# Summary

• Groups and rings, including Sylow theorems,
• Classifying small groups,
• Finitely generated abelian groups,
• Jordan-Holder theorem,
• Solvable groups,
• Simplicity of the alternating group,
• Euclidean domains,
• Principal ideal domains,
• Unique factorization domains,
• Noetherian rings,
• Hilbert basis theorem,
• Zorn’s lemma, and
• Existence of maximal ideals and vector space bases.

Previous course web pages:

# Thursday August 15th

We’ll be using Hungerford’s Algebra text.

## Definitions

The following definitions will be useful to know by heart:

• The order of a group
• Cartesian product
• Relations
• Equivalence relation
• Partition
• Binary operation
• Group
• Isomorphism
• Abelian group
• Cyclic group
• Subgroup
• Greatest common divisor
• Least common multiple
• Permutation
• Transposition
• Orbit
• Cycle
• The symmetric group $$S_{n}$$
• The alternating group $$A_{n}$$
• Even and odd permutations
• Cosets
• Index
• The direct product of groups
• Homomorphism
• Image of a function
• Inverse image of a function
• Kernel
• Normal subgroup
• Factor group
• Simple group

Here is a rough outline of the course:

• Group Theory
• Groups acting on sets
• Sylow theorems and applications
• Classification
• Free and free abelian groups
• Solvable and simple groups
• Normal series
• Galois Theory
• Field extensions
• Splitting fields
• Separability
• Finite fields
• Cyclotomic extensions
• Galois groups
• Module theory
• Free modules
• Homomorphisms
• Projective and injective modules
• Finitely generated modules over a PID
• Linear Algebra
• Matrices and linear transformations
• Rank and determinants
• Canonical forms
• Characteristic polynomials
• Eigenvalues and eigenvectors

## Preliminaries

Definition: A group is an ordered pair $$(G, {\,\cdot\,}: G\times G \to G)$$ where $$G$$ is a set and $${\,\cdot\,}$$ is a binary operation, which satisfies the following axioms:

1. Associativity: $$(g_1 g_2)g_3 = g_1(g_2 g_3)$$,

2. Identity: $$\exists e\in G {~\mathrel{\Big|}~}ge = eg = g$$,

3. Inverses: $$g\in G \implies \exists h\in G {~\mathrel{\Big|}~}gh = gh = e$$.

Examples of groups:

• $$({\mathbb{Z}}, +)$$

• $$({\mathbb{Q}}, +)$$

• $$({\mathbb{Q}}^{\times}, \times)$$

• $$({\mathbb{R}}^{\times}, \times)$$

• ($$\operatorname{GL}(n, {\mathbb{R}}), \times) = \left\{{A \in \mathrm{Mat}_n {~\mathrel{\Big|}~}\det(A) \neq 0}\right\}$$

• $$(S_n, \circ)$$

Definition: A subset $$S \subseteq G$$ is a subgroup of $$G$$ iff

1. Closure: $$s_1, s_2 \in S \implies s_1 s_2 \in S$$

2. Identity: $$e\in S$$

3. Inverses: $$s\in S \implies s^{-1}\in S$$

We denote such a subgroup $$S \leq G$$.

Examples of subgroups:

• $$({\mathbb{Z}}, +) \leq ({\mathbb{Q}}, +)$$

• $${\text{SL}}(n, {\mathbb{R}}) \leq \operatorname{GL}(n, {\mathbb{R}})$$, where $${\text{SL}}(n, {\mathbb{R}}) = \left\{{A\in \operatorname{GL}(n, {\mathbb{R}}) {~\mathrel{\Big|}~}\det(A) = 1}\right\}$$

## Cyclic Groups

Definition: A group $$G$$ is cyclic iff $$G$$ is generated by a single element.

Exercise: Show \begin{align*} \left\langle{g}\right\rangle = \left\{{g^n {~\mathrel{\Big|}~}n\in{\mathbb{Z}}}\right\} \cong \cap_{g\in G} \left\{{H \mathrel{\Big|}H \leq G \text{ and } g\in H}\right\} .\end{align*}

Theorem: Let $$G$$ be a cyclic group, so $$G = \left\langle{g}\right\rangle$$.

• If $${\left\lvert {G} \right\rvert} = \infty$$, then $$G \cong {\mathbb{Z}}$$.

• If $${\left\lvert {G} \right\rvert} = n < \infty$$, then $$G \cong {\mathbb{Z}}_n$$.

Definition: Let $$H \leq G$$, and define a right coset of $$G$$ by $$aH = \left\{{ah {~\mathrel{\Big|}~}H \in H}\right\}$$.

A similar definition can be made for left cosets.

The “Fundamental Theorem of Cosets”: \begin{align*} aH = bH \iff b^{-1}a \in H \text{ and } Ha = Hb \iff ab^{-1}\in H .\end{align*}

Some facts:

• Cosets partition $$H$$, i.e. \begin{align*} b\not\in H \implies aH \cap bH = \left\{{e}\right\} .\end{align*}

• $${\left\lvert {H} \right\rvert} = {\left\lvert {aH} \right\rvert} = {\left\lvert {Ha} \right\rvert}$$ for all $$a\in G$$.

Theorem (Lagrange): If $$G$$ is a finite group and $$H \leq G$$, then $${\left\lvert {H} \right\rvert} {~\Bigm|~}{\left\lvert {G} \right\rvert}$$.

Definition A subgroup $$N \leq G$$ is normal iff $$gN = Ng$$ for all $$g\in G$$, or equivalently $$gNg^{-1}\subseteq N$$. (I denote this $$N {~\trianglelefteq~}G$$.)

When $$N {~\trianglelefteq~}G$$, the set of left/right cosets of $$N$$ themselves have a group structure. So we define \begin{align*} G/N = \left\{{gN {~\mathrel{\Big|}~}g\in G}\right\} \text{ where } (g_1 N)\cdot (g_2 N) \mathrel{\vcenter{:}}=(g_1 g_2) N .\end{align*}

Given $$H, K \leq G$$, define \begin{align*} HK = \left\{{hk \mathrel{\Big|}h\in H, ~k\in K}\right\} .\end{align*}

We have a general formula, \begin{align*} {\left\lvert {HK} \right\rvert} = \frac{{\left\lvert {H} \right\rvert} {\left\lvert {K} \right\rvert}}{{\left\lvert {H \cap K} \right\rvert}}. \end{align*}

## Homomorphisms

Definition: Let $$G,G'$$ be groups, then $$\varphi: G \to G'$$ is a homomorphism if $$\varphi(ab) = \varphi(a) \varphi(b)$$.

Examples of homomorphisms:

• $$\exp\qty{:} ({\mathbb{R}}, +) \to ({\mathbb{R}}^{> 0}, {\,\cdot\,})$$ since \begin{align*} \exp\qty{(}a+b) \mathrel{\vcenter{:}}= e^{a+b} = e^a e^b \mathrel{\vcenter{:}}=\exp\qty{(}a) \exp\qty{(}b) .\end{align*}

• $$\det: (\operatorname{GL}(n, {\mathbb{R}}), \times) \to ({\mathbb{R}}^{\times}, \times)$$ since \begin{align*}\det(AB) = \det(A) \det(B).\end{align*}

• Let $$N {~\trianglelefteq~}G$$ and define

\begin{align*} \varphi: G &\to G/N \\ g &\mapsto gN .\end{align*}

• Let $$\varphi: {\mathbb{Z}}\to {\mathbb{Z}}_n$$ where $$\phi(g) = [g] = g \mod n$$ where $${\mathbb{Z}}_n \cong {\mathbb{Z}}/n{\mathbb{Z}}$$

Definition: Let $$\varphi: G \to G'$$. Then $$\varphi$$ is a monomorphism iff it is injective, an epimorphism iff it is surjective, and an isomorphism iff it is bijective.

## Direct Products

Let $$G_1, G_2$$ be groups, then define \begin{align*} G_1 \times G_2 = \left\{{(g_1, g_2) {~\mathrel{\Big|}~}g_1 \in G, g_2 \in G_2}\right\} \text{ where } (g_1, g_2)(h_1, h_2) = (g_1 h_1, g_2 ,h_2). \end{align*}

We have the formula $${\left\lvert {G_1 \times G_2} \right\rvert} = {\left\lvert {G_1} \right\rvert} {\left\lvert {G_2} \right\rvert}$$.

## Finitely Generated Abelian Groups

Definition: We say a group is abelian if $$G$$ is commutative, i.e. $$g_1, g_2 \in G \implies g_1 g_2 = g_2 g_1$$.

Definition: A group is finitely generated if there exist $$\left\{{g_1, g_2, \cdots g_n}\right\} \subseteq G$$ such that $$G = \left\langle{g_1, g_2, \cdots g_n}\right\rangle$$.

This generalizes the notion of a cyclic group, where we can simply intersect all of the subgroups that contain the $$g_i$$ to define it.

We know what cyclic groups look like – they are all isomorphic to $${\mathbb{Z}}$$ or $${\mathbb{Z}}_n$$. So now we’d like a structure theorem for abelian finitely generated groups.

Theorem: Let $$G$$ be a finitely generated abelian group.

Then \begin{align*}G \cong {\mathbb{Z}}^r \times \displaystyle\prod_{i=1}^s {\mathbb{Z}}_{p_i^{\alpha _i}}\end{align*} for some finite $$r,s \in {\mathbb{N}}$$ where the $$p_i$$ are (not necessarily distinct) primes.

Example: Let $$G$$ be a finite abelian group of order 4.

Then $$G \cong {\mathbb{Z}}_4$$ or $${\mathbb{Z}}_2^2$$, which are not isomorphic because every element in $${\mathbb{Z}}_2^2$$ has order 2 where $${\mathbb{Z}}_4$$ contains an element of order 4.

## Fundamental Homomorphism Theorem

Let $$\varphi: G \to G'$$ be a group homomorphism and define \begin{align*} \ker \varphi \mathrel{\vcenter{:}}=\left\{{g\in G {~\mathrel{\Big|}~}\varphi(g) = e'}\right\} .\end{align*}

### The First Homomorphism Theorem

Theorem: There exists a map $$\varphi': G/\ker \varphi \to G'$$ such that the following diagram commutes:

That is, $$\varphi = \varphi' \circ \eta$$, and $$\varphi'$$ is an isomorphism onto its image, so $$G/\ker \varphi = \operatorname{im}({\varphi})$$.

This map is given by \begin{align*} \varphi'(g(\ker \varphi)) = \varphi(g) .\end{align*}

Exercise: Check that $$\varphi$$ is well-defined.

### The Second Theorem

Theorem: Let $$K, N \leq G$$ where $$N {~\trianglelefteq~}G$$. Then \begin{align*} \frac K {N \cap K} \cong \frac {NK} N \end{align*}

Proof: Define a map

\begin{align*} K &\xrightarrow{\varphi} NK/N \\ k &\mapsto kN .\end{align*}

You can show that $$\varphi$$ is onto, then look at $$\ker \varphi$$; note that \begin{align*} kN = \varphi(k) = N \iff k \in N ,\end{align*}

and so $$\ker \varphi = N \cap K$$.

$$\hfill\blacksquare$$

# Tuesday August 20th

## The Fundamental Homomorphism Theorems

Theorem 1: Let $$\varphi: G \to G'$$ be a homomorphism. Then there is a canonical homomorphism $$\eta: G \to G/\ker \varphi$$ such that the usual diagram commutes.

Moreover, this map induces an isomorphism $$G /\ker \varphi \cong \operatorname{im}({\varphi})$$.

Theorem 2: Let $$K, N \leq G$$ and suppose $$N {~\trianglelefteq~}G$$. Then there is an isomorphism \begin{align*} \frac K {K \cap N} \cong \frac {NK} {N} \end{align*}

Proof Sketch: Show that $$K \cap N {~\trianglelefteq~}G$$, and $$NK$$ is a subgroup exactly because $$N$$ is normal.

Theorem 3: Let $$H, K {~\trianglelefteq~}G$$ such that $$H \leq K$$.

Then

1. $$H/K$$ is normal in $$G/K$$.

2. The quotient $$(G/K) / (H/K) \cong G/H$$.

Proof: We’ll use the first theorem.

Define a map \begin{align*} \phi: G/K &\to G/H \\ gk &\mapsto gH .\end{align*}

Exercise: Show that $$\phi$$ is surjective, and that $$\ker \phi \cong H/K$$.

$$\hfill\blacksquare$$

## Permutation Groups

Let $$A$$ be a set, then a permutation on $$A$$ is a bijective map $$A {\circlearrowleft}$$. This can be made into a group with a binary operation given by composition of functions. Denote $$S_{A}$$ the set of permutations on $$A$$.

Theorem: $$S_{A}$$ is in fact a group.

Proof: Exercise. Follows from checking associativity, inverses, identity, etc.

$$\hfill\blacksquare$$

In the special case that $$A = \left\{{1, 2, \cdots n}\right\}$$, then $$S_{n} \mathrel{\vcenter{:}}= S_{A}$$.

Recall two line notation \begin{align*} \left(\begin{matrix} 1 & 2 & \cdots & n\\ \sigma(1) & \sigma(2) & \cdots & \sigma(n) \end{matrix}\right) \end{align*}

Moreover, $${\left\lvert {S_{n}} \right\rvert} = n!$$ by a combinatorial counting argument.

Example: $$S_{3}$$ is the symmetries of a triangle.

Example: The symmetries of a square are not given by $$S_{4}$$, it is instead $$D_{4}$$.

## Orbits and the Symmetric Group

Permutations $$S_{A}$$ act on $$A$$, and if $$\sigma \in S_{A}$$, then $$\left\langle{\sigma}\right\rangle$$ also acts on $$A$$.

Define $$a \sim b$$ iff there is some $$n$$ such that $$\sigma^{n}(a) = b$$. This is an equivalence relation, and thus induces a partition of $$A$$. See notes for diagram. The equivalence classes under this relation are called the orbits under $$\sigma$$.

Example: \begin{align*} \left(\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 8 & 2 & 6 & 3 & 7 & 4 & 5 & 1 \end{matrix}\right) = (1 8)(2)(3 6 4)(5 7). \end{align*}

Definition: A permutation $$\sigma \in S_{n}$$ is a cycle iff it contains at most one orbit with more than one element.

The length of a cycle is the number of elements in the largest orbit.

Recall cycle notation: $$\sigma = (\sigma(1) \sigma(2) \cdots \sigma(n))$$.

Note that this is read right-to-left by convention!

Theorem: Every permutation $$\sigma \in S_{n}$$ can be written as a product of disjoint cycles.

Definition: A transposition is a cycle of length 2.

Proposition: Every permutation is a product of transpositions.

Proof: \begin{align*} (a_{1} a_{2} \cdots a_{n}) = (a_{1} a_{n}) (a_{1} a_{n-1}) \cdots (a_{1} a_{2}) .\end{align*}

$$\hfill\blacksquare$$

This is not a unique decomposition, however, as e.g. $$\text{id}= (1 2)^{2} = (3 4)^{2}$$.

Theorem: Any $$\sigma \in S_{n}$$ can be written as either

• An even number of transpositions, or

• An odd number of transpositions.

Proof:

Define \begin{align*} A_{n} = \left\{{\sigma \in S_{n} {~\mathrel{\Big|}~}\sigma\text{ is even}}\right\} .\end{align*}

We claim that $$A_{n} {~\trianglelefteq~}S_{n}$$.

1. Closure: If $$\tau_{1}, \tau_{2}$$ are both even, then $$\tau_{1}\tau_{2}$$ also has an even number of transpositions.

2. The identity has an even number of transpositions, since zero is even.

3. Inverses: If $$\sigma = \prod_{i=1}^{s} \tau_{i}$$ where $$s$$ is even, then $$\sigma^{-1}= \prod_{i=1}^{s} \tau_{s-i}$$. But each $$\tau$$ is order 2, so $$\tau^{-1}= \tau$$, so there are still an even number of transpositions.

So $$A_{n}$$ is a subgroup.

It is normal because it is index 2, or the kernel of a homomorphism, or by a direct computation.

## Groups Acting on Sets

Think of this as a generalization of a $$G{\hbox{-}}$$module.

Definition: A group $$G$$ is said to act on a set $$X$$ if there exists a map $$G\times X \to X$$ such that

1. $$e\curvearrowright x = x$$

2. $$(g_{1} g_{2})\curvearrowright x = g_{1} \curvearrowright(g_{2} \curvearrowright x)$$.

Examples:

1. $$G = S_{A} \curvearrowright A$$

2. $$H \leq G$$, then $$G \curvearrowright X = G/H$$ where $$g \curvearrowright xH = (gx)H$$.

3. $$G \curvearrowright G$$ by conjugation, i.e. $$g\curvearrowright x = gxg^{-1}$$.

Definition: Let $$x\in X$$, then define the stabilizer subgroup \begin{align*} G_{x} = \left\{{g\in G {~\mathrel{\Big|}~}g\curvearrowright x = x}\right\} \leq G \end{align*}

We can also look at the dual notion, \begin{align*} X_{g} = \left\{{x\in X {~\mathrel{\Big|}~}g\curvearrowright x = x}\right\}. \end{align*}

We then define the orbit of an element $$x$$ as \begin{align*} Gx = \left\{{g\curvearrowright x {~\mathrel{\Big|}~}g\in G}\right\} \end{align*} and we have a similar result where $$x\sim y \iff x\in Gy$$, and the orbits partition $$X$$.

Theorem: Let $$G$$ act on $$X$$. We want to know the number of elements in an orbit, and it turns out that

\begin{align*} {\left\lvert {Gx} \right\rvert} = [G: G_{x}] \end{align*}

Proof: Construct a map $$Gx \xrightarrow{\psi} G/Gx$$ where $$\psi(g\curvearrowright x) = g Gx$$.

Exercise: Show that this is well-defined, so if 2 elements are equal then they go to the same coset.

Exercise: Show that this is surjective.

Injectivity: $$\psi(g_{1} x) = \psi(g_{2} x)$$, so $$g_{1} Gx = g_{2} Gx$$ and $$(g_{2}^{-1}g_{1}) Gx = Gx$$ so \begin{align*} g_{2}^{-1}g_{1} \in Gx \iff g_{2}^{-1}g_{1} \curvearrowright x = x \iff g_{1}x = g_{2} x .\end{align*} $$\hfill\blacksquare$$

Next time: Burnside’s theorem, proving the Sylow theorems.

# Thursday August 22nd

## Group Actions

Let $$G$$ be a group and $$X$$ be a set; we say $$G$$ acts on $$X$$ (or that $$X$$ is a $$G{\hbox{-}}$$ set) when there is a map $$G\times X \to X$$ such that $$ex = x$$ and \begin{align*} (gh) \curvearrowright x = g \curvearrowright(h \curvearrowright x) .\end{align*}

We then define the stabilizer of $$x$$ as \begin{align*} \mathrm{Stab}_G(x) = G_x \mathrel{\vcenter{:}}=\left\{{g\in G {~\mathrel{\Big|}~}g\curvearrowright x = x}\right\} \leq G, \end{align*}

and the orbit \begin{align*} G.x = \mathcal O_x \mathrel{\vcenter{:}}=\left\{{g\curvearrowright x {~\mathrel{\Big|}~}x\in X}\right\} \subseteq X. \end{align*}

When $$G$$ is finite, we have \begin{align*} {\left\lvert {G.x} \right\rvert} = \frac{{\left\lvert {G} \right\rvert}}{{\left\lvert {G_x} \right\rvert}}. \end{align*}

We can also consider the fixed points of $$X$$, \begin{align*} X_g = \left\{{x\in X \mathrel{\Big|}g\curvearrowright x = x ~~\forall g\in G}\right\} \subseteq X \end{align*}

## Burnside’s Theorem

Theorem (Burnside): Let $$X$$ be a $$G{\hbox{-}}$$set and $$v \mathrel{\vcenter{:}}={\left\lvert {X/G} \right\rvert}$$ be the number of orbits. Then \begin{align*} v {\left\lvert {G} \right\rvert} = \sum_{g\in G} {\left\lvert {X_g} \right\rvert}. \end{align*}

Proof: Define \begin{align*} N = \left\{{(g,x) \mathrel{\Big|}g\curvearrowright x = x}\right\} \subseteq G \times X ,\end{align*}

we then have

\begin{align*} {\left\lvert {N} \right\rvert} &= \sum_{g\in G} {\left\lvert {X_g} \right\rvert} \\ &= \sum_{x\in X} {\left\lvert {G_x} \right\rvert} \\ &= \sum_{x\in X} \frac{{\left\lvert {G} \right\rvert}}{{\left\lvert {G.x} \right\rvert}} \quad\text{by Orbit-Stabilizer} \\ &= {\left\lvert {G} \right\rvert} \left( \sum_{x\in X} \frac{1}{{\left\lvert {G.x} \right\rvert}} \right) \\ &= {\left\lvert {G} \right\rvert} \sum_{G.x ~\in~ X/G} \left( \sum_{y ~\in~ G.x} \frac{1}{{\left\lvert {G.x} \right\rvert}} \right) \\ &= {\left\lvert {G} \right\rvert} \sum_{G.x ~\in~ X/G} \left( {\left\lvert {G.x} \right\rvert} \frac{1}{{\left\lvert {G.x} \right\rvert}} \right) \\ &= {\left\lvert {G} \right\rvert} \sum_{G.x ~\in~ X/G} 1 \\ &= {\left\lvert {G} \right\rvert} v .\end{align*}

The last two equalities follow from the following fact: since the orbits partition $$X$$, say into $$X = \displaystyle{\coprod}_{i=1}^v \sigma_i$$, so let $$\sigma = \left\{{\sigma_i \mathrel{\Big|}1 \leq i \leq v}\right\}$$.

By abuse of notation, replace each orbit in $$\sigma$$ with a representative element $$x_i\in \sigma_i \subset X$$.

We then have \begin{align*} \sum_{x \in \sigma} \frac{1}{{\left\lvert {G.x} \right\rvert}} = \frac{1}{{\left\lvert {G.x} \right\rvert}} {\left\lvert {\sigma} \right\rvert} = 1. \end{align*}

$$\hfill\blacksquare$$

Application: Consider seating 10 people around a circular table. How many distinct seating arrangements are there?

Let $$X$$ be the set of configurations, $$G = S_{10}$$, and let $$G\curvearrowright X$$ by permuting configurations. Then $$v$$, the number of orbits under this action, yields the number of distinct seating arrangements.

By Burnside, we have \begin{align*} v = \frac{1}{{\left\lvert {G} \right\rvert}} \sum_{g\in G} {\left\lvert {X_g} \right\rvert} = \frac{1}{10} (10!) = 9! \end{align*}

since $$X_g = \left\{{x\in X \mathrel{\Big|}g\curvearrowright x = x}\right\} = \emptyset$$ unless $$g = e$$, and $$X_e = X$$.

## Sylow Theory

Recall Lagrange’s theorem:

If $$H \leq G$$ and $$G$$ is finite, then $${\left\lvert {H} \right\rvert}$$ divides $${\left\lvert {G} \right\rvert}$$.

Consider the converse: if $$n$$ divides $${\left\lvert {G} \right\rvert}$$, does there exist a subgroup of size $$n$$?

The answer is no in general, and a counterexample is $$A_4$$ which has $$4!/2 = 12$$ elements but no subgroup of order 6.

### Class Functions

Let $$X$$ be a $$G{\hbox{-}}$$set, and choose orbit representatives $$x_1 \cdots x_v$$.

Then \begin{align*} {\left\lvert {X} \right\rvert} = \sum_{i=1}^v {\left\lvert {G .x_i} \right\rvert}. \end{align*}

We can then separately count all orbits with exactly one element, which is exactly \begin{align*} X_G = \left\{{x\in G \mathrel{\Big|}g\curvearrowright x = x ~ \forall g\in G}\right\} \end{align*}.

We then have \begin{align*} {\left\lvert {X} \right\rvert} = {\left\lvert {X_G} \right\rvert} + \sum_{i=j}^v {\left\lvert {G. x_i} \right\rvert} \end{align*} for some $$j$$ where $${\left\lvert {G.x_i} \right\rvert} > 1$$ for all $$i \geq j$$.

Theorem: Let $$G$$ be a group of order $$p^n$$ for $$p$$ a prime.

Then \begin{align*} {\left\lvert {X} \right\rvert} = {\left\lvert {X_G} \right\rvert} \mod p .\end{align*}

Proof: We know that \begin{align*} {\left\lvert {G.x_i} \right\rvert} = [G : G_{x_i}] \text{ for } j \leq i \leq v \text{ and } {\left\lvert {Gx_i} \right\rvert} > 1 \implies G.x_i \neq G ,\end{align*} and thus $$p$$ divides $$[G: G x_i]$$. The result follows.

$$\hfill\blacksquare$$

Application: If $${\left\lvert {G} \right\rvert} = p^n$$, then the center $$Z(G)$$ is nontrivial.

Let $$X=G$$ act on itself by conjugation, so $$g\curvearrowright x = gxg^{-1}$$. Then \begin{align*} X_G = \left\{{x\in G {~\mathrel{\Big|}~}gxg^{-1}= x}\right\} = \left\{{x\in G {~\mathrel{\Big|}~}gx = xg}\right\} = Z(G) \end{align*}

But then, by the previous theorem, we have \begin{align*} {\left\lvert {Z(G)} \right\rvert} \equiv {\left\lvert {X} \right\rvert}\equiv {\left\lvert {G} \right\rvert} \mod p ,\end{align*} but since $$Z(G) \leq G$$ we have $${\left\lvert {Z(G)} \right\rvert} \cong 0 \mod p$$. So in particular, $$Z(G) \neq \left\{{e}\right\}$$.

Definition: A group $$G$$ is a $$p{\hbox{-}}$$group iff every element in $$G$$ has order $$p^k$$ for some $$k$$. A subgroup is a $$p{\hbox{-}}$$group exactly when it is a $$p{\hbox{-}}$$group in its own right.

### Cauchy’s Theorem

Theorem (Cauchy): Let $$G$$ be a finite group, where $$p$$ is prime and divides $${\left\lvert {G} \right\rvert}$$. Then $$G$$ has an element (and thus a subgroup) of order $$p$$.

Proof: Consider \begin{align*} X = \left\{{(g_1, g_2, \cdots , g_p) \in G^{\oplus p} \mathrel{\Big|}g_1g_2\cdots g_p = e}\right\} .\end{align*}

Given any $$p-1$$ elements, say $$g_1 \cdots g_{p-1}$$, the remaining element is completely determined by $$g_p = (g_1 \cdots g_{p-1})^{-1}$$.

So $${\left\lvert {X} \right\rvert} = {\left\lvert {G} \right\rvert}^{p-1}$$.and since $$p {~\Bigm|~}{\left\lvert {G} \right\rvert}$$, we have $$p {~\Bigm|~}{\left\lvert {X} \right\rvert}$$.

Now let $$\sigma \in S_p$$ the symmetric group act on $$X$$ by index permutation, i.e. \begin{align*} \sigma \curvearrowright(g_1, g_2 \cdots g_p) = (g_{\sigma(1)}, g_{\sigma(2)}, \cdots, g_{\sigma(p)}) .\end{align*}

Exercise: Check that this gives a well-defined group action.

Let $$\sigma = (1~2~\cdots~p) \in S_p$$, and note $$\left\langle{\sigma}\right\rangle \leq S_p$$ also acts on $$X$$ where $${\left\lvert {\left\langle{\sigma}\right\rangle} \right\rvert} = p$$. Therefore we have \begin{align*} {\left\lvert {X} \right\rvert} = {\left\lvert {X_{\left\langle{\sigma}\right\rangle}} \right\rvert} \mod p. \end{align*}

Since $$p{~\Bigm|~}{\left\lvert {X} \right\rvert}$$, it follows that $${\left\lvert {X_{\left\langle{\sigma}\right\rangle}} \right\rvert} = 0 \mod p$$, and thus $$p {~\Bigm|~}{\left\lvert {X_{\left\langle{\sigma}\right\rangle}} \right\rvert}$$.

If $$\left\langle{\sigma}\right\rangle$$ fixes $$(g_1, g_2, \cdots g_p)$$, then $$g_1 = g_2 = \cdots g_p$$.

Note that $$(e, e, \cdots) \in X_{\left\langle{\sigma}\right\rangle}$$, as is $$(a, a, \cdots a)$$ since $$p {~\Bigm|~}{\left\lvert {X_{\left\langle{\sigma}\right\rangle}} \right\rvert}$$. So there is some $$a\in G$$ such that $$a^p = 1$$. Moreover, $$\left\langle{a}\right\rangle \leq G$$ is a subgroup of size $$p$$.

$$\hfill\blacksquare$$

### Normalizers

Let $$G$$ be a group and $$X = S$$ be the set of subgroups of $$G$$. Let $$G$$ act on $$X$$ by $$g\curvearrowright H = gHg^{-1}$$. What is the stabilizer?

\begin{align*} G_x = G_H = \left\{{g\in G \mathrel{\Big|}gHg^{-1}= H}\right\} ,\end{align*} making $$G_H$$ the largest subgroup such that $$H {~\trianglelefteq~}G_H$$.

So we define $$N_G(H) \mathrel{\vcenter{:}}= G_H$$.

Lemma: Let $$H$$ be a $$p{\hbox{-}}$$subgroup of $$G$$ of order $$p^n$$. Then \begin{align*} [N_G(H) : H] = [G : H] \mod p .\end{align*}

Proof: Let $$S = G/H$$ be the set of left $$H{\hbox{-}}$$cosets in $$G$$. Now let $$H$$ act on $$S$$ by \begin{align*} H\curvearrowright x + H \mathrel{\vcenter{:}}=(hx) + H .\end{align*}

By a previous theorem, $${\left\lvert {G/H} \right\rvert} = {\left\lvert {S} \right\rvert} = {\left\lvert {S_H} \right\rvert} \mod p$$, where $${\left\lvert {G/H} \right\rvert} = [G: H]$$. What is $$S_H$$?

This is given by \begin{align*} S_H = \left\{{x + H \in S {~\mathrel{\Big|}~}xHx^{-1}\in H \forall h\in H}\right\} .\end{align*}

Therefore $$x\in N_G(H)$$.

$$\hfill\blacksquare$$

Corollary: Let $$H \leq G$$ be a subgroup of order $$p^n$$. If $$p {~\Bigm|~}[G: H]$$ then $$N_G(H) \neq H$$.

Proof: Exercise. $$\hfill\blacksquare$$

Theorem: Let $$G$$ be a finite group, then $$G$$ is a $$p{\hbox{-}}$$group $$\iff {\left\lvert {G} \right\rvert} = p^n$$ for some $$n\geq 1$$.

Proof: Suppose $${\left\lvert {G} \right\rvert} = p^n$$ and $$a \in G$$. Then $${\left\lvert {\left\langle{a}\right\rangle} \right\rvert} = p^\alpha$$ for some $$\alpha$$.

Conversely, suppose $$G$$ is a $$p{\hbox{-}}$$group. Factor $${\left\lvert {G} \right\rvert}$$ into primes and suppose $$\exists q$$ such that $$q {~\Bigm|~}{\left\lvert {G} \right\rvert}$$ but $$q \neq p$$.

By Cauchy, we can then get a subgroup $$\left\langle{c}\right\rangle$$ such that $${\left\lvert {\left\langle{c}\right\rangle} \right\rvert} {~\Bigm|~}q$$, but then $${\left\lvert {G} \right\rvert} \neq p^n$$.

$$\hfill\blacksquare$$

# Tuesday August 27th

Let $$G$$ be a finite group and $$p$$ a prime. TFAE:

• $${\left\lvert {H} \right\rvert} = p^n$$ for some $$n$$
• Every element of $$H$$ has order $$p^\alpha$$ for some $$\alpha$$.

If either of these are true, we say $$H$$ is a $$p{\hbox{-}}$$group.

Let $$H$$ be a $$p{\hbox{-}}$$group, last time we proved that if $$p {~\Bigm|~}[G : H]$$ then $$N_G(H) \neq H$$.

## Sylow Theorems

Let $$G$$ be a finite group and suppose $${\left\lvert {G} \right\rvert} = p^n m$$ where $$(m, n) = 1$$. Then

### Sylow 1

Idea: take a prime factorization of $${\left\lvert {G} \right\rvert}$$, then there are subgroups of order $$p^i$$ for every prime power appearing, up to the maximal power.

1. $$G$$ contains a subgroup of order $$p^i$$ for every $$1 \leq i \leq n$$.
2. Every subgroup $$H$$ of order $$p^i$$ where $$i < n$$ is a normal subgroup in a subgroup of order $$p^{i+1}$$.

Proof: By induction on $$i$$. For $$i=1$$, we know this by Cauchy’s theorem. If we show (2), that shows (1) as a consequence.

So suppose this holds for $$i < n$$. Let $$H \leq G$$ where $${\left\lvert {H} \right\rvert} = p^i$$, we now want a subgroup of order $$p^{i+1}$$. Since $$p{~\Bigm|~}[G: H]$$, by the previous theorem, $$H < N_G(H)$$ is a proper subgroup (?).

Now consider the canonical projection $$N_G(H) \to N_G(H) /H$$. Since \begin{align*} p {~\Bigm|~}[N_G(H) : H] = {\left\lvert {N_G(H)/ H} \right\rvert} ,\end{align*} by Cauchy there is a subgroup of order $$p$$ in this quotient. Call it $$K$$. Then $$\pi^{-1}(K) \leq N_G(H)$$.

Exercise: Show that $${\left\lvert {\phi^{-1}(K)} \right\rvert} = p^{i+1}$$.

It now follows that $$H {~\trianglelefteq~}\phi^{-1}(K)$$. $$\hfill\blacksquare$$

Definition: For $$G$$ a finite group and $${\left\lvert {G} \right\rvert} = p^n m$$ where $$p$$ does not divide $$m$$.

Then a subgroup of order $$p^n$$ is called a Sylow $$p{\hbox{-}}$$subgroup.

Note: by Sylow 1, these exist.

### Sylow 2

If $$P_1, P_2$$ are Sylow $$p{\hbox{-}}$$subgroups of $$G$$, then $$P_1$$ and $$P_2$$ are conjugate.

Proof: Let $$\mathcal L$$ be the left cosets of $$P_1$$, i.e. $$\mathcal L = G/P_1$$.

Let $$P_2$$ act on $$\mathcal L$$ by \begin{align*} p_2 \curvearrowright(g + P_1) \mathrel{\vcenter{:}}=(p_2g) + P_1 .\end{align*}

By a previous theorem about orbits and fixed points, we have \begin{align*} {\left\lvert {\mathcal L_{P_2}} \right\rvert} = {\left\lvert {\mathcal L} \right\rvert} \mod p. \end{align*}

Since $$p$$ does not divide $${\left\lvert {\mathcal L} \right\rvert}$$, we have $$p$$ does not divide $${\left\lvert {\mathcal L_{P_2}} \right\rvert}$$. So $$\mathcal L_{P_2}$$ is nonempty.

So there exists a coset $$xP_1$$ such that $$xP_1 \in \mathcal L_{P_2}$$, and thus \begin{align*} yxP_1 = xP_1 \text{ for all } y\in P_2 .\end{align*}

Then $$x^{-1}y x P_1 = P_1$$ for all $$y\in P_2$$, and so $$x^{-1}P_2 x = P_1$$. So $$P_1$$ and $$P_2$$ are conjugate.

$$\hfill\blacksquare$$

### Sylow 3

Let $$G$$ be a finite group, and $$p{~\Bigm|~}{\left\lvert {G} \right\rvert}$$. Let $$r_p$$ be the number of Sylow $$p{\hbox{-}}$$subgroups of $$G$$.

Then

• $$r_p \cong 1 \mod p$$.

• $$r_p {~\Bigm|~}{\left\lvert {G} \right\rvert}$$.

• $$r_p = [G : N_G(P)]$$

Proof:

Let $$X = \mathcal S$$ be the set of Sylow $$p{\hbox{-}}$$subgroups, and let $$P \in X$$ be a fixed Sylow $$p{\hbox{-}}$$subgroup.

Let $$P \curvearrowright\mathcal S$$ by conjugation, so for $$\overline P \in \mathcal S$$ let $$x \curvearrowright\overline P = x \overline P x^{-1}$$.

By a previous theorem, we have \begin{align*} {\left\lvert {\mathcal S} \right\rvert} = \mathcal{S}_P \mod p \end{align*}

What are the fixed points $$\mathcal{S}_P$$? \begin{align*} \mathcal{S}_P = \left\{{T \in \mathcal S {~\mathrel{\Big|}~}xTx^{-1}= T \quad \forall x\in P}\right\}. \end{align*}

Let $$\mathcal T \in \mathcal{S}_P$$, so $$xTx^{-1}= T$$ for all $$x\in P$$.

Then $$P \leq N_G(T)$$, so both $$P$$ and $$T$$ are Sylow $$p{\hbox{-}}$$ subgroups in $$N_G(H)$$ as well as $$G$$.

So there exists a $$f\in N_G(T)$$ such that $$T = gPg^{-1}$$. But the point is that in the normalizer, there is only one Sylow $$p{\hbox{-}}$$ subgroup.

But then $$T$$ is the unique largest normal subgroup of $$N_G(T)$$, which forces $$T = P$$.

Then $$\mathcal{S}_P = \left\{{P}\right\}$$, and using the formula, we have $$r_p \cong 1 \mod p$$.

Now modify this slightly by letting $$G$$ act on $$\mathcal S$$ (instead of just $$P$$) by conjugation.

Since all Sylows are conjugate, by Sylow (1) there is only one orbit, so $$\mathcal S = GP$$ for $$P \in \mathcal S$$. But then \begin{align*} r_p = {\left\lvert {\mathcal S} \right\rvert} = {\left\lvert {GP} \right\rvert} = [G: G_p] {~\Bigm|~}{\left\lvert {G} \right\rvert}. \end{align*}

Note that this gives a precise formula for $$r_p$$, although the theorem is just an upper bound of sorts, and $$G_p = N_G(P)$$.

## Applications of Sylow Theorems

Of interest historically: classifying finite simple groups, where a group $$G$$ is simple If $$N {~\trianglelefteq~}G$$ and $$N \neq \left\{{e}\right\}$$, then $$N=G$$.

Example: Let $$G = {\mathbb{Z}}_p$$, any subgroup would need to have order dividing $$p$$, so $$G$$ must be simple.

Example: $$G = A_n$$ for $$n\geq 5$$ (see Galois theory)

One major application is proving that groups of a certain order are not simple.

Applications:

Proposition: Let $${\left\lvert {G} \right\rvert} = p^n q$$ with $$p > q$$. Then $$G$$ is not simple.

Proof:

Strategy: Find a proper normal nontrivial subgroup using Sylow theory. Can either show $$r_p = 1$$, or produce normal subgroups by intersecting distinct Sylow p-subgroups.

Consider $$r_p$$, then $$r_p = p^\alpha q^\beta$$ for some $$\alpha, \beta$$. But since $$r_p \cong 1 \mod p$$, $$p$$ does not divide $$r_p$$, we must have $$r_p = 1, q$$.

But since $$q < p$$ and $$q\neq 1 \mod p$$, this forces $$r_p = 1$$.

So let $$P$$ be a sylow $$p{\hbox{-}}$$subgroup, then $$P < G$$. Then $$gPg^{-1}$$ is also a sylow, but there’s only 1 of them, so $$P$$ is normal.

$$\hfill\blacksquare$$

Proposition: Let $${\left\lvert {G} \right\rvert} = 45$$, then $$G$$ is not simple.

Proof: Exercise.

$$\hfill\blacksquare$$

Proposition: Let $${\left\lvert {G} \right\rvert} = p^n$$, then $$G$$ is not simple if $$n > 1$$.

Proof: By Sylow (1), there is a normal subgroup of order $$p^{n-1}$$ in $$G$$.

$$\hfill\blacksquare$$

Proposition: Let $${\left\lvert {G} \right\rvert} = 48$$, then $$G$$ is not simple.

Proof:

Note $$48 = 2^4 3$$, so consider $$r_2$$, the number of Sylow 2-subgroups. Then $$r_2 \cong 1 \mod 2$$ and $$r_2 {~\Bigm|~}48$$. So $$r_2 = 1, 3$$. If $$r_2 = 1$$, we’re done, otherwise suppose $$r_2 = 3$$.

Let $$H \neq K$$ be Sylow 2-subgroups, so $${\left\lvert {H} \right\rvert} = {\left\lvert {K} \right\rvert} = 2^4 = 16$$. Now consider $$H \cap K$$, which is a subgroup of $$G$$. How big is it?

Since $$H\neq K, {\left\lvert {H \cap K} \right\rvert} < 16$$. The order has to divides 16, so we in fact have $${\left\lvert {H \cap K} \right\rvert} \leq 8$$. Suppose it is less than 4, towards a contradiction. Then \begin{align*} {\left\lvert {HK} \right\rvert} = \frac{{\left\lvert {H} \right\rvert} {\left\lvert {K} \right\rvert}}{{\left\lvert {H \cap K} \right\rvert}} \geq \frac{(16)(16)}{4} = 64 > {\left\lvert {G} \right\rvert} = 48. \end{align*}

So we can only have $${\left\lvert {H \cap K} \right\rvert} = 8$$. Since this is an index 2 subgroup in both $$H$$ and $$K$$, it is in fact normal. But then \begin{align*} H, K \subseteq N_G(H \cap K) \mathrel{\vcenter{:}}= X .\end{align*}

But then $${\left\lvert {X} \right\rvert}$$ must be a multiple of 16 and divide 48, so it’s either 16 or 28. But $${\left\lvert {X} \right\rvert} > 16$$, because $$H \subseteq X$$ and $$K \subseteq X$$. So then \begin{align*} N_G(H \cap K) = G \text{ and so } H \cap K {~\trianglelefteq~}G .\end{align*}

$$\hfill\blacksquare$$

# Thursday August 29th

## Classification of Groups of Certain Orders

We have a classification of some finite abelian groups.

Order of G Number of Groups List of Distinct Groups
1 1 $$\left\{{e}\right\}$$
2 1 $${\mathbb{Z}}_2$$
3 1 $${\mathbb{Z}}_3$$
4 2 $${\mathbb{Z}}_4, {\mathbb{Z}}_2^2$$
5 1 $${\mathbb{Z}}_5$$
6 2 $${\mathbb{Z}}_6, S_3$$ (*)
7 1 $${\mathbb{Z}}_7$$
8 5 $${\mathbb{Z}}_8,{\mathbb{Z}}_4 \times{\mathbb{Z}}_2, {\mathbb{Z}}_2^3, D_4, Q$$
9 2 $${\mathbb{Z}}_9, {\mathbb{Z}}_3^2$$
10 2 $${\mathbb{Z}}_{10}, D_5$$
11 1 $${\mathbb{Z}}_{11}$$

Exercise: show that groups of order $$p^2$$ are abelian.

We still need to justify $$S_3, D_4, Q, D_5$$.

Recall that for any group $$A$$, we can consider the free group on the elements of $$A$$ given by $$F[A]$$.

Note that we can also restrict $$A$$ to just its generators.

There is then a homomorphism $$F[A] \to A$$, where the kernel is the relations.

Example: \begin{align*} {\mathbb{Z}}\ast {\mathbb{Z}}= \left\langle{x, y \mathrel{\Big|}xyx^{-1}y^{-1}= e}\right\rangle \text{ where } x = (1, 0),~ y = (0, 1) .\end{align*}

## Groups of Order 6

Let $$G$$ be nonabelian of order $$6$$.

Idea: look at subgroups of index 2.

Let $$P$$ be a Sylow 3-subgroup of $$G$$, then $$r_3 = 1$$ so $$P{~\trianglelefteq~}G$$. Moreover, $$P$$ is cyclic since it is order 3, so $$P = \left\langle{a}\right\rangle$$.

But since $${\left\lvert {G/P} \right\rvert} = 2$$, it is also cyclic, so $$G/P = \left\langle{bP}\right\rangle$$.

Note that $$b\not\in P$$, but $$b^2 \in P$$ since $$(bP)^2 = P$$, so $$b^2 \in \left\{{e, a, a^2}\right\}$$.

If $$b=a, a^2$$ then $$b$$ has order 6, but this would make $$G = \left\langle{b}\right\rangle$$ cyclic and thus abelian. So $$b^2=1$$.

Since $$P {~\trianglelefteq~}G$$, we have $$bPb^{-1}= P$$, and in particular $$bab^{-1}$$ has order 3.

So either $$bab^{-1}= a$$, or $$bab^{-1}= a^2$$. If $$bab^{-1}= a$$, then $$G$$ is abelian, so $$bab^{-1}= a^2$$. So \begin{align*} G = \left\langle{a, b \mathrel{\Big|}a^3 = e, b^2 = e, bab^{-1}= a^2}\right\rangle .\end{align*}

We’ve shown that if there is such a nonabelian group, then it must satisfy these relations – we still need to produce some group that actually realizes this.

Consider the symmetries of the triangle:

You can check that $$a,b$$ satisfy the appropriate relations.

## Groups of Order 10

For order 10, a similar argument yields

\begin{align*} G = \left\langle{a, b \mathrel{\Big|}a^5 = 1, b^2 = 1, ba=a^4b}\right\rangle ,\end{align*} and this is realized by symmetries of the pentagon where $$a = (1~2~3~4~5), b=(1~4)(2~3)$$.

## Groups of Order 8

Assume $$G$$ is nonabelian of order 8. $$G$$ has no elements of order 8, so the only possibilities for orders of elements are 1, 2, or 4.

Assume all elements have order 1 or 2. Let $$a,b\in G$$, consider \begin{align*} (ab)^2 = abab \implies ab=b^{-1}a^{-1}= ba ,\end{align*} and thus $$G$$ is abelian. So there must be an element of order 4.

So suppose $$a\in G$$ has order 4, which is an index 2 subgroup, and so $$\left\langle{a}\right\rangle {~\trianglelefteq~}G$$.

But $${\left\lvert {G/\left\langle{a}\right\rangle} \right\rvert} = 2$$ is cyclic, so $$G/\left\langle{a}\right\rangle = \left\langle{bH}\right\rangle$$.

Note that $$b^2 \in H = \left\langle{a}\right\rangle$$.

If $$b^2=a, a^3$$ then $$b$$ will have order 8, making $$G$$ cyclic. So $$b^2 = 1, a^2$$. These are both valid possibilities.

Since $$H {~\trianglelefteq~}G$$, we have $$b\left\langle{a}\right\rangle b^{-1}= \left\langle{a}\right\rangle$$, and since $$a$$ has order 4, so does $$bab^{-1}$$.

So $$bab^{-1}= a, a^3$$, but $$a$$ is not an option because this would make $$G$$ abelian.

So we have two options:

\begin{align*} G_1 &= \left\langle{a, b \mathrel{\Big|}a^4 = 1, b^2=1, bab^{-1}= a^3}\right\rangle \\ G_2 &= \left\langle{a, b \mathrel{\Big|}a^4 = 1, b^2 = a^2, bab^{-1}= a^3}\right\rangle .\end{align*}

Exercise: prove $$G_1 \not\cong G_2$$.

Now to realize these groups:

• $$G_1$$ is the group of symmetries of the square, where $$a = (1~2~3~4), b=(1~3)$$.

• $$G_2 \cong Q$$, the quaternions, where $$Q = \left\{{\pm 1,\pm i, \pm j, \pm k}\right\}$$, and there are relations (add picture here).

## Some Nice Facts

• If and $$\phi: G\to G'$$, then
• $$N {~\trianglelefteq~}G \implies N {~\trianglelefteq~}\phi(G)$$, although it is not necessarily normal in $$G$$.
• $$N' {~\trianglelefteq~}G' \implies \phi^{-1}(N') {~\trianglelefteq~}G$$

Definition: A maximal normal subgroup is a normal subgroup $$M {~\trianglelefteq~}G$$ that is properly contained in $$G$$, and if $$M \leq N {~\trianglelefteq~}G$$ (where $$N$$ is proper) then $$M = N$$.

Theorem: $$M$$ is a maximal normal subgroup of $$G$$ iff $$G/M$$ is simple.

## Simple Groups

Definition: A group $$G$$ is simple iff $$N {~\trianglelefteq~}G \implies N = \left\{{e}\right\}, G$$.

Note that if an abelian group has any subgroups, then it is not simple, so $$G = {\mathbb{Z}}_p$$ is the only simple abelian group. Another example of a simple group is $$A_n$$ for $$n\geq 5$$.

Theorem (Feit-Thompson, 1964): Every finite nonabelian simple group has even order.

Note that this is a consequence of the “odd order theorem”.

## Series of Groups

A composition series is a descending series of pairwise normal subgroups such that each successive quotient is simple:

\begin{align*} G_0 {~\trianglelefteq~}G_1 {~\trianglelefteq~}G_2 \cdots {~\trianglelefteq~}\left\{{e}\right\} \\ G_i/G_{i+1}~\text{ simple} .\end{align*}

Example:

\begin{align*} {\mathbb{Z}}_9 {~\trianglelefteq~}{\mathbb{Z}}_3 {~\trianglelefteq~}\left\{{e}\right\} \\ {\mathbb{Z}}_9 / {\mathbb{Z}}_3 = {\mathbb{Z}}_3,\\ {\mathbb{Z}}_3 / \left\{{e}\right\} = {\mathbb{Z}}_3 .\end{align*}

Example:

\begin{align*} {\mathbb{Z}}_6 {~\trianglelefteq~}{\mathbb{Z}}_3 {~\trianglelefteq~}\left\{{e}\right\} \\ {\mathbb{Z}}_6 / {\mathbb{Z}}_3 = {\mathbb{Z}}_2 \\ {\mathbb{Z}}_2 / \left\{{e}\right\} = {\mathbb{Z}}_2 .\end{align*}

but also

\begin{align*} {\mathbb{Z}}_6 {~\trianglelefteq~}{\mathbb{Z}}_2 {~\trianglelefteq~}\left\{{e}\right\} \\ {\mathbb{Z}}_6 / {\mathbb{Z}}_2 = {\mathbb{Z}}_3 \\ {\mathbb{Z}}_3 / \left\{{e}\right\} = {\mathbb{Z}}_3 .\end{align*}

Theorem (Jordan-Holder): Any two composition series are “isomorphic” in the sense that the same quotients appear in both series, up to a permutation.

Definition: A group is solvable iff it has a composition series where all factors are abelian.

Exercise: Show that any abelian group is solvable.

Example: $$S_n$$ is not solvable for $$n\geq 5$$, since

\begin{align*} S_n &{~\trianglelefteq~}A_n {~\trianglelefteq~}\left\{{e}\right\} \\ S_n / A_n &= {\mathbb{Z}}_2~\text{simple} \\ A_n / \left\{{e}\right\} &= A_n~\text{simple} \iff n\geq 5 .\end{align*}

Example:

\begin{align*} S_4 &{~\trianglelefteq~}A_4 {~\trianglelefteq~}G {~\trianglelefteq~}\left\{{e}\right\} \quad \text{where } {\left\lvert {H} \right\rvert} = 4 \\ S_4 / A_4 &= {\mathbb{Z}}_2 \\ A_4 / H &= {\mathbb{Z}}_3 \\ H / \left\{{e}\right\} &= \left\{{a, b}\right\}? .\end{align*}

# August 30th

Recall the Sylow theorems:

• $$p$$ groups exist for every $$p^i$$ dividing $${\left\lvert {G} \right\rvert}$$, and $$H(p) {~\trianglelefteq~}H(p^2) {~\trianglelefteq~}\cdots H(p^n)$$.

• All Sylow $$p{\hbox{-}}$$subgroups are conjugate.

• Numerical constraints

• $$r_p \cong 1 \mod p$$,

• $$r_p {~\Bigm|~}{\left\lvert {G} \right\rvert}$$ and $$r_p {~\Bigm|~}m$$,

## Internal Direct Products

Suppose $$H, K \leq G$$, and consider the smallest subgroup containing both $$H$$ and $$K$$. Denote this $$H \vee K$$.

If either $$H$$ or $$K$$ is normal in $$G$$, then we have $$H\vee K = HK$$.

There is a “recipe” for proving you have a direct product of groups:

Theorem (Recognizing Direct Products): Let $$G$$ be a group, $$H {~\trianglelefteq~}G$$ and $$K{~\trianglelefteq~}G$$, and

1. $$H\vee K = HK = G$$,

2. $$H \cap K = \left\{{e}\right\}$$.

Then $$G \cong H \times K$$.

Proof: We first want to show that $$hk = kh ~\forall k\in K, h\in H$$. We then have \begin{align*} hkh^{-1}k^{-1}= (hkh^{-1})k^{-1}\in K = h(kh^{-1}k^{-1}) \in H \implies hkh^{-1}k^{-1}\in H\cap K = \left\{{e}\right\}. \end{align*}

So define

\begin{align*} \phi: H \times K \to G \\ (h, k) \mapsto hk ,\end{align*}

Exercise: check that this is a homomorphism, it is surjective, and injective.

$$\hfill\blacksquare$$

Applications:

Theorem: Every group of order $$p^2$$ is abelian.

Proof: If $$G$$ is cyclic, then it is abelian and $$G \cong {\mathbb{Z}}_{p^2}$$. So suppose otherwise. By Cauchy, there is an element of order $$p$$ in $$G$$. So let $$H = \left\langle{a}\right\rangle$$, for which we have $${\left\lvert {H} \right\rvert} = p$$.

Then $$H {~\trianglelefteq~}G$$ by Sylow 1, since it’s normal in $$H(p^2)$$, which would have to equal $$G$$.

Now consider $$b\not\in H$$. By Lagrange, we must have $$o(b) = 1, p$$, and since $$e\in H$$, we must have $$o(b) = p$$. This uses fact that $$G$$ is not cyclic.

Now let $$K = \left\langle{b}\right\rangle$$. Then $${\left\lvert {K} \right\rvert} = p$$, and $$K {~\trianglelefteq~}G$$ by the same argument.

$$\hfill\blacksquare$$

Theorem: Let $${\left\lvert {G} \right\rvert} = pq$$ where $$q\neq 1 \mod p$$ and $$p < q$$. Then $$G$$ is cyclic (and thus abelian).

Proof: Use Sylow 1. Let $$P$$ be a sylow $$p{\hbox{-}}$$subgroup. We want to show that $$P {~\trianglelefteq~}G$$ to apply our direct product lemma, so it suffices to show $$r_p = 1$$.

We know $$r_p = 1 \mod p$$ and $$r_p {~\Bigm|~}{\left\lvert {G} \right\rvert} = pq$$, and so $$r_p = 1,q$$. It can’t be $$q$$ because $$p < q$$.

Now let $$Q$$ be a sylow $$q{\hbox{-}}$$subgroup. Then $$r_q \cong 1 \mod 1$$ and $$r_q {~\Bigm|~}pq$$, so $$r_q = 1, q$$. But since $$p< q$$, we must have $$r_q = 1$$. So $$Q {~\trianglelefteq~}G$$ as well.

We now have $$P \cap Q = \emptyset$$ (why?) and \begin{align*} {\left\lvert {PQ} \right\rvert} = \frac{{\left\lvert {P} \right\rvert} {\left\lvert {Q} \right\rvert}}{ {\left\lvert {P \cap Q} \right\rvert} } = {\left\lvert {P} \right\rvert} {\left\lvert {Q} \right\rvert} = pq, \end{align*}

and so $$G = PQ$$, and $$G \cong {\mathbb{Z}}_p \times{\mathbb{Z}}_q \cong {\mathbb{Z}}_{pq}$$.

$$\hfill\blacksquare$$

Example: Every group of order $$15 = 5^1 3^1$$ is cyclic.

## Determination of groups of a given order

Order of G Number of Groups List of Distinct Groups
1 1 $$\left\{{e}\right\}$$
2 1 $${\mathbb{Z}}_2$$
3 1 $${\mathbb{Z}}_3$$
4 2 $${\mathbb{Z}}_4, {\mathbb{Z}}_2^2$$
5 1 $${\mathbb{Z}}_5$$
6 2 $${\mathbb{Z}}_6, S_3$$ (*)
7 1 $${\mathbb{Z}}_7$$
8 5 $${\mathbb{Z}}_8,{\mathbb{Z}}_4 \times{\mathbb{Z}}_2, {\mathbb{Z}}_2^3, D_8,Q$$
9 2 $${\mathbb{Z}}_9, {\mathbb{Z}}_3^2$$
10 2 $${\mathbb{Z}}_{10}, D_5$$
11 1 $${\mathbb{Z}}_{11}$$

We still need to justify 6, 8, and 10.

## Free Groups

Define an alphabet $$A = \left\{{a_1, a_2, \cdots a_n}\right\}$$, and let a syllable be of the form $$a_i^m$$ for some $$m$$. A word is any expression of the form $$\prod_{n_i} a_{n_i}^{m_i}$$.

We have two operations,

• Concatenation, i.e. $$(a_1 a_2) \star (a_3^2 a_5) = a_1 a_2 a_3^2 a_5$$.

• Contraction, i.e. $$(a_1 a_2^2) \star (a_2^{-1}a_5) = a_1 a_2^2 a_2^{-1}a_5 = a_1 a_2 a_5$$.

If we’ve contracted a word as much as possible, we say it is reduced.

We let $$F[A]$$ be the set of reduced words and define a binary operation

\begin{align*} f: F[A] \times F[A] &\to F[A] \\ (w_1, w_2) &\mapsto w_1 w_2 \text{ (reduced) } .\end{align*}

Theorem: $$(A, f)$$ is a group.

Proof: Exercise.

$$\hfill\blacksquare$$

Definition: $$F[A]$$ is called the free group generated by $$A$$. A group $$G$$ is called free on a subset $$A \subseteq G$$ iff $$G \cong F[A]$$.

Examples:

1. $$A = \left\{{x}\right\} \implies F[A] = \left\{{x^n {~\mathrel{\Big|}~}n \in {\mathbb{Z}}}\right\} \cong {\mathbb{Z}}$$.

2. $$A = \left\{{x, y}\right\} \implies F[A] = {\mathbb{Z}}\ast {\mathbb{Z}}$$ (not defined yet!).

Note that there are not relations, i.e. $$xyxyxy$$ is reduced. To abelianize, we’d need to introduce the relation $$xy = yx$$.

Properties:

1. If $$G$$ is free on $$A$$ and free on $$B$$ then we must have $${\left\lvert {A} \right\rvert} = {\left\lvert {B} \right\rvert}$$.

2. Any (nontrivial) subgroup of a free group is free.

(See Fraleigh or Hungerford for possible Algebraic proofs!)

Theorem: Let $$G$$ be generated by some (possibly infinite) subset $$A = \left\{{A_i\mathrel{\Big|}i\in I}\right\}$$ and $$G'$$ be generated by some $$A_i' \subseteq A_i$$.

Then

1. There is at most one homomorphism $$a_i \to a_i'$$.

2. If $$G \cong F[A]$$, there is exactly one homomorphism.

Corollary: Every group $$G'$$ is a homomorphic image of a free group.

Proof: Let $$A$$ be the generators of $$G'$$ and $$G = F[A]$$, then define

\begin{align*} \phi: F[A] &\to G' \\ a_i &\mapsto a_i .\end{align*}

This is onto exactly because $$G' = \left\langle{a_i}\right\rangle$$, and using the theorem above we’re done.

$$\hfill\blacksquare$$

## Generators and Relations

Let $$G$$ be a group and $$A \subseteq G$$ be a generating subset so $$G = \left\langle{a \mathrel{\Big|}a\in A}\right\rangle$$. There exists a $$\phi: F[A] \twoheadrightarrow G$$, and by the first isomorphism theorem, we have $$F[A] / \ker \phi \cong G$$.

Let $$R = \ker \phi$$, these provide the relations.

Examples:

Let $$G = {\mathbb{Z}}_3 = \left\langle{[1]_3}\right\rangle$$. Let $$x = [1]_3$$, then define $$\phi: F[\left\{{x}\right\}] \twoheadrightarrow{\mathbb{Z}}_3$$.

Then since $$[1] + [1] + [1] = [0] \mod 3$$, we have $$\ker \phi = \left\langle{x^3}\right\rangle$$.

Let $$G = {\mathbb{Z}}\oplus {\mathbb{Z}}$$, then $$G \cong \left\langle{x,y \mathrel{\Big|}[x,y] = 1}\right\rangle$$.

We’ll use this for groups of order 6 – there will be only one presentation that is nonabelian, and we’ll exhibit such a group.

# September 9th

## Series of Groups

Recall that a simple group has no nontrivial normal subgroups.

Example:

\begin{align*} {\mathbb{Z}}_6 {~\trianglelefteq~}\left\langle{[3]}\right\rangle {~\trianglelefteq~}\left\langle{[0]}\right\rangle \\ {\mathbb{Z}}_6 / \left\langle{[3]}\right\rangle &= {\mathbb{Z}}_3 \\ \left\langle{[3]}\right\rangle / \left\langle{[0]}\right\rangle &= {\mathbb{Z}}_2 .\end{align*}

Definition: A normal series (or an invariant series) of a group $$G$$ is a finite sequence $$H_i \leq G$$ such that $$H_i {~\trianglelefteq~}H_{i+1}$$ and $$H_n = G$$, so we obtain \begin{align*} H_1 {~\trianglelefteq~}H_2 {~\trianglelefteq~}\cdots {~\trianglelefteq~}H_n = G .\end{align*}

Definition: A normal series $$\left\{{K_i}\right\}$$ is a refinement of $$\left\{{H_i}\right\}$$ if $$K_i \leq H_i$$ for each $$i$$.

Definition: We say two normal series of the same group $$G$$ are isomorphic if there is a bijection from \begin{align*} \left\{{H_i / H_{i+1}}\right\} \iff \left\{{K_j / K_{j+1}}\right\} \end{align*}

Theorem (Schreier): Any two normal series of $$G$$ has isomorphic refinements.

Definition: A normal series of $$G$$ is a composition series iff all of the successive quotients $$H_i / H_{i+1}$$ are simple.

Note that every finite group has a composition series, because any group is a maximal normal subgroup of itself.

Theorem (Jordan-Holder): Any two composition series of a group $$G$$ are isomorphic.

Proof: Apply Schreier’s refinement theorem.

$$\hfill\blacksquare$$

Example: Consider $$S_n {~\trianglelefteq~}A_n {~\trianglelefteq~}\left\{{e}\right\}$$. This is a composition series, with quotients $$Z_2, A_n$$, which are both simple.

Definition: A group $$G$$ is solvable iff it has a composition series in which all of the successive quotients are abelian.

Examples:

• Any abelian group is solvable.

• $$S_n$$ is not solvable for $$n\geq 5$$, since $$A_n$$ is not abelian for $$n\geq 5$$.

Recall Feit-Thompson: Any nonabelian simple group is of even order.

Consequence: Every group of odd order is solvable.

## The Commutator Subgroup

Let $$G$$ be a group, and let $$[G, G] \leq G$$ be the subgroup of $$G$$ generated by elements $$aba^{-1}b^{-1}$$, i.e. every element is a product of commutators. So $$[G, G]$$ is called the commutator subgroup.

Theorem: Let $$G$$ be a group, then

1. $$[G,G] \leq G$$

2. $$[G,G]$$ is a normal subgroup

3. $$G/[G, G]$$ is abelian.

4. $$[G,G]$$ is the smallest normal subgroup such that the quotient is abelian,

I.e., $$H {~\trianglelefteq~}G$$ and if $$G/N$$ is abelian $$\implies [G, G] \leq N$$.

Proof of 1:

$$[G, G]$$ is a subgroup:

• Closure is clear from definition as generators.
• The identity is $$e = e e^{-1}e e^{-1}$$.
• So it suffices to show that $$(aba^{-1}b^{-1})^{-1}\in [G, G]$$, but this is given by $$bab^{-1}a^{-1}$$ which is of the correct form.

$$\hfill\blacksquare$$

Proof of 2:

$$[G, G]$$ is normal.

Let $$x_i \in [G, G]$$, then we want to show $$g \prod x_i g^{-1}\in [G, G]$$, but this reduces to just showing $$gx g^{-1}\in [G, G]$$ for a single $$x\in [G, G]$$.

Then,

\begin{align*} g(aba^{-1}b^{-1}) g^{-1}&= (g^{-1}aba^{-1}) e (b^{-1}g) \\ &= (g^{-1}aba^{-1})(gb^{-1}b g^{-1})(b^{-1}g) \\ &= [(g^{-1}a)b (g^{-1}a)^{-1}b^{-1}] [bg^{-1}b^{-1}g] \\ &\in [G, G] .\end{align*}

$$\hfill\blacksquare$$

Proof of 3:

$$G/[G, G]$$ is abelian.

Let $$H = [G, G]$$. We have $$aH bH = (ab) H$$ and $$bH aH = (ba)H$$.

But $$abH = baH$$ because $$(ba)^{-1}(ab) = a^{-1}b^{-1}a b \in [G, G]$$.

$$\hfill\blacksquare$$

Proof of 4:

$$H {~\trianglelefteq~}G$$ and if $$G/N$$ is abelian $$\implies [G, G] \leq N$$.

Suppose $$G/N$$ is abelian. Let $$aba^{-1}b^{-1}\in [G, G]$$.

Then $$abN = baN$$, so $$aba^{-1}b^{-1}\in N$$ and thus $$[G, G] \subseteq N$$.

$$\hfill\blacksquare$$

## Free Abelian Groups

Example: $${\mathbb{Z}}\times{\mathbb{Z}}$$.

Take $$e_1 = (1, 0), e_2 = (0, 1)$$. Then $$(x,y) \in {\mathbb{Z}}^2$$ can be written $$x(1,0) + y(0, 1)$$, so $$\left\{{e_i}\right\}$$ behaves like a basis for a vector space.

Definition: A group $$G$$ is free abelian if there is a subset $$X\subseteq G$$ such that every $$g \in G$$ can be represented as \begin{align*} g = \sum_{i=1}^r n_i x_i,\quad x_i \in X,~n_i \in {\mathbb{Z}}. \end{align*}

Equivalently, $$X$$ generates $$G$$, so $$G = \left\langle{X}\right\rangle$$, and if $$\sum n_i x_i = 0 \implies n_i = 0~~\forall i$$.

If this is the case, we say $$X$$ is a basis for $$G$$.

Examples:

• $${\mathbb{Z}}^n$$ is free abelian

• $${\mathbb{Z}}_n$$ is not free abelian, since $$n [1] = 0$$ and $$n\neq 0$$.

In general, you can replace $${\mathbb{Z}}_n$$ by any finite group and replace $$n$$ with the order of the group.

Theorem: If $$G$$ is free abelian on $$X$$ where $${\left\lvert {X} \right\rvert} = r$$, then $$G \cong {\mathbb{Z}}^r$$.

Theorem: If $$X = \left\{{x_i}\right\}_{i=1}^r$$, then a basis for $${\mathbb{Z}}^r$$ is given by \begin{align*} \left\{{(1, 0, 0, \cdots), (0, 1, 0, \cdots), \cdots, (0, \cdots, 0, 1)}\right\} \mathrel{\vcenter{:}}=\left\{{e_1, e_2, \cdots, e_r}\right\} \end{align*}

Proof: Use the map $$\phi: G \to {\mathbb{Z}}^r$$ where $$x_i \mapsto e_i$$, and check that this is an isomorphism of groups.

Theorem: Let $$G$$ be free abelian with two bases $$X, X'$$, then $${\left\lvert {X} \right\rvert} = {\left\lvert {X} \right\rvert}'$$.

Definition: Let $$G$$ be free abelian, then if $$X$$ is a basis then $${\left\lvert {X} \right\rvert}$$ is called the rank of of $$G$$.

# Thursday September 5th

## Rings

Recall the definition of a ring: A ring $$(R, +, \times)$$ is a set with binary operations such that

1. $$(R, +)$$ is a group,
2. $$(R, \times)$$ is a monoid.

Examples: $$R = {\mathbb{Z}}, {\mathbb{Q}}, {\mathbb{R}}, {\mathbb{C}}$$, or the ring of $$n\times n$$ matrices, or $${\mathbb{Z}}_n$$.

A ring is commutative iff $$ab = ba$$ for every $$a,b\in R$$, and a ring with unity is a ring such that $$\exists 1 \in R$$ such that $$a1 = 1a = a$$.

Exercise: Show that $$1$$ is unique if it exists.

In a ring with unity, an element $$a\in R$$ is a unit iff $$\exists b\in R$$ such that $$ab = ba = 1$$.

Definition: A ring with unity is a division ring $$\iff$$ every nonzero element is a unit.

Definition: A division ring is a field $$\iff$$ it is commutative.

Definition: Suppose that $$a,b \neq 0$$ with $$ab = 0$$. Then $$a,b$$ are said to be zero divisors.

Definition: A commutative ring without zero divisors is an integral domain.

Example: In $${\mathbb{Z}}_n$$, an element $$a$$ is a zero divisor iff $$\gcd(a, n) \neq 1$$.

Fact: In a ring with no zero divisors, we have \begin{align*} ab = ac \text{ and } a\neq 0 \implies b=c .\end{align*}

Theorem: Every field is an integral domain.

Proof: Let $$R$$ be a field. If $$ab=0$$ and $$a\neq 0$$, then $$a^{-1}$$ exists and so $$b=0$$.

$$\hfill\blacksquare$$

Theorem: Any finite integral domain is a field.

Proof:

Idea: Similar to the pigeonhole principle.

Let $$D = \left\{{0, 1, a_1, \cdots, a_n}\right\}$$ be an integral domain. Let $$a_j \neq 0, 1$$ be arbitrary, and consider $$a_j D = \left\{{a_j x \mathrel{\Big|}x\in D\setminus\left\{{0}\right\}}\right\}$$.

Then $$a_j D = D\setminus\left\{{0}\right\}$$ as sets. But \begin{align*} a_j D = \left\{{a_j, a_j a_1, a_j a_2, \cdots, a_j a_n}\right\}. \end{align*}

Since there are no zero divisors, $$0$$ does not occur among these elements, so some $$a_j a_k$$ must be equal to 1.

$$\hfill\blacksquare$$

## Field Extensions

If $$F \leq E$$ are fields, then $$E$$ is a vector space over $$F$$, for which the dimension turns out to be important.

Definition: We can consider \begin{align*} \operatorname{Aut}(E/F) \mathrel{\vcenter{:}}=\left\{{\sigma: E {\circlearrowleft}\mathrel{\Big|}f\in F \implies \sigma(f) = f}\right\}, \end{align*} i.e. the field automorphisms of $$E$$ that fix $$F$$.

Examples of field extensions: $${\mathbb{C}}\to {\mathbb{R}}\to {\mathbb{Q}}$$.

Let $$F(x)$$ be the smallest field containing both $$F$$ and $$x$$. Given this, we can form a diagram

Let $$F[x]$$ the polynomials with coefficients in $$F$$.

Theorem: Let $$F$$ be a field and $$f(x) \in F[x]$$ be a non-constant polynomial. Then there exists an $$F \to E$$ and some $$\alpha \in E$$ such that $$f(\alpha) = 0$$.

Proof: Since $$F[x]$$ is a unique factorization domain, given $$f(x)$$ we can find an irreducible $$p(x)$$ such that $$f(x) = p(x) g(x)$$ for some $$g(x)$$. So consider $$E = F[x] / (p)$$.

Since $$p$$ is irreducible, $$(p)$$ is a prime ideal, but in $$F[x]$$ prime ideals are maximal and so $$E$$ is a field.

Then define \begin{align*} \psi: F &\to E \\ a &\mapsto a + (p) .\end{align*}

Then $$\psi$$ is a homomorphism of rings: supposing $$\psi(\alpha) = 0$$, we must have $$\alpha \in (p)$$. But all such elements are multiples of a polynomial of degree $$d \geq 1$$, and $$\alpha$$ is a scalar, so this can only happen if $$\alpha = 0$$.

Then consider $$\alpha = x + (p)$$; the claim is that $$p(\alpha) = 0$$ and thus $$f(\alpha) = 0$$. We can compute

\begin{align*} p(x + (p)) &= a_0 + a_1(x + (p)) + \cdots + a_n(x + (p))^n \\ &= p(x) + (p) = 0 .\end{align*}

$$\hfill\blacksquare$$

Example: $${\mathbb{R}}[x] / (x^2 + 1)$$ over $${\mathbb{R}}$$ is isomorphic to $${\mathbb{C}}$$ as a field.

## Algebraic and Transcendental Elements

Definition: An element $$\alpha \in E$$ with $$F \to E$$ is algebraic over $$F$$ iff there is a nonzero polynomial in $$f \in F[x]$$ such that $$f(\alpha) = 0$$.

Otherwise, $$\alpha$$ is said to be transcendental.

Examples:

• $$\sqrt 2 \in {\mathbb{R}}\leftarrow{\mathbb{Q}}$$ is algebraic, since it satisfies $$x^2 - 2$$.

• $$\sqrt{-1} \in {\mathbb{C}}\leftarrow{\mathbb{Q}}$$ is algebraic, since it satisfies $$x^2 + 1$$.

• $$\pi, e \in {\mathbb{R}}\leftarrow{\mathbb{Q}}$$ are transcendental

This takes some work to show.

An algebraic number $$\alpha \in {\mathbb{C}}$$ is an element that is algebraic over $${\mathbb{Q}}$$.

Fact: The set of algebraic numbers forms a field.

Definition: Let $$F \leq E$$ be a field extension and $$\alpha \in E$$. Define a map

\begin{align*} \phi_\alpha: F[x] &\to E \\ \phi_\alpha(f) &= f(\alpha) .\end{align*}

This is a homomorphism of rings and referred to as the evaluation homomorphism.

Theorem: Then $$\phi_\alpha$$ is injective iff $$\alpha$$ is transcendental.

Note: otherwise, this map will have a kernel, which will be generated by a single element that is referred to as the minimal polynomial of $$\alpha$$.

## Minimal Polynomials

Theorem: Let $$F\leq E$$ be a field extension and $$\alpha \in E$$ algebraic over $$F$$. Then

1. There exists a polynomial $$p\in F[x]$$ of minimal degree such that $$p(\alpha) = 0$$.

2. $$p$$ is irreducible.

3. $$p$$ is unique up to a constant.

Proof:

Since $$\alpha$$ is algebraic, $$f(\alpha) = 0$$. So write $$f$$ in terms of its irreducible factors, so $$f(x) = \prod p_j(x)$$ with each $$p_j$$ irreducible. Then $$p_i(\alpha) = 0$$ for some $$i$$ because we are in a field and thus don’t have zero divisors.

So there exists at least one $$p_i(x)$$ such that $$p(\alpha) = 0$$, so let $$q$$ be one such polynomial of minimal degree.

Suppose that $$\deg q < \deg p_i$$. Using the Euclidean algorithm, we can write $$p(x) = q(x) c(x) + r(x)$$ for some $$c$$, and some $$r$$ where $$\deg r < \deg q$$.

But then $$0 = p(\alpha) = q(\alpha)c(\alpha) + r(\alpha)$$, but if $$q(\alpha) = 0$$, then $$r(\alpha) = 0$$. So $$r(x)$$ is identically zero, and so $$p(x) - q(x) = c(x) = c$$, a constant.

$$\hfill\blacksquare$$

Definition: Let $$\alpha \in E$$ be algebraic over $$F$$, then the unique monic polynomial $$p\in F[x]$$ of minimal degree such that $$p(\alpha) = 0$$ is the minimal polynomial of $$\alpha$$.

Example: $$\sqrt{1 + \sqrt 2}$$ has minimal polynomial $$x^4 + x^2 - 1$$, which can be found by raising it to the 2nd and 4th power and finding a linear combination that is constant.

# Tuesday September 10th

## Vector Spaces

Definition: Let $${\mathbb{F}}$$ be a field. A vector space is an abelian group $$V$$ with a map $${\mathbb{F}}\times V \to V$$ such that

• $$\alpha(\beta \mathbf{v}) = (\alpha \beta) \mathbf{v}$$

• $$(\alpha + \beta)\mathbf{v} = \alpha \mathbf{v} + \beta \mathbf{v}$$,

• $$\alpha(\mathbf{v} + \mathbf{w}) = \alpha \mathbf{v} + \alpha \mathbf{w}$$

• $$1\mathbf{v} = \mathbf{v}$$

Examples: $${\mathbb{R}}^n, {\mathbb{C}}^n , F[x] = \mathrm{span}(\left\{{1, x, x^2, \cdots}\right\}), L^2({\mathbb{R}})$$

Definition: Let $$V$$ be a vector space over $${\mathbb{F}}$$; then a set $$W \subseteq V$$ spans $$V$$ iff for every $$\mathbf{v}\in V$$, one can write $$\mathbf{v} = \sum \alpha_i \mathbf{w}_i$$ where $$\alpha_i \in {\mathbb{F}},~\mathbf{w}_i \in W$$.

Definition: $$V$$ is finite dimensional if there exists a finite spanning set.

Definition: A set $$W \subseteq V$$ is linearly independent iff \begin{align*} \sum \alpha_i \mathbf{w}_i = \mathbf{0} \implies \alpha_i = 0 \text{ for all } i .\end{align*}

Definition: A basis for $$V$$ is a set $$W \subseteq V$$ such that

1. $$W$$ is linearly independent, and

2. $$W$$ spans $$V$$.

A basis is a midpoint between a spanning set and a linearly independent set.

We can add vectors to a set until it is spanning, and we can throw out vectors until the remaining set is linearly independent. This is encapsulated in the following theorems:

Theorem: If $$W$$ spans $$V$$, then some subset of $$W$$ spans $$V$$.

Theorem: If $$W$$ is a set of linearly independent vectors, then some superset of $$W$$ is a basis for $$V$$.

Fact: Any finite-dimensional vector spaces has a finite basis.

Theorem: If $$W$$ is a linearly independent set and $$B$$ is a basis, then $${\left\lvert {B} \right\rvert} \leq {\left\lvert {W} \right\rvert}$$.

Corollary: Any two bases have the same number of elements.

So we define the dimension of $$V$$ to be the number of elements in any basis, which is a unique number.

## Algebraic Extensions

Definition: $$E \geq F$$ is an algebraic extension iff every $$\alpha \in E$$ is algebraic of $$F$$.

Definition: $$E \geq F$$ is a finite extension iff $$E$$ is finite-dimensional as an $$F{\hbox{-}}$$vector space.

Notation: $$[E: F] = \dim_F E$$, the dimension of $$E$$ as an $$F{\hbox{-}}$$vector space.

Observation: If $$E = F(\alpha)$$ where $$\alpha$$ is algebraic over $$F$$, then $$E$$ is an algebraic extension of $$F$$.

Observation: If $$E\geq F$$ and $$[E: F] = 1$$, then $$E=F$$.

Theorem: If $$E \geq F$$ is a finite extension, then $$E$$ is algebraic over $$F$$.

Proof: Let $$\beta \in E$$. Then the set $$\left\{{1, \beta, \beta^2, \cdots}\right\}$$ is not linearly independent. So $$\sum_{i=0}^n c_i \beta^i = 0$$ for some $$n$$ and some $$c_i$$. But then $$\beta$$ is algebraic. $$\hfill\blacksquare$$

Note that the converse is not true in general. Example: Let $$E = \overline {\mathbb{R}}$$ be the algebraic numbers. Then $$E \geq {\mathbb{Q}}$$ is algebraic, but $$[E : {\mathbb{Q}}] = \infty$$.

Theorem: Let $$K \geq E \geq F$$, then $$[K: F] = [K: E] [E: F]$$.

Proof: Let $$\left\{{\alpha_i}\right\}^m$$ be a basis for $$E/F$$ Let $$\left\{{\beta_i}\right\}^n$$ be a basis for $$K / E$$. Then the RHS is $$mn$$.

Claim: $$\left\{{\alpha_i \beta_j}\right\}^{m, n}$$ is a basis for $$K/ F$$.

Linear independence:

\begin{align*} \sum_{i, j} c_{ij} \alpha _i \beta_j &= 0 \\ \implies \sum_j \sum_i c_{ij} \alpha_i \beta_j &= 0 \\ \implies \sum_i c_{ij} \alpha_i &= 0 \quad \text{since \beta form a basis} \\ \implies \sum c_{ij} &= 0 \quad \text{since \alpha form a basis} .\end{align*}

Exercise: Show this is also a spanning set.

$$\hfill\blacksquare$$

Corollary: Let $$E_r \geq E_{r-1} \geq \cdots \geq E_1 \geq F$$, then \begin{align*} [E_r: F]= [E_r: E_{r-1}][E_{r-1}:E_{r-2}] \cdots [E_2: E_1][E_1 : F] .\end{align*}

Observation: If $$\alpha \in E \geq F$$ and $$\alpha$$ is algebraic over $$F$$ where $$E \geq F(\alpha) \geq F$$, then $$F(\alpha)$$ is algebraic (since $$[F(\alpha): F] < \infty$$) and $$[F(\alpha): F]$$ is the degree of the minimal polynomial of $$\alpha$$ over $$F$$.

Corollary: Let $$E = F(\alpha) \geq F$$ where $$\alpha$$ is algebraic. Then \begin{align*}\beta \in F(\alpha) \implies \deg \min(\beta, F) {~\Bigm|~}\deg \min(\alpha, F) .\end{align*}

Proof: Since $$F(\alpha) \geq F(\beta) \geq F$$, we have $$[F(\alpha): F] = [F(\alpha): F(\beta)][F(\beta): F]$$. But just note that \begin{align*} [F(\alpha): F] &= \deg \min (\alpha, F) \text{ and } \\ [F(\beta): F] &= \deg \min (\beta, F) .\end{align*}

$$\hfill\blacksquare$$

Theorem: Let $$E \geq F$$ be algebraic, then \begin{align*} [E: F] < \infty \iff E = F(\alpha_1, \cdots, \alpha_n) \text{ for some } \alpha_n \in E .\end{align*}

## Algebraic Closures

Definition: Let $$E \geq F$$, and define \begin{align*} \overline{F_E} = \left\{{\alpha \in E \mathrel{\Big|}\alpha \text{ is algebraic over } F}\right\} \end{align*} to be the algebraic closure of $$F$$ in $$E$$.

Example: $${\mathbb{Q}}\hookrightarrow{\mathbb{C}}$$, while $$\overline {\mathbb{Q}}= \Bbb{A}$$ is the field of algebraic numbers, which is a dense subfield of $${\mathbb{C}}$$.

Proposition: $$\overline{F_E}$$ is a always field.

Proof: Let $$\alpha, \beta \in \overline{F_E}$$, so $$[F(\alpha, \beta): F] < \infty$$. Then $$F(\alpha, \beta) \subseteq \overline{F_E}$$ is algebraic over $$F$$ and \begin{align*} \alpha\pm \beta, \quad \alpha\beta,\quad \frac \alpha \beta \quad \in F(\alpha, \beta) .\end{align*}

So $$\overline{F_E}$$ is a subfield of $$E$$ and thus a field.

Definition: A field $$F$$ is algebraically closed iff every non-constant polynomial in $$F[x]$$ is a root in $$F$$. Equivalently, every polynomial in $$F[x]$$ can be factored into linear factors.

If $$F$$ is algebraically closed and $$E\geq F$$ and $$E$$ is algebraic, then $$E=F$$.

### The Fundamental Theorem of Algebra

Theorem (Fundamental Theorem of Algebra): $${\mathbb{C}}$$ is an algebraically closed field.

Proof:

Liouville’s theorem: A bounded entire function $$f: {\mathbb{C}}{\circlearrowleft}$$ is constant.

• Bounded means $$\exists M {~\mathrel{\Big|}~}z\in {\mathbb{C}}\implies {\left\lvert {f(z)} \right\rvert} \leq M$$.

• Entire means analytic everywhere.

Let $$f(z) \in {\mathbb{C}}[z]$$ be a polynomial without a zero which is non-constant.

Then $$\frac 1 {f(z)}: {\mathbb{C}}{\circlearrowleft}$$ is analytic and bounded, and thus constant, and contradiction.

$$\hfill\blacksquare$$

## Geometric Constructions:

Given the tools of a straightedge and compass, what real numbers can be constructed? Let $$\mathcal C$$ be the set of such numbers.

Theorem: $$C$$ is a subfield of $${\mathbb{R}}$$.

# Thursday September 12th

## Geometric Constructions

Definition: A real number $$\alpha$$ is said to be constructible iff $${\left\lvert {\alpha} \right\rvert}$$ is constructible using a ruler and compass. Let $$\mathcal C$$ be the set of constructible numbers.

Note that $$\pm 1$$ is constructible, and thus so is $${\mathbb{Z}}$$.

Theorem: $$\mathcal{C}$$ is a field.

Proof: It suffices to construct $$\alpha \pm \beta,~~ \alpha\beta,~~ \alpha / \beta$$.

Showing $$\pm$$ and inverses: Relatively easy.

Showing closure under products:

Corollary: $${\mathbb{Q}}\leq \mathcal C$$ is a subfield.

Can we get all of $${\mathbb{R}}$$ with $$\mathcal C$$? The operations we have are

1. Intersect 2 lines (gives nothing new)

2. Intersect a line and a circle

3. Intersect 2 circles

Operation (3) reduces to (2) by subtracting two equations of a circle ($$x^2 + y^2 + ax + by + c$$) to get an equation of a line.

Operation (2) reduces to solving quadratic equations.

Theorem: $$\mathcal C$$ contains precisely the real numbers obtained by adjoining finitely many square roots of elements in $${\mathbb{Q}}$$.

Proof: Need to show that $$\alpha \in \mathcal C \implies \sqrt \alpha \in \mathcal C$$.

• Bisect $$PA$$ to get $$B$$.

• Draw a circle centered at $$B$$.

• Let $$Q$$ be intersection of circle with $$y$$ axis and $$O$$ be the origin.

• Note triangles 1 and 2 are similar, so \begin{align*} \frac{OQ}{OA} = \frac{PO}{OQ} \implies (OQ)^2 = (PO)(OA) = 1\alpha .\end{align*}

$$\hfill\blacksquare$$

Corollary: Let $$\gamma \in \mathcal{C}$$ be constructible. Then there exist $$\left\{{\alpha_i}\right\}_{i=1}^n$$ such that \begin{align*} \gamma = \prod_{i=1}^n \alpha_i \quad\text{and}\quad [{\mathbb{Q}}(\alpha_1, \cdots, \alpha_j): {\mathbb{Q}}(\alpha_1, \cdots, \alpha_{j-1})] = 2 ,\end{align*} and $$[{\mathbb{Q}}(\alpha): {\mathbb{Q}}] = 2^d$$ for some $$d$$.

Applications:

Doubling the cube: Given a cube of size 1, can we construct one of size 2? To do this, we’d need $$x^3 = 2$$. But note that $$\min(\sqrt[3]{2}, {\mathbb{Q}}) = x^3 - 2 = f(x)$$ is irreducible over $${\mathbb{Q}}$$. So $$[{\mathbb{Q}}(\sqrt[3]{2}): {\mathbb{Q}}] = 3 \neq 2^d$$ for any $$d$$, so this can not be constructible.

Trisections of angles: We want to construct regular polygons, so we’ll need to construct angles. We can get some by bisecting known angles, but can we get all of them?

Example: Attempt to construct $$20^\circ$$ by trisecting the known angle $$60^\circ$$, which is constructible using a triangle of side lengths $$1,2,\sqrt 3$$.

If $$20^\circ$$ were constructible, $$\cos 20^\circ$$ would be as well. There is an identity \begin{align*} \cos 3\theta = 4\cos^3 \theta - 3\cos \theta .\end{align*}

Letting $$\theta = 20^\circ$$ so $$3\theta = 60^\circ$$, we obtain \begin{align*} \frac 1 2 = 4(\cos 20^\circ)^3 - 3\cos 20^\circ, \end{align*}

so if we let $$x = \cos 20^\circ$$ then $$x$$ satisfies the polynomial $$f(x) = 8x^3 - 6x - 1$$, which is irreducible. But then $$[{\mathbb{Q}}(20^\circ):{\mathbb{Q}}] = 3 \neq 2^d$$, so $$\cos 20^\circ \not\in\mathcal C$$.

## Finite Fields

Definition: The characteristic of $$F$$ is the smallest $$n\geq 0$$ such that $$n1 = 0$$, or $$0$$ if such an $$n$$ does not exist.

Exercise: For a field $$F$$, show that $$\operatorname{char}~F = 0$$ or $$p$$ a prime.

Note that if $$\operatorname{char}~F = 0$$, then $${\mathbb{Z}}\in F$$ since $$1,~ 1+1,~ 1+1+1, \cdots$$ are all in $$F$$. Since inverses must also exist in $$F$$, we must have $${\mathbb{Q}}\in F$$ as well. So $$\operatorname{char}~F = 0 \iff F$$ is infinite.

If $$\operatorname{char}~F = p$$, it follows that $${\mathbb{Z}}_p \subset F$$.

Theorem: \begin{align*} \text{For } E \geq F \text{ where } [E: F] = n \text{ and } F \text{ finite }, \quad {\left\lvert {F} \right\rvert} = q \implies {\left\lvert {E} \right\rvert} = q^n .\end{align*}

Proof: $$E$$ is a vector space over $$F$$. Let $$\left\{{v_i}\right\}^n$$ be a basis. Then $$\alpha \in E \implies \alpha = \sum_{i=1}^n a_i v_i$$ where each $$a_i \in F$$. There are $$q$$ choices for each $$a_i$$, and $$n$$ coefficients, yielding $$q^n$$ distinct elements.

$$\hfill\blacksquare$$

Corollary: Let $$E$$ be a finite field where $$\operatorname{char}~E = p$$. Then $${\left\lvert {E} \right\rvert} = p^n$$ for some $$n$$.

Theorem: Let $${\mathbb{Z}}_p \leq E$$ with $${\left\lvert {E} \right\rvert} = p^n$$. If $$\alpha \in E$$, then $$\alpha$$ satisfies \begin{align*} x^{p^n} - x \in {\mathbb{Z}}_p[x]. \end{align*}

Proof: If $$\alpha = 0$$, we’re done. So suppose $$\alpha \neq 0$$, then $$\alpha \in E^{\times}$$, which is a group of order $$p^n - 1$$. So $$\alpha^{p^n - 1} = 1$$, and thus $$\alpha \alpha^{p^n - 1} = \alpha 1 \implies \alpha^{p^n} = \alpha$$.

$$\hfill\blacksquare$$

Definition: $$\alpha \in F$$ is an $$n$$th root of unity iff $$\alpha^n = 1$$. It is a primitive root of unity of $$n$$ iff $$k\leq n \implies \alpha^k \neq 1$$ (so $$n$$ is the smallest power for which this holds).

Fact: If $$F$$ is a finite field, then $$F^{\times}$$ is a cyclic group.

Corollary: If $$E \geq F$$ with $$[E: F] = n$$, then $$E = F(\alpha)$$ for just a single element $$\alpha$$.

Proof: Choose $$\alpha \in E^{\times}$$ such that $$\left\langle{\alpha }\right\rangle= E^{\times}$$. Then $$E = F(\alpha)$$.

$$\hfill\blacksquare$$

Next time: Showing the existence of a field with $$p^n$$ elements.

For now: derivatives.

Let $$f(x) \in F[x]$$ by a polynomial with a multiple zero $$\alpha \in E$$ for some $$E \geq F$$.

If it has multiplicity $$m \geq 2$$, then note that \begin{align*} f(x) = (x-\alpha)^m g(x) \implies f'(x) m(x-\alpha)^{m-1}g(x) + g'(x)(x-\alpha)^m \implies f'(\alpha) = 0. \end{align*} So \begin{align*} \alpha \text{ a multiple zero of } f \implies f'(\alpha) = 0 .\end{align*}

The converse is also useful.

Application: Let $$f(x) = x^{p^n} - x$$, then $$f'(x) = p^n x^{p^n - 1} - 1 = -1 \neq 0$$, so all of the roots are distinct.

# Tuesday September 17th

## Finite Fields and Roots of Polynomials

Recall from last time:

Let $${\mathbb{F}}$$ be a finite field. Then $${\mathbb{F}}^{\times}= {\mathbb{F}}\setminus\left\{{0}\right\}$$ is cyclic (this requires some proof).

Let $$f \in {\mathbb{F}}[x]$$ with $$f(\alpha) = 0$$. Then $$\alpha$$ is a multiple root if $$f'(\alpha) = 0$$.

Lemma: Let $${\mathbb{F}}$$ be a finite field with characteristic $$p > 0$$. Then \begin{align*} f(x) = x^{p^n} - x \in {\mathbb{F}}[x] \end{align*}

has $$p^n$$ distinct roots.

Proof: \begin{align*} f'(x) = p^n x^{p^n-1}-1 = -1 ,\end{align*} since we are in char $$p$$.

This is identically -1, so $$f'(x) \neq 0$$ for any $$x$$. So there are no multiple roots. Since there are at most $$p^n$$ roots, this gives exactly $$p^n$$ distinct roots.

$$\hfill\blacksquare$$

Theorem: A field with $$p^n$$ elements exists (denoted $$\mathbb{GF}(p^n)$$) for every prime $$p$$ and every $$n > 0$$.

Proof: Consider $${\mathbb{Z}}_p \subseteq K \subseteq \overline{{\mathbb{Z}}}_p$$ where $$K$$ is the set of zeros of $$x^{p^n}-x$$. Then we claim $$K$$ is a field.

Suppose $$\alpha, \beta \in K$$. Then $$(\alpha \pm \beta)^{p^n} = \alpha^{p^n} \pm \beta^{p^n}$$.

We also have \begin{align*} (\alpha\beta)^{p^n} = \alpha^{p^n}\beta^{p^n} - \alpha\beta \text{ and } \alpha^{-p^n} = \alpha^{-1} .\end{align*}

So $$K$$ is a field and $${\left\lvert {K} \right\rvert} = p^n$$.

$$\hfill\blacksquare$$

Corollary: Let $$F$$ be a finite field. If $$n\in{\mathbb{N}}^+$$, then there exists an $$f(x) \in F[x]$$ that is irreducible of degree $$n$$.

Proof: Let $$F$$ be a finite field, so $${\left\lvert {F} \right\rvert} = p^r$$. By the previous lemma, there exists a $$K$$ such that $${\mathbb{Z}}_p \subseteq k \subseteq \overline F$$.

$$K$$ is defined as \begin{align*} K \mathrel{\vcenter{:}}=\left\{{\alpha \in F \mathrel{\Big|}\alpha^{p^n} - \alpha = 0}\right\} .\end{align*}

We also have \begin{align*} F = \left\{{\alpha \in \overline F {~\mathrel{\Big|}~}\alpha^{p^n} - \alpha = 0}\right\}.\end{align*}

Moreover, $$p^{rs} = p^r p^{r(s-1)}$$. So let $$\alpha \in F$$, then $$\alpha^{p^r} - \alpha = 0$$.

Then \begin{align*} \alpha^{p^{rn}} = \alpha^{p^r p^{r(n-1)}} = (\alpha^{p^r})^{p^{r(n-1)}} = \alpha^{p^{r(n-1)}} ,\end{align*}

and we can continue reducing this way to show that this is yields to $$\alpha^{p^r} = \alpha$$.

So $$\alpha \in K$$, and thus $$F \leq K$$. We have $$[K:F] = n$$ by counting elements. Now $$K$$ is simple, because $$K^{\times}$$ is cyclic. Let $$\beta$$ be the generator, then $$K = F(\beta)$$. This the minimal polynomial of $$\beta$$ in $$F$$ has degree $$n$$, so take this to be the desired $$f(x)$$.

$$\hfill\blacksquare$$

## Simple Extensions

Let $$F \leq E$$ and \begin{align*} \phi_\alpha: F[x] &\to E \\ f &\mapsto f(\alpha) .\end{align*}

denote the evaluation map.

Case 1: Suppose $$\alpha$$ is algebraic over $$F$$.

There is a kernel for this map, and since $$F[x]$$ is a PID, this ideal is generated by a single element – namely, the minimal polynomial of $$\alpha$$.

Thus (applying the first isomorphism theorem), we have $$F(\alpha) \supseteq E$$ isomorphic to $$F[x] / \min(\alpha, F)$$. Moreover, $$F(\alpha)$$ is the smallest subfield of $$E$$ containing $$F$$ and $$\alpha$$.

Case 2: Suppose $$\alpha$$ is transcendental over $$F$$.

Then $$\ker \phi_\alpha = 0$$, so $$F[x] \hookrightarrow E$$. Thus $$F[x] \cong F[\alpha]$$.

Definition: $$E \geq F$$ is a simple extension if $$E = F(\alpha)$$ for some $$\alpha \in E$$.

Theorem: Let $$E = F(\alpha)$$ be a simple extension of $$F$$ where $$\alpha$$ is algebraic over $$F$$.

Then every $$\beta \in E$$ can be uniquely expressed as \begin{align*} \beta = \sum_{i=0}^{n-1} c_i \alpha^i \text{ where } n = \deg \min(\alpha, F) .\end{align*}

Proof:

Existence: We have \begin{align*} F(\alpha) = \left\{{\sum_{i=1}^r \beta_i \alpha^i {~\mathrel{\Big|}~}\beta_i \in F}\right\} ,\end{align*} so all elements look like polynomials in $$\alpha$$.

Using the minimal polynomial, we can reduce the degree of any such element by rewriting $$\alpha^n$$ in terms of lower degree terms:

\begin{align*} f(x) = \sum_{i=0}^n a_i x^i, \quad f(\alpha) &= 0 \\ \implies \sum_{i=0}^n a_i \alpha^i &= 0 \\ \implies \alpha^n &= -\sum_{i=0}^{n-1} a_i \alpha^i .\end{align*}

Uniqueness: Suppose $$\sum c_i \alpha^i = \sum^{n-1} d_i \alpha^i$$. Then $$\sum^{n-1} (c_i - d_i) \alpha^i = 0$$. But by minimality of the minimal polynomial, this forces $$c_i - d_i = 0$$ for all $$i$$.

$$\hfill\blacksquare$$

Note: if $$\alpha$$ is algebraic over $$F$$, then $$\left\{{1, \alpha, \cdots \alpha^{n-1}}\right\}$$ is a basis for $$F(\alpha)$$ over $$F$$ where $$n = \deg \min(\alpha, F)$$. Moreover, \begin{align*} [F(\alpha):F] = \dim_F F(\alpha) = \deg\min(\alpha, F) .\end{align*}

Note: adjoining any root of a minimal polynomial will yield isomorphic (usually not identical) fields. These are distinguished as subfields of the algebraic closure of the base field.

Theorem: Let $$F \leq E$$ with $$\alpha \in E$$ algebraic over $$F$$.

If $$\deg\min(\alpha, F) = n$$, then $$F(\alpha)$$ has dimension $$n$$ over $$F$$, and $$\left\{{1, \alpha, \cdots, \alpha^{n-1}}\right\}$$ is a basis for $$F(\alpha)$$ over $$F$$.

Moreover, any $$\beta \in F(\alpha)$$, is also algebraic over $$F$$,and $$\deg\min(\beta, F) {~\Bigm|~}\deg \min(\alpha, F)$$.

Proof of first part: Exercise.

Proof of second part: We want to show that $$\beta$$ is algebraic over $$F$$.

We have \begin{align*} [F(\alpha):F] = [F(\alpha): F(\beta)][F(\beta): F] ,\end{align*} so $$[F(\beta) : F]$$ is less than $$n$$ since this is a finite extension, and the division of degrees falls out immediately.

$$\hfill\blacksquare$$

## Automorphisms and Galois Theory

Let $$F$$ be a field and $$\overline F$$ be its algebraic closure. Consider subfields of the algebraic closure, i.e. $$E$$ such that $$F \leq E \leq \overline F$$. Then $$E \geq F$$ is an algebraic extension.

Definition: $$\alpha, \beta \in E$$ are conjugates iff $$\min(\alpha, F) = \min(\beta, F)$$.

Examples:

• $$\sqrt[3]{3}, \sqrt[3]{3}\zeta, \sqrt[3]{3}\zeta^2$$ are all conjugates, where $$\zeta = e^{2\pi i/3}$$.

• $$\alpha = a+bi \in {\mathbb{C}}$$ has conjugate $$\mkern 1.5mu\overline{\mkern-1.5mu\alpha \mkern-1.5mu}\mkern 1.5mu= a-bi$$, and \begin{align*} \min(\alpha, {\mathbb{R}}) = \min(\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu, {\mathbb{R}}) = x^2 - 2ax + (a^2 + b^2) .\end{align*}

# Thursday September 19th

## Conjugates

Let $$E \geq F$$ be a field extension. Then $$\alpha,\beta \in E$$ are conjugate $$\iff \min(\alpha, F) = \min(\beta, F)$$ in $$F[x]$$.

Example: $$a + bi, a-bi$$ are conjugate in $${\mathbb{C}}/{\mathbb{R}}$$, since they both have minimal polynomial $$x^2 - 2ax + (a^2 + b^2)$$ over $${\mathbb{R}}$$.

Theorem: Let $$F$$ be a field and $$\alpha, \beta \in E \geq F$$ with $$\deg \min (\alpha, F) = \deg \min(\beta, F)$$, i.e. \begin{align*} [F(\alpha): F] = [F(\beta): F] .\end{align*}

Then $$\alpha, \beta$$ are conjugates $$\iff F(\alpha) \cong F(\beta)$$ under the map

\begin{align*} \phi: F(\alpha) &\to F(\beta)\\ \sum_i a_i \alpha^i &\mapsto \sum_i a_i \beta^i .\end{align*}

Proof: Suppose $$\phi$$ is an isomorphism.

Let \begin{align*} f \mathrel{\vcenter{:}}=\min (\alpha, F) = \sum c_i x^i \text{ where } c_i \in F ,\end{align*} so $$f(\alpha) = 0$$.

Then \begin{align*} 0 = f(\alpha) = f(\sum c_i \alpha^i) = \sum c_i \beta^i ,\end{align*}

so $$\beta$$ satisfies $$f$$ as well, and thus $$f = \min(\alpha, F) {~\Bigm|~}\min(\beta, F)$$.

But we can repeat this argument with $$f^{-1}$$ and $$g(x) \mathrel{\vcenter{:}}=\min(\beta, F)$$, and so we get an equality. Thus $$\alpha, \beta$$ are conjugates.

Conversely, suppose $$\alpha, \beta$$ are conjugates so that $$f = g$$. Check that $$\phi$$ is a homomorphism of fields, so that \begin{align*} \phi(x + y) = \phi(x) + \phi(y) \text{ and } \phi(xy) = \phi(x) \phi(y) .\end{align*} Then $$\phi$$ is clearly surjective, so it remains to check injectivity.

To see that $$\phi$$ is injective, suppose $$f(z) = 0$$. Then $$\sum a_i \beta^i = 0$$. But by linear independence, this forces $$a_i = 0$$ for all $$i$$, which forces $$z=0$$. $$\hfill\blacksquare$$

Corollary: Let $$\alpha \in \overline F$$ be algebraic over $$F$$.

Then

1. $$\phi: F(\alpha) \hookrightarrow\overline F$$ for which $$\phi(f) = f$$ for all $$f\in F$$ maps $$\alpha$$ to one of its conjugates.

2. If $$\beta \in \overline F$$ is a conjugate of $$\alpha$$, then there exists one isomorphism $$\psi: F(\alpha) \to F(\beta)$$ such that $$\psi(f) = f$$ for all $$f\in F$$.

Corollary: Let $$f \in {\mathbb{R}}[x]$$ and suppose $$f(a+bi) = 0$$. Then $$f(a - bi) = 0$$ as well.

Proof: We know $$i, -i$$ are conjugates since they both have minimal polynomial $$f(x) = x^2 + 1$$. By (2), we have an isomorphism $${\mathbb{R}}[i] \xrightarrow{\psi} {\mathbb{R}}[-i]$$. We have $$\psi(a+bi) = a-bi$$, and $$f(a+bi) = 0$$.

This isomorphism commutes with $$f$$, so we in fact have \begin{align*} 0 = \psi(f(a+bi)) = f(\psi(a-bi)) = f(a-bi) .\end{align*}

$$\hfill\blacksquare$$

## Fixed Fields and Automorphisms

Definition: Let $$F$$ be a field and $$\psi: F {\circlearrowleft}$$ is an automorphism iff $$\psi$$ is an isomorphism.

Definition: Let $$\sigma: E{\circlearrowleft}$$ be an automorphism. Then $$\sigma$$ is said to fix $$a\in E$$ iff $$\sigma(a) = a$$. For any subset $$F \subseteq E$$, $$\sigma$$ fixes $$F$$ iff $$\sigma$$ fixes every element of $$F$$.

Example: Let $$E = {\mathbb{Q}}(\sqrt 2, \sqrt 5) \supseteq {\mathbb{Q}}= F$$.

A basis for $$E/F$$ is given by $$\left\{{1, \sqrt 2, \sqrt 5, \sqrt {10}}\right\}$$. Suppose $$\psi: E{\circlearrowleft}$$ fixes $${\mathbb{Q}}$$. By the previous theorem, we must have $$\psi(\sqrt 2) = \pm \sqrt 2$$ and $$\psi(\sqrt 5) = \pm \sqrt 5$$.

What is fixed by $$\psi$$? Suppose we define $$\psi$$ on generators, $$\psi(\sqrt 2) = -\sqrt 2$$ and $$\psi(\sqrt 5) = \sqrt 5$$.

Then \begin{align*} f(c_0 + c_1 \sqrt 2 + c_2 \sqrt 5 + c_3 \sqrt{10}) = c_0 - c_1\sqrt 2 + c_2 \sqrt 5 - c_3\sqrt{10} .\end{align*}

This forces $$c_1 = 0, c_3 = 0$$, and so $$\psi$$ fixes $$\left\{{ c_0 + c_2 \sqrt 5 }\right\} = {\mathbb{Q}}(\sqrt 5)$$.

Theorem: Let $$I$$ be a set of automorphisms of $$E$$ and define \begin{align*} E_I = \left\{{\alpha \in E {~\mathrel{\Big|}~}\sigma(a) = a ~\forall \sigma \in I}\right\} \end{align*}

Then $$E_I \leq E$$ is a subfield.

Proof: Let $$a,b \in E_i$$. We need to show $$a \pm b, ab, b\neq 0 \implies b^{-1}\in I$$.

We have $$\sigma(a\pm b) = \sigma(a) \pm \sigma(b) = a + b \in I$$ since $$\sigma$$ fixes everything in $$I$$. Moreover \begin{align*} \sigma(ab) = \sigma(a)\sigma(b) = ab \in I \quad \text{ and } \quad \sigma(b^{-1}) = \sigma(b)^{-1}= b^{-1}\in I .\end{align*}

$$\hfill\blacksquare$$

Definition: Given a set $$I$$ of automorphisms of $$F$$, $$E_I$$ is called the fixed field of $$E$$ under $$I$$.

Theorem: Let $$E$$ be a field and $$A = \left\{{\sigma:E {\circlearrowleft}{~\mathrel{\Big|}~}\sigma\text{ is an automorphism }}\right\}$$. Then $$A$$ is a group under function composition.

Theorem: Let $$E/F$$ be a field extension, and define \begin{align*} G(E/F) = \left\{{\sigma:E{\circlearrowleft}{~\mathrel{\Big|}~}f\in F \implies \sigma(f) = f}\right\} .\end{align*} Then $$G(E/F) \leq A$$ is a subgroup which contains $$F$$.

Proof: This contains the identity function.

Now if $$\sigma(f) = f$$ then $$f = \sigma^{-1}(f)$$, and \begin{align*} \sigma, \tau \in G(E/F) \implies (\sigma \circ \tau)(f) = \sigma(\tau(f)) = \sigma(f) = f .\end{align*}

$$\hfill\blacksquare$$

Note $$G(E/F)$$ is called the group of automorphisms of $$E$$ fixing $$F$$, i.e. the Galois Group.

Theorem (Isomorphism Extension): Suppose $$F \leq E \leq \overline F$$, so $$E$$ is an algebraic extension of $$F$$.

Suppose similarly that we have $$F' \leq E' \leq \overline F'$$, where we want to find $$E'$$.

Then any $$\sigma: F \to F'$$ that is an isomorphism can be lifted to some $$\tau: E \to E'$$, where $$\tau(f) = \sigma(f)$$ for all $$f\in F$$.

# Tuesday October 1st

## Isomorphism Extension Theorem

Suppose we have $$F\leq E \leq \overline F$$ and $$F' \leq E' \leq \overline{F}'$$. Supposing also that we have an isomorphism $$\sigma: F \to F'$$, we want to extend this to an isomorphism from $$E$$ to some subfield of $$\overline{F}'$$ over $$F'$$.

Theorem: Let $$E$$ be an algebraic extension of $$F$$ and $$\sigma: F \to F'$$ be an isomorphism of fields. Let $$\overline{F}'$$ be the algebraic closure of $$F'$$.

Then there exists a $$\tau: E \to E'$$ where $$E' \leq F'$$ such that $$\tau(f) = \sigma(f)$$ for all $$f \in F$$.

Proof: See Fraleigh. Uses Zorn’s lemma.

$$\hfill\blacksquare$$

Corollary: Let $$F$$ be a field and $$\overline F, \overline F'$$ be algebraic closures of $$F$$. Then $$\overline F \cong \overline F'$$.

Proof: Take the identity $$F \to F$$ and lift it to some $$\tau: \overline F \to E = \tau(\overline F)$$ inside $$\overline F '$$.

Then $$\tau(\overline F)$$ is algebraically closed, and $$\overline F' \geq \tau(\overline F)$$ is an algebraic extension. But then $$\overline F' = \tau(\overline F)$$. $$\hfill\blacksquare$$

Corollary: Let $$E \geq F$$ be an algebraic extension with $$\alpha, \beta \in E$$ conjugates. Then the conjugation isomorphism that sends $$\alpha \to \beta$$ can be extended to $$E$$.

Proof:

Note: Any isomorphism needs to send algebraic elements to algebraic elements, and even more strictly, conjugates to conjugates.

Counting the number of isomorphisms:

Let $$E \geq F$$ be a finite extension. We want to count the number of isomorphisms from $$E$$ to a subfield of $$\overline F$$ that leave $$F$$ fixed.

I.e., how many ways can we fill in the following diagram?

Let $$G(E/F) \mathrel{\vcenter{:}}=\operatorname{Gal}(E/F)$$; this will be a finite group if $$[E: F] < \infty$$.

Theorem: Let $$E \geq F$$ with $$[E: F] < \infty$$ and $$\sigma: F \to F'$$ be an isomorphism.

Then the number of isomorphisms $$\tau: E \to E'$$ extending $$\sigma$$ is finite.

Proof: Since $$[E: F]$$ is finite, we have $$F_0 \mathrel{\vcenter{:}}= F(\alpha_1, \alpha_2, \cdots, \alpha_t)$$ for some $$t\in {\mathbb{N}}$$. Let $$\tau: F_0 \to E'$$ be an isomorphism extending $$\sigma$$.

Then $$\tau(\alpha_i)$$ must be a conjugate of $$\alpha_i$$, of which there are only finitely many since $$\deg \min(\alpha_j, F)$$ is finite. So there are at most $$\prod_i \deg\min(\alpha_i, F)$$ isomorphisms.

Example: $$f(x) = x^3 - 2$$, which has roots $$\sqrt[3] 2, \sqrt[3] 2 \zeta, \sqrt[3] \zeta^2$$.

• Separability (multiple roots)

• Splitting Fields (containing all roots)

Definition: Let \begin{align*} \left\{{E: F}\right\} \mathrel{\vcenter{:}}= {\left\lvert { \left\{{\sigma: E \to E' {~\mathrel{\Big|}~}\sigma \text{ is an isomorphism extending } \text{id}: F \to F}\right\}} \right\rvert} ,\end{align*}

and define this to be the index.

Theorem: Suppose $$F \leq E \leq K$$, then \begin{align*} \left\{{K: F}\right\} = \left\{{K: E}\right\} \left\{{E: F}\right\}. \end{align*}

Proof: Exercise.

$$\hfill\blacksquare$$

Example: $${\mathbb{Q}}(\sqrt 2, \sqrt 5)/{\mathbb{Q}}$$, which is an extension of degree 4. It also turns out that \begin{align*} \left\{{{\mathbb{Q}}(\sqrt 2, \sqrt 5) : {\mathbb{Q}}}\right\} = 4. \end{align*}

Questions:

1. When does $$[E: F] = \left\{{E: F}\right\}$$? (This is always true in characteristic zero.)

2. When is $$\left\{{E: F}\right\} = {\left\lvert {\operatorname{Gal}(E/F)} \right\rvert}$$?

Note that in this example, $$\sqrt 5 \mapsto \pm \sqrt 5$$ and likewise for $$\sqrt 2$$, so any isomorphism extending the identity must in fact be an automorphism.

We have automorphisms \begin{align*} \sigma_1: (\sqrt 2, \sqrt 5) &\mapsto (-\sqrt 2,\sqrt 5) \\ \sigma_2: (\sqrt 2, \sqrt 5) &\mapsto (\sqrt 2, -\sqrt 5) ,\end{align*}

as well as $$\text{id}$$ and $$\sigma_1 \circ \sigma_2$$. Thus $$\operatorname{Gal}(E/F) \cong {\mathbb{Z}}_2^2$$.

## Separable Extensions

Goal: When is $$\left\{{E: F}\right\} = [E: F]$$? We’ll first see what happens for simple extensions.

Definition: Let $$f \in F[x]$$ and $$\alpha$$ be a zero of $$f$$ in $$\overline F$$.

The maximum $$\nu$$ such that $$(x-\alpha)^\nu {~\Bigm|~}f$$ is called the multiplicity of $$f$$.

Theorem: Let $$f$$ be irreducible.

Then all zeros of $$f$$ in $$\overline F$$ have the same multiplicity.

Proof: Let $$\alpha, \beta$$ satisfy $$f$$, where $$f$$ is irreducible. Then consider the following lift:

This induces a map \begin{align*} F(\alpha)[x] &\xrightarrow{\tau} F(\beta)[x] \\ \sum c_i x^i &\mapsto \sum \psi(c_i) x^i ,\end{align*}

so $$x\mapsto x$$ and $$\alpha \mapsto \beta$$, so $$x\mapsto x$$ and $$\alpha \mapsto \beta$$.

Then $$\tau(f(x)) = f(x)$$ and \begin{align*} \tau((x-\alpha)^\nu) = (x-\beta)^\nu .\end{align*}

So write $$f(x) = (x-\alpha)^\nu h(x)$$, then \begin{align*} \tau(f(x)) = \tau((x-\alpha)^\nu) \tau(h(x)) .\end{align*}

Since $$\tau(f(x)) = f(x)$$, we then have \begin{align*} f(x) = (x-\beta)^\nu \tau(h(x)) .\end{align*}

So we get $$\mathrm{mult}(\alpha) \leq \mathrm{mult}(\beta)$$. But repeating the argument with $$\alpha, \beta$$ switched yields the reverse inequality, so they are equal.

$$\hfill\blacksquare$$

Observation: If $$F(\alpha) \to E'$$ extends the identity on $$F$$, then $$E' = F(\beta)$$ where $$\beta$$ is a root of $$f \mathrel{\vcenter{:}}=\min(\alpha, F)$$. Thus we have \begin{align*} \left\{{F(\alpha): F}\right\} = {\left\lvert {\left\{{\text{distinct roots of } f}\right\}} \right\rvert} .\end{align*}

Moreover, \begin{align*} [F(\alpha): F] = \left\{{F(\alpha) : F}\right\} \nu \end{align*} where $$\nu$$ is the multiplicity of a root of $$\min(\alpha, F)$$.

Theorem: Let $$E \geq F$$, then $$\left\{{E: F}\right\} {~\Bigm|~}[E: F]$$.

# Thursday October 3rd

When can we guarantee that there is a $$\tau: E{\circlearrowleft}$$ lifting the identity?

If $$E$$ is separable, then we have $${\left\lvert { \operatorname{Gal}(E/F) } \right\rvert} = \left\{{E: F}\right\} [E: F]$$.

Fact: $$\{F(\alpha): F \}$$ is equal to number of distinct zeros of $$\mathrm{min}(\alpha, F)$$.

If $$F$$ is algebraic, then $$[F(\alpha): F]$$ is the degree of the extension, and $$\left\{{F(\alpha): F}\right\} {~\Bigm|~}[F(\alpha): F]$$.

Theorem: Let $$E \geq F$$ be finite, then $$\left\{{E: F}\right\} {~\Bigm|~}[E:F]$$.

Proof: If $$E \geq F$$ is finite, $$E = F(\alpha_1, \cdots, \alpha_n)$$.

So $$\mathrm{min}(\alpha_i, F)$$ has $$a_j$$ as a root, so let $$n_j$$ be the number of distinct roots, and $$v_j$$ the respective multiplicities.

Then \begin{align*} [F: F(\alpha_1, \cdots, \alpha_{n-1})] = n_j v_j = v_j \left\{{F: F(\alpha_1, \cdots, \alpha_{n-1})}\right\} .\end{align*}

So $$[E: F] = \prod_j n_j v_j$$ and $$\left\{{E:F}\right\} = \prod_j n_j$$, and we obtain divisibility.

$$\hfill\blacksquare$$

Definitions:

1. An extension $$E \geq F$$ is separable iff $$[E:F] = \{E: F\}$$

2. An element $$\alpha \in E$$ is separable iff $$F(\alpha) \geq F$$ is a separable extension.

3. A polynomial $$f(x) \in F[x]$$ is separable iff $$f(\alpha) = 0 \implies \alpha$$ is separable over $$F$$.

Lemma:

1. $$\alpha$$ is separable over $$F$$ iff $$\min(\alpha, F)$$ has zeros of multiplicity one.

2. Any irreducible polynomial $$f(x) \in F[x]$$ is separable iff $$f(x)$$ has zeros of multiplicity one.

Proof of (1): Note that $$[F(\alpha): F] = \deg \min(\alpha, F)$$, and $$\left\{{F(\alpha): F}\right\}$$ is the number of distinct zeros of $$\min(\alpha, F)$$.

Since all zeros have multiplicity 1, we have $$[F(\alpha): F] = \left\{{F(\alpha): F}\right\}$$.

$$\hfill\blacksquare$$

Proof of (2): If $$f(x) \in F[x]$$ is irreducible and $$\alpha\in \overline F$$ a root, then $$\min(\alpha, F) {~\Bigm|~}f(\alpha)$$.

But then $$f(x) = \ell \min(\alpha, F)$$ for some constant $$\ell \in F$$, since $$\min(\alpha, F)$$ was monic and only had zeros of multiplicity one.

$$\hfill\blacksquare$$

Theorem: If $$K \geq E \geq F$$ and $$[K:F] < \infty$$, then $$K$$ is separable over $$F$$ iff $$K$$ is separable over $$E$$ and $$E$$ is separable over $$F$$.

Proof:

\begin{align*} [K: F] &= [K:E] [E: F] \\ &= \{K:E\} \{E: F\} \\ &= \{K: F\} .\end{align*}

Corollary: Let $$E \geq F$$ be a finite extension. Then \begin{align*} E \text{ is separable over } F \iff \text{ Every } \alpha \in E \text{ is separable over } F .\end{align*}

Proof:

$$\implies$$: Suppose $$E \geq F$$ is separable.

Then $$E \geq F(\alpha) \geq F$$ implies that $$F(\alpha)$$ is separable over $$F$$ and thus $$\alpha$$ is separable.

$$\impliedby$$: Suppose every $$\alpha \in E$$ is separable over $$F$$.

Since $$E = F(\alpha_1, \cdots, \alpha_n)$$, build a tower of extensions over $$F$$. For the first step, consider $$F(\alpha_1, \alpha_2) \to F(\alpha_1) \to F$$.

We know $$F(\alpha_1)$$ is separable over $$F$$. To see that $$F(\alpha_1, \alpha_2)$$ is separable over $$F(\alpha_1)$$, consider $$\alpha_2$$.

$$\alpha_2$$ is separable over $$F \iff \min(\alpha_2, F)$$ has roots of multiplicity one.

Then $$\min(\alpha_2, F(\alpha_1)) {~\Bigm|~}\min(\alpha_2, F)$$, so $$\min(\alpha_2, F(\alpha))$$ has roots of multiplicity one.

Thus $$F(\alpha_1, \alpha_2)$$ is separable over $$F(\alpha_1)$$.

$$\hfill\blacksquare$$

## Perfect Fields

Lemma: $$f(x) \in F[x]$$ has a multiple root $$\iff f(x), f'(x)$$ have a nontrivial (multiple) common factor.

Proof:

$$\implies$$: Let $$K\geq F$$ be an extension field of $$F$$.

Suppose $$f(x), g(x)$$ have a common factor in $$K[x]$$; then $$f,g$$ also have a common factor in $$F[x]$$.

If $$f, g$$ do not have a common factor in $$F[x]$$, then $$\gcd(f, g) = 1$$ in $$F[x]$$, and we can find $$p(x), q(x) \in F[x]$$ such that $$f(x)p(x) + g(x)q(x) = 1$$.

But this equation holds in $$K[x]$$ as well, so $$\gcd(f, g) = 1$$ in $$K[x]$$.

We can therefore assume that the roots of $$f$$ lie in $$F$$. Let $$\alpha\in F$$ be a root of $$f$$. Then

\begin{align*} f(x) &= (x-\alpha)^m g(x) \\ f'(x) &= m(x-\alpha)^{m-1} g(x) + (x-\alpha)^m g'(x) .\end{align*}

If $$\alpha$$ is a multiple root, $$m > 2$$, and thus $$(x-\alpha) {~\Bigm|~}f'$$.

$$\impliedby$$: Suppose $$f$$ does not have a multiple root.

We can assume all of the roots are in $$F$$, so we can split $$f$$ into linear factors.

So

\begin{align*} f(x) = \prod_{i=1}^n (x-\alpha_i) \\ f'(x) = \sum_{i=1}^n \prod_{j\neq i} (x-\alpha_j) .\end{align*}

But then $$f'(\alpha_k) = \prod{j\neq k} (x - \alpha_j) \neq 0$$. Thus $$f, f'$$ can not have a common root.

$$\hfill\blacksquare$$

Moral: we can thus test separability by taking derivatives.

Definition: A field $$F$$ is perfect if every finite extension of $$F$$ is separable.

Theorem: Every field of characteristic zero is perfect.

Proof: Let $$F$$ be a field with $$\mathrm{char}(F) = 0$$, and let $$E \geq F$$ be a finite extension.

Let $$\alpha \in E$$, we want to show that $$\alpha$$ is separable. Consider $$f = \min(\alpha, F)$$. We know that $$f$$ is irreducible over $$F$$, and so its only factors are $$1, f$$. If $$f$$ has a multiple root, then $$f, f'$$ have a common factor in $$F[x]$$. By irreducibility, $$f {~\Bigm|~}f'$$, but $$\deg f' < \deg f$$, which implies that $$f'(x) = 0$$. But this forces $$f(x) = c$$ for some constant $$c\in F$$, which means $$f$$ has no roots – a contradiction.

So $$\alpha$$ separable for all $$\alpha \in E$$, so $$E$$ is separable over $$F$$, and $$F$$ is thus perfect.

$$\hfill\blacksquare$$

Theorem: Every finite field is perfect.

Proof: Let $$F$$ be a finite field with $$\mathrm{char} F = p > 0$$ and let $$E \geq F$$ be finite. Then $$E = F(\alpha)$$ for some $$\alpha\in E$$, since $$E$$ is a simple extension (look at $$E^*$$?) So $$E$$ is separable over $$F$$ iff $$\min(\alpha, F)$$ has distinct roots.

So $$E^{\times}= E\setminus\left\{{0}\right\}$$, and so $${\left\lvert {E} \right\rvert} = p^n \implies {\left\lvert {E} \right\rvert} = p^{n-1}$$. Thus all elements of $$E$$ satisfy \begin{align*} f(x) \mathrel{\vcenter{:}}= x^{p^n} - x \in {\mathbb{Z}}_p[x] .\end{align*}

So $$\min(\alpha, F) {~\Bigm|~}f(x)$$. One way to see this is that every element of $$E$$ satisfies $$f$$, since there are exactly $$p^n$$ distinct roots.

Another way is to note that \begin{align*} f'(x) = p^nx^{p^n - 1} - 1 = -1 \neq 0 .\end{align*}

Since $$f(x)$$ has no multiple roots, $$\min(\alpha, F)$$ can not have multiple roots either.

$$\hfill\blacksquare$$

Note that $$[E: F] < \infty \implies F(\alpha_1, \cdots, \alpha_n)$$ for some $$\alpha_i \in E$$ that are algebraic over $$F$$.

## Primitive Elements

Theorem (Primitive Element): Let $$E\geq F$$ be a finite extension and separable.

Then there exists an $$\alpha \in E$$ such that $$E = F(\alpha)$$.

Proof: See textbook.

Corollary: Every finite extension of a field of characteristic zero is simple.

# Tuesday October 8th

## Splitting Fields

For $$\overline F \geq E \geq F$$, we can use the lifting theorem to get a $$\tau: E \to E'$$. What conditions guarantee that $$E = E'$$?

If $$E = F(\alpha)$$, then $$E' = F(\beta)$$ for some $$\beta$$ a conjugate of $$\alpha$$. Thus we need $$E$$ to contain conjugates of all of its elements.

Definition: Let $$\left\{{f_i(x) \in F[x] {~\mathrel{\Big|}~}i\in I}\right\}$$ be any collection of polynomials. We way that $$E$$ is a splitting field $$\iff E$$ is the smallest subfield of $$\overline F$$ containing all roots of the $$f_i$$.

Examples:

• $${\mathbb{Q}}(\sqrt 2, \sqrt 3)$$ is a splitting field for $$\left\{{x^-2, x^2 - 5}\right\}$$.

• $${\mathbb{C}}$$ is a splitting field for $$\left\{{x^2 + 1}\right\}$$.

• $${\mathbb{Q}}(\sqrt[3] 2)$$ is not a splitting field for any collection of polynomials.

Theorem: Let $$F \leq E \leq \overline F$$. Then $$E$$ is a splitting field over $$F$$ for some set of polynomials $$\iff$$ every isomorphism of $$E$$ fixing $$F$$ is in fact an automorphism.

Proof:

$$\implies:$$ Let $$E$$ be a splitting field of $$\left\{{f_i(x) {~\mathrel{\Big|}~}f_i(x) \in F[x], i\in I}\right\}$$.

Then $$E = \left\langle{\alpha_j \mathrel{\Big|}j\in J}\right\rangle$$ where $$\alpha_j$$ are the roots of all of the $$f_i$$.

Suppose $$\sigma: E \to E'$$ is an isomorphism fixing $$F$$. Then consider $$\sigma(\alpha_j)$$ for some $$j \in J$$. We have \begin{align*} \min(\alpha, F) = p(x) = a_0 + a_1 x + \cdots a_{n-1}x^{n-1} + a_n x^n ,\end{align*} and so \begin{align*} p(x) = 0,~~ 0\in F \implies 0 = \sigma(p(\alpha_j)) = \sum_i a_i \sigma(\alpha_j)^i .\end{align*} Thus $$\sigma(\alpha_j)$$ is a conjugate, and thus a root of some $$f_i(x)$$.

$$\impliedby:$$ Suppose any isomorphism of $$E$$ leaving $$F$$ fixed is an automorphism.

Let $$g(x)$$ be an irreducible polynomial and $$\alpha \in E$$ a root.

Using the lifting theorem, where $$F(\alpha \leq E$$, we get a map $$\tau: E \to E'$$ lifting the identity and the conjugation homomorphism. But this says that $$E'$$ must contain every conjugate of $$\alpha$$.

Therefore we can take the collection \begin{align*} S = \left\{{g_i(x) \in F[x] {~\mathrel{\Big|}~}g_i \text{ irreducible and has a root in } E}\right\} .\end{align*}

This defines a splitting field for $$\left\{{g_j}\right\}$$, and we’re done.

$$\hfill\blacksquare$$

Examples:

1. $$x^2 + 1 \in {\mathbb{R}}[x]$$ splits in $${\mathbb{C}}$$, i.e. $$x^2 + 1 = (x+i)(x-i)$$.
2. $$x^2 - 2 \in {\mathbb{Q}}[x]$$ splits in $${\mathbb{Q}}(\sqrt 2)$$.

Corollary: Let $$E$$ be a splitting field over $$F$$. Then every irreducible polynomial in $$F[x]$$ with a root $$\alpha \in E$$ splits in $$E[x]$$.

Corollary: The index $$\{ E: F \}$$ (the number of distinct lifts of the identity). If $$E$$ is a splitting field and $$\tau:E \to E'$$ lifts the identity on $$F$$, then $$E = E'$$. Thus $$\{ E : F \}$$ is the number of automorphisms, i.e. $${\left\lvert {\operatorname{Gal}(E/F)} \right\rvert}$$.

Question: When is it the case that \begin{align*} [E: F] = \{E: F\} = {\left\lvert {\operatorname{Gal}(E/F)} \right\rvert}? \end{align*}

• The first equality occurs when $$E$$ is separable.

• The second equality occurs when $$E$$ is a splitting field.

Characteristic zero implies separability

Definition: If $$E$$ satisfies both of these conditions, it is said to be a Galois extension.

Some cases where this holds:

• $$E \geq F$$ a finite algebraic extension with $$E$$ characteristic zero.

• $$E$$ a finite field, since it is a splitting field for $$x^{p^n} - x$$.

Example 1: $${\mathbb{Q}}(\sqrt 2, \sqrt 5)$$ is

1. A degree 4 extension,

2. The number of automorphisms was 4, and

3. The Galois group was $${\mathbb{Z}}_2^2$$, of size 4.

Example 2: $$E$$ the splitting field of $$x^3 - 3$$ over $${\mathbb{Q}}$$.

This polynomial has roots $$\sqrt[3] 3,~ \zeta_3 \sqrt[3] 3,~ \zeta_3^2 \sqrt[3] 3$$ where $$\zeta_3^3 = 1$$.

Then $$E = {\mathbb{Q}}(\sqrt[3] 3, \zeta_3)$$, where \begin{align*} \min(\sqrt[3] 3, {\mathbb{Q}}) &= x^3 - 3 \\ \min(\zeta_3, {\mathbb{Q}}) &= x^2 + x + 1 ,\end{align*}

so this is a degree 6 extension.

Since $$\operatorname{char}~{\mathbb{Q}}= 0$$, we have $$[E: {\mathbb{Q}}] = \{E: {\mathbb{Q}}\}$$ for free.

We know that any automorphism has to map \begin{align*} \sqrt[3] 3 &\mapsto \sqrt[3] 3,~ \sqrt[3] 3 \zeta_3,~ \sqrt[3] 3 \zeta_3^2 \\ \zeta_3 &\mapsto \zeta_3,~ \zeta_3^2 .\end{align*}

You can show this is nonabelian by composing a few of these, thus the Galois group is $$S^3$$.

Example 3 If $$[E: F] = 2$$, then $$E$$ is automatically a splitting field.

Since it’s a finite extension, it’s algebraic, so let $$\alpha \in E\setminus F$$.

Then $$\min(\alpha, F)$$ has degree 2, and thus $$E = F(\alpha)$$ contains all of its roots, making $$E$$ a splitting field.

## The Galois Correspondence

There are three key players here: \begin{align*} [E: F],\quad \{E: F\},\quad \operatorname{Gal}(E/F) .\end{align*}

How are they related?

Definition: Let $$E \geq F$$ be a finite extension. $$E$$ is normal (or Galois) over $$F$$ iff $$E$$ is a separable splitting field over $$F$$.

Examples:

1. $${\mathbb{Q}}(\sqrt 2, \sqrt 3)$$ is normal over $${\mathbb{Q}}$$.

2. $${\mathbb{Q}}(\sqrt[3] 3)$$ is not normal (not a splitting field of any irreducible polynomial in $${\mathbb{Q}}[x]$$).

3. $${\mathbb{Q}}(\sqrt[3] 3, \zeta_3)$$ is normal

Theorem: Let $$F \leq E \leq K \leq \overline F$$, where $$K$$ is a finite normal extension of $$F$$. Then

1. $$K$$ is a normal extension of $$E$$ as well,

2. $$\operatorname{Gal}(K/E) \leq \operatorname{Gal}(K/F)$$.

3. For $$\sigma, \tau \in \operatorname{Gal}(K/F)$$, \begin{align*} \sigma\mathrel{\Big|}_E = \tau\mathrel{\Big|}_E \iff \sigma, \tau \text{ are in the same left coset of }~ \frac{\operatorname{Gal}(K/F)}{\operatorname{Gal}(K/E)} .\end{align*}

Proof of (1): Since $$K$$ is separable over $$F$$, we have $$K$$ separable over $$E$$.

Then $$K$$ is a splitting field for polynomials in $$F[x] \subseteq E[x]$$. Thus $$K$$ is normal over $$E$$.

$$\hfill\blacksquare$$

Proof of (2):

So this follows by definition.

$$\hfill\blacksquare$$

Proof of (3): Let $$\sigma, \tau \in \operatorname{Gal}(K/F)$$ be in the same left coset. Then \begin{align*} \tau^{-1}\sigma \in \operatorname{Gal}(K/E) ,\end{align*} so let $$\mu \mathrel{\vcenter{:}}=\tau^{-1}\sigma$$.

Note that $$\mu$$ fixes $$E$$ by definition.

So $$\sigma = \tau \mu$$, and thus \begin{align*} \sigma(e) = \tau(\mu(e)) = \tau(e) \text{ for all } e\in E .\end{align*}

$$\hfill\blacksquare$$

Note: We don’t know if the intermediate field $$E$$ is actually a normal extension of $$F$$.

Standard example: $$K \geq E \geq F$$ where \begin{align*} K = {\mathbb{Q}}(\sqrt[3] 3, \zeta_3)\quad E = {\mathbb{Q}}(\sqrt[3] 3) \quad F = {\mathbb{Q}} .\end{align*} Then $$K {~\trianglelefteq~}E$$ and $$K{~\trianglelefteq~}F$$, since $$\operatorname{Gal}(K/F) = S_3$$ and $$\operatorname{Gal}(K/E) = {\mathbb{Z}}_2$$. But $$E \ntrianglelefteq F$$, since $${\mathbb{Z}}_2 \ntrianglelefteq S_3$$.

# Thursday October 10th

## Computation of Automorphisms

Setup:

• $$F \leq E \leq K \leq \overline F$$

• $$[K: F] < \infty$$

• $$K$$ is a normal extension of $$F$$

Facts:

• $$\operatorname{Gal}(K/E) = \left\{{\sigma \in \operatorname{Gal}(K/F) {~\mathrel{\Big|}~}\sigma(e) = e ~\forall e\in E}\right\}$$.

• $$\sigma, \tau \in \operatorname{Gal}(K/F)$$ and $${\left.{\sigma}\right|_{E}} = {\left.{\tau}\right|_{E}} \iff \sigma, \tau$$ are in the same left coset of $$\operatorname{Gal}(K/F) / \operatorname{Gal}(K/E)$$.

Example: $$K = {\mathbb{Q}}(\sqrt 2, \sqrt 5)$$.

Then $$\operatorname{Gal}(K/{\mathbb{Q}}) \cong {\mathbb{Z}}_2^2$$, given by the following automorphisms:

\begin{align*} \text{id}: \sqrt 2 &\mapsto \sqrt 2, \quad& \sqrt 5 &\mapsto \sqrt 5 \\ \rho_1: \sqrt 2 &\mapsto \sqrt 2, \quad& \sqrt 5 &\mapsto -\sqrt 5 \\ \rho_2: \sqrt 2 &\mapsto -\sqrt 2, \quad& \sqrt 5 &\mapsto \sqrt 5 \\ \rho_1 \circ \rho_2: \sqrt 2 &\mapsto -\sqrt 2, \quad& \sqrt 5 &\mapsto -\sqrt 5 .\end{align*}

We then get the following subgroup/subfield correspondence:

## Fundamental Theorem of Galois Theory

Recall that $$\mathrel{\vcenter{:}}=\operatorname{Gal}(K/E)$$.

Theorem (Fundamental Theorem of Galois Theory):

Let $$\mathcal D$$ be the collection of subgroups of $$\operatorname{Gal}(K/F)$$ and $$\mathcal C$$ be the collection of subfields $$E$$ such that $$F \leq E \leq K$$.

Define a map \begin{align*} \lambda: \mathcal C &\to \mathcal D \\ \lambda(E) &\coloneqq\left\{{\sigma \in \operatorname{Gal}(K/F) \mathrel{\Big|}\sigma(e) = e ~\forall e\in E}\right\} .\end{align*}

Then $$\lambda$$ is a bijective map, and

1. $$\lambda(E) = \operatorname{Gal}(K/E)$$

2. $$E = K_{\lambda(E)}$$

3. If $$H \leq \operatorname{Gal}(K/F)$$ then \begin{align*} \lambda(K_H) = H \end{align*}

4. $$[K: E] = {\left\lvert {\lambda(E)} \right\rvert}$$ and \begin{align*} [E: F] = [\operatorname{Gal}(K/F): \lambda(E)] \end{align*}

5. $$E$$ is normal over $$F \iff \lambda(E) {~\trianglelefteq~}\operatorname{Gal}(K/F)$$, and in this case \begin{align*} \operatorname{Gal}(E/F) \cong \operatorname{Gal}(K/F) / \operatorname{Gal}(K/E) .\end{align*}

6. $$\lambda$$ is order-reversing, i.e. \begin{align*} E_1 \leq E_2 \implies \lambda(E_2) \leq \lambda(E_1) .\end{align*}

Proof of 1: Proved earlier.

$$\hfill\blacksquare$$

Proof of 2: We know that $$E \leq L_{\operatorname{Gal}(K/E)}$$. Let $$\alpha \in K\setminus E$$; we want to show that $$\alpha$$ is not fixed by all automorphisms in $$\operatorname{Gal}(K/E)$$.

We build the following tower:

This uses the isomorphism extension theorem, and the fact that $$K$$ is normal over $$F$$.

If $$\beta\neq \alpha$$, then $$\beta$$ must be a conjugate of $$\alpha$$, so $$\tau'(\alpha) \neq \alpha$$ while $$\tau' \in \operatorname{Gal}(K/E)$$.

$$\hfill\blacksquare$$

Claim: $$\lambda$$ is injective.

Proof: Suppose $$\lambda(E_1) = \lambda(E_2)$$. Then by (2), $$E_1 = K_{\lambda(E_1)} = K_{\lambda(E_2)} = E_2$$. $$\hfill\blacksquare$$

Proof of 3: We want to show that if $$H\leq \operatorname{Gal}(K/F)$$ then $$\lambda(K_H) = H$$.

We know $$H \leq \lambda(K_H) = \operatorname{Gal}(K/K_H) \leq \operatorname{Gal}(K/F)$$, so suppose $$H \lneq \lambda(K_H)$$.

Since $$K$$ is a finite, separable extension, $$K = K_H(\alpha)$$ for some $$\alpha \in K$$.

Let \begin{align*} n = [K: K_H] = {K: K_H} = {\left\lvert {\operatorname{Gal}(K/K_H)} \right\rvert} .\end{align*}

Since $$H \lneq \lambda(K_H)$$, we have $${\left\lvert {H} \right\rvert} < n$$. So denote $$H = \left\{{\sigma, \sigma_2, \cdots}\right\}$$ and let define \begin{align*} f(x) = \prod_i (x - \sigma_i(\alpha)) .\end{align*}

We then have

• $$\deg f = {\left\lvert {H} \right\rvert}$$

• The coefficients of $$f$$ are symmetric polynomials in the $$\sigma_i(\alpha)$$ and are fixed under any $$\sigma\in H$$

• $$f(x) \in K_H(\alpha)[x]$$

• $$f(\alpha) = 0$$ since $$\sigma_i(\alpha) = \alpha$$ for every $$i$$.

This is a contradiction, so we must have \begin{align*} [K_H: K] = n = \deg \min(\alpha, K_H) \leq \deg f = {\left\lvert {H} \right\rvert} .\end{align*}

$$\hfill\blacksquare$$

Assuming (3), $$\lambda$$ is surjective, so suppose $$H < \operatorname{Gal}(K/F)$$. Then $$\lambda(K_H) = H \implies \lambda$$ is surjective.

Proof of 4: \begin{align*} {\left\lvert {\lambda(E)} \right\rvert} &= {\left\lvert {\operatorname{Gal}(K/E)} \right\rvert} =_{\text{splitting field}} [K: E] \\ [E: F] &=_{\text{separable}} \{E: F\} =_{\text{previous part}} [\operatorname{Gal}(K/F): \lambda(E)] .\end{align*}

Proof of 5:

We have $$F\leq E \leq K$$ and $$E$$ is separable over $$F$$, so $$E$$ is normal over $$F \iff E$$ is a splitting field over $$F$$.

That is, every extension $$E'/E$$ maps $$K$$ to itself, since $$K$$ is normal.

So $$E$$ is normal over $$F \iff$$ for all $$\sigma \in \operatorname{Gal}(K/F), \sigma(\alpha) \in E$$ for all $$\alpha \in E$$.

By a previous property, $$E = K_{\operatorname{Gal}(K/E)}$$, and so \begin{align*} \sigma(\alpha) \in E &\iff \tau(\sigma(\alpha)) = \sigma(\alpha) &\quad \forall \tau \in \operatorname{Gal}(K/E) \\ &\iff (\sigma^{-1}\tau \sigma) (\alpha) = \alpha S&\quad \forall \tau \in \operatorname{Gal}(K/E) \\ &\iff \sigma^{-1}\tau\sigma \in \operatorname{Gal}(K/E) \\ &\iff \operatorname{Gal}(K/E) {~\trianglelefteq~}\operatorname{Gal}(K/F) .\end{align*}

Now assume $$E$$ is a normal extension of $$F$$, and let \begin{align*} \phi: \operatorname{Gal}(K/F) &\to \operatorname{Gal}(E/F) \\ \sigma &\mapsto {\left.{\sigma}\right|_{E}} .\end{align*}

Then $$\phi$$ is well-defined precisely because $$E$$ is normal over $$F$$, and we can apply the extension theorem:

$$\phi$$ is surjective by the extension theorem, and $$\phi$$ is a homomorphism, so consider $$\ker \phi$$.

Let $$\phi(\sigma) = {\left.{\sigma}\right|_{E}} = \text{id}$$. Then $$\phi$$ fixes elements of $$E \iff \sigma \in \operatorname{Gal}(K/E)$$, and thus $$\ker \phi = \operatorname{Gal}(K/E)$$.

$$\hfill\blacksquare$$

Proof of 6: \begin{align*} E_1 \leq E_2 \iff &\operatorname{Gal}(K/E_2) \leq &\operatorname{Gal}(K/E_1) \\ & \shortparallel &\shortparallel \\ &\lambda(E_2) \leq &\lambda(E_1) .\end{align*}

Example: $$K = {\mathbb{Q}}(\sqrt[3] 2, \zeta_3)$$. Then $$\min(\zeta, {\mathbb{Q}}) = x^2 + x + 1$$ and $$\operatorname{Gal}(K/{\mathbb{Q}}) = S_3$$. There is a subgroup of order 2, $$E = \operatorname{Gal}(K/{\mathbb{Q}}(\sqrt[3] 2)) \leq \operatorname{Gal}(K/{\mathbb{Q}})$$, but $$E$$ doesn’t correspond to a normal extension of $$F$$, so this subgroup is not normal. On the other hand, $$\operatorname{Gal}({\mathbb{Q}}(\zeta_3), {\mathbb{Q}}) {~\trianglelefteq~}\operatorname{Gal}(K/ {\mathbb{Q}})$$.

# Tuesday October 15th

## Cyclotomic Extensions

Definition: Let $$K$$ denote the splitting field of $$x^n-1$$ over $$F$$. Then $$K$$ is called the $$n$$th cyclotomic extension of $$F$$.

If we set $$f(x) = x^n-1$$, then $$f'(x) = nx^{n-1}$$.

So if $$\operatorname{char}~F$$ does not divide $$n$$, then the splitting field is separable. So this splitting field is in fact normal.

Suppose that $$\operatorname{char}~F$$ doesn’t divide $$n$$, then $$f(x)$$ has $$n$$ zeros, and let $$\zeta_1, \zeta_2$$ be two zeros. Then $$(\zeta_1 \zeta_2)^n = \zeta_1^n \zeta_2^n = 1$$, so the product is a zero as well, and the roots of $$f$$ form a subgroup in $$K^{\times}$$.

So let’s specialize to $$F = {\mathbb{Q}}$$.

The roots of $$f$$ are the $$n$$th roots of unity, i.e. $$\zeta_n = e^{2\pi i / n}$$, and are given by $$\left\{{\zeta_n, \zeta_n^2, \zeta_n^3, \cdots, \zeta_n^{n-1}}\right\}$$.

The primitive roots of unity are given by $$\left\{{\zeta_n^m {~\mathrel{\Big|}~}\gcd(m, n) = 1}\right\}$$.

Definition: Let \begin{align*} \Phi_n(x) = \prod_{i=1}^{\varphi(n)} (x-\alpha_i) ,\end{align*} where this product runs over all of the primitive $$n$$th roots of unity.

Let $$G$$ be $$\operatorname{Gal}(K/{\mathbb{Q}})$$. Then any $$\sigma\in G$$ will permute the primitive $$n$$th roots of unity. Moreover, it only permutes primitive roots, so every $$\sigma$$ fixes $$\Phi_n(x)$$. But this means that the coefficients must lie in $${\mathbb{Q}}$$.

Since $$\zeta$$ generates all of the roots of $$\Phi_n$$, we in fact have $$K = {\mathbb{Q}}(\zeta)$$. But what is the group structure of $$G$$?

Since any automorphism is determined by where it sends a generator, we have automorphisms $$\tau_m(\zeta) = \zeta^m$$ for each $$m$$ such that $$\gcd(m, n) = 1$$.

But then $$\tau_{m_1} \circ \tau_{m_2} = \tau_{m_1 + m_2}$$, and so $$G \cong G_m \leq {\mathbb{Z}}_n$$ as a ring, where

\begin{align*} G_m = \left\{{[m] {~\mathrel{\Big|}~}\gcd(m, n) = 1}\right\} \end{align*} and $${\left\lvert {G} \right\rvert} = \varphi(n)$$.

Note that as a set, there are the units $${\mathbb{Z}}_n^{\times}$$.

Theorem: The Galois group of the $$n$$th cyclotomic extension over $${\mathbb{Q}}$$ has $$\varphi(n)$$ elements and is isomorphic to $$G_m$$.

Special case: $$n=p$$ where $$p$$ is a prime.

Then $$\phi(p) = p-1$$, and \begin{align*} \Phi_p(x) = \frac{x^p - 1}{x-1} = x^{p-1} + x^{p-2} + \cdots + x + 1 .\end{align*}

Note that $${\mathbb{Z}}_p^{\times}$$ is in fact cyclic, although this may not always happen. In this case, we have $$\operatorname{Gal}(K/{\mathbb{Q}}) \cong {\mathbb{Z}}_p^{\times}$$.

## Construction of n-gons

To construct the vertices of an n-gon, we will need to construct the angle $$2\pi/n$$, or equivalently, $$\zeta_n$$. Note that if $$[{\mathbb{Q}}(\zeta_n) : {\mathbb{Q}}] \neq 2^\ell$$ for some $$\ell\in{\mathbb{N}}$$, then the $$n{\hbox{-}}$$gon is not constructible.

Example: An 11-gon. Noting that $$[{\mathbb{Q}}(\zeta_{11}) : {\mathbb{Q}}] = 10 \neq 2^\ell$$, the 11-gon is not constructible.

Since this is only a sufficient condition, we’ll refine this.

Definition: A prime of the form $$p = 2^{2^k}+1$$ are called Fermat primes.

Theorem: The regular $$n{\hbox{-}}$$gon is constructible $$\iff$$ all odd primes dividing $$n$$ are Fermat primes $$p$$ where $$p^2$$ does not divide $$n$$.

Example: Consider \begin{align*} \Phi_5(x) = x^4 + x^3 + x^2 + x + 1 .\end{align*}

Then take $$\zeta = \zeta_5$$; we then obtain the roots as $$\left\{{1, \zeta, \zeta^2, \zeta^3, \zeta^4}\right\}$$ and $${\mathbb{Q}}(\zeta)$$ is the splitting field.

Any automorphism is of the form $$\sigma_r: \zeta \mapsto \zeta^r$$ for $$r=1,2,3,4$$. So $${\left\lvert {\operatorname{Gal}(K/{\mathbb{Q}})} \right\rvert} = 4$$, and is cyclic and thus isomorphic to $${\mathbb{Z}}_4$$. Corresponding to $$0 \to {\mathbb{Z}}_2 \to {\mathbb{Z}}_4$$, we have the extensions \begin{align*} {\mathbb{Q}}\to {\mathbb{Q}}(\zeta^2) \to {\mathbb{Q}}(\zeta) .\end{align*}

How can we get a basis for the degree 2 extension $${\mathbb{Q}}(\zeta^2)/{\mathbb{Q}}$$? Let \begin{align*} \lambda(E) = \left\{{\sigma \in \operatorname{Gal}({\mathbb{Q}}(\zeta)/{\mathbb{Q}}) {~\mathrel{\Big|}~}\sigma(e) = e ~\forall e\in E }\right\} ,\end{align*} $$\lambda(K_H) = H$$ where $$H$$ is a subgroup of $$\operatorname{Gal}({\mathbb{Q}}(\zeta)/{\mathbb{Q}})$$, and \begin{align*} K_H = \left\{{x\in K {~\mathrel{\Big|}~}\sigma(x) = x ~\forall \sigma\in H}\right\} .\end{align*}

Note that if $${\mathbb{Z}}_4 = \left\langle{\psi}\right\rangle$$, then $${\mathbb{Z}}_2 \leq {\mathbb{Z}}_4$$ is given by $${\mathbb{Z}}_2 = \left\langle{\psi^2}\right\rangle$$.

We can compute that if $$\psi(\zeta) = \zeta^2$$, then

\begin{align*} \psi^2(\zeta) &= \zeta^{-1}\\ \psi^2(\zeta^2) &= \zeta^{-2}\\ \psi^2(\zeta^3) &= \zeta^{-3} .\end{align*}

Noting that $$\zeta_4$$ is a linear combination of the other $$\zeta$$s, we have a basis $$\left\{{1, \zeta, \zeta^2, \zeta^3}\right\}$$.

Then you can explicitly compute the fixed field by writing out \begin{align*} \sigma(a + b\zeta + c\zeta^2 + d\zeta^3) = a + b\sigma(\zeta) + c\sigma(\zeta^2) + \cdots ,\end{align*} gathering terms, and seeing how this restricts the coefficients.

In this case, it yields $${\mathbb{Q}}(\zeta^2 + \zeta^3)$$.

## The Frobenius Automorphism

Definition: Let $$p$$ be a prime and $$F$$ be a field of characteristic $$p>0$$. Then

\begin{align*} \sigma_p: F &\to F \\ \sigma_p(x) &= x^p \end{align*}

is denoted the Frobenius map.

Theorem: Let $$F$$ be a finite field of characteristic $$p > 0$$. Then

1. $$\phi_p$$ is an automorphism, and
2. $$\phi_p$$ fixes $$F_{\sigma_p} = {\mathbb{Z}}_p$$.

Proof of part 1: Since $$\sigma_p$$ is a field homomorphism, we have \begin{align*} \sigma_p(x+y) = (x+y)^p = x^p + y^p \text{ and } \sigma(xy) = (xy)^p = x^p y^p \end{align*}

Note that $$\sigma_p$$ is injective, since $$\sigma_p(x) =0 \implies x^p=0 \implies x=0$$ since we are in a field. Since $$F$$ is finite, $$\sigma_p$$ is also surjective, and is thus an automorphism.

Proof of part 2: If $$\sigma(x) = x$$, then \begin{align*} x^p = x \implies x^p-x = 0 ,\end{align*} which implies that $$x$$ is a root of $$f(x) = x^p - x$$. But these are exactly the elements in the prime ring $${\mathbb{Z}}_p$$.

$$\hfill\blacksquare$$

# Thursday October 17th

## Example Galois Group Computation

Example: What is the Galois group of $$x^4-2$$ over $${\mathbb{Q}}$$?

First step: find the roots. We can find directly that there are 4 roots given by \begin{align*} \left\{{\pm \sqrt[4] 2, \pm i \sqrt[4] 2}\right\} \mathrel{\vcenter{:}}=\left\{{r_i}\right\} .\end{align*}

The splitting field will then be $${\mathbb{Q}}(\sqrt[4] 2, i)$$, which is separable because we are in characteristic zero. So this is a normal extension.

We can find some automorphisms: \begin{align*} \sqrt[4] 2 \mapsto r_i, \quad i \mapsto \pm i .\end{align*}

So $${\left\lvert {G} \right\rvert} = 8$$, and we can see that $$G$$ can’t be abelian because this would require every subgroup to be abelian and thus normal, which would force every intermediate extension to be normal.

But the intermediate extension $${\mathbb{Q}}(\sqrt[4] 2)/{\mathbb{Q}}$$ is not a normal extension since it’s not a splitting field.

So the group must be $$D_4$$.

$$\hfill\blacksquare$$

## Insolubility of the Quintic

### Symmetric Functions

Let $$F$$ be a field, and let \begin{align*} F(y_1, \cdots , y_n) = \left\{{\frac{f(y_1, \cdots, y_n)}{g(y_1, \cdots, y_n)} {~\mathrel{\Big|}~}f, g \in F[y_1, \cdots, y_n]}\right\} \end{align*} be the set of rational functions over $$F$$.

Then $$S_n \curvearrowright F(y_1, \cdots, y_n)$$ by permuting the $$y_i$$, i.e.

\begin{align*} \sigma \left(\frac{ f(y_1, \cdots, y_n) }{ g(y_1, \cdots, y_n) }\right) = \frac{ f(\sigma(y_1), \cdots, \sigma(y_n)) }{ g(\sigma(y_1), \cdots, \sigma(y_n)) } .\end{align*}

Definition: A function $$f \in F(\alpha_1, \cdots, \alpha_n)$$ is symmetric $$\iff$$ under this action, $$\sigma\curvearrowright f = f$$ for all $$\sigma \in S_n$$.

Examples:

1. $$f(y_1, \cdots, y_n) = \prod y_i$$
2. $$f(y_1, \cdots, y_n) = \sum y_i$$.

### Elementary Symmetric Functions

Consider $$f(x) \in F(y_1, \cdots, y_n)[x]$$ given by $$\prod (x-y_i)$$. Then $$\sigma f = f$$, so $$f$$ is a symmetric function. Moreover, all coefficients are fixed by $$S_n$$. So the coefficients themselves are symmetric functions.

Concretely, we have

Coefficient Term
1 $$(-1)^n$$
$$x^{n-1}$$ $$-y_1 - y_2 - \cdots - y_n$$
$$x^{n-2}$$ $$y_1y_2 + y_1y_3 + \cdots + y_2y_3 + \cdots$$

The coefficient of $$x^{n-i}$$ is referred to as the $$i$$th elementary symmetric function.

Consider an intermediate extension $$E$$ given by joining all of the elementary symmetric functions:

Let $$K$$ denote the base field with all symmetric functions adjoined; then $$K$$ is an intermediate extension, and we have the following results:

Theorem:

1. $$E \leq K$$ is a field extension.

2. $$E \leq F(y_1, \cdots, y_n)$$ is a finite, normal extension since it is the splitting field of $$f(x) = \prod (x-y_i)$$, which is separable.

We thus have \begin{align*} [F(y_1, \cdots, y_n): E] \leq n! < \infty .\end{align*}

Proof:

We’ll show that in fact $$E = K$$, so all symmetric functions are generated by the elementary symmetric functions.

By definition of symmetric functions, $$K$$ is exactly the fixed field $$F(y_1, \cdots, y_n)_{S_n}$$, and $${\left\lvert {S} \right\rvert}_n = n!$$.

So we have

\begin{align*} n! &= {\left\lvert { \operatorname{Gal}(F(y_1, \cdots, y_n / K))} \right\rvert} \\ & \leq \{F(y_1, \cdots, y_n) : K\} \\ & \leq [F(y_1, \cdots, y_n): K] .\end{align*}

But now we have \begin{align*} n! \leq [F(y_1, \cdots, y_n):K] \leq [F(y_1, \cdots, y_n) : E] \leq n! \end{align*} which forces $$K=E$$.

$$\hfill\blacksquare$$

Theorem:

1. Every symmetric function can be written as a combination of sums, products, and possibly quotients of elementary symmetric functions.

2. $$F(y_1, \cdots, y_n)$$ is a finite normal extension of $$F(s_1, \cdots, s_n)$$ of degree $$n!$$.

3. $$\operatorname{Gal}(F(y_1, \cdots, y_n) / F(s_1, \cdots, s_n)) \cong S_n$$.

We know that every group $$G \hookrightarrow S_n$$ by Cayley’s theorem. So there exists an intermediate extension \begin{align*} F(s_1, \cdots, s_n) \leq L \leq F(y_1, \cdots, y_n) \end{align*} such that $$G = \operatorname{Gal}(F(y_1, \cdots, y_n) / L)$$.

Open question: which groups can be realized as Galois groups over $${\mathbb{Q}}$$? Old/classic question, possibly some results in the other direction (i.e. characterizations of which groups can’t be realized as such Galois groups).

Let $$p(x) = \sum a_i x^i \in {\mathbb{Q}}[x]$$ be a polynomial of degree $$n$$. Can we find a formula for the roots as a function of the coefficients, possibly involving radicals?

• For $$n = 1$$ this is clear

• For $$n=2$$ we have the quadratic formula.

• For $$n = 3$$, there is a formula by work of Cardano.

• For $$n = 4$$, this is true by work of Ferrari.

• For $$n \geq 5$$, there can not be a general equation.

Definition: Let $$K \geq F$$ be a field extension. Then $$K$$ is an extension of $$F$$ by radicals (or a radical extension) $$\iff$$ $$K = \alpha_1, \cdots, \alpha_n$$ for some $$\alpha_i$$ such that

1. Each $$\alpha_i^{m_i} \in F$$ for some $$m_i > 0$$.

2. For each $$i$$, $$\alpha_i^{\ell_i} \in F(\alpha_1, \cdots, \alpha_{i-1})$$ for some $$\ell_i < m_i$$ (?).

Definition: A polynomial $$f(x) \in F[x]$$ is solvable by radicals over $$F$$ $$\iff$$ the splitting field of $$f$$ is contained in some radical extension.

Example: Over $${\mathbb{Q}}$$, the polynomials $$x^5-1$$ and $$x^3-2$$ are solvable by radicals.

Recall that $$G$$ is solvable if there exists a normal series \begin{align*} 1 {~\trianglelefteq~}H_1 {~\trianglelefteq~}H_2 \cdots {~\trianglelefteq~}H_n {~\trianglelefteq~}G \text{ such that } H_n/H_{n-1} \text{ is abelian } \forall n .\end{align*}

### The Splitting Field of $$x^n-a$$ is Solvable

Lemma: Let $$\operatorname{char}~F = 0$$ and $$a\in F$$. If $$K$$ is the splitting field of $$p(x) = x^n-a$$, then $$\operatorname{Gal}(K/F)$$ is a solvable group.

Example: Let $$p(x) = x^4-2 / {\mathbb{Q}}$$, which had Galois group $$D_4$$.

Proof: Suppose that $$F$$ contains all $$n$$th roots of unity, $$\left\{{1, \zeta, \zeta^2, \cdots, \zeta^[n-1]}\right\}$$ where $$\zeta$$ is a primitive $$n$$th root of unity. If $$\beta$$ is any root of $$p(x)$$, then $$\zeta^i\beta$$ is also a root for any $$1\leq i \leq n-1$$. This in fact yields $$n$$ distinct roots, and is thus all of the them. Since the splitting field $$K$$ is of the form $$F(\beta)$$, then if $$\sigma \in \operatorname{Gal}(K/F)$$, then $$\sigma(\beta) = \zeta^i \beta$$ for some $$i$$. Then if $$\tau \in \operatorname{Gal}(K/F)$$ is any other automorphism, then $$\tau(\beta) = \zeta^k \beta$$ and thus (exercise) the Galois group is abelian and thus solvable.

Suppose instead that $$F$$ does not contain all $$n$$th roots of unity. So let $$F' = F(\zeta)$$, so $$F \leq F(\zeta) = F' \leq K$$. Then $$F \leq F(\zeta)$$ is a splitting field (of $$x^n-1$$) and separable since we are in characteristic zero and this is a finite extension. Thus this is a normal extension.

We thus have $$\operatorname{Gal}(K/F) / \operatorname{Gal}(K/F(\zeta)) \cong \operatorname{Gal}(F(\zeta)/ F)$$. We know that $$\operatorname{Gal}(F(\zeta)/ F)$$ is abelian since this is a cyclotomic extension, and so is $$\operatorname{Gal}(K/F(\zeta))$$. We thus obtain a normal series \begin{align*} 1 {~\trianglelefteq~}\operatorname{Gal}(K/F(\zeta)) {~\trianglelefteq~}\operatorname{Gal}(K/F) \end{align*} Thus we have a solvable group.

$$\hfill\blacksquare$$

# Tuesday October 22nd

## Certain Radical Extensions are Solvable

Recall the definition of an extension being radical (see above).

We say that a polynomial $$f(x) \in K[x]$$ is solvable by radicals iff its splitting field $$L$$ is a radical extension of $$K$$.

Lemma: Let $$F$$ be a field of characteristic zero.

If $$K$$ is a splitting field of $$f(x) = x^n - a \in F[x]$$, then $$\operatorname{Gal}(K/F)$$ is a solvable group.

Theorem: Let $$F$$ be characteristic zero, and suppose $$F \leq E \leq K \leq \overline F$$ be algebraic extension where $$E/F$$ is normal and $$K$$ a radical extension of $$F$$. Moreover, suppose $$[K:F] < \infty$$.

Then $$\operatorname{Gal}(E/F)$$ is solvable.

Proof: The claim is that $$K$$ is contained in some $$L$$ where $$F \subset L$$, $$L$$ is a finite normal radical extension, and $$\operatorname{Gal}{L/F}$$ is solvable.

Since $$K$$ is a radical extension of $$F$$, we have $$F = K(\alpha_1, \cdots, \alpha_n)$$ and $$\alpha_i^{n_i} \in K(\alpha_1, \cdots, \alpha_{i-1})$$ for each $$i$$ and some $$n_i \in {\mathbb{N}}$$.

Let $$L_1$$ be the splitting field of $$f_1(x) = x^{n_1} - \alpha_1^{n_1}$$, then by the previous lemma, $$L_1$$ is a normal extension and $$\operatorname{Gal}(L_1/F)$$ is a solvable group.

Inductively continue this process, and letting \begin{align*} f_2(x) = \prod_{\sigma \in \operatorname{Gal}(L_1/F)} x^{n_2} - \sigma(\alpha_2)^{n_2} \in F[x] .\end{align*} Note that the action of the Galois group on this polynomial is stable. Let $$L_2$$ be the splitting field of $$f_2$$, then $$L_2$$ is a finite normal radical extension.

Then \begin{align*} \frac{ \operatorname{Gal}(L_2/F) }{ \operatorname{Gal}(L_2/L_1) } \cong \operatorname{Gal}(L_1/F) ,\end{align*} which is solvable, and the denominator in this quotient is solvable, so the total group must be solvable as well. $$\hfill\blacksquare$$

## Proof: Insolubility of the Quintic

Theorem (Insolubility of the quintic): Let $$y_1, \cdots, y_n$$ be independent transcendental elements in $${\mathbb{R}}$$, then the polynomial $$f(x) = \prod (x-y_i)$$ is not solvable by radicals over $${\mathbb{Q}}(s_1, \cdots, s_n)$$ where the $$s_i$$ are the elementary symmetric polynomials in $$y_i$$.

So there are no polynomial relations between the transcendental elements.

Proof:

Let $$n\geq 5$$ and suppose $$y_i$$ are transcendental over $${\mathbb{R}}$$ and linearly independent over $${\mathbb{Q}}$$. Then consider

\begin{align*} s_1 &= \sum y_i \\ s_2 &= \sum_{i\leq j} y_i y_j \\ \cdots \\ s_n &= \prod_i y_i .\end{align*}

Then $${\mathbb{Q}}(y_1, \cdots, y_n)/ {\mathbb{Q}}(s_1, \cdots, s_n)$$ would be a normal extension precisely if $$A_n {~\trianglelefteq~}S_n$$ (by previous theorem). For $$n\geq 5$$, $$A_n$$ is simple, and thus $$S_n$$ is not solvable in this range.

Thus the polynomial is not solvable by radicals, since the splitting field of $$f(x)$$ is $${\mathbb{Q}}(y_1, \cdots, y_n)$$. $$\hfill\blacksquare$$

## Rings and Modules

Recall that a ring is given by $$(R, +, \cdot)$$, where

1. $$(R, +)$$ is an abelian group,
2. $$(R, \cdot)$$ is a monoid,
3. The distributive laws hold.

An ideal is certain type of subring that allows taking quotients, and is defined by $$I {~\trianglelefteq~}R \iff I\leq R$$ and $$RI, IR \subseteq I$$. The quotient is given by $$R/I = \left\{{r + I {~\mathrel{\Big|}~}r\in R}\right\}$$, and the ideal property is what makes this well-defined.

Much like groups, we have some notion of homomorphism $$\phi: R\to R'$$, where $$\phi(ax+y) = \phi(a)\phi(x) + \phi(y)$$.

### Modules

We want to combine the following two notions:

• Groups acting on sets, and

• Vector spaces

Definition: Let $$R$$ be a ring and $$M$$ an abelian group. Then if there is a map \begin{align*} R\times M &\to M \\ (r,m) &\mapsto rm .\end{align*}

such that $$\forall s,r_1,r_2 \in R$$ and $$m_1,m_2 \in M$$ we have

• $$(sr_1 + r_2)(m_1 + m_2) = sr_1m_1 + sr_1m_2 + r_2m_1 + r_2 m_2$$
• $$1\in R \implies 1m = m$$.

then $$M$$ is said to be an $$R{\hbox{-}}$$module.

Think of $$R$$ like the group acting by scalar multiplication, and $$M$$ the set of vectors with vector addition.

Examples:

1. $$R = k$$ a field, then a $$k{\hbox{-}}$$module is a vector space.

2. $$R = G$$ an abelian group, then $$R$$ is a $${\mathbb{Z}}{\hbox{-}}$$module where \begin{align*} n\curvearrowright a \mathrel{\vcenter{:}}=\sum_{i=1}^n a .\end{align*}

(In fact, these two notions are equivalent.)

1. $$I {~\trianglelefteq~}R$$, then $$M \mathrel{\vcenter{:}}= R/I$$ is an ring, which has an underlying abelian group, so $$M$$ is an $$R{\hbox{-}}$$module where \begin{align*} M\curvearrowright R = r\curvearrowright(s+I) \mathrel{\vcenter{:}}=(rs) + I .\end{align*}

2. For $$M$$ an abelian group, $$R \mathrel{\vcenter{:}}=\mathrm{End}(M) = \hom_{\text{AbGrp}}(M, M)$$ is a ring, and $$M$$ is a left $$R{\hbox{-}}$$module given by \begin{align*} f\curvearrowright m \mathrel{\vcenter{:}}= f(m) .\end{align*}

Definition: Let $$M, N$$ be left $$R{\hbox{-}}$$modules. Then $$f: M \to N$$ is an $$R{\hbox{-}}$$module homomorphism $$\iff$$ \begin{align*} f(rm_1 + m_2) = rf(m_1) + f(m_2) .\end{align*}

Definition: Monomorphisms are injective maps, epimorphisms are surjections, and isomorphisms are both.

Definition: A submodule $$N\leq M$$ is a subset that is closed under all module operations.

We can consider images, kernels, and inverse images, so we can formulate homomorphism theorems analogous to what we saw with groups/rings:

Theorem:

1. If $$M \xrightarrow{f} N$$ in $$R{\hbox{-}}$$mod, then \begin{align*} M / \ker(f) \cong \operatorname{im}({(})f) .\end{align*}

2. Let $$M, N \leq L$$, then $$M+N \leq L$$ as well, and \begin{align*} \frac{M}{M\cap N} \cong \frac{M+N}{N} .\end{align*}

3. If $$M\leq M\leq L$$, then \begin{align*} \frac{M}{N} \cong \frac{L/M}{L/N} \end{align*}

Note that we can always quotient, since there’s an underlying abelian group, and thus the “normality”/ideal condition is always satisfied for submodules. Just consider \begin{align*} M/N \mathrel{\vcenter{:}}=\left\{{m + N {~\mathrel{\Big|}~}m\in M}\right\} ,\end{align*} then $$R\curvearrowright(M/N)$$ in a well-defined way that gives $$M/N$$ the structure of an $$R{\hbox{-}}$$module as well.

# Thursday October 24

## Conjugates

Let $$E\geq F$$. Then $$\alpha, \beta \in E$$ are conjugate iff $$\min(\alpha, F) = \min(\beta, F)$$.

Example: $$\alpha \pm bi \in {\mathbb{C}}$$.

Theorem: Let $$F$$ be a field and $$\alpha, \beta \in F$$ with $$\deg \min (\alpha, F) = \deg \min (\beta, F)$$, so \begin{align*} [F(\alpha): F] = [F(\beta): F] .\end{align*}

Then $$\alpha, \beta$$ are conjugates $$\iff$$ $$F(\alpha) \cong F(\beta)$$ under the conjugation map, \begin{align*} \psi: F(\alpha) &\to F(\beta) \\ \sum_{i=1}^{n-1} a_i \alpha^i &\mapsto \sum_{i=1}^{n-1} a_i \beta^i .\end{align*}

Proof:

$$\impliedby$$:

Suppose that $$\psi$$ is an isomorphism. Let $$\min(\alpha, F) = p(x) = \sum c_i x^i$$ where each $$c_i \in F$$. Then \begin{align*} 0 = \psi(0) = \psi(p(\alpha)) = p(\beta) \implies \min(\beta, F) {~\Bigm|~}\min(\alpha, F) .\end{align*}

Applying the same argument to $$q(x) = \min(\beta, F)$$ yields $$\min(\beta, F) = \min(\alpha, F)$$.

$$\implies$$:

Suppose $$\alpha, \beta$$ are conjugates.

Exercise: Check that $$\psi$$ is surjective and \begin{align*} \psi(x+y) = \psi(x) + \psi(y) \\ \psi(xy) = \psi(x) \psi(y) .\end{align*}

Let $$z = \sum a_i \alpha^i$$. Supposing that $$\psi(z) = 0$$, we have $$\sum a_i \beta^i = 0$$. By linear independence, this forces $$a_i = 0$$ for all $$i$$, and thus $$z=0$$. So $$\psi$$ is injective.

$$\hfill\blacksquare$$

Corollary: Let $$\alpha \in \overline F$$ be algebraic. Then

1. Any $$\phi: F(\alpha) \hookrightarrow\overline F$$ such that $$\phi(f) = f$$ for all $$f\in F$$ must map $$\alpha$$ to a conjugate.

2. If $$\beta \in \overline F$$ is a conjugate of $$\alpha$$, then there exists an isomorphism $$\phi: F(\alpha) \to F(\beta) \subseteq \overline F$$ such that $$\phi(f) = f$$ for all $$f\in F$$.

Proof of 1:

Let $$\min(\alpha, F) = p(x) = \sum a_i x^i$$. Note that $$0 = \psi(p(\alpha)) = p(\psi(\alpha))$$, and since $$p$$ was irreducible, $$p$$ must also be the minimal polynomial of $$\psi(\alpha)$$. Thus $$\psi(\alpha)$$ is a conjugate of $$\alpha$$.

$$\hfill\blacksquare$$

Proof of 2:

$$F(\alpha)$$ is generated by $$F$$ and $$\alpha$$, and $$\psi$$ is completely determined by where it sends $$F$$ and $$\alpha$$. This shows uniquness.

$$\hfill\blacksquare$$

Corollary: Let $$f(x) \in {\mathbb{R}}[x]$$ and suppose $$f(a+bi)= 0$$. Then $$f(a-bi) = 0$$.

Proof: Both $$i, -i$$ are conjugates and $$\min(i, {\mathbb{R}}) = \min(-i, {\mathbb{R}}) = x^2 + 1 \in {\mathbb{R}}[x]$$. We then have a map \begin{align*} \psi: {\mathbb{R}}[i] &\to {\mathbb{R}}[-i] \\ \psi(a+bi) = a + b(-i) .\end{align*}

So if $$f(a+bi) = 0$$, then $$0 = \psi(f(a+bi)) = f(\psi(a+bi)) = f(a-bi)$$.

$$\hfill\blacksquare$$

# October 27th

## Modules

Let $$R$$ be a ring and $$M$$ be an $$R{\hbox{-}}$$module.

Definition: For a subset $$X\subseteq M$$, we can define the submodule generated by $$X$$ as \begin{align*} \left\langle{X}\right\rangle \mathrel{\vcenter{:}}=\cap_{X \subseteq N \leq M} N \subseteq M .\end{align*}

Then $$M$$ is generated by $$X$$ iff $$M = \left\langle{X}\right\rangle$$.

As a special case, when $$X = \left\{{m}\right\}$$ consists of a single element, we write \begin{align*} \left\langle{m}\right\rangle = Rm \mathrel{\vcenter{:}}=\left\{{rm {~\mathrel{\Big|}~}r\in R}\right\} .\end{align*}

In general, we have \begin{align*} \left\langle{X}\right\rangle = \left\{{\sum r_i x_i \mathrel{\Big|}r_i \in R, x_i \in X}\right\} .\end{align*}

## Direct Products and Direct Sums

Definition: Let $$\left\{{M_i}\right\}$$ be a finite collection of $$R{\hbox{-}}$$modules, and let \begin{align*} N = \bigoplus M_i = \left\{{\sum m_i \mathrel{\Big|}m_i \in M_i}\right\} \end{align*}

with multiplication given by $$\gamma \sum m_i = \sum \gamma m_i$$ denote the direct sum.

For an infinite collection, we require that all but finitely many terms are zero.

Definition: Define $$N = \prod M_i$$ denote the direct product, where we now drop the condition that finitely many terms are zero.

When the indexing set is finite, $$\bigoplus M_i \cong \prod M_i$$. In general, $$\bigoplus M_i \hookrightarrow\prod M_i$$.

Note that the natural inclusions \begin{align*} \iota_j: M_j \hookrightarrow\prod M_i \end{align*}

and projections \begin{align*} \pi_j: \prod M_i \twoheadrightarrow M_j \end{align*}

are both $$R{\hbox{-}}$$module homomorphisms.

Theorem: $$M \cong \bigoplus M_i$$ iff there exist maps $$\pi_j: M \to M_j$$ and $$\iota_j: M_j \to M$$ such that

1. \begin{align*} \pi_j \circ \iota_k = \begin{cases} 1m & j=k \\ 0 & \text{else}\end{cases} \end{align*}

2. $$\sum_j \iota_j \circ \pi_j = \text{id}_M$$

Remark: Let $$M, N$$ be $$R{\hbox{-}}$$modules. Then $$\hom_{R-\text{mod}}(M, N)$$ is an abelian group.

## Internal Direct Sums

For a collection of submodules of $$M$$ given by $$\left\{{M_i}\right\}$$, denote the internal direct sum \begin{align*} \sum M_i \coloneqq\left\{{m_1 + m_2 + \cdots \mathrel{\Big|}m_i \in M_i}\right\} \end{align*}

iff it satisfies the following conditions:

1. $$M = \sum_i M_i$$

2. $$M_i \cap M_j = \left\{{0}\right\}$$ for $$i\neq j$$.

## Exact Sequences

Definition: A sequence of the form \begin{align*} 0 \to M_1 \xrightarrow{i} M_2 \xrightarrow{p} M_3 \to 0 \end{align*}

where

• $$i$$ is a monomorphism

• $$p$$ is an empimorphism

• $$\operatorname{im}({i}) = \ker p$$

is said to be short exact.

Examples:

• \begin{align*} 0 \to 2{\mathbb{Z}}\hookrightarrow{\mathbb{Z}}\twoheadrightarrow{\mathbb{Z}}/2{\mathbb{Z}}\to 0 \end{align*}

• For any epimorphism $$\pi: M\to N$$, \begin{align*} 0 \to \ker \pi \to M \to N \to 0 \end{align*}

• \begin{align*} 0 \to M_1 \to M_1 \oplus M_2 \to M_2 \to 0 \end{align*}

In general, any sequence \begin{align*} \cdots \to M_i \xrightarrow{f_i} M_{i+1} \xrightarrow{f_{i+1}} \cdots \end{align*}

is exact iff $$\operatorname{im}({f})_i = \ker f_{i+1}$$.

1. If $$\alpha, \gamma$$ are monomorphisms then $$\beta$$ is a monomorphism.

# Tuesday October 29th

## Exact Sequences

Lemma (Short Five):

Consider a diagram of the following form:

1. $$\alpha, \gamma$$ monomorphisms implies $$\beta$$ is a monomorphism.

2. $$\alpha, \gamma$$ epimorphisms implies $$\beta$$ is an epimorphism.

3. $$\alpha, \gamma$$ isomorphisms implies $$\beta$$ is an isomorphism.

Moreover, (1) and (2) together imply (3).

Proof: Exercise.

Example proof of (2): Suppose $$\alpha, \gamma$$ are monomorphisms.

• Let $$n\in N$$ with $$\beta(n) = 0$$, then $$g' \circ \beta(n) = 0$$.
• $$\implies \gamma \circ g (n) = 0$$.
• $$\implies g(n) = 0$$
• $$\implies \exists m\in M$$ such that $$f(m) = n$$
• $$\implies \beta \circ f (m) = \beta(n)$$
• $$\implies f' \alpha(m) = \beta (n) = 0$$
• $$\implies \alpha(m) = 0$$
• $$\implies f'$$ is injective, so $$m=0$$ and $$n=f(m) = 0$$.

$$\hfill\blacksquare$$

Definition: Two exact sequences are isomorphic iff in the following diagram, $$f,g,h$$ are all isomorphisms:

Theorem: Let $$0 \to M_1 \xrightarrow{f} M_2 \xrightarrow{g} M_3 \to 0$$ be a SES. Then TFAE:

• There exists an $$R{\hbox{-}}$$module homomorphisms $$h: M_3 \to M_2$$ such that $$g\circ h = \text{id}_{M_3}$$.

• There exists an $$R{\hbox{-}}$$module homomorphisms $$k: M_2 \to M_1$$ such that $$k\circ f = \text{id}_{M_1}$$.

• The sequence is isomorphic to $$0 \to M_1 \to M_1 \oplus M_3 \to M_3 \to 0$$.

Proof: Define $$\phi: M_1 \oplus M_3 \to M_2$$ by $$\phi(m_1 + m_2) = f(m_1) + h(m_2)$$. We need to show that the following diagram commutes:

We can check that \begin{align*} (g\circ \phi)(m_1 + m_2) = g( f(m_1)) + g(h(m_2)) = m_2 = \pi(m_1 + m_2).\end{align*}

This yields $$1 \implies 3$$, and $$2 \implies 3$$ is similar.

To see that $$3 \implies 1, 2$$, we attempt to define $$k, h$$ in the following diagram:

So define $$k = \pi_1 \circ \phi^{-1}$$ and $$h = \phi \circ \iota_2$$. It can then be checked that \begin{align*} g \circ h = g \circ \phi \circ \iota_2 = \pi_2 \circ \iota_2 = \text{id}_{M_3} .\end{align*}

$$\hfill\blacksquare$$

## Free Modules

Moral: A free module is a module with a basis.

Definition: A subset $$X = \left\{{x_i}\right\}$$ is linearly independent iff \begin{align*} \sum r_i x_i = 0 \implies r_i = 0 ~\forall i .\end{align*}

Definition: A subset $$X$$ spans $$M$$ iff \begin{align*} m\in M \implies m = \sum_{i=1}^n r_i x_i \quad \text{ for some }r_i \in R,~x_i \in X .\end{align*}

Definition: A subset $$X$$ is a basis $$\iff$$ it is a linearly independent spanning set.

Example: $${\mathbb{Z}}_6$$ is an abelian group and thus a $${\mathbb{Z}}{\hbox{-}}$$module, but not free because $$3 \curvearrowright[2] = [6] = 0$$, so there are torsion elements. This contradicts linear independence for any subset.

Theorem (Characterization of Free Modules): Let $$R$$ be a unital ring and $$M$$ a unital $$R{\hbox{-}}$$module (so $$1\curvearrowright m = m$$).

TFAE:

• There exists a nonempty basis of $$M$$.

• $$M = \oplus_{i\in I} R$$ for some index set $$I$$.

• There exists a non-empty set $$X$$ and a map $$\iota: X \hookrightarrow M$$ such that given $$f: X \to N$$ for $$N$$ any $$R{\hbox{-}}$$ module, $$\exists! \tilde f: M \to N$$ such that the following diagram commutes.

Definition: An $$R{\hbox{-}}$$module is free iff any of 1,2, or 3 hold.

Proof of $$1 \implies 2$$:

Let $$X$$ be a basis for $$M$$, then define $$M \to \oplus_{x\in X} Rx$$ by $$\phi(m) = \sum r_i x_i$$.

It can be checked that

• This is an $$R{\hbox{-}}$$module homomorphism,

• $$\phi(m) = 0 \implies r_j = 0 ~\forall j \implies m = 0$$, so $$\phi$$ is injective,

• $$\phi$$ is surjective, since $$X$$ is a spanning set.

So $$M \cong \bigoplus_{x\in X} Rx$$, so it only remains to show that $$Rx \cong R$$. We can define the map \begin{align*} \pi_x: R &\to Rx \\ r &\mapsto rx .\end{align*}

Then $$\pi_x$$ is onto, and is injective exactly because $$X$$ is a linearly independent set. Thus $$M \cong \oplus R$$.

$$\hfill\blacksquare$$

Proof of $$1 \implies 3$$:

Let $$X$$ be a basis, and suppose there are two maps $$X \xrightarrow{\iota} M$$ and $$X \xrightarrow{f} M$$. Then define

\begin{align*} \tilde f: M &\to N \\ \sum_i r_i x_i &\mapsto \sum_i r_i f(x_i) .\end{align*}

This is clearly an $$R{\hbox{-}}$$module homomorphism, and the diagram commutes because $$(\tilde f \circ \iota)(x) = f(x)$$.

This is unique because $$\tilde f$$ is determined precisely by $$f(X)$$.

$$\hfill\blacksquare$$

Proof of $$3 \implies 2$$:

We use the usual “2 diagram” trick to produce maps \begin{align*} \tilde f: M \to \bigoplus_{x\in X} R \\ \tilde g: \bigoplus_{x\in X}R \to M .\end{align*}

Then commutativity forces \begin{align*} \tilde f \circ \tilde g = \tilde g \circ \tilde f = \text{id} .\end{align*}

$$\hfill\blacksquare$$

Proof of $$2 \implies 1$$:

We have $$M = \oplus_{i\in I} R$$ by (2). So there exists a map \begin{align*} \psi: \oplus_{i\in I} R \to M ,\end{align*}

so let $$X \mathrel{\vcenter{:}}=\left\{{\psi(1_i) \mathrel{\Big|}i\in I}\right\}$$, which we claim is a basis.

To see that $$X$$ is a basis, suppose $$\sum r_i \psi(1_i) = 0$$. Then $$\psi(\sum r_i 1_i) = 0$$ and thus $$\sum r_i 1_i = 0$$ and $$r_i = 0$$ for all $$i$$.

Checking that it’s a spanning set: Exercise.

$$\hfill\blacksquare$$

Corollary: Every $$R{\hbox{-}}$$module is the homomorphic image of a free module.

Proof: Let $$M$$ be an $$R{\hbox{-}}$$module, and let $$X$$ be any set of generators of $$R$$. Then we can make a map \begin{align*} M \to \bigoplus_{x\in X} R \end{align*} and there is a map $$X \hookrightarrow M$$, so the universal property provides a map \begin{align*} \tilde f: \bigoplus_{x\in X} R \to M .\end{align*}

Moreover, $$\bigoplus_{x\in X} R$$ is free.

$$\hfill\blacksquare$$

Examples:

• $${\mathbb{Z}}_n$$ is not a free $${\mathbb{Z}}{\hbox{-}}$$module for any $$n$$.

• If $$V$$ is a vector space over a field $$k$$, then $$V$$ is a free $$k{\hbox{-}}$$module (even if $$V$$ is infinite dimensional).

• Every nonzero submodule of a free module over a PID is free.

Some facts:

Let $$R = k$$ be a field (or potentially a division ring).

1. Every maximal linearly independent subset is a basis for $$V$$.

2. Every vector space has a basis.

3. Every linearly independent set is contained in a basis

4. Every spanning set contains a basis.

5. Any two bases of a vector space have the same cardinality.

Theorem (Invariant Dimension): Let $$R$$ be a commutative ring and $$M$$ a free $$R{\hbox{-}}$$module.

If $$X_1, X_2$$ are bases for $$R$$, then $${\left\lvert {X_1} \right\rvert} = {\left\lvert {X_2} \right\rvert}$$.

Any ring satisfying this condition is said to have the invariant dimension property.

Note that it’s difficult to say much more about generic modules. For example, even a finitely generated module may not have an invariant number of generators.

# Tuesday November 5th

## Free vs Projective Modules

Let $$R$$ be a PID. Then any nonzero submodule of a free module over a PID is free, and any projective module over $$R$$ is free.

Recall that a module $$M$$ is projective $$\iff M$$ is a direct summand of a free module.

In general,

• Free $$\implies$$ projective, but

• Projective $$\centernot\implies$$ free.

Example:

Consider $${\mathbb{Z}}_6 = {\mathbb{Z}}_2 \oplus {\mathbb{Z}}_3$$ as a $${\mathbb{Z}}{\hbox{-}}$$module. Is this free as a $${\mathbb{Z}}{\hbox{-}}$$module?

Note that $${\mathbb{Z}}_2$$ is a submodule and thus projective, but $${\mathbb{Z}}_2$$ is not free since it is not a free module over $${\mathbb{Z}}$$. What fails here is that $${\mathbb{Z}}_6$$ is not a PID, since it is not a domain.

## Annihilators

Definition: Let $$m\in M$$ a module, then define \begin{align*} \mathrm{Ann}_m \mathrel{\vcenter{:}}=\left\{{r\in R {~\mathrel{\Big|}~}r.m = 0 }\right\} {~\trianglelefteq~}R. \end{align*} We can then define a map

\begin{align*} \phi: R \to R.m \\ r \mapsto r.m .\end{align*}

Then $$\ker \phi = \mathrm{Ann}_m$$, and $$R/\mathrm{Ann} \cong R.m$$.

We can also define \begin{align*} M_t \mathrel{\vcenter{:}}=\left\{{m\in M {~\mathrel{\Big|}~}\mathrm{Ann}_m \neq 0}\right\} \leq M. \end{align*}

Lemma: Let $$R$$ be a PID and $$p$$ a prime element. Then

• If $$p^i m = 0$$ then $$\mathrm{Ann}_m = (p^j)$$ where $$0\leq j\leq i$$.
• If $$\mathrm{Ann}_m = (p^i)$$, then $$p^jm \neq 0$$ for any $$j < m$$.

Proof of (1): Since we are in a PID and the annihilator is an ideal, we have $$\mathrm{Ann}_m \mathrel{\vcenter{:}}=(r)$$ for some $$r\in M$$. Then $$p^i \in (r)$$, so $$r {~\Bigm|~}p^i$$. But $$p$$ was prime, to up to scaling by units, we have $$r = p^j$$ for some $$j \leq i$$.

$$\hfill\blacksquare$$

Proof of (2): Towards a contradiction, suppose that $$\mathrm{Ann}_m = (p^i)$$ and $$p^jm = 0$$ for some $$j < i$$. Then $$p^j \in \mathrm{Ann}_m$$, so $$p^j {~\Bigm|~}p^i$$. But this forces $$j \leq i$$, a contradiction.

$$\hfill\blacksquare$$

Some terminology:

• $$\mathrm{Ann}_m$$ is the order ideal of $$m$$.

• $$M_t$$ is the torsion submodule of $$M$$.

• $$M$$ is torsion iff $$M = M_t$$.

• $$M$$ is torsion free iff $$M_t = 0$$.

• $$\mathrm{Ann}_m = (r)$$ is said to have order $$r$$.

• $$Rm$$ is the cyclic module generated by $$m$$.

Theorem: A finitely generated torsion-free module over a PID is free.

Proof: Let $$M = \left\langle{X}\right\rangle$$ for some finite generating set.

We can assume $$M \neq (0)$$. If $$m\neq 0 \in M$$, with $$rm = 0$$ iff $$r=0$$.

So choose $$S = \left\{{x_1, \cdots , x_n}\right\} \subseteq X$$ to be a maximal linearly independent subset of generators, so \begin{align*} \sum r_i x_i = 0 \implies r_i = 0 ~\forall i .\end{align*}

Consider the submodule $$F \mathrel{\vcenter{:}}=\left\langle{x_1, \cdots, x_n}\right\rangle \leq M$$; then $$S$$ is a basis for $$F$$ and thus $$F$$ is free.

The claim is that $$M \cong F$$. Supposing otherwise, let $$y\in X\setminus S$$. Then $$S \cup\left\{{y}\right\}$$ can not be linearly independent, so there exists $$r_y, r_i \in R$$ such that \begin{align*} r_y y + \sum r_i x^i = 0 .\end{align*} Thus $$r_y y = - \sum r_i x^i$$, where $$r_y \neq 0$$.

Since $${\left\lvert {X} \right\rvert} < \infty$$, let \begin{align*} r = \prod_{y \in X\setminus S} r_y .\end{align*}

Then $$rX = \left\{{rx {~\mathrel{\Big|}~}x\in X}\right\} \subseteq F$$, and $$rM \leq F$$.

Now using the particular $$r$$ we’ve just defined, define a map \begin{align*} f: M &\to M \\ m &\mapsto rm .\end{align*}

Then $$\operatorname{im}({f}) = r.M$$, and since $$M$$ is torsion-free, $$\ker f = (0)$$. So $$M \cong rM \subseteq F$$ and $$M$$ is free.

$$\hfill\blacksquare$$

Theorem: Let $$M$$ be a finitely generated module over a PID $$R$$. Then $$M$$ can be decomposed as \begin{align*} M \cong M_t \oplus F \end{align*} where $$M_t$$ is torsion and $$F$$ is free of finite rank, and $$F \cong M/M_t$$.

Note: we also have $$M/F \cong F_t$$ since this is a direct sum.

Proof:

Part 1: $$M/M_t$$ is torsion free.

Suppose that $$r(m + M_t) = M_t$$, so that $$r$$ acting on a coset is the zero coset. Then $$rm + M_t = M_t$$, so $$rm \in M_t$$, so there exists some $$r'$$ such that $$r'(rm) = 0$$ by definition of $$M_t$$. But then $$(r'r)m = 0$$, so in fact $$m\in M_t$$ and thus $$m + M_t = M_t$$, making $$M/M_t$$ torsion free.

Part 2: $$F \cong M/M_t$$.

We thus have a SES

\begin{align*} 0 \to M_t \to M \to M/M_t \coloneqq F \to 0 ,\end{align*}

and since we’ve shown that $$F$$ is torsion-free, by the previous theorem $$F$$ is free. Moreover, every SES with a free module in the right-hand slot splits:

For $$X = \left\{{x_j}\right\}$$ a generating set of $$F$$, we can choose elements $$\left\{{y_i}\right\} \in \pi^{-1}(\iota(X))$$ to construct a set map $$f: X \to M$$. By the universal property of free modules, we get a map $$h: F \to M$$.

It remains to check that this is actually a splitting, but we have

\begin{align*} \pi \circ h (x_j) = \pi(h(\iota(x_j))) = \pi(f(x_j)) = \pi(y_j) = x_j. \end{align*} Lemma: Let $$R$$ be a PID, and $$r\in R$$ factor as $$r = \prod p_i^{k_i}$$ as a prime factorization. Then \begin{align*} R/(r) \cong \bigoplus R/(p_i^{k_i}). \end{align*}

Since $$R$$ is a UFD, suppose that $$\gcd(s ,t) = 1$$. Then the claim is that \begin{align*} R/(st) = R/(s) \oplus R/(t) ,\end{align*} which will prove the lemma by induction.

Define a map \begin{align*} \alpha: R/(s) \oplus R/(t) &\to R/(st) \\ (x + (s), y+(t)) &\mapsto tx + sy + (st) .\end{align*}

Exercise: Show that this map is well-defined.

Since $$\gcd(s, t) = 1$$, there exist $$u, v$$ such that $$su + vt = 1$$. Then for any $$r\in R$$, we have \begin{align*} rsu + rvt = r ,\end{align*} so for any given $$r\in R$$ we can pick $$x =tv$$ and $$y=su$$ so that this holds. As a result, the map $$\alpha$$ is onto.

Now suppose $$tx + sy \in (st)$$; then $$tx + sy = stz$$. We have $$su + vt = 1$$, and thus \begin{align*} utx + usy = ustz \implies utx + (y-tvy) = ustz .\end{align*}

We can thus write \begin{align*} y = ustv - utx + tvy \in (t) .\end{align*}

Similarly, $$x\in (t)$$, so $$\ker \alpha = 0$$.

$$\hfill\blacksquare$$

## Classification of Finitely Generated Modules Over a PID

Theorem (Classification of Finitely Generated Modules over a PID):

Let $$M$$ be a finitely generated $$R{\hbox{-}}$$module where $$R$$ is a PID. Then

1. \begin{align*} M \cong F \bigoplus_{i=1}^t R/(r_i) \end{align*} where $$F$$ is free of finite rank and $$r_1 {~\Bigm|~}r_2 {~\Bigm|~}\cdots {~\Bigm|~}r_t$$. The rank and list of ideals occurring is uniquely determined by $$M$$. The $$r_i$$ are referred to as the invariant factors.
1. \begin{align*} M \cong F \bigoplus_{i=1}^k R/(p_i^{s_i}) \end{align*} where $$F$$ is free of finite rank and $$p_i$$ are primes that need not be distinct. The rank and ideals are uniquely determined by $$M$$. The $$p_i^{s_i}$$ are referred to as elementary divisors.

# Thursday November 7th

## Projective Modules

Definition: A projective module $$P$$ over a ring $$R$$ is an $$R{\hbox{-}}$$module such that the following diagram commutes:

i.e. for every surjective map $$g:M \twoheadrightarrow N$$ and every map $$f: P \to N$$ there exists a lift $$\phi: P \to M$$ such that that $$g \circ \phi = f$$.

Theorem: Every free module is projective.

Proof: Suppose $$M \twoheadrightarrow N \to 0$$ and $$F \xrightarrow{f} N$$, so we have the following situation:

For every $$x\in X$$, there exists an $$m_x \in M$$ such that $$g(m_x) = f(i(x))$$. By freeness, there exists a $$\phi: F \to M$$ such that this diagram commutes.

$$\hfill\blacksquare$$

Corollary: Every $$R{\hbox{-}}$$module is the homomorphic image of a projective module.

Proof: If $$M$$ is an $$R{\hbox{-}}$$module, then $$F \twoheadrightarrow M$$ where $$F$$ is free, but free modules are surjective.

$$\hfill\blacksquare$$

Theorem: Let $$P$$ be an $$R{\hbox{-}}$$module. Then TFAE:

1. $$P$$ is projective.

2. Every SES $$0 \to M \to N \to P \to 0$$ splits.

3. There exists a free module $$F$$ such that $$F = P \oplus K$$ for some other module $$K$$.

Proof:

$$a \implies b$$:

We set up the following situation, where $$s$$ is produced by the universal property:

$$\hfill\blacksquare$$

$$b \implies c$$:

Suppose we have $$0 \to M \to N \to P \to 0$$ a SES which splits, then $$N \cong M \oplus P$$ by a previous theorem.

$$\hfill\blacksquare$$

$$c\implies a$$:

We have the following situation:

By the previous argument, there exists an $$h: F\to M$$ such that $$g\circ h = f \circ \pi$$. Set $$\phi = h\circ \iota$$.

Exercise: Check that $$g\circ \phi = f$$.

$$\hfill\blacksquare$$

Theorem: $$\bigoplus P_i$$ is projective $$\iff$$ each $$P_i$$ is projective.

Proof:

$$\implies$$: Suppose $$\oplus P_i$$ is projective.

Then there exists some $$F = K \oplus \bigoplus P_i$$ where $$F$$ is free. But then $$P_i$$ is a direct summand of $$F$$, and is thus projective.

$$\impliedby$$: Suppose each $$P_i$$ is projective.

Then there exists $$F_i = P_i \oplus K_i$$, so $$F \mathrel{\vcenter{:}}=\bigoplus F_i = \bigoplus (P_i \oplus K_i) = \bigoplus P_i \oplus \bigoplus K_i$$. So $$\bigoplus P_i$$ is a direct summand of a free module, and thus projective.

$$\hfill\blacksquare$$

Note that a direct sum has finitely many nonzero terms. Can use the fact that a direct sum of free modules is still free by taking a union of bases.

Example of a projective module that is not free:

Take $$R = {\mathbb{Z}}_6$$, which is not a PID and not a domain. Then $${\mathbb{Z}}_6 = {\mathbb{Z}}_2 \oplus {\mathbb{Z}}_3$$, and $${\mathbb{Z}}_2, {\mathbb{Z}}_3$$ are projective $$R{\hbox{-}}$$modules. By previous statements, we know these are torsion as $${\mathbb{Z}}{\hbox{-}}$$modules, and thus not free.

## Endomorphisms as Matrices

See section 7.1 in Hungerford

Let $$M_{m, n}({\mathbb{R}})$$ denote $$m\times n$$ matrices with coefficients in $$R$$. This is an $$R{\hbox{-}}R$$ bimodule, and since $$R$$ is not necessarily a commutative ring, these two module actions may not be equivalent.

If $$m=n$$, then $$M_{n,n}(R)$$ is a ring under the usual notions of matrix addition and multiplication.

Theorem: Let $$V, W$$ be vector spaces where $$\dim V = m$$ and $$\dim W = n$$. Let $$\hom_k(V, W)$$ be the set of linear transformations between them.

Then $$\hom_k(V, W) \cong M_{m, n}(k)$$ as $$k{\hbox{-}}$$vector spaces.

Proof: Choose bases of $$V, W$$. Then consider

\begin{align*} T: V \to W \\ v_1 \mapsto \sum_{i=1}^n a_{1, i} ~w_i \\ v_2 \mapsto \sum_{i=1}^n a_{2, i} ~w_i \\ \vdots \end{align*}

This produces a map \begin{align*} f: \hom_k(V, W) &\to M_{m, n}(k) \\ T &\mapsto (a_{i, j}) ,\end{align*}

which is a matrix.

Exercise: Check that this is bijective.

$$\hfill\blacksquare$$

Theorem: Let $$M, N$$ be free left $$R{\hbox{-}}$$modules of rank $$m, n$$ respectively. Then $$\hom_R(M, N) \cong M_{m, n}(R)$$ as $$R{\hbox{-}}R$$ bimodules.

Notation: Suppose $$M, N$$ are free $$R{\hbox{-}}$$modules, then denote $$\beta_m, \beta_n$$ be fixed respective bases. We then write $$[T]_{\beta_m, \beta_n} \mathrel{\vcenter{:}}=(a_{i, j})$$ to be its matrix representation.

Theorem: Let $$R$$ be a ring and let $$V, W, Z$$ be three free left $$R{\hbox{-}}$$modules with bases $$\beta_v, \beta_w, \beta_z$$ respectively. If $$T: V \to W, S: W\to Z$$ are $$R{\hbox{-}}$$module homomorphisms, then $$S \circ T: V \to Z$$ exists and \begin{align*} [S \circ T]_{\beta_v, \beta_z} = [T]_{\beta_v, \beta_w} [S]_{\beta_w, \beta_z} \end{align*}

Proof: Exercise.

Show that \begin{align*} (S \circ T)(v_i) = \sum_j^t \sum_k^m a_{ik} b_{kj} z_j .\end{align*}

$$\hfill\blacksquare$$

## Matrices and Opposite Rings

Suppose $$\Gamma: \hom_R(V, V) \to M_n(R)$$ and $$V$$ is a free left $$R{\hbox{-}}$$module. By the theorem, we have $$\Gamma(T \circ S) = \Gamma(S) \Gamma(T)$$. We say that $$\Gamma$$ is an anti-homomorphism.

To address this mixup, given a ring $$R$$ we can define $$R^{op}$$ which has the same underlying set of $$R$$ but with the modified multiplication \begin{align*} x \cdot y \mathrel{\vcenter{:}}= yx \in R .\end{align*}

If $$R$$ is commutative, then $$R \cong R^{op}$$.

$$\hfill\blacksquare$$

Theorem: Let $$R$$ be a unital ring and $$V$$ an $$R{\hbox{-}}$$module.

Then $$\hom_R(V, V) \cong M_n(R^{op})$$ as rings.

Proof: Since $$\Gamma(S \circ T) = \Gamma(T) \Gamma(S)$$, define a map \begin{align*} \Theta: M_{n, n}(R) &\to M_{n, n}(R^{op}) \\ A &\mapsto A^t .\end{align*}

Then \begin{align*} \Theta(AB) = (AB)^t = B^t A^t = \Theta(B) \Theta(A) ,\end{align*} so $$\Theta$$ is an anti-isomorphism.

Thus $$\Theta\circ \Gamma$$ is an anti-anti-homomorphism, i.e. a usual homomorphism.

$$\hfill\blacksquare$$

Definition: A matrix $$A$$ is invertible iff there exists a $$B$$ such that $$AB = BA = \text{id}_n$$.

Proposition: Let $$R$$ be a unital ring and $$V, W$$ free $$R{\hbox{-}}$$modules with $$\dim V = n, \dim W = m$$. Then

1. $$T \in \hom_R(V, W)$$ is an isomorphisms iff $$[T]_{\beta_v, \beta_w}$$ is invertible.

2. $$[T^{-1}]_{\beta_v, \beta_w} = [T]_{\beta_v, \beta_w}^{-1}$$.

Definition: We’ll say that two matrices $$A, B$$ are equivalent iff there exist $$P, Q$$ invertible such that $$PAQ = B$$.

# Tuesday November 12th

## Equivalence and Similarity

Recall from last time:

If $$V, W$$ are free left $$R{\hbox{-}}$$modules of ranks $$m,n$$ respectively with bases $$\beta_v, \beta_w$$ respectively, then \begin{align*} \hom_R(V, W) \cong M_{m, n}(R) .\end{align*}

Definition: Two matrices $$A, B \in M_{m \times n}(R)$$ are equivalent iff \begin{align*} \exists P \in \operatorname{GL}(m, R),~ \exists Q \in \operatorname{GL}(n, R) \quad \text{ such that } \quad A = PBQ .\end{align*}

Definition: Two matrices $$A, B \in M_m(R)$$ are similar iff \begin{align*} \exists P \in \operatorname{GL}(m, R) \quad \text{ such that } \quad A = P^{-1}B P .\end{align*}

Theorem: Let $$T: V\to W$$ be an $$R{\hbox{-}}$$module homomorphism.

Then $$T$$ has an $$m\times n$$ matrix relative to other bases for $$V, W$$ $$\iff$$ \begin{align*} B = P [T]_{\beta_v, \beta_w} Q .\end{align*}

Proof: $$\implies$$:

Let $$\beta_v', \beta_w'$$ be other bases. Then we want $$B = [T]_{\beta_v', \beta_w'}$$, so just let

\begin{align*} P = [\text{id}]_{\beta_v', \beta_v} \quad Q = [\text{id}]_{\beta_w, \beta_w'} .\end{align*}

$$\hfill\blacksquare$$

$$\impliedby$$:

Suppose $$B = P [T]_{\beta_v, \beta_w} Q$$ for some $$P, Q$$.

Let $$g: V\to V$$ be the transformation associated to $$P$$, and $$h: W \to W$$ associated to $$Q^{-1}$$.

Then \begin{align*} P &= [\text{id}]_{g(\beta_v), \beta_v} \\ \implies Q^{-1}&= [\text{id}]_{h(\beta_w), \beta_w} \\ \implies Q &= [\text{id}]_{\beta_w, h(\beta_w)} \\ \implies B &= [T]_{g(\beta_v), h(\beta_w)} .\end{align*}

$$\hfill\blacksquare$$

Corollary: Let $$V$$ be a free $$R{\hbox{-}}$$module and $$\beta_v$$ a basis of size $$n$$.

Then $$T: V\to V$$ has an $$n\times n$$ matrix relative to $$\beta_v$$ relative to another basis $$\iff$$ \begin{align*} B = P [T]_{\beta_v, \beta_v} P^{-1} .\end{align*}

Note how this specializes to the case of linear transformations, particularly when $$B$$ is diagonalizable.

## Review of Linear Algebra:

Let $$D$$ be a division ring. Recall the notions of rank and nullity, and the statement of the rank-nullity theorem.

Note that we can always factor a linear transformation $$\phi: E\to F$$ as the following short exact sequence:

\begin{align*} 0 \to \ker \phi \to E \xrightarrow{\phi} \operatorname{im}({\phi })\to 0, \end{align*}

and since every module over a division ring is free, this sequence splits and $$E \cong \ker\phi \oplus \operatorname{im}({\phi})$$. Taking dimensions yields the rank-nullity theorem.

Let $$A\in M_{m, n}(D)$$ and define

• $$R(A) \in D^n$$ is the span of the rows of $$A$$, and

• $$C(A) \in D^m$$ is the span of the columns of $$A$$.

Recall that finding a basis of the row space involves doing Gaussian Elimination and taking the rows which have nonzero pivots.

For a basis of the column space, you take the corresponding columns in the original matrix.

Note that in this case, $$\dim R(A) = \dim C(A)$$, and in fact these are always equal.

Theorem (Rank and Equivalence): Let $$\phi: V\to W$$ be a linear transformation and $$A$$ be the matrix of $$\phi$$ relative to $$\beta_v, \beta_v'$$.

Then $$\dim \operatorname{im}({\pi })= \dim C(A) = \dim R(A)$$.

Proof: Construct the matrix $$A = [\phi]_{\beta_v, \beta_w}$$.

Then $$\phi: V \to W$$ descends to a map $$A: D^m \to D^n$$. Writing the matrix $$A$$ out and letting $$v\in D^m$$ a row vector act on $$A$$ from the left yields a column vector $$Av \in D^n$$.

But then $$\operatorname{im}({\phi})$$ corresponds to $$R(A)$$, and so \begin{align*} \dim \operatorname{im}({\phi })= \dim R(A) = \dim C(A) .\end{align*}

$$\hfill\blacksquare$$

## Canonical Forms

Let $$1 \leq r \leq \min(m, n)$$, and define $$E_r$$ to be the $$m\times n$$ matrix with the $$r\times r$$ identity matrix in the top-left block.

Theorem: Let $$A, B \in M_{m,n}(D)$$. Then

1. $$A$$ is equivalent to $$E_r \iff \operatorname{rank}A = r$$

• That is, $$\exists P,Q$$ such that $$E_r = PAQ$$
2. $$A$$ is equivalent to $$B$$ iff $$\operatorname{rank}A = \operatorname{rank}B$$.

3. $$E_r$$ for $$r = 0, 1, \cdots, \min(m,n)$$ is a complete set of representatives for the relation of matrix equivalence on $$M_{m, n}(D)$$.

Let $$X = M_{m, n}(D)$$ and $$G = \operatorname{GL}_m(D) \times\operatorname{GL}_n(D)$$, then \begin{align*} G \curvearrowright X \text{ by } (P, Q) \curvearrowright A \mathrel{\vcenter{:}}= PAQ^{-1} .\end{align*} Then the orbits under this action are exactly $$\left\{{E_r \mathrel{\Big|}0 \leq r \leq \min(m, n)}\right\}$$.

Proof: Note that 2 and 3 follow from 1, so we’ll show 1.

$$\implies$$:

Let $$A$$ be an $$m\times n$$ matrix for some linear transformation $$\phi: D^m \to D^n$$ relative to some basis. Assume $$\operatorname{rank}A = \dim \operatorname{im}({\phi })= r$$. We can find a basis such that $$\phi(u_i) = v_i$$ for $$1 \leq i \leq r$$, and $$\phi(u_i) = 0$$ otherwise. Relative to this basis, $$[\phi] = E_r$$. But then $$A$$ is equivalent to $$E_r$$.

$$\impliedby$$:

If $$A = PE_r Q$$ with $$P, Q$$ invertible, then $$\dim \operatorname{im}({A}) = \dim \operatorname{im}({E})_r$$, and thus $$\operatorname{rank}A = \operatorname{rank}E_r = r$$.

How do we do this? Recall the row operations:

• Interchange rows

• Multiply a row by a unit

• Add one row to another

But each corresponds to left-multiplication by an elementary matrix, each of which is invertible. If you proceed this way until the matrix is in RREF, you produce $$P \prod P_i A$$. You can now multiply on the right by elementary matrices to do column operations and move all pivots to the top-left block, which yields $$E_r$$.

$$\hfill\blacksquare$$

Theorem: Let $$A \in M_{m, n}(R)$$ where $$R$$ is a PID.

Then $$A$$ is equivalent to a matrix with $$L_r$$ in the top-left block, where $$L_r$$ is a diagonal matrix with $$L_{ii} = d_i$$ such that $$d_1 {~\Bigm|~}d_2 {~\Bigm|~}\cdots {~\Bigm|~}d_r$$. Each $$(d_i)$$ is uniquely determined by $$A$$.

# Thursday November 14th

## Equivalence to Canonical Forms

Let $$D$$ be a division ring and $$k$$ a field.

Recall that a matrix $$A$$ is equivalent to $$B \iff \exists P, Q$$ such that $$PBQ=A$$. From a previous theorem, if $$\operatorname{rank}(A) = r$$, then $$A$$ is equivalent to a matrix with $$I_r$$ in the top-left block.

Theorem: Let $$A$$ be a matrix over a PID $$R$$. Then $$A$$ is equivalent to a matrix with $$L_r$$ in the top-left corner, where $$L_r = \mathrm{diag}(d_1, d_2, \cdots, d_r)$$ and $$d_1 {~\Bigm|~}d_2 {~\Bigm|~}\cdots {~\Bigm|~}d_r$$, and the $$d_i$$ are uniquely determined.

Theorem: Let $$A$$ be an $$n\times n$$ matrix over a division ring $$D$$. TFAE:

1. $$\operatorname{rank}A = n$$.

2. $$A$$ is equivalent to $$I_n$$.

3. $$A$$ is invertible.

$$1\implies 2$$: Use Gaussian elimination.

$$2\implies 3$$: $$A = PI_n Q = PQ$$ where $$P, Q$$ are invertible, so $$PQ = A$$ is invertible.

$$3\implies 1$$: If $$A$$ is invertible, then $$A: D^n \to D^n$$ is bijective and thus surjective, so $$\dim \operatorname{im}({A}) = n$$.

Note: the image is now row space because we are taking left actions.

$$\hfill\blacksquare$$

## Determinants

Definition: Let $$M_1, \cdots, M_n$$ be $$R{\hbox{-}}$$modules, and then $$f: \prod M_i \to R$$ is $$n{\hbox{-}}$$linear iff

\begin{align*} f( m_1, m_2, \cdots, rm_k + sm_k', \cdots, m_n ) = \\ r f( m_1, \cdots, m_k, \cdots m_k) + sf(m_1, \cdots, m_k', \cdots, m_n ) .\end{align*}

Example: The inner product is a 2-linear form.

Definition: $$f$$ is symmetric iff \begin{align*} f(m_1, \cdots, m_n) = f(m_{\sigma(1)}, \cdots, m_{\sigma(n)}) ~~\forall \sigma \in S_n .\end{align*}

Definition: $$f$$ is skew-symmetric iff \begin{align*} f(m_1, \cdots, m_n) = \mathrm{sgn}(\sigma) f(m_{\sigma(1)}, \cdots, m_{\sigma(n)}) ~~\forall \sigma \in S_n ,\end{align*}

where

\begin{align*} \mathrm{sgn}(\sigma) = \begin{cases} 1 & \sigma \text{ is even } \\ -1 & \sigma \text{ is odd } \end{cases} .\end{align*}

Definition: $$f$$ is alternating iff

\begin{align*} m_i = m_j \text{ for some pair } (i, j) \implies f(m_1, \cdots, m_n) = 0 .\end{align*}

Theorem: Let $$f$$ be an $$n{\hbox{-}}$$linear form. If $$f$$ is alternating, then $$f$$ is skew-symmetric.

Proof: It suffices to show the $$n=2$$ case. We have

\begin{align*} 0 &= f(m+1 + m_2, m_1 + m_2) \\ &= f(m_1, m_1) + f(m_1, m_2) + f(m_2, m_1) + f(m_2, m_2) \\ &= f(m_1, m_2) + f(m_2, m_1)\\ \implies f(m_1, m_2) &= - f(m_2, m_1) .\end{align*}

$$\hfill\blacksquare$$

Theorem: Let $$R$$ be a unital commutative ring and let $$r\in R$$ be arbitrary.

Then \begin{align*} \exists! f: \bigoplus_{i=1}^n R^n \to R ,\end{align*}

where $$f$$ is an alternating $$R{\hbox{-}}$$form such that $$f(\mathbf{e}_i) = r$$ for all $$i$$, where $$\mathbf{e}_i = [0, 0, \cdots, 0, 1, 0, \cdots, 0, 0]$$.

$$R^n$$ is a free module, so $$f$$ can be identified with a matrix once a basis is chosen.

Proof:

Existence: Let $$x_i = [a_{i1}, a_{i2}, \cdots, a_{in}]$$ and define \begin{align*} f(x_1, \cdots, x_n) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) r \prod_i a_{i \sigma(i)} .\end{align*}

Exercise: Check that $$f(\mathbf{e}_1, \cdots, \mathbf{e}_n) = r$$ and $$f$$ is $$n{\hbox{-}}$$linear.

Moreover, $$f$$ is alternating. Consider $$f(x_1, \cdots, x_n)$$ where $$x_i = x_j$$ for some $$i\neq j$$.

Letting $$\phi = (i, j)$$, we can write $$S_n = A_n {\coprod}A_n \rho$$.

If $$\sigma$$ is even, then the summand is \begin{align*} (+1)r a_{1\sigma(1)} \cdots a_{n\sigma(n)} .\end{align*}

Since $$x_i = x_j$$, we’ll have $$\prod_k a_{ik} = \prod a_{jk}$$. Then consider applying $$\sigma \rho$$. We have

\begin{align*} -r \prod a_{i\sigma(i)} &= -r a_{1\sigma(1)} \cdots \mathbf{a}_{j \sigma(j)} \cdots \mathbf{a}_{i \sigma(i)} \cdots a_{n, \sigma(n)} \\ &= -r \prod a_{i\sigma(i)} = -r a_{1\sigma(1)} \cdots \mathbf{a}_{i \sigma(i)} \cdots \mathbf{a}_{j \sigma(j)} \cdots a_{n, \sigma(n)} ,\end{align*}

which permutes the $$i,j$$ terms. So these two terms cancel, the remaining terms are untouched.

Uniqueness: Let $$x_i = \sum_j a_{ij} \mathbf{e}_j$$. Then

\begin{align*} f(x_1, \cdots, x_n) &= f(\sum_{j_1} a^1_j \mathbf{e}_j, \cdots, \sum_{j_n} a^n_j \mathbf{e}_j) \\ &= \sum_{j_1} \cdots \sum_{j_n} f(\mathbf{e}_{j_1}, \cdots, \mathbf{e}_{j_n} ) a_{1, j_1} \cdots a_{n, j_n} \\ &= \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) f(\mathbf{e}_1, \cdots, \mathbf{e}_n) a_{1, \sigma(1)} \cdots a_{n, \sigma(n)} \\ &= \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) r a_{1, \sigma(1)} \cdots a_{n, \sigma(n)} .\end{align*}

$$\hfill\blacksquare$$

Definition: Let $$R$$ be a commutative unital ring and define $$\mathrm{det}: M_n(R) \to R$$ is the unique $$n{\hbox{-}}$$alternating form with $$\det(I) = 1$$, and is called the determinant.

Theorem: Let $$A, B \in M_{n}(R)$$. Then

1. $${\left\lvert {AB} \right\rvert} = {\left\lvert {A} \right\rvert} {\left\lvert {B} \right\rvert}$$

2. $$A$$ is invertible $$\iff {\left\lvert {A} \right\rvert} \in R^{\times}$$

3. $$A \sim B \implies {\left\lvert {A} \right\rvert} = {\left\lvert {B} \right\rvert}$$.

4. $${\left\lvert {A^t} \right\rvert} = {\left\lvert {A} \right\rvert}$$.

5. If $$A$$ is triangular, then $${\left\lvert {A} \right\rvert}$$ is the product of the diagonal entries.

Proof of a: Let $$B$$ be fixed.

Let $$\Delta_B: M_n(R) \to R$$ be defined as $$C \mapsto {\left\lvert {CB} \right\rvert}$$. Then this is an alternating form, so by the theorem, $$\Delta_B = r \mathrm{det}$$. But then $$\Delta_B(C) = r{\left\lvert {C} \right\rvert}$$, so $$r{\left\lvert {C} \right\rvert} = {\left\lvert {CB} \right\rvert}$$. So pick $$C = I$$, then $$r = {\left\lvert {B} \right\rvert}$$.

$$\hfill\blacksquare$$

Proof of b: Suppose $$A$$ is invertible.

Then $$AA^{-1}= I$$, so $${\left\lvert {AA^{-1}} \right\rvert} = {\left\lvert {A} \right\rvert}{\left\lvert {A^{-1}} \right\rvert} = 1$$, which shows that $${\left\lvert {A} \right\rvert}$$ is a unit.

$$\hfill\blacksquare$$

Proof of c: Let $$A = PBP^{-1}$$. Then \begin{align*} {\left\lvert {A} \right\rvert} = {\left\lvert {PBP^{-1}} \right\rvert} = {\left\lvert {P} \right\rvert} {\left\lvert {B} \right\rvert} {\left\lvert {P^{-1}} \right\rvert} = {\left\lvert {P} \right\rvert} {\left\lvert {P^{-1}} \right\rvert} {\left\lvert {B} \right\rvert} = {\left\lvert {B} \right\rvert} .\end{align*}

$$\hfill\blacksquare$$

Proof of d: Let $$A = (a_{ij})$$, so $$B = (b_{ij}) = (a_{ji})$$. Then

\begin{align*} {\left\lvert {A^t} \right\rvert} &= \sum_{\sigma} \mathrm{sgn}(\sigma) \prod_k b_{k \sigma(k)} \\ &= \sum_\sigma \mathrm{sgn}(\sigma) \prod_k a_{\sigma(k) k} \\ &= \sum_{\sigma^{-1}} \mathrm{sgn}(\sigma) \prod_k a_{k \sigma^{-1}(k)} \\ &= \sum_\sigma \mathrm{sgn}(\sigma) \prod_k a_{k \sigma(k)} \\ &= {\left\lvert {A} \right\rvert} .\end{align*}

$$\hfill\blacksquare$$

Proof of e: Let $$A$$ be upper-triangular. Then \begin{align*} {\left\lvert {A} \right\rvert} = \sum_\sigma \mathrm{sgn}(\sigma) \prod_k a_{k \sigma(k)} = a_{11} a_{22} \cdots a_{nn} .\end{align*}

$$\hfill\blacksquare$$

Next time:

• Calculate determinants
• Gaussian elimination
• Cofactors
• Formulas for $$A^{-1}$$
• Cramer’s rule

# Tuesday November 19th

## Determinants

Let $$A\in M_n(R)$$, where $$R$$ is a commutative unital ring.

Given $$A = (a_{ij})$$, recall that \begin{align*} \det A = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod a_{i, \sigma(i)} .\end{align*}

This satisfies a number of properties:

• $$\det(AB) = \det A \det B$$

• $$A$$ invertible $$\implies$$ $$\det A$$ is a unit in $$R$$

• $$A \sim B \implies \det(A) = \det(B)$$

• $$\det A^t = \det A$$

• $$A$$ is triangular $$\implies \det A = \prod a_{ii}$$.

### Calculating Determinants

1. Gaussian Elimination

1. $$B$$ is obtained from $$A$$ by interchanging rows: $$\det B = -\det A$$

2. $$B$$ is obtained from $$A$$ by multiplying $$\det B = r \det A$$

3. $$B$$ is obtained from $$A$$ by adding a scalar multiple of one row to another: $$\det B = \det A$$.

2. Cofactors Let $$A_{ij}$$ be the $$(n-1)\times (n-1)$$ minor obtained by deleting row $$i$$ and column $$j$$, and $$C_{ij} = (-1)^{i+j} \det A_{ij}$$.

Then (theorem) $$\det A = \sum_{j=1}^n a_{ij} C_{ij}$$ by expanding along either a row or column.

Theorem: \begin{align*} A \mathrm{Adj}(A) = \det (A) I_n ,\end{align*} where $$\mathrm{Adj} = (C_{ij})^t$$.

If $$A^{-1}$$ is a unit, then $$A^{-1}= \mathrm{Adj}(A) / \det(A)$$.

### Decomposition of a Linear Transformation:

Let $$\phi: V \to V$$ be a linear transformation of vector spaces. and $$R = \hom_k(V, V)$$. Then $$R$$ is a ring.

Let $$f(x) = \sum a_j x^j \in k[x]$$ be an arbitrary polynomial. Then for $$\phi \in R$$, it makes sense to evaluate $$f(\phi)$$ where $$\phi^n$$ denotes an $$n{\hbox{-}}$$fold composition, and $$f(\phi): V \to V$$.

Lemma:

• There exists a unique monic polynomial $$q_\phi(x) \in k[x]$$ such that $$q_\phi(\phi) = 0$$ and $$f(\phi) = 0 \implies q_\phi {~\Bigm|~}f$$. $$q_\phi$$ is referred to as the minimal polynomial of $$\phi$$.

• The exact same conclusion holds with $$\phi$$ replaced by a matrix $$A$$, yielding $$q_A$$.

• If $$A$$ is the matrix of $$\phi$$ relative to a fixed basis, then $$q_\phi = q_A$$.

Proof of a and b: Fix $$\phi$$, and define

\begin{align*} \Gamma: k[x] &\to \hom_k(V, V) \\ f &\mapsto f(\phi) .\end{align*}

Since $$\dim_k V^\vee= \dim_k V < \infty$$ and $$\dim_k k[x] = \infty$$, we must have $$\ker \Gamma \neq 0$$.

Since $$k[x]$$ is a PID, we have $$\ker \Gamma = (q)$$ for some $$q\in k[x]$$. Then if $$f(\phi) = 0$$, we have $$f(x) \in \ker \Gamma \implies q {~\Bigm|~}f$$. We can then rescale $$q$$ to be monic, which makes it unique.

Note: for (b), just replace $$\phi$$ with $$A$$ everywhere.

$$\hfill\blacksquare$$

Proof of c: Suppose $$A = [\phi]_\mathcal{B}$$ for some fixed basis $$\mathcal B$$.

Then $$\hom_k(V, V) \cong M_n(k)$$, so we have the following commutative diagram:

$$\hfill\blacksquare$$

### Finitely Generated Modules over a PID

Let $$M$$ be a finitely generated module over $$R$$ a PID. Then

\begin{align*} M &\cong F \oplus \bigoplus_{i=1}^n R/(r_i) \quad r_1 {~\Bigm|~}r_2 {~\Bigm|~}\cdots r_n \\ M &\cong F \oplus \bigoplus_{i=1}^n R/(p_i^{s_i}) \quad p_i \text{ not necessarily distinct primes. } .\end{align*}

Letting $$R = k[x]$$ and $$\phi: V\to V$$ with $$\dim_k V < \infty$$, $$V$$ becomes a $$k[x]{\hbox{-}}$$module by defining \begin{align*} f(x) \curvearrowright\mathbf{v} \mathrel{\vcenter{:}}= f(\phi)(\mathbf{v}) \end{align*}

Note that $$W$$ is a $$k[x]{\hbox{-}}$$submodule iff $$\phi: W \to W$$.

Let $$v\in V$$, and $$\left\langle{v}\right\rangle = \left\{{\phi^i(v) {~\mathrel{\Big|}~}i = 0,1,2,\cdots}\right\}$$ is the cyclic submodule generated by $$v$$, and we write $$\left\langle{v}\right\rangle = k[x].v$$.

Theorem: Let $$\phi: V\to V$$ be a linear transformation. Then

1. There exist cyclic $$k[x]{\hbox{-}}$$submodules $$V_i$$ such that $$V = \bigoplus_{i=1}^t V_i$$, where for each $$i$$ there exists a $$q_i: V_i \to V_i$$ such that $$q_1 {~\Bigm|~}q_2 {~\Bigm|~}\cdots {~\Bigm|~}q_t$$.

2. There exist cyclic $$k[x]{\hbox{-}}$$submodules $$V_j$$ such that $$V = \bigoplus_{j=1}^\nu$$ and $$p_j^{m_j}$$ is the minimal polynomial of $$\phi: V_j \to V_j$$.

Proof: Apply the classification theorem to write $$V = \bigoplus R/(r_i)$$ as an invariant factor decomposition.

Then $$R/(q_i) \cong V_i$$, some vector space, and since there is a direct sum decomposition, the invariant factors are minimal polynomials for $$\phi_i: V_i \to V_i$$, and thus $$k[x]/(q_i)$$.

$$\hfill\blacksquare$$

### Canonical Forms for Matrices

We’ll look at

• Rational Canonical Form

• Jordan Canonical Form

Theorem: Let $$\phi: V\to V$$ be linear, then $$V$$ is a cyclic $$k[x]{\hbox{-}}$$module and $$\phi: V\to V$$ has minimal polynomial $$q(x) = \sum_j a_j x^j$$ iff $$\dim V = n$$ and $$V$$ has an ordered basis of the form

\begin{align*} [\phi]_{\mathcal{B}} = \left[ \begin{array}{ccccc} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \end{array} \right] \end{align*}

with ones on the super-diagonal.

Proof:

$$\impliedby$$:

Let $$V = k[x].v = \left\langle{v, \phi(v), \cdots, \phi^{n-1}(v)}\right\rangle$$ where $$\deg q(x) = n$$. The claim is that this is a linearly independent spanning set.

Linear independence: suppose $$\sum_{j=0}^{n-1} k_j \phi^j(v) = 0$$ with some $$k_j \neq 0$$. Then $$f(x) = \sum k_j x^j$$ is a polynomial where $$f(\phi) = 0$$, but this contradicts the minimality of $$q(x)$$.

But then we have $$n$$ linearly independent vectors in $$V$$ which is dimension $$n$$, so this is a spanning set.

$$\implies$$:

We can just check where basis elements are sent. Set $$\mathcal{B} = \left\{{v, \phi(v), \cdots, \phi^{n-1}(v)}\right\}$$. Then

\begin{align*} v &\mapsto \phi(v) \\ \phi(v) &\mapsto \phi^2(v) \\ &\vdots \\ \phi{n-1}(v) &\mapsto \phi^n(v) = -\sum a_i \phi^i(v) \\ .\end{align*}

$$\impliedby$$ Fix a basis $$B = \left\{{v_1, \cdots, v_n}\right\}$$ and $$A = [\phi]_B$$, then

\begin{align*} v_1 &\mapsto v_2 = \phi(v_1) \\ v_1 &\mapsto v_3 = \phi^2(v_1) \\ v_{n-2} &\mapsto v_{n-1} = \phi^2(v_1) .\end{align*}

and \begin{align*} \phi^n(v) = -a_k v_1 \neq -a_1 \phi(v_1), \cdots -a_{n-1} \phi^{n-1}(v_1) .\end{align*}

Thus $$V = k[x].v_1$$, since $$\dim V = n$$ with $$\left\{{v_1, \phi(v_1), \cdots, \phi^{n-1}(v_1)}\right\}$$ as a basis.

$$\hfill\blacksquare$$

# Thursday November 21

## Cyclic Decomposition

Let $$\phi: V\to V$$ be a linear transformation; then $$V$$ is a $$k[x]$$ module under $$f(x) \curvearrowright v \mathrel{\vcenter{:}}= f(\phi)(v)$$.

By the structure theorem, since $$k[x]$$ is a PID, we have an invariant factor decomposition $$V = \bigoplus V_i$$ where each $$V_i$$ is a cyclic $$k[x]{\hbox{-}}$$module. If $$q_i$$ is the minimal polynomial for $$\phi_i: V_i \to V_i$$, then $$q_{i} {~\Bigm|~}q_{i+1}$$ for all $$i$$.

We also have an elementary divisor decomposition where $$p_i^{m_i}$$ are the minimal polynomials for $$\phi_i$$.

Note: one is only for the restriction to the subspaces? Check.

Recall that if $$\phi$$ has minimal polynomial $$q(x)$$. Then if $$\dim V = n$$, there exists a basis of $$B$$ if $$V$$ such that $$[\phi]_B$$ is given by the companion matrix of $$q(x)$$. This is the rational canonical form.

Corollary: Let $$\phi: V\to V$$ be a linear transformation. Then $$V$$ is a cyclic $$k[x]{\hbox{-}}$$module and $$\phi$$ has minimal polynomial $$(x-b)^n \iff \dim V = n$$ and there exists a basis such that

\begin{align*} [\phi]_B = \left[\begin{array}{ccccccc} b & 1 & 0 & \cdots & 0 & 0 & 0\\ 0 & b & 1 & \cdots & 0 & 0 & 0\\ 0 & 0 & b & 1 &\cdots & 0 & 0\\ 0 & 0 & 0 & 0 & \cdots & b & 1 \end{array}\right] .\end{align*}

This is the Jordan Canonical form.

Note that if $$k$$ is not algebraically closed, we can only reduce to RCF. If $$k$$ is closed, we can reduce to JCF, which is slightly nicer.

Proof:

Let $$\delta = \phi - b \cdot\text{id}_V$$. Then

• $$q(x)$$ is the minimal polynomial for $$\phi \iff x^n$$ is the minimal polynomial for $$\delta$$.

• A priori, $$V$$ has two $$k[x]$$ structures – one given by $$\phi$$, and one by $$\delta$$.

• Exercise: $$V$$ is cyclic with respect to the $$\phi$$ structure $$\iff$$ $$V$$ is cyclic with respect to the the $$\delta$$ structure.

Then the matrix $$[\delta]_B$$ relative to an ordered basis for $$\delta$$ is with only zeros on the diagonal and 1s on the super-diagonal, and $$[\phi]_B$$ is the same but with $$b$$ on the diagonal.

$$\hfill\blacksquare$$

Lemma: Let $$\phi: V\to V$$ with $$V = \bigoplus_i^t V_i$$ as $$k[x]{\hbox{-}}$$modules. Then $$M_i$$ is a matrix of $${\left.{\phi}\right|_{V_i}}: V_i \to V_i$$ relative to some basis for $$V_i \iff$$the matrix of $$\phi$$ wrt some ordered basis is given by

\begin{align*} \left[ \begin{array}{cccc} M_1 & & & \\ & M_2 & & \\ & & \ddots & \\ & & & M_t \end{array}\right] .\end{align*}

Proof:

$$\implies$$: Suppose $$B_i$$ is a basis for $$V_i$$ and $$[\phi]_{B_i} = M_i$$. Then let $$B = \cup_i B_i$$; then $$B$$ is a basis for $$V$$ and the matrix is of the desired form.

$$\impliedby$$: Suppose that we have a basis $$B$$ and $$[\phi]_B$$ is given by a block diagonal matrix filled with blocks $$M_i$$. Suppose $$\dim M_i = n_i$$. If $$B = \left\{{v_1, v_2, \cdots, v_n}\right\}$$, then take $$B_1 = \left\{{v_1, \cdots, v_{n_1}}\right\}$$ and so on. Then $$[\phi_i]_{B_i} = M_i$$ as desired.

$$\hfill\blacksquare$$

Application: Let $$V = \bigoplus V_i$$ with $$q_i$$ the minimal polynomials of $$\phi: V_i \to V_i$$ with $$q_i {~\Bigm|~}q_{i+1}$$.

Then there exists a basis where $$[\phi]_B$$ is block diagonal with blocks $$M_i$$, where each $$M_i$$ is in rational canonical form with minimal polynomial $$q_i(x)$$. If $$k$$ is algebraically closed, we can obtain elementary divisors $$p_i(x) = (x - b_i)^{m_i}$$. Then there exists a similar basis where now each $$M_i$$ is a Jordan block with $$b_i$$ on the diagonals and ones on the super-diagonal.

Moreover, in each case, there is a basis such that $$A = P [M_i] P^{-1}$$ (where $$M_i$$ are the block matrices obtained). When $$A$$ is diagonalizable, $$P$$ contains the eigenvectors of $$A$$.

Corollary: Two matrices are similar $$\iff$$ they have the same invariant factors and elementary divisors.

Example: Let $$\phi: V\to V$$ have invariant factors $$q_1(x) = (x-1)$$ and $$q_2(x) = (x-1)(x-2)$$.

Then $$\dim V = 3$$, $$V = V_1 \oplus V_2$$ where $$\dim V_1 = 1$$ and $$\dim V_2 = 2$$. We thus have

\begin{align*} [\phi]_B = \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -2 & 3 \end{array}\right) .\end{align*}

Moreover, we have

\begin{align*} V \cong \frac{k[x]}{(x-1)} \oplus \frac{k[x]}{(x-1)(x-2)} \cong \frac{k[x]}{(x-1)} \oplus \frac{k[x]}{(x-1)} \oplus \frac{k[x]}{(x-2)} ,\end{align*}

so the elementary divisors are $$x-1, x-1, x-2$$.

Invariant factor decompositions should correspond to rational canonical form blocks, and elementary divisors should correspond to Jordan blocks.

Theorem: Let $$A$$ be an $$n\times n$$ matrix over $$k$$. Then the matrix $$xI_n - A \in M_n(k[x])$$ is equivalent in $$k[x]$$ to a diagonal matrix $$D$$ with non-zero entries $$f_1, f_2, \cdots f_t \in k[x]$$ such that the $$f_i$$ are monic and $$f_i {~\Bigm|~}f_{i+1}$$. The non-constant polynomials among the $$f_i$$ are the invariant factors of $$A$$.

Proof (Sketch): Let $$V = k^n$$ and $$\phi: k^n \to k^n$$ correspond to $$A$$ under the fixed standard basis $$\left\{{e_i}\right\}$$. Then $$V$$ has a $$k[x]{\hbox{-}}$$module structure induced by $$\phi$$.

Let $$F$$ be the free $$k[x]$$ module with basis $$\left\{{u_i}\right\}_{i=1}^n$$, and define the maps

\begin{align*} \pi: F &\to k^n \\ u_i &\mapsto e_i \end{align*}

and

\begin{align*} \psi: F &\to F \\ u_i &\mapsto xu_i - \sum_j a_{ij} u_j .\end{align*}

Then $$\psi$$ relative to the basis $$\left\{{u_i}\right\}$$ is $$xI_n - A$$.

Then (exercise) the sequence \begin{align*} F \xrightarrow{\psi} F \xrightarrow{\pi} k^n \to 0 \end{align*} is exact, $$\operatorname{im}({\pi })= k^n$$, and $$\operatorname{im}({\psi })= \ker \pi$$.

We then have $$k^n \cong F/\ker \pi = F / \operatorname{im}({\psi})$$, and since $$k[x]$$ is a PID,

\begin{align*} xI_n - A \sim D \coloneqq \left[\begin{array}{cc} L_r & 0 \\ 0 & 0 \end{array}\right] .\end{align*}

where $$L_r$$ is diagonal with $$f_i$$s where $$f_i {~\Bigm|~}f_{i+1}$$.

However, $$\det(xI_n - A) \neq 0$$ because $$x I_n - A$$ is a monic polynomial of degree $$n$$.

But $$\det{xI_n - A} = \det(D)$$, so this means that $$L_r$$ must take up the entire matrix of $$D$$, so there is no zero in the bottom-right corner. So $$L_r = D$$, and $$D$$ is the matrix of $$\psi$$ with respect to $$B_1 = \left\{{v_i}\right\}$$ and $$B_2 = \left\{{w_i}\right\}$$ with $$\psi(v_i) = f_i w_i$$.

Thus \begin{align*} \operatorname{im}({\psi })= \bigoplus_{i=1}^n k[x] f_i w_i. \end{align*}

But then \begin{align*} V = k^n \cong F/ \operatorname{im}({\psi}) &\cong \frac{k[x] w_1 \oplus \cdots \oplus k[x] w_n} {k[x] f_1 w_1 \oplus \cdots \oplus k[x] f_n w_n} \\ &\cong \bigoplus_{i=1}^n k[x]/(f_i) .\end{align*}

$$\hfill\blacksquare$$

# Tuesday November 26th

## Minimal and Characteristic Polynomials

Theorem

1. ? (Todo)

2. (Cayley Hamilton) If $$p$$ is the minimal polynomial of a linear transformation $$\phi$$, then $$p(\phi) = 0$$

3. For any $$f(x) \in k[x]$$ that is irreducible, $$f(x) {~\Bigm|~}p_\phi(x) \iff f(x) {~\Bigm|~}q_\phi(x)$$.

Proof of (a): ?

$$\hfill\blacksquare$$

Proof of (b):

If $$q_\phi(x) {~\Bigm|~}p_\phi(x)$$ and $$q_\phi(\phi) = 0$$, then $$p_\phi(\phi) = 0$$ as well.

$$\hfill\blacksquare$$

Proof of (c): We have $$f(x) {~\Bigm|~}q_\phi(x) \implies f(x) {~\Bigm|~}p_\phi(x)$$ and $$f(x) {~\Bigm|~}p_\phi(x) \implies f(x) {~\Bigm|~}q_i(x)$$ for some $$i$$, and so $$f(x) {~\Bigm|~}q_\phi(x)$$.

$$\hfill\blacksquare$$

## Eigenvalues and Eigenvectors

Definition: Let $$\phi: V\to V$$ be a linear transformation. Then

1. An eigenvector is a vector $$\mathbf{v} = \mathbf{0}$$ such that $$\phi(\mathbf{v}) = \lambda \mathbf{v}$$ for some $$\lambda \in k$$.

2. If such a $$\mathbf{v}$$ exists, then $$\lambda$$ is called an eigenvalue of $$\phi$$.

Theorem: The eigenvalues of $$\phi$$ are the roots of $$p_\phi(x)$$ in $$k$$.

Proof: Let $$[\phi]_B = A$$, then

\begin{align*} &p_A(\lambda) = p_\phi(\lambda) = \det(\lambda I - A) = 0 \\ &\iff \exists \mathbf{v}\neq \mathbf{0} \text{ such that } (\lambda I - A)\mathbf{v} = \mathbf{0} \\ &\iff \lambda I\mathbf{v} = A \mathbf{v} \\ &\iff A\mathbf{v} = \lambda \mathbf{v} \\ &\iff \lambda \text{ is an eigenvalue and } \mathbf{v} \text{ is an eigenvector} .\end{align*}

$$\hfill\blacksquare$$

# Tuesday December 3rd

## Similarity and Diagonalizability

Recall that $$A \sim B \iff A = PBP^{-1}$$.

Fact: If $$T:V \to V$$ is a linear transformation and $$\mathcal{B}, \mathcal{B}'$$ are bases where $$[T]_{\mathcal{B}} = A$$ and $$[T]_{\mathcal{B}'}$$, then $$A \sim B$$.

Theorem: Let $$A$$ be an $$n\times n$$ matrix. Then

1. $$A$$ is similar to a diagonal matrix / diagonalizable $$\iff A$$ has $$n$$ linearly independent eigenvectors.

2. $$A = PDP^{-1}$$ where $$D$$ is diagonal and $$P = [\mathbf{v_1}, \mathbf{v_2}, \cdots, \mathbf{v_n}]$$ with the $$\mathbf{v_i}$$ linearly independent.

Proof: Consider $$AP = PD$$, then $$AP$$ has columns $$A\mathbf{v_i}$$ and $$PD$$ has columns $$\lambda_i \mathbf{v_i}$$. $$\hfill\blacksquare$$

Corollary: If $$A$$ has distinct eigenvalues, then $$A$$ is diagonalizable.

Examples:

1. Let \begin{align*} A = \left[\begin{array}{ccc} 4 & 0 & 0 \\ -1 & 4 & 0 \\ 0 & 0 & 5 \end{array}\right] \end{align*}

$$A$$ has eigenvalues $$4,5$$, and it turns out that $$A$$ is defective.

Note that $$\dim \Lambda_4 + \dim \Lambda_5 = 2 < 3$$, so the eigenvectors can’t form a basis of $${\mathbb{R}}^3$$.

2. \begin{align*} A = \left[\begin{array}{ccc} 4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \end{array}\right] \end{align*} $$A$$ has eigenvalues $$2, 8$$. $$\Lambda_2 = {\operatorname{span}}_{\mathbb{R}}\left\{{[-1, 1, 0]^t, [-1, 0, 1]^t}\right\}$$ and $$\Lambda_8 = {\operatorname{span}}_{\mathbb{R}}\left\{{[1,1,1]^t}\right\}$$. These vectors become the columns of $$P$$, which is (by no coincidence!) an orthogonal matrix, since $$A$$ was symmetric.

Exercise: \begin{align*} \left[\begin{array}{ccc} 0 & 4 & 2 \\ -1 & -4 & -1 \\ 0 & 0 & -2 \end{array}\right] .\end{align*}

Find $$J = JCF(A)$$ (so $$A = PJP^{-1}$$) and compute $$P$$.

Definition: Let $$A = (a_{ij})$$, then define that trace of $$A$$ by $$\operatorname{Tr}(A) = \sum_i a_{ii}$$.

The trace satisfies several properties:

• $$\operatorname{Tr}(A+B) = \operatorname{Tr}(A) + \operatorname{Tr}(B)$$,

• $$\operatorname{Tr}(kA) = k\operatorname{Tr}(A)$$,

• $$\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$$.

Theorem: Let $$T: V\to V$$ be a linear transformation with $$\dim V < \infty$$, $$A = [T]_{\mathcal{B}}$$ with respect to some basis, and $$p_T(x)$$ be the characteristic polynomial of $$A$$.

Then \begin{align*} p_T(x) &= x^n + c_{n-1}x^{n-1} + \cdots + c_1 x + c_0, \\ c_0 &= (-1)^n \det(A), \\ c_{n-1} &= -\operatorname{Tr}(A) .\end{align*}

Proof: We have $$p_T(0) = \det(0 I_n - A) = \det(-A) = (-1)^n \det(A)$$.

Compute $$p_T(x)$$ by expanding $$\det{xI - A}$$ along the first row. The first term looks like $$\prod (x-a_{ii})$$, and no other term contributes to the coefficient of $$x^{n-1}$$.

$$\hfill\blacksquare$$

Definition: A Lie Algebra is a vector space with an operation $$[{\,\cdot\,}, {\,\cdot\,}]: V\times V \to V$$ satisfying

1. Bilinearity,

2. $$[x, x] = 0$$,

3. The Jacobi identity $$[x, [y, z]] = [y, [z, x]] + [z, [x, y]] = 0$$.

Examples:

1. $$L = {\mathfrak{gl}}(n, {\mathbb{C}}) = n \times n$$ invertible matrices over $${\mathbb{C}}$$ with $$[A, B] = AB - BA$$.

2. $$L = {\mathfrak{sl}}(n, {\mathbb{C}}) = \left\{{A \in {\mathfrak{gl}}(n, {\mathbb{C}}) \mathrel{\Big|}\operatorname{Tr}(A) = 0}\right\}$$ with the same operation, and it can be checked that \begin{align*} \operatorname{Tr}([A, B]) = \operatorname{Tr}(AB - BA) = \operatorname{Tr}(AB) - \operatorname{Tr}(BA) = 0 .\end{align*}

This turns out to be a simple algebra, and simple algebras over $${\mathbb{C}}$$ can be classified using root systems and Dynkin diagrams – this is given by type $$A_{n-1}$$.

# Preface

These are notes live-tex’d from a graduate Algebra course taught by Dan Nakano at the University of Georgia in Fall 2019. As such, any errors or inaccuracies are almost certainly my own.