Let \(H\) be a normal subgroup of a finite group \(G\) where the order of \(H\) and the index of \(H\) in \(G\) are relatively prime. Prove that no other subgroup of \(G\) has the same order as \(H\).

For a finite group \(G\), let \(c(G)\) denote the number of conjugacy classes of \(G\).

Prove that if two elements of \(G\) are chosen uniformly at random,then the probability they commute is precisely \begin{align*} \frac{c(G)}{{\left\lvert {G} \right\rvert}} .\end{align*}

State the class equation for a finite group.

Using the class equation (or otherwise) show that the probability in part (a) is at most \begin{align*} \frac 1 2 + \frac 1 {2[G : Z(G)]} .\end{align*}

Here, as usual, \(Z(G)\) denotes the center of \(G\).

Concepts Used:

- Notation: \(X/G\) is the set of \(G{\hbox{-}}\)orbits
- Notation: \(X^g = \left\{{x\in x{~\mathrel{\Big|}~}g\cdot x = x}\right\}\)
- Burnside’s formula: \({\left\lvert {G} \right\rvert} {\left\lvert {X/G} \right\rvert} = \sum {\left\lvert {X^g} \right\rvert}\).

**Solution**

Strategy: Burnside.

Define a sample space \(\Omega = G \times G\), so \({\left\lvert {\Omega} \right\rvert} = {\left\lvert {G} \right\rvert}^2\).

Identify the event we want to analyze: \(A \mathrel{\vcenter{:}}=\left\{{(g,h) \in G\times G {~\mathrel{\Big|}~}[g,h] = 1}\right\}\).

- Define and note: \begin{align*} A_g \mathrel{\vcenter{:}}=\left\{{(g, h) {~\mathrel{\Big|}~}h\in H, [g, h] = 1}\right\} \implies A = {\coprod}_{g\in G} A_g .\end{align*}

Set \(n\) be the number of conjugacy classes, note we want to show \(P(A) = n / {\left\lvert {G} \right\rvert}\).

Let \(G\) act on itself by conjugation, which partitions \(G\) into conjugacy classes.

What are the orbits? \begin{align*} \mathcal{O}_g = \left\{{hgh^{-1}{~\mathrel{\Big|}~}h\in G}\right\} ,\end{align*} which is the conjugacy class of \(g\).

What are the fixed points? \begin{align*}X^g = \left\{{h\in G {~\mathrel{\Big|}~}hgh^{-1}= g}\right\},\end{align*} which are the elements of \(G\) that commute with \(g\), which is precisely \(A_g\).

Note \({\left\lvert {X/G} \right\rvert} = n\), the number of conjugacy classes.

Note that \begin{align*} {\left\lvert {A} \right\rvert} = {\left\lvert {{\coprod}_{g\in G} A_g} \right\rvert} = \sum_{g\in G} {\left\lvert {A_g} \right\rvert} = \sum_{g\in G}{\left\lvert {X^g} \right\rvert} .\end{align*}

Apply Burnside \begin{align*} {\left\lvert {X / G} \right\rvert} = \frac { 1 } { | G | } \sum _ { g \in G } \left| X ^ { g } \right|, \end{align*}

Rearrange and use definition: \begin{align*} n {\left\lvert {G} \right\rvert} = {\left\lvert {X/G} \right\rvert} {\left\lvert {G} \right\rvert} = \sum _ { g \in G } \left| X ^ { g } \right| \end{align*}

Compute probability: \begin{align*} P(A) = {{\left\lvert {A} \right\rvert} \over {\left\lvert {\Omega} \right\rvert}} = \frac{\sum_{ g \in G } \left| X ^ { g } \right|}{{\left\lvert {G} \right\rvert}^2} = \frac{{\left\lvert {X/G} \right\rvert}{\left\lvert {G} \right\rvert}}{{\left\lvert {G} \right\rvert}^2} = \frac{n {\left\lvert {G} \right\rvert}}{{\left\lvert {G} \right\rvert}^2} = \frac n {{\left\lvert {G} \right\rvert}} .\end{align*}

\(\hfill\blacksquare\)

Class equation: \begin{align*} {\left\lvert {G} \right\rvert} = Z(G) + \sum_{\substack{\text{One $x$ from each} \\ \text{conjugacy class}}}[G: Z(x)] \end{align*}

where \(Z(x) = \left\{{g\in G {~\mathrel{\Big|}~}[g, x] = 1}\right\}\).

Todo: revisit.

As shown in part 1, \begin{align*} \mathcal{O}_x = \left\{{g\curvearrowright x {~\mathrel{\Big|}~}g\in G}\right\} = \left\{{h\in G {~\mathrel{\Big|}~}ghg^{-1}= h}\right\} = C_G(g) ,\end{align*} and by the class equation

\begin{align*} {\left\lvert {G} \right\rvert} = {\left\lvert {Z(G)} \right\rvert} + \sum_{\substack{\text{One $x$ from each} \\ \text{conjugacy class}}}[G: Z(x)] \end{align*}

Now note

Each element of \(Z(G)\) is in its own conjugacy class, contributing \({\left\lvert {Z(G)} \right\rvert}\) classes to \(n\).

Every other class of elements in \(G\setminus Z(G)\) contains at least 2 elements

- Claim: each such class contributes
**at least**\(\frac 1 2 {\left\lvert {G \setminus Z(G)} \right\rvert}\).

- Claim: each such class contributes

Thus \begin{align*} n &\leq {\left\lvert {Z(G)} \right\rvert} + \frac 1 2{\left\lvert {G \setminus Z(G)} \right\rvert} \\ &= {\left\lvert {Z(G)} \right\rvert} + \frac 1 2{\left\lvert {G} \right\rvert} - \frac 1 2 {\left\lvert {Z(G)} \right\rvert} \\ &= \frac 1 2 {\left\lvert {G} \right\rvert} + \frac 1 2 {\left\lvert {Z(G)} \right\rvert} \\ \\ \implies \frac n {{\left\lvert {G} \right\rvert}} &\leq \frac 1 2 \frac{{\left\lvert {G} \right\rvert}}{{\left\lvert {G} \right\rvert}} + \frac 1 2 \frac{{\left\lvert {Z(G)} \right\rvert}}{{\left\lvert {G} \right\rvert}} \\ &= \frac 1 2 + \frac 1 2 \frac 1 {[G: Z(G)]} .\end{align*}

Let \(G\) be a finite group and \(p\) a prime number such that there is a normal subgroup \(H{~\trianglelefteq~}G\) with \({\left\lvert {H} \right\rvert} = p^i > 1\).

Show that \(H\) is a subgroup of any Sylow \(p{\hbox{-}}\)subgroup of \(G\).

Show that \(G\) contains a nonzero abelian normal subgroup of order divisible by \(p\).

Let \(G\) be a finite group and \(\pi: G\to \operatorname{Sym}(G)\) the Cayley representation.

(Recall that this means that for an element \(x\in G\), \(\pi(x)\) acts by left translation on \(G\).)

Prove that \(\pi(x)\) is an odd permutation \(\iff\) the order \({\left\lvert {\pi(x)} \right\rvert}\) of \(\pi(x)\) is even and \({\left\lvert {G} \right\rvert} / {\left\lvert {\pi(x)} \right\rvert}\) is odd.

Let \(G\) be a finite group and \(s, t\in G\) be two distinct elements of order 2. Show that subgroup of \(G\) generated by \(s\) and \(t\) is a dihedral group.

Recall that the dihedral groups of order \(2m\) for \(m\geq 2\) are of the form \begin{align*} D_{2m} = \left\langle{\sigma, \tau {~\mathrel{\Big|}~}\sigma^m = 1 = \tau^2, \tau \sigma = \sigma^{-1}\tau}\right\rangle .\end{align*}

Let \(G\) be a group containing a subgroup \(H\) not equal to \(G\) of finite index. Prove that \(G\) has a normal subgroup which is contained in every conjugate of \(H\) which is of finite index.

For a prime \(p\), let \(G\) be a finite \(p{\hbox{-}}\)group and let \(N\) be a normal subgroup of \(G\) of order \(p\). Prove that \(N\) is contained in the center of \(G\).

Let \(G\) be a group and \(H, K < G\) be subgroups of finite index. Show that \begin{align*} [G: H\cap K] \leq [G: H] ~ [G:K] .\end{align*}

Let \(P\) be a finite \(p{\hbox{-}}\)group. Prove that every nontrivial normal subgroup of \(P\) intersects the center of \(P\) nontrivially.

Let \(G\) be a group of order \(p^2q\) for \(p, q\) prime. Show that \(G\) has a nontrivial normal subgroup.

Let \(p\) be a prime. Show that \(S_p = \left\langle{\tau, \sigma}\right\rangle\) where \(\tau\) is a transposition and \(\sigma\) is a \(p{\hbox{-}}\)cycle.

Let \(G\) be a nonabelian group of order \(p^3\) for \(p\) prime. Show that \(Z(G) = [G, G]\)

Let \(G\) be a finite group with \(n\) distinct conjugacy classes. Let \(g_1 \cdots g_n\) be representatives of the conjugacy classes of \(G\).

Prove that if \(g_i g_j = g_j g_i\) for all \(i, j\) then \(G\) is abelian.

Concepts used:

- Centralizer: \begin{align*} C_G(h) = Z(h) = \left\{{g\in G {~\mathrel{\Big|}~}[g,h] = 1}\right\} \quad\text{Centralizer} \end{align*}
- Class equation: \begin{align*} {\left\lvert {G} \right\rvert} = \sum_{\substack{\text{One $h$ from each } \\ \text{ conjugacy class}}} \frac{{\left\lvert {G} \right\rvert}}{{\left\lvert {Z(h)} \right\rvert}} \end{align*}
- Notation: \begin{align*} h^g &= ghg^{-1}\\ h^G &= \left\{{ h^g {~\mathrel{\Big|}~}g\in G}\right\} \quad\text{Conjugacy Class}\\ H^g &= \left\{{h^g {~\mathrel{\Big|}~}h\in H}\right\} \\ N_G(H) &= \left\{{g\in G {~\mathrel{\Big|}~}H^g = H}\right\} \supseteq H \quad\text{Normalizer} .\end{align*}

**Solution**:

**Claim 1**: \({\left\lvert {h^G} \right\rvert} = [G: Z(h)]\)

**Claim 2**: \({\left\lvert {\left\{{H^g {~\mathrel{\Big|}~}g\in G}\right\}} \right\rvert} = [G: N_G(H)]\)

*Proof:*Let \(G\curvearrowright\left\{{H {~\mathrel{\Big|}~}H \leq G}\right\}\) by \(H \mapsto gHg^{-1}\).- Then the \(\mathcal O_H\) is the set of conjugate subgroups, \(\mathrm{Stab}(H) = N_G(H)\).
- So Orbit-Stabilizer says \(\mathcal O_h \cong G/\mathrm{Stab}(H)\); then just take sizes.

**Claim 3:** \(\cup_{g\in G} H^g = \cup_{g\in G} gHg^{-1}\subsetneq G\) for any proper \(H \leq G\).

*Proof:*By theorem 2, since each coset is of size \({\left\lvert {H} \right\rvert}\), which only intersect at the identity, and there are exactly \([G: N_G(H)]\) of them \begin{align*} {\left\lvert {\cup_{g\in G} H^g} \right\rvert} &= \qty{ {\left\lvert {H} \right\rvert} - 1} [G: N_G(H)] + 1\\ &= {\left\lvert {H} \right\rvert} [G: N_G(H)] - [G:N_G(H)] + 1\\ &= {\left\lvert {G} \right\rvert} \frac{{\left\lvert {G} \right\rvert}}{{\left\lvert {N_G(H)} \right\rvert}} - \frac{{\left\lvert {G} \right\rvert}}{{\left\lvert {N_G(H)} \right\rvert}} + 1 \\ &\leq {\left\lvert {H} \right\rvert} \frac{{\left\lvert {G} \right\rvert}}{{\left\lvert {H} \right\rvert}} - \frac{{\left\lvert {G} \right\rvert}}{{\left\lvert {H} \right\rvert}} + 1 \\ &= {\left\lvert {G} \right\rvert} - ([G: H] - 1) \\ &< {\left\lvert {G} \right\rvert} ,\end{align*} where we use the fact that \(H \subseteq N_G(H) \implies {\left\lvert {H} \right\rvert} \leq {\left\lvert {N_G(H)} \right\rvert} \implies \frac{1}{{\left\lvert {N_G(H)} \right\rvert}} \leq \frac{1}{{\left\lvert {H} \right\rvert}}\), and since \(H < G\) is proper, \([G:H] \geq 2\).Since \([g_i, g_j] = 1\), we have \(g_i \in Z(g_j)\) for every \(i, j\).

Then \begin{align*} g\in G &\implies g = g_i^h \quad \text{ for some } h \\ &\implies g \in Z(g_\mathbf{j})^h \quad\text{for every } j \text{ since }g_i \in Z(g_j) ~\forall j \\ &\implies g \in \cup_{h\in G} Z(g_j)^h \quad\text{for every } j\\ &\implies G \subseteq \cup_{h\in G} Z(g_j)^h \quad\text{for every } j ,\end{align*}

which by Theorem 3, if \(Z(g_j) < G\) were proper, then the RHS is properly contained in \(G\).

So it must be the case that that \(Z(g_j)\) is not proper and thus equal to \(G\) for every \(j\).

But \(Z(g_i) = G \iff g_i \in Z(G)\), and so each conjugacy class is size one.

So for every \(g\in G\), we have \(g = g_j\) for some \(j\), and thus \(g = g_j \in Z(g_j) = Z(G)\), so \(g\) is central.

Then \(G\subseteq Z(G)\) and \(G\) is abelian.

Let \(G\) be a finite group and let \(P\) be a sylow \(p{\hbox{-}}\)subgroup for \(p\) prime. Show that \(N(N(P)) = N(P)\) where \(N\) is the normalizer in \(G\).

Let \(G\) be a group of order 30.

Show that \(G\) has a subgroup of order 15.

Show that every group of order 15 is cyclic.

Show that \(G\) is isomorphic to some semidirect product \({\mathbb{Z}}_{15} \rtimes{\mathbb{Z}}_2\).

Exhibit three nonisomorphic groups of order 30 and prove that they are not isomorphic. You are not required to use your answer to (c).

Let \(G\subset S_9\) be a Sylow-3 subgroup of the symmetric group on 9 letters.

Show that \(G\) contains a subgroup \(H\) isomorphic to \({\mathbb{Z}}_3 \times{\mathbb{Z}}_3 \times{\mathbb{Z}}_3\) by exhibiting an appropriate set of cycles.

Show that \(H\) is normal in \(G\).

Give generators and relations for \(G\) as an abstract group, such that all generators have order 3.

Also exhibit elements of \(S_9\) in cycle notation corresponding to these generators.

Without appealing to the previous parts of the problem, show that \(G\) contains an element of order 9.

Let \(G\) be a group of order 96.

Show that \(G\) has either one or three 2-Sylow subgroups.

Show that either \(G\) has a normal subgroup of order 32, or a normal subgroup of order 16.

State the three Sylow theorems.

Prove that any group of order 1225 is abelian.

Write down exactly one representative in each isomorphism class of abelian groups of order 1225.

How many isomorphism classes of abelian groups of order 56 are there? Give a representative for one of each class.

Prove that if \(G\) is a group of order 56, then either the Sylow-2 subgroup or the Sylow-7 subgroup is normal.

Give two non-isomorphic groups of order 56 where the Sylow-7 subgroup is normal and the Sylow-2 subgroup is *not* normal. Justify that these two groups are not isomorphic.

Classify the abelian groups of order 36.

For the rest of the problem, assume that \(G\) is a non-abelian group of order 36.

You may assume that the only subgroup of order 12 in \(S_4\) is \(A_4\) and that \(A_4\) has no subgroup of order 6.

Prove that if the 2-Sylow subgroup of \(G\) is normal, \(G\) has a normal subgroup \(N\) such that \(G/N\) is isomorphic to \(A_4\).

Show that if \(G\) has a normal subgroup \(N\) such that \(G/N\) is isomorphic to \(A_4\) and a subgroup \(H\) isomorphic to \(A_4\) it must be the direct product of \(N\) and \(H\).

Show that the dihedral group of order 36 is a non-abelian group of order 36 whose Sylow-2 subgroup is not normal.

Let \(G\) be a group of order 30.

Show that \(G\) contains normal subgroups of orders 3, 5, and 15.

Give all possible presentations and relations for \(G\).

Determine how many groups of order 30 there are up to isomorphism.

Let \(G\) be a finite group whose order is divisible by a prime number \(p\). Let \(P\) be a normal \(p{\hbox{-}}\)subgroup of \(G\) (so \({\left\lvert {P} \right\rvert} = p^c\) for some \(c\)).

Show that \(P\) is contained in every Sylow \(p{\hbox{-}}\)subgroup of \(G\).

Let \(M\) be a maximal proper subgroup of \(G\). Show that either \(P \subseteq M\) or \(|G/M | = p^b\) for some \(b \leq c\).

Concepts Used:

- Sylow 2: All Sylow \(p{\hbox{-}}\)subgroups are conjugate.
- \({\left\lvert {HK} \right\rvert} = {\left\lvert {H} \right\rvert} {\left\lvert {K} \right\rvert} / {\left\lvert {H\cap K} \right\rvert}\).
- Lagrange’s Theorem: \(H\leq G \implies {\left\lvert {H} \right\rvert} \divides {\left\lvert {G} \right\rvert}\)

**Solution**

Every \(p{\hbox{-}}\)subgroup is contained in some Sylow \(p{\hbox{-}}\)subgroup, so \(P \subseteq S_p^i\) for some \(S_p^i \in \mathrm{Syl}_p(G)\).

\(P {~\trianglelefteq~}G \iff gPg^{-1}= P\) for all \(g\in G\).

Let \(S_p^j\) be any other Sylow \(p{\hbox{-}}\)subgroup,

Since Sylow \(p{\hbox{-}}\)subgroups are all conjugate \(gS_p^i g^{-1}= S_p^j\) for

*some*\(g\in G\).Then \begin{align*} P = gPg^{-1}\subseteq gS_p^i g^{-1}= S_p^j .\end{align*}

If \(P\) is not contained in \(M\), then \(M < MP\) is a proper subgroup

By maximality of \(M\), \(MP = G\)

Note that \(M\cap P \leq P\) and \({\left\lvert {P} \right\rvert} = p^c\) implies \({\left\lvert {M\cap P} \right\rvert} = p^a\) for some \(a\leq c\) by Lagrange

Then write \begin{align*} G = MP &\iff {\left\lvert {G} \right\rvert} = \frac{{\left\lvert {M} \right\rvert} {\left\lvert {P} \right\rvert}}{{\left\lvert {M\cap P} \right\rvert}} \\ \\ &\iff { {\left\lvert {G} \right\rvert} \over {\left\lvert {M} \right\rvert}} = {{\left\lvert {P} \right\rvert} \over {\left\lvert {M\cap P} \right\rvert}} = {p^c \over p^a} = p^{c-a} \mathrel{\vcenter{:}}= p^b \end{align*}

where \(a\leq c \implies 0 \leq c-b \leq c\) so \(0\leq b \leq c\).

Let \(G\) be a group of order 105 and let \(P, Q, R\) be Sylow 3, 5, 7 subgroups respectively.

Prove that at least one of \(Q\) and \(R\) is normal in \(G\).

Prove that \(G\) has a cyclic subgroup of order 35.

Prove that both \(Q\) and \(R\) are normal in \(G\).

Prove that if \(P\) is normal in \(G\) then \(G\) is cyclic.

Relevant Concepts:

The \(pqr\) theorem.

Sylow 3: \({\left\lvert {G} \right\rvert} = p^n m\) implies \(n_p \divides m\) and \(n_p \cong 1 \pmod p\).

**Theorem**: If \(H, K \leq G\) and any of the following conditions hold, \(HK\) is a subgroup:- \(H{~\trianglelefteq~}G\) (wlog)
- \([H, K] = 1\)
- \(H \leq N_G(K)\)

**Theorem**: For a positive integer \(n\), all groups of order \(n\) are cyclic \(\iff n\) is squarefree and, for each pair of distinct primes \(p\) and \(q\) dividing \(n\), \(q - 1 \neq 0 \pmod p\).**Theorem:**\begin{align*} A_i{~\trianglelefteq~}G, \quad G = A_1 \cdots A_k,\quad A_k \cap\prod_{i\neq k} A_i = \emptyset \implies G = \prod A_i .\end{align*}The intersection of subgroups is a again a subgroup.

Any subgroups of coprime order intersect trivially?

**Solution**

We have

- \(n_3 \divides 5\cdot 7, \quad n_3 \cong 1 \pmod 3 \implies n_3 \in \left\{{1, 5, 7, 35}\right\} \setminus \left\{{5, 35}\right\}\)
- \(n_5 \divides 3\cdot 7, \quad n_5 \cong 1 \pmod 5 \implies n_5 \in \left\{{1, 3, 7, 21}\right\}\setminus \left\{{3, 7}\right\}\)
- \(n_7 \divides 3\cdot 5, \quad n_7 \cong 1 \pmod 7 \implies n_7 \in \left\{{1, 3, 5, 15}\right\}\setminus\left\{{3, 5}\right\}\)

Thus \begin{align*} n_3 \in \left\{{1, 7}\right\} \quad n_5 \in \left\{{1, 21}\right\} \quad n_7 \in \left\{{1, 15}\right\} .\end{align*}

Toward a contradiction, if \(n_5\neq 1\) and \(n_7 \neq 1\), then \begin{align*} {\left\lvert {{\operatorname{Syl}}(5) \cup{\operatorname{Syl}}(7)} \right\rvert} = (5-1)n_5 + (7-1)n_7 + 1 &= 4(21) + 6(15) = 174 > 105 \text{ elements} \end{align*} using the fact that Sylow \(p{\hbox{-}}\)subgroups for distinct primes \(p\) intersect trivially (?).

By (a), either \(Q\) or \(R\) is normal. Thus \(QR \leq G\) is a subgroup, and it has order \({\left\lvert {Q} \right\rvert} \cdot {\left\lvert {R} \right\rvert} = 5\cdot 7 = 35\).

By the \(pqr\) theorem, since \(5\) does not divide \(7-1=6\), \(QR\) is cyclic.

We want to show \(Q, R{~\trianglelefteq~}G\), so we proceed by showing \(\textbf{not }\qty{n_5 = 21 \text{ or } n_7 = 15}\), which is equivalent to \(\qty{n_5 = 1 \text{ and } n_7 = 1}\) by the previous restrictions.

Note that we can write \begin{align*} G = \left\{{\text{elements of order } n}\right\} {\coprod}\left\{{\text{elements of order not } n}\right\} .\end{align*} for any \(n\), so we count for \(n=5, 7\):

- Elements in \(QR\) of order
**not**equal to 5: \({\left\lvert {QR - Q\left\{{\operatorname{id}}\right\} + \left\{{\operatorname{id}}\right\}} \right\rvert} = 35 - 5 + 1 = 31\) - Elements in \(QR\) of order
**not**equal to 7: \({\left\lvert {QR - \left\{{\operatorname{id}}\right\}R + \left\{{\operatorname{id}}\right\}} \right\rvert} = 35 - 7 + 1 = 29\)

Since \(QR \leq G\), we have

- Elements in \(G\) of order
**not**equal to 5 \(\geq 31\). - Elements in \(G\) of order
**not**equal to 7 \(\geq 29\).

Now both cases lead to contradictions:

\(n_5 = 21\): \begin{align*} {\left\lvert {G} \right\rvert} &= {\left\lvert {\left\{{\text{elements of order } 5}\right\} {\coprod}\left\{{\text{elements of order not } 5}\right\}} \right\rvert} \\ &\geq n_5(5-1) + 31 = 21(4) + 31 = 115 > 105 = {\left\lvert {G} \right\rvert} .\end{align*}

\(n_7 = 15\): \begin{align*} {\left\lvert {G} \right\rvert} &= {\left\lvert {\left\{{\text{elements of order } 7}\right\} {\coprod}\left\{{\text{elements of order not } 7}\right\}} \right\rvert} \\ &\geq n_7(7-1) + 29 = 15(6) + 29 = 119 > 105 = {\left\lvert {G} \right\rvert} .\end{align*}

Suppose \(P\) is normal and recall \({\left\lvert {P} \right\rvert} = 3, {\left\lvert {Q} \right\rvert} = 5, {\left\lvert {R} \right\rvert} = 7\).

- \(P\cap QR = \left\{{e}\right\}\) since \((3, 35) = 1\)
- \(R\cap PQ = \left\{{e}\right\}\) since \((5, 21) = 1\)
- \(Q\cap RP = \left\{{e}\right\}\) since \((7, 15) = 1\)

We also have \(PQR = G\) since \({\left\lvert {PQR} \right\rvert} = {\left\lvert {G} \right\rvert}\) (???).

We thus have an internal direct product \begin{align*} G \cong P\times Q \times R \cong {\mathbb{Z}}_3 \times{\mathbb{Z}}_5 \times{\mathbb{Z}}_7 \cong {\mathbb{Z}}_{105} .\end{align*} by the Chinese Remainder Theorem, which is cyclic.

Let \(G\) be a finite group and \(X\) a set on which \(G\) acts.

Let \(x\in X\) and \(G_x \mathrel{\vcenter{:}}=\left\{{g\in G {~\mathrel{\Big|}~}g\cdot x = x}\right\}\). Show that \(G_x\) is a subgroup of \(G\).

Let \(x\in X\) and \(G\cdot x \mathrel{\vcenter{:}}=\left\{{g\cdot x {~\mathrel{\Big|}~}g\in G}\right\}\). Prove that there is a bijection between elements in \(G\cdot x\) and the left cosets of \(G_x\) in \(G\).

Let \(G\) be a finite group, \(H\) a \(p{\hbox{-}}\)subgroup, and \(P\) a sylow \(p{\hbox{-}}\)subgroup for \(p\) a prime. Let \(H\) act on the left cosets of \(P\) in \(G\) by left translation.

Prove that this is an orbit under this action of length 1.

Prove that \(xP\) is an orbit of length 1 \(\iff H\) is contained in \(xPx^{-1}\).

Let \(G\) be a finite group acting on a set \(X\). For \(x\in X\), let \(G_x\) be the stabilizer of \(x\) and \(G\cdot x\) be the orbit of \(x\).

Prove that there is a bijection between the left cosets \(G/G_x\) and \(G\cdot x\).

Prove that the center of every finite \(p{\hbox{-}}\)group \(G\) is nontrivial by considering that action of \(G\) on \(X=G\) by conjugation.

Suppose the group \(G\) acts on the set \(A\). Assume this action is faithful (recall that this means that the kernel of the homomorphism from \(G\) to \(\operatorname{Sym}(A)\) which gives the action is trivial) and transitive (for all \(a, b\) in \(A\), there exists \(g\) in \(G\) such that \(g \cdot a = b\).)

For \(a \in A\), let \(G_a\) denote the stabilizer of \(a\) in \(G\). Prove that for any \(a \in A\), \begin{align*} \bigcap_{\sigma\in G} \sigma G_a \sigma^{-1}= \left\{{1}\right\} .\end{align*}

Suppose that \(G\) is abelian. Prove that \(|G| = |A|\). Deduce that every abelian transitive subgroup of \(S_n\) has order \(n\).

Suppose the group \(G\) acts on the set \(X\) . Show that the stabilizers of elements in the same orbit are conjugate.

Let \(G\) be a finite group and let \(H\) be a proper subgroup. Show that the union of the conjugates of \(H\) is strictly smaller than \(G\), i.e. \begin{align*} \bigcup_{g\in G} gHg^{-1}\subsetneq G \end{align*}

Suppose \(G\) is a finite group acting transitively on a set \(S\) with at least 2 elements. Show that there is an element of \(G\) with no fixed points in \(S\).

Concepts used:

- Orbit: \(G\cdot x \mathrel{\vcenter{:}}=\left\{{g\cdot x {~\mathrel{\Big|}~}g\in G}\right\} \subseteq X\)
- Stabilizer: \(G_x \mathrel{\vcenter{:}}=\left\{{g\in G{~\mathrel{\Big|}~}g\cdot x = x}\right\} \leq G\)
- Orbit-Stabilizer: \(G\cdot x \simeq G/G_x\).
- \(abc\in H \iff b\in a^{-1}H c^{-1}\)
- Set of orbits for \(G\curvearrowright X\), notated \(X/G\).
- Set of fixed points for \(G\curvearrowright X\), notated \(X^g\).
- Burnside’s Lemma: \({\left\lvert {X/G} \right\rvert} \cdot {\left\lvert {G} \right\rvert} = \sum_{g\in G} {\left\lvert {X^g} \right\rvert}\)
- Number of orbits equals average number of fixed points.

**Solution**

- Fix \(x\) and let \(y\in G_x\) be another element in the orbit of \(x\).
- Then there exists a \(g\in G\) such that \(g\cdot x = y\), so \(x = g^{-1}\cdot y\)
- Then \begin{align*} h \in G\cdot x &\iff h\cdot x = x && \text{by being in the stabilizer} \\ &\iff h\cdot (g^{-1}\cdot y) = g^{-1}\cdot y && \text{using that $x, y$ are in the same orbit} \\ &\iff (g h g^{-1}) \cdot y = y \\ &\iff ghg^{-1}\in G_y && \text{by the defn of the stabilizer}\\ &\iff h\in g ^{-1} G_y g ,\end{align*}

so every \(h\in G\cdot x\) is conjugate to some element in \(G_y\).

Let \(G\) act on its subgroups by conjugation,

The orbit \(G\cdot H\) is the set of all subgroups conjugate to \(H\), and

The stabilizer of \(H\) is \(G_H = N_G(H)\).

By orbit-stabilizer, \begin{align*} G\cdot H = [G: G_H] = [G: N_G(H)] .\end{align*}

Since \({\left\lvert {H} \right\rvert} = n\), and all of its conjugate also have order \(n\).

Note that \begin{align*} H\leq N_G(H) \implies {\left\lvert {H} \right\rvert} \leq {\left\lvert {N_G(H)} \right\rvert} \implies {1\over {\left\lvert {N_G(H)} \right\rvert}} \leq {1\over {\left\lvert {H} \right\rvert}} ,\end{align*}

Now

*strictly*bound the size of the union by overcounting their intersections at the identity: \begin{align*} {\left\lvert {\bigcup_{g\in G}gHg^{-1}} \right\rvert} &< (\text{Number of Conjugates of } H) \cdot (\text{Size of each conjugate}) \\ & \text{strictly overcounts since they intersect in at least the identity} \\ &= [G: N_G(H)] {\left\lvert {H} \right\rvert} \\ &= {{\left\lvert {G} \right\rvert} \over {\left\lvert {N_G(H)} \right\rvert}} {\left\lvert {H} \right\rvert} \\ & \text{since $G$ is finite} \\ &\leq {{\left\lvert {G} \right\rvert} \over {\left\lvert {H} \right\rvert}} {\left\lvert {H} \right\rvert} \\ &= {\left\lvert {G} \right\rvert} .\end{align*}

- Let \(G\curvearrowright X\) transitively where \({\left\lvert {X} \right\rvert} \geq 2\)
- An action is transitive iff there is only one orbit, so \({\left\lvert {X/G} \right\rvert} = 1\).
- Apply Burnside’s Lemma \begin{align*} 1 = {\left\lvert {X/G} \right\rvert} = \frac{1}{{\left\lvert {G} \right\rvert}} \sum_{g\in G} {\left\lvert {X^g} \right\rvert} \implies {\left\lvert {G} \right\rvert} = \sum_{g\in G} {\left\lvert {X^g} \right\rvert} \end{align*}
- Note that \(X^e = X\), since the identity must fix every element, so \({\left\lvert {X^e} \right\rvert} \geq 2\).
- Not
*every*other term in the sum can be greater than 1, otherwise the RHS is greater than the size of \(G\) - Thus we must have \({\left\lvert {X^g} \right\rvert} = 0\) for some \(g\in G\), i.e. \(g\) has no fixed points in \(X\).

Show that any group of order 2020 is solvable.

Give (without proof) a classification of all abelian groups of order 2020.

Describe one nonabelian group of order 2020.

How many isomorphism classes are there of groups of order 45?

Describe a representative from each class.

Concepts used:

- Sylow theorems:
- \(n_p \cong 1 \pmod p\)
- \(n_p \divides m\).

**Solution**

It turns out that \(n_3 = 1\) and \(n_5 = 1\), so \(G \cong S_3 \times S_5\) since both subgroups are normal.

There is only one possibility for \(S_5\), namely \(S_5\cong {\mathbb{Z}}/(5)\).

There are two possibilities for \(S_3\), namely \(S_3 \cong {\mathbb{Z}}/(3^2)\) and \({\mathbb{Z}}/(3)^2\).

Thus

- \(G \cong {\mathbb{Z}}/(9) \times{\mathbb{Z}}/(5)\), or
- \(G \cong {\mathbb{Z}}/(3)^2 \times{\mathbb{Z}}/(5)\).

Let \(G\) be a group of order 70.

Show that \(G\) is not simple.

Exhibit 3 nonisomorphic groups of order 70 and prove that they are not isomorphic.

How many groups are there up to isomorphism of order \(pq\) where \(p<q\) are prime integers?

Use the Class Equation (equivalently, the conjugation action of a group on itself) to prove that any \(p{\hbox{-}}\)group (a group whose order is a positive power of a prime integer \(p\)) has a nontrivial center.

Prove that any group of order \(p^2\) (where \(p\) is prime) is abelian.

Prove that any group of order \(5^2 \cdot 7^2\) is abelian.

Write down exactly one representative in each isomorphism class of groups of order \(5^2 \cdot 7^2\).

Concepts Used:

Centralizer: \(C_G(x) = \left\{{g\in G {~\mathrel{\Big|}~}[gx] = 1}\right\}\).

Class Equation: \({\left\lvert {G} \right\rvert} = {\left\lvert {Z(G)} \right\rvert} + \sum [G: C_G(x_i)]\)

\(G/Z(G)\) cyclic \(\iff G\) is abelian.

*Proof:*\begin{align*} G/Z(G) = \left\langle{xZ}\right\rangle &\iff g\in G \implies gZ = x^mZ \\ &\iff g(x^m)^{-1}\in Z \\ &\iff g = x^m z {\quad \operatorname{for some} \quad}z\in Z\\ &\implies gh = x^mz_1 x^n z_2 = x^n z_2 x^m z_1 = hg .\end{align*}Every group of order \(p^2\) is abelian.

Classification of finite abelian groups.

Strategy: get \(p\) to divide \({\left\lvert {Z(G)} \right\rvert}\).

Apply the class equation: \begin{align*} {\left\lvert {G} \right\rvert} = {\left\lvert {Z(G)} \right\rvert} + \sum [G: C_G(x_i)] .\end{align*}

Since \(C_G(x_i) \leq G\) and \({\left\lvert {G} \right\rvert} = p^k\), by Lagrange \({\left\lvert {C_G(x_i)} \right\rvert} = p^\ell\) for some \(0\leq \ell \leq k\).

Since \({\left\lvert {G} \right\rvert} = p^k\) for some \(k\) and \(Z(G), C_G(x_i) \leq G\) are subgroups, their orders are powers of \(p\).

Use \begin{align*}[G: C_G(x_i)] = 1 \iff C_G(x_i) = G \iff \left\{{g\in G{~\mathrel{\Big|}~}gx_ig^{-1}= x_i}\right\} = G \iff x_i \in Z(G).\end{align*}

- Thus every index appearing in the sum is greater than 1, and thus equal to \(p^{\ell_i}\) for some \(1\leq \ell_i \leq k\)
- So \(p\) divides every term in the sum

Rearrange \begin{align*} {\left\lvert {G} \right\rvert} - \sum [G: C_G(x_i)] = {\left\lvert {Z(G)} \right\rvert} .\end{align*}

\(p\) divides both terms on the LHS, so must divide the RHS, so \({\left\lvert {Z(G)} \right\rvert} \geq p\).

Strategy: examine \({\left\lvert {G/Z(G)} \right\rvert}\) by cases.

- \(1\): Then \(G = Z(G)\) and \(G\) is abelian.
- \(p\): Then \(G/Z(G)\) is cyclic so \(G\) is abelian
- \(p^2\): Not possible, since \({\left\lvert {Z(G)} \right\rvert} > 1\) by (a).

By Sylow

- \(n_5 \divides 7^2,\quad n_5\cong 1\pmod 5 \implies n_5\in\left\{{1, 7, 49}\right\}\setminus\left\{{7, 49}\right\} = \left\{{1}\right\} \implies n_5 = 1\)
- \(n_7 \divides 5^2, \quad n_7 \cong 1 \pmod 7 \implies n_7 \in \left\{{1, 5, 25}\right\}\setminus\left\{{5, 25}\right\} =\left\{{1}\right\} \implies n_7 = 1\)

By recognition of direct products, \(G = S_5 \times S_7\)

- By above, \(S_5, S_7{~\trianglelefteq~}G\)
- Check \(S_5\cap S_7 = \left\{{e}\right\}\) since they have coprime order.
- Check \(S_5S_7 = G\) since \({\left\lvert {S_5 S_7} \right\rvert} = 5^2 7^2 = {\left\lvert {G} \right\rvert}\)

By (b), \(S_5, S_7\) are abelian since they are groups of order \(p^2\)

The direct product of abelian groups is abelian.

- \({\mathbb{Z}}_{5^2} \times{\mathbb{Z}}_{7^2}\)
- \({\mathbb{Z}}_{5}^2 \times{\mathbb{Z}}_{7^2}\)
- \({\mathbb{Z}}_{5^2} \times{\mathbb{Z}}_{7}^2\)
- \({\mathbb{Z}}_{5}^2 \times{\mathbb{Z}}_{7}^2\)

Define what it means for a group \(G\) to be *solvable*.

Show that every group \(G\) of order 36 is solvable.

Hint: you can use that \(S_4\) is solvable.

Let \(N\) be a positive integer, and let \(G\) be a finite group of order \(N\).

Let \(\operatorname{Sym}G\) be the set of all bijections from \(G\to G\) viewed as a group under composition. Note that \(\operatorname{Sym}G \cong S_N\). Prove that the Cayley map \begin{align*} C: G&\to \operatorname{Sym}G\\ g &\mapsto (x\mapsto gx) \end{align*} is an injective homomorphism.

Let \(\Phi: \operatorname{Sym}G\to S_N\) be an isomorphism. For \(a\in G\) define \(\varepsilon(a) \in \left\{{\pm 1}\right\}\) to be the sign of the permutation \(\Phi(C(a))\). Suppose that \(a\) has order \(d\). Prove that \(\varepsilon(a) = -1 \iff d\) is even and \(N/d\) is odd.

Suppose \(N> 2\) and \(n\equiv 2 \pmod 4\). Prove that \(G\) is not simple.

Hint: use part (b).

Let \(p, n\) be integers such that \(p\) is prime and \(p\) does not divide \(n\). Find a real number \(k = k (p, n)\) such that for every integer \(m\geq k\), every group of order \(p^m n\) is not simple.

Let \(p, q\) be distinct primes.

Let \(\mkern 1.5mu\overline{\mkern-1.5muq\mkern-1.5mu}\mkern 1.5mu \in {\mathbb{Z}}_p\) be the class of \(q\pmod p\) and let \(k\) denote the order of \(\mkern 1.5mu\overline{\mkern-1.5muq\mkern-1.5mu}\mkern 1.5mu\) as an element of \({\mathbb{Z}}_p^{\times}\).

Prove that no group of order \(pq^k\) is simple.

Let \(G\) be a group of order \(pq\), and prove that \(G\) is not simple.

Define a *simple group*.

Prove that a group of order 56 can not be simple.

Show that there exist no simple groups of order 148.

Let \(R\) be a ring and \(f: M\to N\) and \(g: N\to M\) be \(R{\hbox{-}}\)module homomorphisms such that \(g\circ f = \operatorname{id}_M\). Show that \(N \cong \operatorname{im}f \oplus \ker g\).

Let \(R\) be a ring with the property that for every \(a \in R, a^2 = a\).

Prove that \(R\) has characteristic 2.

Prove that \(R\) is commutative.

Just fiddling with computations. Context hints that we should be considering things like \(x^2\) and \(a+b\).

\begin{align*} 2a = (2a)^2 = 4a^2 = 4a \implies 2a = 0 .\end{align*}

Note that this implies \(x = -x\) for all \(x\in R\).

\begin{align*} a+b = (a+b)^2 &= a^2 + ab + ba + b^2 = a + ab + ba + b \\ &\implies ab + ba = 0 \\ &\implies ab = -ba \\ &\implies ab = ba \quad\text{by (a)} .\end{align*}

Let \(R\) be a commutative ring with multiplicative identity. Assume Zorn’s Lemma.

Show that \begin{align*} N = \{r \in R \mathrel{\Big|}r^n = 0 \text{ for some } n > 0\} \end{align*} is an ideal which is contained in any prime ideal.

Let \(r\) be an element of \(R\) not in \(N\). Let \(S\) be the collection of all proper ideals of \(R\) not containing any positive power of \(r\). Use Zorn’s Lemma to prove that there is a prime ideal in \(S\).

Suppose that \(R\) has exactly one prime ideal \(P\) . Prove that every element \(r\) of \(R\) is either nilpotent or a unit.

Prime ideal: \(\mathfrak{p}\) is prime iff \(ab \in \mathfrak{p} \implies a\in \mathfrak{p}\) or \(b\in \mathfrak{p}\). Silly fact: 0 is in every ideal!

Zorn’s Lemma:Given a poset, if every chain has an upper bound, then there is a maximal element. (Chain: totally ordered subset.)

Corollary:If \(S\subset R\) is multiplicatively closed with \(0\not\in S\) then \(\left\{{I {~\trianglelefteq~}R {~\mathrel{\Big|}~}J\cap S = \emptyset}\right\}\) has a maximal element. (TODO: PROVE)

Theorem:If \(R\) is commutative, maximal \(\implies\) prime for ideals. (TODO: PROVE)

Theorem:Non-units are contained in a maximal ideal. (See HW?)

Let \(\mathfrak{p}\) be prime and \(x\in N\). Then \(x^k = 0 \in \mathfrak{p}\) for some \(k\), and thus \(x^k = x x^{k-1} \in \mathfrak p\). Since \(\mathfrak p\) is prime, inductively we obtain \(x\in\mathfrak p\).

Let \(S = \left\{{r^k \mathrel{\Big|}k\in {\mathbb{N}}}\right\}\) be the set of positive powers of \(r\). Then \(S^2 \subseteq S\), since \(r^{k_1}r^{k_2} = r^{k_1+k_2}\) is also a positive power of \(r\), and \(0\not\in S\) since \(r\neq 0\) and \(r\not\in N\).

By the corollary, \(\left\{{I {~\trianglelefteq~}R {~\mathrel{\Big|}~}I\cap S = \emptyset}\right\}\) has a maximal element \(\mathfrak{p}\).

Since \(R\) is commutative, \(\mathfrak{p}\) is prime.

Suppose \(R\) has a unique prime ideal \(\mathfrak{p}\).

Suppose \(r\in R\) is not a unit, and toward a contradiction, suppose that \(r\) is also not nilpotent.

Since \(r\) is not a unit, \(r\) is contained in some maximal (and thus prime) ideal, and thus \(r \in \mathfrak{p}\).

Since \(r\not\in N\), by (b) there is a maximal ideal \(\mathfrak{m}\) that avoids all positive powers of \(r\). Since \(\mathfrak{m}\) is prime, we must have \(\mathfrak{m} = \mathfrak{p}\). But then \(r\not\in \mathfrak{p}\), a contradiction.

Let \(R\) be a commutative ring with 1.

Recall that \(x \in R\) is nilpotent iff \(xn = 0\) for some positive integer \(n\).

Show that every proper ideal of \(R\) is contained within a maximal ideal.

Let \(J(R)\) denote the intersection of all maximal ideals of \(R\).

Show that \(x \in J(R) \iff 1 + rx\) is a unit for all \(r \in R\).

Suppose now that \(R\) is finite. Show that in this case \(J(R)\) consists precisely of the nilpotent elements in R.

Define the set of proper ideals \begin{align*} S = \left\{{J {~\mathrel{\Big|}~}I \subseteq J < R}\right\} ,\end{align*}

which is a poset under set inclusion.

Given a chain \(J_1 \subseteq \cdots\), there is an upper bound \(J \mathrel{\vcenter{:}}=\cup J_i\), so Zorn’s lemma applies.

\(\implies\):

We will show that \(x\in J(R) \implies 1+x \in R^{\times}\), from which the result follows by letting \(x=rx\).

Let \(x\in J(R)\), so it is in every maximal ideal, and suppose toward a contradiction that \(1+x\) is **not** a unit.

Then consider \(I = \left\langle{1+x}\right\rangle {~\trianglelefteq~}R\). Since \(1+x\) is not a unit, we can’t write \(s(1+x) = 1\) for any \(s\in R\), and so \(1 \not\in I\) and \(I\neq R\)

So \(I < R\) is proper and thus contained in some maximal proper ideal \(\mathfrak{m} < R\) by part (1), and so we have \(1+x \in \mathfrak{m}\). Since \(x\in J(R)\), \(x\in \mathfrak{m}\) as well.

But then \((1+x) - x = 1 \in \mathfrak{m}\) which forces \(\mathfrak{m} = R\).

\(\impliedby\)

Fix \(x\in R\), and suppose \(1+rx\) is a unit for all \(r\in R\).

Suppose towards a contradiction that there is a maximal ideal \(\mathfrak{m}\) such that \(x\not \in \mathfrak{m}\) and thus \(x\not\in J(R)\).

Consider \begin{align*} M' \mathrel{\vcenter{:}}=\left\{{rx + m {~\mathrel{\Big|}~}r\in R,~ m\in M}\right\} .\end{align*}

Since \(\mathfrak{m}\) was maximal, \(\mathfrak{m} \subsetneq M'\) and so \(M' = R\).

So every element in \(R\) can be written as \(rx + m\) for some \(r\in R, m\in M\). But \(1\in R\), so we have \begin{align*} 1 = rx + m .\end{align*}

So let \(s = -r\) and write \(1 = sx - m\), and so \(m = 1 + sx\).

Since \(s\in R\) by assumption \(1+sx\) is a unit and thus \(m \in \mathfrak{m}\) is a unit, a contradiction.

So \(x\in \mathfrak{m}\) for every \(\mathfrak{m}\) and thus \(x\in J(R)\).

- \(\mathfrak N(R) = \left\{{x\in R {~\mathrel{\Big|}~}x^n = 0 \text{ for some } n}\right\}\).
- \(J(R) = \operatorname{Spec}_{\text{max}}(R) = \displaystyle\cap_{{\mathfrak{m}}\text{ maximal}} {\mathfrak{m}}\).

We want to show \(J(R) = \mathfrak N(R)\).

\(\mathfrak N(R) \subseteq J(R)\):

We’ll use the fact \(x\in \mathfrak N(R) \implies x^n = 0 \implies 1 + rx\) is a unit \(\iff x\in J(R)\) by (b): \begin{align*} \sum_{k=1}^{n-1} (-x)^k = \frac{1 - (-x)^n}{1- (-x)} = (1+x)^{-1} .\end{align*}

\(J(R) \subseteq \mathfrak N(R)\):

Let \(x \in J(R) \setminus \mathfrak N(R)\).

Since \(R\) is finite, \(x^m = x\) for some \(m > 0\). Without loss of generality, we can suppose \(x^2 = x\) by replacing \(x^m\) with \(x^{2m}\).

If \(1-x\) is not a unit, then \(\left\langle{1-x}\right\rangle\) is a nontrivial proper ideal, which by (a) is contained in some maximal ideal \({\mathfrak{m}}\). But then \(x\in {\mathfrak{m}}\) and \(1-x \in {\mathfrak{m}}\implies x + (1-x) = 1 \in {\mathfrak{m}}\), a contradiction.

So \(1-x\) is a unit, so let \(u = (1-x)^{-1}\).

Then \begin{align*} (1-x)x &= x - x^2 = x - x = 0 \\ &\implies u (1-x)x = x = 0 \\ &\implies x=0 .\end{align*}

Let \(R\) be a commutative ring.

Let \(r \in R\). Show that the map \begin{align*} r\bullet : R &\to R \\ x &\mapsto r x .\end{align*} is an \(R{\hbox{-}}\)module endomorphism of \(R\).

We say that \(r\) is a **zero-divisor** if r\(\bullet\) is not injective. Show that if \(r\) is a zero-divisor and \(r \neq 0\), then the kernel and image of \(R\) each consist of zero-divisors.

Let \(n \geq 2\) be an integer. Show: if \(R\) has exactly \(n\) zero-divisors, then \(\#R \leq n^2\) .

Show that up to isomorphism there are exactly two commutative rings \(R\) with precisely 2 zero-divisors.

You may use without proof the following fact: every ring of order 4 is isomorphic to exactly one of the following: \begin{align*} \frac{ {\mathbb{Z}}}{ 4{\mathbb{Z}}}, \quad \frac{ \frac{ {\mathbb{Z}}}{ 2{\mathbb{Z}}} [t]}{(t^2 + t + 1)}, \quad \frac{ \frac{ {\mathbb{Z}}}{ 2{\mathbb{Z}}} [t]}{ (t^2 - t)}, \quad \frac{ \frac{ {\mathbb{Z}}}{2{\mathbb{Z}}}[t]}{(t^2 )} .\end{align*}

Let \(\phi\) denote the map in question, it suffices to show that \(\phi\) is \(R{\hbox{-}}\)linear, i.e. \(\phi(s\mathbf{x} + \mathbf{y}) = s\phi(\mathbf{x}) + \phi(\mathbf{y})\):

\begin{align*} \phi(s\mathbf{x} + \mathbf{y}) &= r(s\mathbf{x} + \mathbf{y}) \\ &= rs\mathbf{x} + r\mathbf{y} \\ &= s(r\mathbf{x}) + (r\mathbf{y}) \\ &= s\phi(\mathbf{x}) + \phi(\mathbf{y}) .\end{align*}

We identify \(\ker \phi = \left\{{x\in R {~\mathrel{\Big|}~}rx = 0}\right\}\), and since \(r\neq 0\) by assumption, this implies each such \(x\) is a zero divisor by definition (and \(\ker \phi\) is nonempty by assumption).

Similarly, we identify \(\operatorname{im}\phi = \left\{{y = rx {~\mathrel{\Big|}~}x\in R}\right\}\). So let \(y\in \operatorname{im}\phi\). Since \(r\) is a zero divisor, there exists some \(z\in R\) such that \(rz = 0\).

But then \begin{align*} yz = rxz = xrz = x\cdot 0 = 0 \end{align*} since \(R\) is commutative, so \(y\) is a zero divisor.

See 1964 Annals “Properties of rings with a finite number of zero divisors”

Let \(Z \mathrel{\vcenter{:}}=\left\{{z_i}\right\}_{i=1}^n\) be the set of \(n\) zero divisors in \(R\). Let \(\phi_i\) be the \(n\) maps \(x \mapsto z_i x\), and let \(K_i = \ker \phi_i\) be the corresponding kernels.

Fix an \(i\). By (b), \(K_i\) consists of zero divisors, so \begin{align*} {\left\lvert {K_i} \right\rvert} \leq n < \infty \quad \text{for each } i .\end{align*}

Now consider \(R/K_i \mathrel{\vcenter{:}}=\left\{{r + K_i}\right\}\). By the first isomorphism theorem, \(R/K_i \cong \operatorname{im}\phi\), and by (b) every element in the image is a zero divisor, so \begin{align*} [R: K_i] = {\left\lvert {R/K_i} \right\rvert} = {\left\lvert {\operatorname{im}\phi_i} \right\rvert} \leq n .\end{align*}

But then \begin{align*} {\left\lvert {R} \right\rvert} = [R:K_i]\cdot {\left\lvert {K_i} \right\rvert} \leq n\cdot n = n^2 .\end{align*}

By (c), if there are exactly 2 zero divisors then \({\left\lvert {R} \right\rvert} \leq 4\). Since every element in a finite ring is either a unit or a zero divisor, and \({\left\lvert {R^{\times}} \right\rvert} \geq 2\) since \(\pm 1\) are always units, we must have \({\left\lvert {R} \right\rvert} = 4\).

Since the characteristic of a ring must divide its size, we have \(\operatorname{ch}R = 2\) or \(4\).

Using the hint, we see that only \({\mathbb{Z}}/(4)\) has characteristic 4, which has exactly 2 zero divisors given by \([0]_4\) and \([2]_4\).

If \(R\) has characteristic 2, we can check the other 3 possibilities.

We can write \({\mathbb{Z}}/(2)[t]/(t^2) = \left\{{a + bt {~\mathrel{\Big|}~}a,b\in {\mathbb{Z}}/(2)}\right\}\), and checking the multiplication table we have

\begin{align*} \begin{array}{c|cccc} & 0 & 1 & t & 1+t \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & t & 1+t \\ t & 0 & t & \mathbf{0} & t \\ 1 + t & 0 & 1+t & t & 1 \\ \end{array} ,\end{align*}

and so we find that \(t, 0\) are the zero divisors.

In \({\mathbb{Z}}/(2)[t]/(t^2 - t)\), we can check that \(t^2 = t \implies t t^2 = t^2 \implies t(t^2 + 1) = 0 \implies t(t+1) = 0\), so both \(t\) and \(t+1\) are zero divisors, along with zero, so this is not a possibility.

Similarly, in \({\mathbb{Z}}/(2)[t]/(t^2 + t + 1)\), we can check the bottom-right corner of the multiplication table to find \begin{align*} \left[\begin{array}{c|cc} & t & 1 +t \\ \hline t & 1+t & 1 \\ t & 1 & t \\ \end{array}\right] ,\end{align*}

and so this ring only has one zero divisor.

Thus the only possibilities are:

\begin{align*} R &\cong {\mathbb{Z}}/(4) \\ R &\cong {\mathbb{Z}}/(2)[t] / (t^2) .\end{align*}

Let $ \begin{align*} M=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right) \quad \text{and} \quad N=\left(\begin{array}{cc}{x} & {u} \\ {-y} & {-v}\end{array}\right) \end{align*}

over a commutative ring \(R\), where \(b\) and \(x\) are units of \(R\). Prove that \begin{align*} M N=\left(\begin{array}{ll}{0} & {0} \\ {0} & {*}\end{array}\right) \implies MN = 0 .\end{align*}

Let \(R = C[0, 1]\) be the ring of continuous real-valued functions on the interval \([0, 1]\). Let I be an ideal of \(R\).

Show that if \(f \in I, a \in [0, 1]\) are such that \(f (a) \neq 0\), then there exists \(g \in I\) such that \(g(x) \geq 0\) for all \(x \in [0, 1]\), and \(g(x) > 0\) for all \(x\) in some open neighborhood of \(a\).

If \(I \neq R\), show that the set \(Z(I) = \{x \in [0, 1] {~\mathrel{\Big|}~}f(x) = 0 \text{ for all } f \in I\}\) is nonempty.

Show that if \(I\) is maximal, then there exists \(x_0 \in [0, 1]\) such that \(I = \{ f \in R {~\mathrel{\Big|}~}f (x_0 ) = 0\}\).

A ring \(R\) is called *simple* if its only two-sided ideals are \(0\) and \(R\).

Suppose \(R\) is a commutative ring with 1. Prove \(R\) is simple if and only if \(R\) is a field.

Let \(k\) be a field. Show the ring \(M_n (k)\), \(n \times n\) matrices with entries in \(k\), is a simple ring.

For a ring \(R\), let \(U(R)\) denote the multiplicative group of units in \(R\). Recall that in an integral domain \(R\), \(r \in R\) is called *irreducible* if \(r\) is not a unit in R, and the only divisors of \(r\) have the form \(ru\) with \(u\) a unit in \(R\).

We call a non-zero, non-unit \(r \in R\) *prime* in \(R\) if \(r \divides ab \implies r \divides a\) or \(r \divides b\). Consider the ring \(R = \{a + b \sqrt{-5}{~\mathrel{\Big|}~}a, b \in Z\}\).

Prove \(R\) is an integral domain.

Show \(U(R) = \{\pm1\}\).

Show \(3, 2 + \sqrt{-5}\), and \(2 - \sqrt{-5}\) are irreducible in \(R\).

Show 3 is not prime in \(R\).

Conclude \(R\) is not a PID.

Let \(R\) be a commutative ring with 1. Suppose that \(M\) is a free \(R{\hbox{-}}\)module with a finite basis \(X\).

Let \(I {~\trianglelefteq~}R\) be a proper ideal. Prove that \(M/IM\) is a free \(R/I{\hbox{-}}\)module with basis \(X'\), where \(X'\) is the image of \(X\) under the canonical map \(M\to M/IM\).

Prove that any two bases of \(M\) have the same number of elements. You may assume that the result is true when \(R\) is a field.

Let \(R\) be an integral domain with quotient field \(F\). Suppose that \(p(x), a(x), b(x)\) are monic polynomials in \(F[x]\) with \(p(x) = a(x) b(x)\) and with \(p(x) \in R[x]\), \(a(x)\) not in \(R[x]\), and both \(a(x), b(x)\) not constant.

Prove that \(R\) is not a UFD.

(You may assume Gauss’ lemma)

Prove that \({\mathbb{Z}}[2\sqrt{2}]\) is not a UFD.

Hint: let \(p(x) = x^2-2\).

Let \(R\) be a simple rng (a nonzero ring which is not assume to have a 1, whose only two-sided ideals are \((0)\) and \(R\)) satisfying the following two conditions:

- \(R\) has no zero divisors, and
- If \(x\in R\) with \(x\neq 0\) then \(2x\neq 0\), where \(2x\mathrel{\vcenter{:}}= x+x\).

Prove the following:

For each \(x\in R\) there is one and only one element \(y\in R\) such that \(x = 2y\).

Suppose \(x,y\in R\) such that \(x\neq 0\) and \(2(xy) = x\), then \(yz = zy\) for all \(z\in R\).

You can get partial credit for (b) by showing it in the case \(R\) has a 1.

Let \(R\) be a rng (a ring without 1) which contains an element \(u\) such that for all \(y\in R\), there exists an \(x\in R\) such that \(xu=y\).

Prove that \(R\) contains a maximal left ideal.

Hint: imitate the proof (using Zorn’s lemma) in the case where \(R\) does have a 1.

Let \(R\) be a PID and \((a_1) < (a_2) < \cdots\) be an ascending chain of ideals in \(R\). Prove that for some \(n\), we have \((a_j) = (a_n)\) for all \(j\geq n\).

Let \(R\) be a commutative ring, and \(S\subset R\) be a nonempty subset that does not contain 0 such that for all \(x, y\in S\) we have \(xy\in S\). Let \({\mathcal{I}}\) be the set of all ideals \(I{~\trianglelefteq~}R\) such that \(I\cap S = \emptyset\).

Show that for every ideal \(I\in {\mathcal{I}}\), there is an ideal \(J\in {\mathcal{I}}\) such that \(I\subset J\) and \(J\) is not properly contained in any other ideal in \({\mathcal{I}}\).

Prove that every such ideal \(J\) is prime.

Give a careful proof that \({\mathbb{C}}[x, y]\) is not a PID.

Let \(R\) be a nonzero commutative ring without unit such that \(R\) does not contain a proper maximal ideal. Prove that for all \(x\in R\), the ideal \(xR\) is proper.

You may assume the axiom of choice.

Let \(R\) be a commutative ring and \(a\in R\). Prove that \(a\) is not nilpotent \(\iff\) there exists a commutative ring \(S\) and a ring homomorphism \(\phi: R\to S\) such that \(\phi(a)\) is a unit.

Note: by definition, \(a\) is nilpotent \(\iff\) there is a natural number \(n\) such that \(a^n = 0\).

Let \(R\) be a commutative ring with identity and let \(n\) be a positive integer.

Prove that every surjective \(R{\hbox{-}}\)linear endomorphism \(T: R^n \to R^n\) is injective.

Show that an injective \(R{\hbox{-}}\)linear endomorphism of \(R^n\) need not be surjective.

Define *prime ideal*, give an example of a nontrivial ideal in the ring \({\mathbb{Z}}\) that is not prime, and prove that it is not prime.

Define *maximal ideal*, give an example of a nontrivial maximal ideal in \({\mathbb{Z}}\) and prove that it is maximal.

Let \(R\) be a commutative ring with \(1\neq 0\). Recall that \(x\in R\) is *nilpotent* iff \(x^n = 0\) for some positive integer \(n\).

Show that the collection of nilpotent elements in \(R\) forms an ideal.

Show that if \(x\) is nilpotent, then \(x\) is contained in every prime ideal of \(R\).

Suppose \(x\in R\) is not nilpotent and let \(S = \left\{{x^n {~\mathrel{\Big|}~}n\in {\mathbb{N}}}\right\}\). There is at least on ideal of \(R\) disjoint from \(S\), namely \((0)\).

By Zorn’s lemma the set of ideals disjoint from \(S\) has a maximal element with respect to inclusion, say \(I\). In other words, \(I\) is disjoint from \(S\) and if \(J\) is any ideal disjoint from \(S\) with \(I\subseteq J \subseteq R\) then \(J=I\) or \(J=R\).

Show that \(I\) is a prime ideal.

Deduce from (a) and (b) that the set of nilpotent elements of \(R\) is the intersection of all prime ideals of \(R\).

Let \(R\) be a commutative ring.

Define a *maximal ideal* and prove that \(R\) has a maximal ideal.

Show than an element \(r\in R\) is not invertible \(\iff r\) is contained in a maximal ideal.

Let \(M\) be an \(R{\hbox{-}}\)module, and recall that for \(0\neq \mu \in M\), the *annihilator* of \(\mu\) is the set
\begin{align*}
\operatorname{Ann}(\mu) = \left\{{r\in R {~\mathrel{\Big|}~}r\mu = 0}\right\}
.\end{align*}
Suppose that \(I\) is an ideal in \(R\) which is maximal with respect to the property that there exists an element \(\mu \in M\) such that \(I = \operatorname{Ann}(\mu)\) for some \(\mu \in M\). In other words, \(I = \operatorname{Ann}(\mu)\) but there does not exist \(\nu\in M\) with \(J = \operatorname{Ann}(\nu) \subsetneq R\) such that \(I\subsetneq J\).

Prove that \(I\) is a prime ideal.

Define a *Euclidean domain*.

Define a *unique factorization domain*.

Is a Euclidean domain an UFD? Give either a proof or a counterexample with justification.

Is a UFD a Euclidean domain? Give either a proof or a counterexample with justification.

How many monic irreducible polynomials over \({\mathbb{F}}_p\) of prime degree \(\ell\) are there? Justify your answer.

Let \(F = {\mathbb{F}}_2\) and let \(\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\) denote its algebraic closure.

Show that \(\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\) is not a finite extension of \(F\).

Suppose that \(\alpha \in \mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\) satisfies \(\alpha^{17} = 1\) and \(\alpha\neq 1\). Show that \(F(\alpha)/F\) has degree 8.

Let \(F\) be a finite field with \(q\) elements.

Let \(n\) be a positive integer relatively prime to \(q\) and let \(\omega\) be a primitive \(n\)th root of unity in an extension field of \(F\).

Let \(E = F [\omega]\) and let \(k = [E : F]\).

Prove that \(n\) divides \(q^{k}-1\).

Let \(m\) be the order of \(q\) in \({\mathbb{Z}}/n{\mathbb{Z}}^{\times}\). Prove that \(m\) divides \(k\).

Prove that \(m = k\).

Concepts used:

- Theorem: \(F^{\times}\) is always cyclic for \(F\) a field

**Solution**:

Since \({\left\lvert {F} \right\rvert} = q\) and \([E:F] = k\), we have \({\left\lvert {E} \right\rvert} = q^k\) and \({\left\lvert {E^{\times}} \right\rvert} = q^k-1\).

Noting that \(\zeta \in E^{\times}\) we must have \(n = o(\zeta) \divides {\left\lvert {E^{\times}} \right\rvert} = q^k-1\) by Lagrange’s theorem.

- Rephrasing (a), we have \begin{align*} n \divides q^k-1 &\iff q^k-1 \cong 0 \pmod n \\ &\iff q^k \cong 1 \pmod n \\ &\iff m \mathrel{\vcenter{:}}= o(q) \divides k .\end{align*}

Since \(m\divides k \iff k = \ell m\), (

**claim**) there is an intermediate subfield \(M\) such that \begin{align*} E \leq M \leq F \quad k = [F:E] = [F:M] [M:E] = \ell m ,\end{align*}so \(M\) is a degree \(m\) extension of \(E\).

Now consider \(M^{\times}\).

By the argument in (a), \(n\) divides \(q^m - 1 = {\left\lvert {M^{\times}} \right\rvert}\), and \(M^{\times}\) is cyclic, so it contains a cyclic subgroup \(H\) of order \(n\).

But then \(x\in H \implies p(x)\mathrel{\vcenter{:}}= x^n-1 = 0\), and since \(p(x)\) has at most \(n\) roots in a field.

So \(H = \left\{{x \in M {~\mathrel{\Big|}~}x^n-1 = 0}\right\}\), i.e. \(H\) contains all solutions to \(x^n-1\) in \(E[x]\).

But \(\zeta\) is one such solution, so \(\zeta \in H \subset M^{\times}\subset M\).

Since \(F[\zeta]\) is the smallest field extension containing \(\zeta\), we must have \(F = M\), so \(\ell = 1\), and \(k = m\).

Let \(\zeta_n\) denote a primitive \(n\)th root of 1 \(\in {\mathbb{Q}}\). You may assume the roots of the minimal polynomial \(p_n(x)\) of \(\zeta_n\) are exactly the primitive \(n\)th roots of 1.

Show that the field extension \({\mathbb{Q}}(\zeta_n )\) over \({\mathbb{Q}}\) is Galois and prove its Galois group is \(({\mathbb{Z}}/n{\mathbb{Z}})^{\times}\).

How many subfields are there of \({\mathbb{Q}}(\zeta_{20} )\)?

Concepts Used:

**Galois**= normal + separable.**Separable**: Minimal polynomial of every element has distinct roots.**Normal (if separable)**: Splitting field of an irreducible polynomial.Definition: \(\zeta\) is a primitive root of unity iff \(o(\zeta) = n\) in \(F^{\times}\).

\(\phi(p^k) = p^{k-1}(p-1)\)

The lattice:

**Solution**:

Let \(K = {\mathbb{Q}}(\zeta)\). Then \(K\) is the splitting field of \(f(x) = x^n - 1\), which is irreducible over \({\mathbb{Q}}\), so \(K/{\mathbb{Q}}\) is normal. We also have \(f'(x) = nx^{n-1}\) and \(\gcd(f, f') = 1\) since they can not share any roots.

Or equivalently, \(f\) splits into distinct linear factors \(f(x) = \prod_{k\leq n}(x-\zeta^k)\).

Since it is a Galois extension, \({\left\lvert {\operatorname{Gal}(K/{\mathbb{Q}})} \right\rvert} = [K: {\mathbb{Q}}] = \phi(n)\) for the totient function.

We can now define maps \begin{align*} \tau_j: K &\to K \\ \zeta &\mapsto \zeta^j \end{align*}

and if we restrict to \(j\) such that \(\gcd(n, j) = 1\), this yields \(\phi(n)\) maps. Noting that if \(\zeta\) is a primitive root, then \((n, j) = 1\) implies that that \(\zeta^j\) is also a primitive root, and hence another root of \(\min(\zeta, {\mathbb{Q}})\), and so these are in fact automorphisms of \(K\) that fix \({\mathbb{Q}}\) and thus elements of \(\operatorname{Gal}(K/{\mathbb{Q}})\).

So define a map \begin{align*} \theta: {\mathbb{Z}}_n^{\times}&\to K \\ [j]_n &\mapsto \tau_j .\end{align*}

from the *multiplicative* group of units to the Galois group.

The claim is that this is a surjective homomorphism, and since both groups are the same size, an isomorphism.

**Surjectivity:**

Letting \(\sigma \in K\) be arbitrary, noting that \([K: {\mathbb{Q}}]\) has a basis \(\left\{{1, \zeta, \zeta^2, \cdots, \zeta^{n-1}}\right\}\), it suffices to specify \(\sigma(\zeta)\) to fully determine the automorphism. (Since \(\sigma(\zeta^k) = \sigma(\zeta)^k\).)

In particular, \(\sigma(\zeta)\) satisfies the polynomial \(x^n - 1\), since \(\sigma(\zeta)^n = \sigma(\zeta^n) = \sigma(1) = 1\), which means \(\sigma(\zeta)\) is another root of unity and \(\sigma(\zeta) = \zeta^k\) for some \(1\leq k \leq n\).

Moreover, since \(o(\zeta) = n \in K^{\times}\), we must have \(o(\zeta^k) = n \in K^{\times}\) as well. Noting that \(\left\{{\zeta^i}\right\}\) forms a cyclic subgroup \(H\leq K^{\times}\), then \(o(\zeta^k) = n \iff (n, k) = 1\) (by general theory of cyclic groups).

Thus \(\theta\) is surjective.

**Homomorphism:**

\begin{align*} \tau_j \circ \tau_k (\zeta) =\tau_j(\zeta^k) = \zeta^{jk} \implies \tau_{jk} = \theta(jk) = \tau_j \circ \tau_k .\end{align*}

**Part 2:**

We have \(K \cong {\mathbb{Z}}_{20}^{\times}\) and \(\phi(20) = 8\), so \(K \cong {\mathbb{Z}}_8\), so we have the following subgroups and corresponding intermediate fields:

- \(0 \sim {\mathbb{Q}}(\zeta_{20})\)
- \({\mathbb{Z}}_2 \sim {\mathbb{Q}}(\omega_1)\)
- \({\mathbb{Z}}_4 \sim {\mathbb{Q}}(\omega_2)\)
- \({\mathbb{Z}}_8 \sim {\mathbb{Q}}\)

For some elements \(\omega_i\) which exist by the primitive element theorem.

Let \(F = {\mathbb{F}}_p\) , where \(p\) is a prime number.

Show that if \(\pi(x) \in F[x]\) is irreducible of degree \(d\), then \(\pi(x)\) divides \(x^{p^d} - x\).

Show that if \(\pi(x) \in F[x]\) is an irreducible polynomial that divides \(x^{p^n} - x\), then \(\deg \pi(x)\) divides \(n\).

Go to a field extension. Orders of multiplicative groups for finite fields are known.

We can consider the quotient \(K = \displaystyle{\frac{{\mathbb{F}}_p[x]}{\left\langle{\pi(x)}\right\rangle}}\), which since \(\pi(x)\) is irreducible is an extension of \({\mathbb{F}}_p\) of degree \(d\) and thus a field of size \(p^d\) with a natural quotient map of rings \(\rho: {\mathbb{F}}_p[x] \to K\).

Since \(K^{\times}\) is a group of size \(p^d-1\), we know that for any \(y \in K^{\times}\), we have by Lagrange’s theorem that the order of \(y\) divides \(p^d-1\) and so \(y^{p^d} = y\).

So every element in \(K\) is a root of \(q(x) = x^{p^d}-x\).

Since \(\rho\) is a ring morphism, we have

\begin{align*} \rho(q(x)) = \rho(x^{p^d} - x) = \rho(x)^{p^d} - \rho(x) &= 0 \in K \\ &\iff q(x) \in \ker \rho \\ &\iff q(x) \in \left\langle{\pi(x)}\right\rangle \\ &\iff \pi(x) \divides q(x) = x^{p^d}-x \quad\text{"to contain is to divide"} .\end{align*}

\(\hfill\blacksquare\)

Some potentially useful facts:

- \({\mathbb{GF}}(p^n)\) is the splitting field of \(x^{p^n} - x \in {\mathbb{F}}_p[x]\).
- \(x^{p^d} - x \divides x^{p^n} - x \iff d \divides n\)
- \({\mathbb{GF}}(p^d) \leq {\mathbb{GF}}(p^n) \iff d\divides n\)
- \(x^{p^n} - x = \prod f_i(x)\) over all irreducible monic \(f_i\) of degree \(d\) dividing \(n\).

Claim: \(\pi(x)\) divides \(x^{p^n}-x \iff \deg \pi\) divides \(n\).

\(\implies\): Let \(L \cong {\mathbb{GF}}(p^n)\) be the splitting field of \(\phi_n(x) \mathrel{\vcenter{:}}= x^{p^n}-x\); then since \(\pi \divides \phi_n\) by assumption, \(\pi\) splits in \(L\). Let \(\alpha \in L\) be any root of \(\pi\); then there is a tower of extensions \({\mathbb{F}}_p \leq {\mathbb{F}}_p(\alpha) \leq L\).

Then \({\mathbb{F}}_p \leq {\mathbb{F}}_p(\alpha) \leq L\), and so \begin{align*} n &= [L: {\mathbb{F}}_p] \\ &= [L: {\mathbb{F}}_p(\alpha)]~[{\mathbb{F}}_p(\alpha): {\mathbb{F}}_p] \\ &= \ell d ,\end{align*}

for some \(\ell \in {\mathbb{Z}}^{\geq 1}\), so \(d\) divides \(n\).

\(\impliedby\): If \(d\divides n\), use the fact (claim) that \(x^{p^n} - x = \prod f_i(x)\) over all irreducible monic \(f_i\) of degree \(d\) dividing \(n\). So \(f = f_i\) for some \(i\).

Let \(\zeta = e^{2\pi i/8}\).

What is the degree of \({\mathbb{Q}}(\zeta)/{\mathbb{Q}}\)?

How many quadratic subfields of \({\mathbb{Q}}(\zeta)\) are there?

What is the degree of \({\mathbb{Q}}(\zeta, \sqrt[4] 2)\) over \({\mathbb{Q}}\)?

Concepts used:

- \(\zeta_n \mathrel{\vcenter{:}}= e^{2\pi i \over n}\), and \(\zeta_n^k\) is a primitive \(n\)th root of unity \(\iff \gcd(n, k) = 1\)
- In general, \(\zeta_n^k\) is a primitive \({n \over \gcd(n, k)}\)th root of unity.

- \(\deg \Phi_n(x) = \phi(n)\)
- \(\phi(p^k) = p^k - p^{k-1} = p^{k-1}(p-1)\) (proof: for a nontrivial gcd, the possibilities are \(p, 2p, 3p, 4p, \cdots, p^{k-2}p, p^{k-1}p\).)
- \(\operatorname{Gal}({\mathbb{Q}}(\zeta)/{\mathbb{Q}}) \cong {\mathbb{Z}}/(n)^{\times}\)

**Solution**:

Let \(K = {\mathbb{Q}}(\zeta)\)

- \(\zeta \mathrel{\vcenter{:}}= e^{2\pi i / 8}\) is a primitive \(8\)th root of unity
- The minimal polynomial of an \(n\)th root of unity is the \(n\)th cyclotomic polynomial \(\Phi_n\)
- The degree of the field extension is the degree of \(\Phi_8\), which is \begin{align*} \phi(8) = \phi(2^3) = 2^{3-1} \cdot (2-1) = 4 .\end{align*}
- So \([{\mathbb{Q}}(\zeta): {\mathbb{Q}}] = 4\).

- \(\operatorname{Gal}({\mathbb{Q}}(\zeta)/{\mathbb{Q}}) \cong {\mathbb{Z}}/(8)^{\times}\cong {\mathbb{Z}}/(4)\) by general theory
- \({\mathbb{Z}}/(4)\) has exactly one subgroup of index 2.
- Thus there is exactly
**one**intermediate field of degree 2 (a quadratic extension).

Let \(L = {\mathbb{Q}}(\zeta, \sqrt[4] 2)\).

Note \({\mathbb{Q}}(\zeta) = {\mathbb{Q}}(i, \sqrt 2)\)

- \({\mathbb{Q}}(i, \sqrt{2})\subseteq {\mathbb{Q}}(\zeta)\)
- \(\zeta_8^2 = i\), and \(\zeta_8 = \sqrt{2}^{-1}+ i\sqrt{2}^{-1}\) so \(\zeta_8 + \zeta_8 ^{-1}= 2/\sqrt{2} = \sqrt{2}\).

- \({\mathbb{Q}}(\zeta) \subseteq {\mathbb{Q}}(i, \sqrt{2})\):
- \(\zeta = e^{2\pi i / 8} = \sin(\pi/4) + i\cos(\pi/4) = {\sqrt 2 \over 2}\qty{1+i}\).

- \({\mathbb{Q}}(i, \sqrt{2})\subseteq {\mathbb{Q}}(\zeta)\)
Thus \(L = {\mathbb{Q}}(i, \sqrt{2})(\sqrt[4]{2}) = {\mathbb{Q}}(i, \sqrt 2, \sqrt[4] 2) = {\mathbb{Q}}(i, \sqrt[4]{2})\).

- Uses the fact that \({\mathbb{Q}}(\sqrt 2) \subseteq {\mathbb{Q}}(\sqrt[4] 2)\) since \(\sqrt[4]{2}^2 = \sqrt{2}\)

Conclude \begin{align*} [L: {\mathbb{Q}}] = [L: {\mathbb{Q}}(\sqrt[4] 2)] ~[{\mathbb{Q}}(\sqrt[4] 2): {\mathbb{Q}}] = 2 \cdot 4 = 8 \end{align*} using the fact that the minimal polynomial of \(i\) over any subfield of \({\mathbb{R}}\) is always \(x^2 + 1\), so \(\min_{{\mathbb{Q}}(\sqrt[4] 2)}(i) = x^2 + 1\) which is degree 2.

Let \(F \subset K \subset L\) be finite degree field extensions. For each of the following assertions, give a proof or a counterexample.

If \(L/F\) is Galois, then so is \(K/F\).

If \(L/F\) is Galois, then so is \(L/K\).

If \(K/F\) and \(L/K\) are both Galois, then so is \(L/F\).

Let \(L/K/F\).

**False**: Take \(L/K/F = {\mathbb{Q}}(\zeta_2, \sqrt[3] 2) \to {\mathbb{Q}}(\sqrt[3] 2) \to {\mathbb{Q}}\).

Then \(L/F\) is Galois, since it is the splitting field of \(x^3 - 2\) and \({\mathbb{Q}}\) has characteristic zero.

But \(K/F\) is not Galois, since it is not the splitting field of any irreducible polynomial.

**True**: If \(L/F\) is Galois, then \(L/K\) is normal and separable:

\(L/K\) is normal, since if \(\sigma: L \hookrightarrow\overline K\) lifts the identity on \(K\) and fixes \(L\), i-t also lifts the identity on \(F\) and fixes \(L\) (and \(\overline K = \overline F\)).

\(L/K\) is separable, since \(F[x] \subseteq K[x]\), and so if \(\alpha \in L\) where \(f(x) \mathrel{\vcenter{:}}=\min(\alpha, F)\) has no repeated factors, then \(f'(x) \mathrel{\vcenter{:}}=\min(\alpha, K)\) divides \(f\) and thus can not have repeated factors.

**False**: Use the fact that every quadratic extension is Galois, and take \(L/K/F = {\mathbb{Q}}(\sqrt[4] 2) \to {\mathbb{Q}}(\sqrt 2) \to {\mathbb{Q}}\).

Then each successive extension is quadratic (thus Galois) but \({\mathbb{Q}}(\sqrt[4] 2)\) is not the splitting field of any polynomial (noting that it does not split \(x^4 - 2\) completely.)

Let \(f(x) = x^4 - 4x^2 + 2 \in {\mathbb{Q}}[x]\).

Find the splitting field \(K\) of \(f\), and compute \([K: {\mathbb{Q}}]\).

Find the Galois group \(G\) of \(f\), both as an explicit group of automorphisms, and as a familiar abstract group to which it is isomorphic.

Exhibit explicitly the correspondence between subgroups of \(G\) and intermediate fields between \({\mathbb{Q}}\) and \(k\).

Note that \(g(x) = x^2 - 4x + 2\) has roots \(\beta = 2 \pm \sqrt{2}\), and so \(f\) has roots \begin{align*} \alpha_1 &= \sqrt{2 + \sqrt 2} \\ \alpha_2 &= \sqrt{2 - \sqrt 2} \\ \alpha_3 &= -\alpha_1 \\ \alpha_4 &= -\alpha_2 .\end{align*}

and splitting field \(K = {\mathbb{Q}}(\left\{{\alpha_i}\right\})\).

\(K\) is the splitting field of a separable polynomial and thus Galois over \({\mathbb{Q}}\). Moreover, Since \(f\) is irreducible by Eisenstein with \(p=2\), the Galois group is a transitive subgroup of \(S^4\), so the possibilities are:

- \(S_4\)
- \(A_4\)
- \(D_4\)
- \({\mathbb{Z}}/(2) \times{\mathbb{Z}}/(2)\)
- \({\mathbb{Z}}/(4)\)

We can note that \(g\) splits over \(L \mathrel{\vcenter{:}}={\mathbb{Q}}(\sqrt 2)\), an extension of degree 2.

We can now note that \(\min(\alpha, L)\) is given by \(p(x) = x^2 - (2 + \sqrt 2)\), and so \([K: L] = 2\).

We then have \begin{align*} [K: {\mathbb{Q}}] = [K: L] [L : {\mathbb{Q}}] = (2)(2) = 4 .\end{align*}

This \({\left\lvert {\operatorname{Gal}(K/{\mathbb{Q}})} \right\rvert} = 4\), which leaves only two possibilities:

- \({\mathbb{Z}}/(2) \times{\mathbb{Z}}/(2)\)
- \({\mathbb{Z}}/(4)\)

We can next check orders of elements. Take \begin{align*} \sigma &\in \operatorname{Gal}(K/{\mathbb{Q}}) \\ \alpha_1 &\mapsto \alpha_2 .\end{align*}

Computations show that

- \(\alpha_1^2 \alpha_2^2 = 2\), so \(\alpha_1 \alpha_2 = \sqrt 2\)
- \(\alpha_1^2 = 2 + \sqrt 2 \implies \sqrt 2 = \alpha_1^2 - 2\)

and thus \begin{align*} \sigma^2(\alpha_1) &= \sigma(\alpha_2) \\ &= \sigma\left(\frac{\sqrt 2}{\alpha_1}\right) \\ &= \frac{\sigma(\sqrt 2)}{\sigma(\alpha_1)} \\ &= \frac{\sigma(\alpha_1^2 - 2)}{\alpha_2} \\ &= \frac{\alpha_2^2 - 2}{\alpha_2} \\ &= \alpha_2 -2\alpha_2^{-1}\\ &= \alpha_2 - \frac{2\alpha_1}{\sqrt 2} \\ &= \alpha_2 -\alpha_1 \sqrt 2 \\ &\neq \alpha_1 ,\end{align*}

and so the order of \(\sigma\) is strictly greater than 2, and thus 4, and thus \(\operatorname{Gal}(K/{\mathbb{Q}}) = \left\{{\sigma^k {~\mathrel{\Big|}~}1\leq k \leq 4}\right\} \cong {\mathbb{Z}}/(4)\).

?? The subgroup of index 2 \(\left\langle{\sigma^2}\right\rangle\) corresponds to the field extension \(Q(\sqrt 2) / {\mathbb{Q}}\).

Let \(K\) be a Galois extension of \({\mathbb{Q}}\) with Galois group \(G\), and let \(E_1 , E_2\) be intermediate fields of \(K\) which are the splitting fields of irreducible \(f_i (x) \in {\mathbb{Q}}[x]\).

Let \(E = E_1 E_2 \subset K\).

Let \(H_i = \operatorname{Gal}(K/E_i)\) and \(H = \operatorname{Gal}(K/E)\).

Show that \(H = H_1 \cap H_2\).

Show that \(H_1 H_2\) is a subgroup of \(G\).

Show that \begin{align*} \operatorname{Gal}(K/(E_1 \cap E_2 )) = H_1 H_2 .\end{align*}

Moral: \(H_1 \cap H_2 \iff E_1 E_2\), \(H_1 H_2 \iff E_1 \cap E_2\).

By the Galois correspondence, it suffices to show that the fixed field of \(H_1 \cap H_2\) is \(E_1 E_2\).

Let \(\sigma \in H_1 \cap H_2\); then \(\sigma \in {\operatorname{Aut}}(K)\) fixes both \(E_1\) and \(E_2\).

Not sure if this works – compositum is not literally product..?

Writing \(x \in E_1E_2\) as \(x=e_1 e_2\), we have \begin{align*} \sigma(x) = \sigma(e_1 e_2) = \sigma(e_1) \sigma(e_2) = e_1 e_2 =x, \end{align*}

so \(\sigma\) fixes \(E_1 E_2\).

That \(H_1 H_2 \subseteq G\) is clear, since if \(\sigma = \tau_1 \tau_2 \in H_1 H_2\), then each \(\tau_i\) is an automorphism of \(K\) that fixes \(E_i \supseteq {\mathbb{Q}}\), so each \(\tau_i\) fixes \({\mathbb{Q}}\) and thus \(\sigma\) fixes \({\mathbb{Q}}\).

That it is a subgroup follows from the fact that elements commute. (?)

To see this, let \(\sigma = \sigma_1 \sigma_2 \in H_1 H_2\).

Note that \(\sigma_1(e) = e\) for all \(e\in E_1\) by definition, since \(H_1\) fixes \(E_1\), and \(\sigma_2(e) \in E_1\) (?).

Then \begin{align*} \sigma_1(e) = e \quad \forall e \in E_1 \implies \sigma_1(\sigma_2(e)) = \sigma_2(e) \end{align*}

and substituting \(e = \sigma_1(e)\) on the RHS yields

\begin{align*} \sigma_1 \sigma_2(e) = \sigma_2 \sigma_1(e) ,\end{align*}

where a similar proof holds for \(e\in E_2\) and thus for arbitrary \(x\in E_1 E_2\).

By the Galois correspondence, the subgroup \(H_1H_2 \leq G\) will correspond to an intermediate field \(E\) such that \(K/E/{\mathbb{Q}}\) and \(E\) is the fixed field of \(H_1 H_2\).

But if \(\sigma \in H_1 H_2\), then \(\sigma = \tau_1 \tau_2\) where \(\tau_i\) is an automorphism of \(K\) that fixes \(E_i\), and so \(\sigma(x) = x \iff \tau_1\tau_2(x) = x \iff \tau_2(x) = x ~\&~ \tau_1(x) = x \iff x \in E_1 \cap E_2\).

Let \(f(x) = x^4-2 \in {\mathbb{Q}}[x]\).

Define what it means for a finite extension field \(E\) of a field \(F\) to be a Galois extension.

Determine the Galois group \(\operatorname{Gal}(E/{\mathbb{Q}})\) for the polynomial \(f(x)\), and justify your answer carefully.

Exhibit a subfield \(K\) in \((b)\) such that \({\mathbb{Q}}\leq K \leq E\) with \(K\) not a Galois extension over \({\mathbb{Q}}\). Explain.

Let \(E\) be an extension field of \(F\) and \(\alpha\in E\) be algebraic of odd degree over \(F\).

Show that \(F(\alpha) = F(\alpha^2)\).

Prove that \(\alpha^{2020}\) is algebraic of odd degree over \(F\).

Let \(f (x)\) be an irreducible polynomial of degree 4 in \({\mathbb{Q}}[x]\) whose splitting field \(K\) over \({\mathbb{Q}}\) has Galois group \(G = S_4\).

Let \(\theta\) be a root of \(f(x)\). Prove that \({\mathbb{Q}}[\theta]\) is an extension of \({\mathbb{Q}}\) of degree 4 and that there are no intermediate fields between \({\mathbb{Q}}\) and \({\mathbb{Q}}[\theta]\).

Prove that if \(K\) is a Galois extension of \({\mathbb{Q}}\) of degree 4, then there is an intermediate subfield between \(K\) and \({\mathbb{Q}}\).

Let \(F\) be a field. Let \(f(x)\) be an irreducible polynomial in \(F[x]\) of degree \(n\) and let \(g(x)\) be any polynomial in \(F[x]\). Let \(p(x)\) be an irreducible factor (of degree \(m\)) of the polynomial \(f(g(x))\).

Prove that \(n\) divides \(m\). Use this to prove that if \(r\) is an integer which is not a perfect square, and \(n\) is a positive integer then every irreducible factor of \(x^{2n} - r\) over \({\mathbb{Q}}[x]\) has even degree.

Let \(F\) be a field and let \(f(x) \in F[x]\).

Define what a splitting field of \(f(x)\) over \(F\) is.

Let \(F\) now be a finite field with \(q\) elements. Let \(E/F\) be a finite extension of degree \(n>0\). Exhibit an explicit polynomial \(g(x) \in F[x]\) such that \(E/F\) is a splitting field of \(g(x)\) over \(F\). Fully justify your answer.

Show that the extension \(E/F\) in (b) is a Galois extension.

Let \(K\) denote the splitting field of \(x^5 - 2\) over \({\mathbb{Q}}\). Show that the Galois group of \(K/{\mathbb{Q}}\) is isomorphic to the group of invertible matrices \begin{align*} \left(\begin{array}{ll} a & b \\ 0 & 1 \end{array}\right) {\quad \operatorname{where} \quad} a\in {\mathbb{F}}_5^{\times}\text{ and } b\in {\mathbb{F}}_5 .\end{align*}

Determine all intermediate fields between \(K\) and \({\mathbb{Q}}\) which are Galois over \({\mathbb{Q}}\).

Set \(f(x) = x^3 - 5 \in {\mathbb{Q}}[x]\).

Find the splitting field \(K\) of \(f(x)\) over \({\mathbb{Q}}\).

Find the Galois group \(G\) of \(K\) over \({\mathbb{Q}}\).

Exhibit explicitly the correspondence between subgroups of \(G\) and intermediate fields between \({\mathbb{Q}}\) and \(K\).

Let \(K = {\mathbb{Q}}[\sqrt 2 + \sqrt 5]\).

Find \([K: {\mathbb{Q}}]\).

Show that \(K/{\mathbb{Q}}\) is Galois, and find the Galois group \(G\) of \(K/{\mathbb{Q}}\).

Exhibit explicitly the correspondence between subgroups of \(G\) and intermediate fields between \({\mathbb{Q}}\) and \(K\).

Let \(K\) be a Galois extension of a field \(F\) with \([K: F] = 2015\). Prove that \(K\) is an extension by radicals of the field \(F\).

Let \(u = \sqrt{2 + \sqrt{2}}\), \(v = \sqrt{2 - \sqrt{2}}\), and \(E = {\mathbb{Q}}(u)\).

Find (with justification) the minimal polynomial \(f(x)\) of \(u\) over \({\mathbb{Q}}\).

Show \(v\in E\), and show that \(E\) is a splitting field of \(f(x)\) over \({\mathbb{Q}}\).

Determine the Galois group of \(E\) over \({\mathbb{Q}}\) and determine all of the intermediate fields \(F\) such that \({\mathbb{Q}}\subset F \subset E\).

Let \(G\) be a finite group. Show that there exists a field extension \(K/F\) with \(\operatorname{Gal}(K/F) = G\).

You may assume that for any natural number \(n\) there is a field extension with Galois group \(S_n\).

Let \(K\) be a Galois extension of \(F\) with \({\left\lvert {\operatorname{Gal}(K/F)} \right\rvert} = 12\). Prove that there exists an intermediate field \(E\) of \(K/F\) with \([E: F] = 3\).

With \(K/F\) as in (b), does an intermediate field \(L\) necessarily exist satisfying \([L: F] = 2\)? Give a proof or counterexample.

Let \({\mathbb{F}}\) be a finite field.

Give (with proof) the decomposition of the additive group \(({\mathbb{F}}, +)\) into a direct sum of cyclic groups.

The *exponent* of a finite group is the least common multiple of the orders of its elements. Prove that a finite abelian group has an element of order equal to its exponent.

Prove that the multiplicative group \(({\mathbb{F}}^{\times}, \cdot)\) is cyclic.

Let \(f(x) = x^4 - 5 \in {\mathbb{Q}}[x]\).

Compute the Galois group of \(f\) over \({\mathbb{Q}}\).

Compute the Galois group of \(f\) over \({\mathbb{Q}}(\sqrt{5})\).

Let \(f\in {\mathbb{Q}}[x]\) be an irreducible polynomial and \(L\) a finite Galois extension of \({\mathbb{Q}}\). Let \(f(x) = g_1(x)g_2(x)\cdots g_r(x)\) be a factorization of \(f\) into irreducibles in \(L[x]\).

Prove that each of the factors \(g_i(x)\) has the same degree.

Give an example showing that if \(L\) is not Galois over \({\mathbb{Q}}\), the conclusion of part (a) need not hold.

Consider the polynomial \(f(x) = x^4 - 7 \in {\mathbb{Q}}[x]\) and let \(E/{\mathbb{Q}}\) be the splitting field of \(f\).

What is the structure of the Galois group of \(E/{\mathbb{Q}}\)?

Give an explicit description of all of the intermediate subfields \({\mathbb{Q}}\subset K \subset E\) in the form \(K = {\mathbb{Q}}(\alpha), {\mathbb{Q}}(\alpha, \beta), \cdots\) where \(\alpha, \beta\), etc are complex numbers. Describe the corresponding subgroups of the Galois group.

Let \(F\subset C\) be a field extension with \(C\) algebraically closed.

Prove that the intermediate field \(C_{\text{alg}} \subset C\) consisting of elements algebraic over \(F\) is algebraically closed.

Prove that if \(F\to E\) is an algebraic extension, there exists a homomorphism \(E\to C\) that is the identity on \(F\).

Let \(E\subset {\mathbb{C}}\) denote the splitting field over \({\mathbb{Q}}\) of the polynomial \(x^3 - 11\).

Prove that if \(n\) is a squarefree positive integer, then \(\sqrt{n}\not\in E\).

Hint: you can describe all quadratic extensions of \({\mathbb{Q}}\) contained in \(E\).

Find the Galois group of \((x^3 - 11)(x^2 - 2)\) over \({\mathbb{Q}}\).

Prove that the minimal polynomial of \(11^{1/3} + 2^{1/2}\) over \({\mathbb{Q}}\) has degree 6.

Let \(L/K\) be a finite extension of fields.

Define what it means for \(L/K\) to be *separable*.

Show that if \(K\) is a finite field, then \(L/K\) is always separable.

Give an example of a finite extension \(L/K\) that is not separable.

Let \(K\) be the splitting field of \(x^4-2\) over \({\mathbb{Q}}\) and set \(G = \operatorname{Gal}(K/{\mathbb{Q}})\).

Show that \(K/{\mathbb{Q}}\) contains both \({\mathbb{Q}}(i)\) and \({\mathbb{Q}}(\sqrt[4]{2})\) and has degree 8 over \({\mathbb{Q}}\)/

Let \(N = \operatorname{Gal}(K/{\mathbb{Q}}(i))\) and \(H = \operatorname{Gal}(K/{\mathbb{Q}}(\sqrt[4]{2}))\). Show that \(N\) is normal in \(G\) and \(NH = G\).

Hint: what field is fixed by \(NH\)?

Show that \(\operatorname{Gal}(K/{\mathbb{Q}})\) is generated by elements \(\sigma, \tau\), of orders 4 and 2 respectively, with \(\tau \sigma\tau^{-1}= \sigma^{-1}\).

Equivalently, show it is the dihedral group of order 8.

How many distinct quartic subfields of \(K\) are there? Justify your answer.

Let \(f(x) = g(x) h(x) \in {\mathbb{Q}}[x]\) and \(E,B,C/{\mathbb{Q}}\) be the splitting fields of \(f,g,h\) respectively.

Prove that \(\operatorname{Gal}(E/B)\) and \(\operatorname{Gal}(E/C)\) are normal subgroups of \(\operatorname{Gal}(E/{\mathbb{Q}})\).

Prove that \(\operatorname{Gal}(E/B) \cap\operatorname{Gal}(E/C) = \left\{{1}\right\}\).

If \(B\cap C = {\mathbb{Q}}\), show that \(\operatorname{Gal}(E/B) \operatorname{Gal}(E/C) = \operatorname{Gal}(E/{\mathbb{Q}})\).

Under the hypothesis of (c), show that \(\operatorname{Gal}(E/{\mathbb{Q}}) \cong \operatorname{Gal}(E/B) \times \operatorname{Gal}(E/C)\).

Use (d) to describe \(\operatorname{Gal}({\mathbb{Q}}[\alpha]/{\mathbb{Q}})\) where \(\alpha = \sqrt 2 + \sqrt 3\).

Let \(F\) be the field with 2 elements and \(K\) a splitting field of \(f(x) = x^6 + x^3 + 1\) over \(F\). You may assume that \(f\) is irreducible over \(F\).

Show that if \(r\) is a root of \(f\) in \(K\), then \(r^9 = 1\) but \(r^3\neq 1\).

Find \(\operatorname{Gal}(K/F)\) and express each intermediate field between \(F\) and \(K\) as \(F(\beta)\) for an appropriate \(\beta \in K\).

Let \(f(x) \in {\mathbb{Q}}[x]\) be an irreducible polynomial of degree 5. Assume that \(f\) has all but two roots in \({\mathbb{R}}\). Compute the Galois group of \(f(x)\) over \({\mathbb{Q}}\) and justify your answer.

Let \(f(x) \in {\mathbb{Q}}[x]\) be a polynomial and \(K\) be a splitting field of \(f\) over \({\mathbb{Q}}\). Assume that \([K:{\mathbb{Q}}] = 1225\) and show that \(f(x)\) is solvable by radicals.

Suppose that \(F\subset E\) are fields such that \(E/F\) is Galois and \({\left\lvert {\operatorname{Gal}(E/F)} \right\rvert} = 14\).

Show that there exists a unique intermediate field \(K\) with \(F\subset K \subset E\) such that \([K: F] = 2\).

Assume that there are at least two distinct intermediate subfields \(F \subset L_1, L_2 \subset E\) with \([L_i: F]= 7\). Prove that \(\operatorname{Gal}(E/F)\) is nonabelian.

Let \(f(x) = x^7 - 3\in {\mathbb{Q}}[x]\) and \(E/{\mathbb{Q}}\) be a splitting field of \(f\) with \(\alpha \in E\) a root of \(f\).

Show that \(E\) contains a primitive 7th root of unity.

Show that \(E\neq {\mathbb{Q}}(\alpha)\).

Compute the Galois group of \(f(x) = x^3-3x -3\in {\mathbb{Q}}[x]/{\mathbb{Q}}\).

Show that a field \(k\) of characteristic \(p\neq 0\) is perfect \(\iff\) for every \(x\in k\) there exists a \(y\in k\) such that \(y^p=x\).

Let \(k\) be a field of characteristic \(p\neq 0\) and \(f\in k[x]\) irreducible. Show that \(f(x) = g(x^{p^d})\) where \(g(x) \in k[x]\) is irreducible and separable.

Conclude that every root of \(f\) has the same multiplicity \(p^d\) in the splitting field of \(f\) over \(k\).

Let \(n\geq 3\) and \(\zeta_n\) be a primitive \(n\)th root of unity. Show that \([{\mathbb{Q}}(\zeta_n + \zeta_n^{-1}): {\mathbb{Q}}] = \phi(n)/2\) for \(\phi\) the totient function. 10.

Let \(L/K\) be a finite normal extension.

Show that if \(L/K\) is cyclic and \(E/K\) is normal with \(L/E/K\) then \(L/E\) and \(E/K\) are cyclic.

Show that if \(L/K\) is cyclic then there exists exactly one extension \(E/K\) of degree \(n\) with \(L/E/K\) for each divisor \(n\) of \([L:K]\).

Let \(R\) be a commutative ring, and let \(M\) be an \(R{\hbox{-}}\)module. An \(R{\hbox{-}}\)submodule \(N\) of \(M\) is maximal if there is no \(R{\hbox{-}}\)module \(P\) with \(N \subsetneq P \subsetneq M\).

Show that an \(R{\hbox{-}}\)submodule \(N\) of \(M\) is maximal \(\iff M /N\) is a simple \(R{\hbox{-}}\)module: i.e., \(M /N\) is nonzero and has no proper, nonzero \(R{\hbox{-}}\)submodules.

Let \(M\) be a \({\mathbb{Z}}{\hbox{-}}\)module. Show that a \({\mathbb{Z}}{\hbox{-}}\)submodule \(N\) of \(M\) is maximal \(\iff \#M /N\) is a prime number.

Let \(M\) be the \({\mathbb{Z}}{\hbox{-}}\)module of all roots of unity in \({\mathbb{C}}\) under multiplication. Show that there is no maximal \({\mathbb{Z}}{\hbox{-}}\)submodule of \(M\).

a

By the correspondence theorem, submodules of \(M/N\) biject with submodules \(A\) of \(M\) containing \(N\).

So

\(M\) is maximal:

\(\iff\) no such (proper, nontrivial) submodule \(A\) exists

\(\iff\) there are no (proper, nontrivial) submodules of \(M/N\)

\(\iff M/N\) is simple.

b

Identify \({\mathbb{Z}}{\hbox{-}}\)modules with abelian groups, then by (a), \(N\) is maximal \(\iff\) \(M/N\) is simple \(\iff\) \(M/N\) has no nontrivial proper subgroups.

By Cauchy’s theorem, if \({\left\lvert {M/N} \right\rvert} = ab\) is a composite number, then \(a\divides ab \implies\) there is an element (and thus a subgroup) of order \(a\). In this case, \(M/N\) contains a nontrivial proper cyclic subgroup, so \(M/N\) is not simple. So \({\left\lvert {M/N} \right\rvert}\) can not be composite, and therefore must be prime.

c

Let \(G = \left\{{x \in {\mathbb{C}}{~\mathrel{\Big|}~}x^n=1 \text{ for some }n\in {\mathbb{N}}}\right\}\), and suppose \(H < G\) is a proper subgroup.

Then there must be a prime \(p\) such that the \(\zeta_{p^k} \not \in H\) for all \(k\) greater than some constant \(m\) – otherwise, we can use the fact that if \(\zeta_{p^k} \in H\) then \(\zeta_{p^\ell} \in H\) for all \(\ell \leq k\), and if \(\zeta_{p^k} \in H\) for all \(p\) and all \(k\) then \(H = G\).

But this means there are infinitely many elements in \(G\setminus H\), and so \(\infty = [G: H] = {\left\lvert {G/H} \right\rvert}\) is not a prime. Thus by (b), \(H\) can not be maximal, a contradiction.

Consider the \({\mathbb{Z}}{\hbox{-}}\)submodule \(N\) of \({\mathbb{Z}}^3\) spanned by \(f_1 = [-1, 0, 1], f_2 = [2,-3,1], f_3 = [0, 3, 1], f_4 = [3,1,5]\). Find a basis for \(N\) and describe \({\mathbb{Z}}^3/N\).

Let \begin{align*} M = \{(w, x, y, z) \in {\mathbb{Z}}^4 {~\mathrel{\Big|}~}w + x + y + z \in 2{\mathbb{Z}}\} ,\end{align*}

and

\begin{align*} N = \{(w, x, y, z) \in {\mathbb{Z}}^4 {~\mathrel{\Big|}~}4\divides (w - x),~ 4\divides (x - y),~ 4\divides ( y - z)\} .\end{align*}

Show that \(N\) is a \({\mathbb{Z}}{\hbox{-}}\)submodule of \(M\) .

Find vectors \(u_1 , u_2 , u_3 , u_4 \in {\mathbb{Z}}^4\) and integers \(d_1 , d_2 , d_3 , d_4\) such that \begin{align*} \{u_1 , u_2 , u_3 , u_4 \} \end{align*} is a free basis for \(M\), and \begin{align*} \{d_1 u_1,~ d_2 u_2,~ d_3 u_3,~ d_4 u_4 \} \end{align*} is a free basis for \(N\) .

Use the previous part to describe \(M/N\) as a direct sum of cyclic \({\mathbb{Z}}{\hbox{-}}\)modules.

Let \(R\) be a PID and \(M\) be an \(R{\hbox{-}}\)module. Let \(p\) be a prime element of \(R\). The module \(M\) is called *\(\left\langle{p}\right\rangle{\hbox{-}}\)primary* if for every \(m \in M\) there exists \(k > 0\) such that \(p^k m = 0\).

Suppose M is \(\left\langle{p}\right\rangle{\hbox{-}}\)primary. Show that if \(m \in M\) and \(t \in R, ~t \not\in \left\langle{p}\right\rangle\), then there exists \(a \in R\) such that \(atm = m\).

A submodule \(S\) of \(M\) is said to be *pure* if \(S \cap r M = rS\) for all \(r \in R\). Show that if \(M\) is \(\left\langle{p}\right\rangle{\hbox{-}}\)primary, then \(S\) is pure if and only if \(S \cap p^k M = p^k S\) for all \(k \geq 0\).

Let \(R\) be a ring and \(f: M\to N\) and \(g: N\to M\) be \(R{\hbox{-}}\)module homomorphisms such that \(g\circ f = \operatorname{id}_M\). Show that \(N\cong \operatorname{im}f \oplus \ker g\).

Let \(R\) be a ring with the following commutative diagram of \(R{\hbox{-}}\)modules, where each row represents a short exact sequence of \(R{\hbox{-}}\)modules:

Prove that if \(\alpha\) and \(\gamma\) are isomorphisms then \(\beta\) is an isomorphism.

Let \(R\) be a PID and \(M\) a finitely generated \(R{\hbox{-}}\)module.

Prove that there are \(R{\hbox{-}}\)submodules \begin{align*} 0 = M_0 \subset M_1 \subset \cdots \subset M_n = M \end{align*} such that for all \(0\leq i \leq n-1\), the module \(M_{i+1}/M_i\) is cyclic.

Is the integer \(n\) in part (a) uniquely determined by \(M\)? Prove your answer.

Let \(R\) be a ring and \(M\) an \(R{\hbox{-}}\)module. Recall that \(M\) is *Noetherian* iff any strictly increasing chain of submodule \(M_1 \subsetneq M_2 \subsetneq \cdots\) is finite. Call a proper submodule \(M' \subsetneq M\) *intersection-decomposable* if it can not be written as the intersection of two proper submodules \(M' = M_1\cap M_2\) with \(M_i \subsetneq M\).

Prove that for every Noetherian module \(M\), any proper submodule \(N\subsetneq M\) can be written as a finite intersection \(N = N_1 \cap\cdots \cap N_k\) of intersection-indecomposable modules.

Let \(A\) be an abelian group, and show \(A\) is a \({\mathbb{Z}}{\hbox{-}}\)module in a unique way.

Let \(R\) be a ring and \(M\) an \(R{\hbox{-}}\)module.

Recall that the set of torsion elements in M is defined by \begin{align*} \operatorname{Tor}(M) = \{m \in M {~\mathrel{\Big|}~}\exists r \in R, ~r \neq 0, ~rm = 0\} .\end{align*}

Prove that if \(R\) is an integral domain, then \(\operatorname{Tor}(M )\) is a submodule of \(M\) .

Give an example where \(\operatorname{Tor}(M )\) is not a submodule of \(M\).

If \(R\) has zero-divisors, prove that every non-zero \(R{\hbox{-}}\)module has non-zero torsion elements.

One-step submodule test.

It suffices to show that \begin{align*} r\in R, ~t_1, t_2\in \operatorname{Tor}(M) \implies rt_1 + t_2 \in \operatorname{Tor}(M) .\end{align*}

We have \begin{align*} t_1 \in \operatorname{Tor}(M) &\implies \exists s_1 \neq 0 \text{ such that } s_1 t_1 = 0 \\ t_2 \in \operatorname{Tor}(M) &\implies \exists s_2 \neq 0 \text{ such that } s_2 t_2 = 0 .\end{align*}

Since \(R\) is an integral domain, \(s_1 s_2 \neq 0\). Then \begin{align*} s_1 s_2(rt_1 + t_2) &= s_1 s_2 r t_1 + s_1 s_2t_2 \\ &= s_2 r (s_1 t_1) + s_1 (s_2 t_2) \quad\text{since $R$ is commutative} \\ &= s_2 r(0) + s_1(0) \\ &= 0 .\end{align*}

Let \(R = {\mathbb{Z}}/6{\mathbb{Z}}\) as a \({\mathbb{Z}}/6{\mathbb{Z}}{\hbox{-}}\)module, which is not an integral domain as a ring.

Then \([3]_6\curvearrowright[2]_6 = [0]_6\) and \([2]_6\curvearrowright[3]_6 = [0]_6\), but \([2]_6 + [3]_6 = [5]_6\), where 5 is coprime to 6, and thus \([n]_6\curvearrowright[5]_6 = [0] \implies [n]_6 = [0]_6\). So \([5]_6\) is *not* a torsion element.

So the set of torsion elements are not closed under addition, and thus not a submodule.

Suppose \(R\) has zero divisors \(a,b \neq 0\) where \(ab = 0\). Then for any \(m\in M\), we have \(b\curvearrowright m \mathrel{\vcenter{:}}= bm \in M\) as well, but then \begin{align*} a\curvearrowright bm = (ab)\curvearrowright m = 0\curvearrowright m = 0_M ,\end{align*} so \(m\) is a torsion element for any \(m\).

Let \(R\) be an integral domain. Recall that if \(M\) is an \(R{\hbox{-}}\)module, the *rank* of \(M\) is defined to be the maximum number of \(R{\hbox{-}}\)linearly independent elements of \(M\) .

Prove that for any \(R{\hbox{-}}\)module \(M\), the rank of \(\operatorname{Tor}(M )\) is 0.

Prove that the rank of \(M\) is equal to the rank of of \(M/\operatorname{Tor}(M )\).

Suppose that M is a non-principal ideal of \(R\).

Prove that \(M\) is torsion-free of rank 1 but not free.

**Part a**

- Suppose toward a contradiction \(\operatorname{Tor}(M)\) has rank \(n \geq 1\).
- Then \(\operatorname{Tor}(M)\) has a linearly independent generating set \(B = \left\{{\mathbf{r}_1, \cdots, \mathbf{r}_n}\right\}\), so in particular \begin{align*} \sum_{i=1}^n s_i \mathbf{r}_i = 0 \implies s_i = 0_R \,\forall i .\end{align*}
- Let \(\mathbf{r}\) be any of of these generating elements.
- Since \(\mathbf{r}\in \operatorname{Tor}(M)\), there exists an \(s\in R\setminus 0_R\) such that \(s\mathbf{r} = 0_M\).
- Then \(s\mathbf{r} = 0\) with \(s\neq 0\), so \(\left\{{\mathbf{r}}\right\} \subseteq B\) is
*not*a linearly independent set, a contradiction.

**Part b**

- Let \(n = \operatorname{rank}M\), and let \(\mathcal B = \left\{{\mathbf{r}_i}\right\}_{i=1}^n \subseteq R\) be a generating set.
- Let \(\tilde M \mathrel{\vcenter{:}}= M/\operatorname{Tor}(M)\) and \(\pi: M \to M'\) be the canonical quotient map.

**Claim:**
\begin{align*}
\tilde {\mathcal{B}}\mathrel{\vcenter{:}}=\pi(\mathcal B) = \left\{{\mathbf{r}_i + \operatorname{Tor}(M)}\right\}
\end{align*}
is a basis for \(\tilde M\).

**Linearly Independent:**Suppose that \begin{align*} \sum_{i=1}^n s_i (\mathbf{r}_i + \operatorname{Tor}(M)) = \mathbf{0}_{\tilde M} .\end{align*}

Then using the definition of coset addition/multiplication, we can write this as \begin{align*} \sum_{i=1}^n \qty { s_i \mathbf{r}_i + \operatorname{Tor}(M)} = \qty{ \sum_{i=1}^n s_i \mathbf{r}_i} + \operatorname{Tor}(M) = 0_{\tilde M} .\end{align*}

Since \(\tilde{\mathbf{x}} = 0 \in \tilde M \iff \tilde{\mathbf{x}} = \mathbf{x} + \operatorname{Tor}(M)\) where \(\mathbf{x} \in \operatorname{Tor}(M)\), this forces \(\sum s_i \mathbf{r}_i \in \operatorname{Tor}(M)\).

Then there exists a scalar \(\alpha\in R^{\bullet}\) such that \(\alpha \sum s_i \mathbf{r}_i = 0_M\).

Since \(R\) is an integral domain and \(\alpha \neq 0\), we must have \(\sum s_i \mathbf{r}_i = 0_M\).

Since \(\left\{{\mathbf{r}_i}\right\}\) was linearly independent in \(M\), we must have \(s_i = 0_R\) for all \(i\).

**Spanning:**Write \(\pi(\mathcal B) = \left\{{\mathbf{r}_i + \operatorname{Tor}(M)}\right\}_{i=1}^n\) as a set of cosets.

Letting \(\mathbf{x} \in M'\) be arbitrary, we can write \(\mathbf{x} = \mathbf{m} + \operatorname{Tor}(M)\) for some \(\mathbf{m} \in M\) where \(\pi(\mathbf{m}) = \mathbf{x}\) by surjectivity of \(\pi\).

Since \(\mathcal B\) is a basis for \(M\), we have \(\mathbf{m} = \sum_{i=1}^n s_i \mathbf{r}_i\), and so \begin{align*} \mathbf{x} &= \pi(\mathbf{m}) \\ &\mathrel{\vcenter{:}}=\pi\qty{ \sum_{i=1}^n s_i \mathbf{r}_i} \\ &= \sum_{i=1}^n s_i \pi(\mathbf{r}_i) \quad\text{since $\pi$ is an $R{\hbox{-}}$module morphism}\\ &\mathrel{\vcenter{:}}=\sum_{i=1}^n s_i \mathbf{(}\mathbf{r}_i + \operatorname{Tor}(M)) ,\end{align*} which expresses \(\mathbf{x}\) as a linear combination of elements in \(\mathcal B'\).

**Part c**

Notation: Let \(0_R\) denote \(0\in R\) regarded as a ring element, and \(\mathbf{0} \in R\) denoted \(0_R\) regarded as a module element (where \(R\) is regarded as an \(R{\hbox{-}}\)module over itself)

**\(M\) is not free:**

**Claim**: If \(I\subseteq R\) is an ideal*and*a free \(R{\hbox{-}}\)module, then \(I\) is principal .Suppose \(I\) is free and let \(I = \left\langle{B}\right\rangle\) for some basis, we will show \({\left\lvert {B} \right\rvert} = 1\)>

Toward a contradiction, suppose \({\left\lvert {B} \right\rvert} \geq 2\) and let \(m_1, m_2\in B\).

Then since \(R\) is commutative, \(m_2 m_1 - m_1 m_2 = 0\) and this yields a linear dependence

So \(B\) has only one element \(m\).

But then \(I = \left\langle{m}\right\rangle = R_m\) is cyclic as an \(R{\hbox{-}}\) module and thus principal as an ideal of \(R\).

Now since \(M\) was assumed to

*not*be principal, \(M\) is not free (using the contrapositive of the claim).

**\(M\) is rank 1**:

For any module, we can take an element \(\mathbf{m}\in M^{\bullet}\) and consider the cyclic submodule \(R\mathbf{m}\).

Since \(M\) is not principle, it is not the zero ideal, and contains at least two elements. So we can consider an element \(\mathbf{m}\in M\).

We have \(\operatorname{rank}_R(M) \geq 1\), since \(R\mathbf{m} \leq M\) and \(\left\{{m}\right\}\) is a subset of some spanning set.

\(R\mathbf{m}\) can not be linearly dependent, since \(R\) is an integral domain and \(M\subseteq R\), so \(\alpha \mathbf{m} = \mathbf{0} \implies \alpha = 0_R\).

Claim: since \(R\) is commutative, \(\operatorname{rank}_R(M) \leq 1\).

- If we take two elements \(\mathbf{m}, \mathbf{n} \in M^{\bullet}\), then since \(m, n\in R\) as well, we have \(nm = mn\) and so \begin{align*} (n)\mathbf{m} + (-m)\mathbf{n} = 0_R = \mathbf{0} \end{align*} is a linear dependence.

**\(M\) is torsion-free**:

Let \(\mathbf{x} \in \operatorname{Tor}M\), then there exists some \(r\neq 0\in R\) such that \(r\mathbf{x} = \mathbf{0}\).

But \(\mathbf{x}\in R\) as well and \(R\) is an integral domain, so \(\mathbf{x}=0_R\), and thus \(\operatorname{Tor}(M) = \left\{{0_R}\right\}\).

Let \(R\) be a ring with unity.

Give a definition for a free module over \(R\).

Define what it means for an \(R{\hbox{-}}\)module to be torsion free.

Prove that if \(F\) is a free module, then any short exact sequence of \(R{\hbox{-}}\)modules of the following form splits: \begin{align*} 0 \to N \to M \to F \to 0 .\end{align*}

Let \(R\) be a PID. Show that any finitely generated \(R{\hbox{-}}\)module \(M\) can be expressed as a direct sum of a torsion module and a free module.

You may assume that a finitely generated torsionfree module over a PID is free.

Let \(R\) be a ring with 1.

An \(R{\hbox{-}}\)module \(M\) is **free** if any of the following conditions hold:

- \(M\) admits an \(R{\hbox{-}}\)linearly independent spanning set \(\left\{{\mathbf{b}_\alpha}\right\}\), so \begin{align*}m\in M \implies m = \sum_\alpha r_\alpha \mathbf{b}_\alpha\end{align*} and \begin{align*}\sum_\alpha r_\alpha \mathbf{b}_\alpha = 0_M \implies r_\alpha = 0_R\end{align*} for all \(\alpha\).
- \(M \cong \bigoplus_{\alpha} R\) are isomorphic as \(R{\hbox{-}}\)modules.
- There is a nonempty set \(X\) and an inclusion \(X\hookrightarrow M\) such that for every \(R{\hbox{-}}\)modules \(N\), every map \(X\to N\) lifts to a unique map \(M\to N\), so the following diagram commutes:
`{=tex} \begin{center} \begin{tikzcd} M \ar[rd, dotted, "\exists ! \tilde f"] & \\ X \ar[u, hook] \ar[r, "f"] & N \end{tikzcd} \end{center}`

\(M\) is **torsionfree** iff \(M_t \mathrel{\vcenter{:}}=\left\{{m\in M {~\mathrel{\Big|}~}\operatorname{Ann}(m) \neq 0}\right\} \leq M\) is the trivial submodule, where \(\operatorname{Ann}(m) \mathrel{\vcenter{:}}=\left\{{r\in R {~\mathrel{\Big|}~}r\cdot m = 0_M}\right\} {~\trianglelefteq~}R\).

Let the following be an SES where \(F\) is a free \(R{\hbox{-}}\)module: \begin{align*} 0 \to N \to M \xrightarrow{\pi} F \to 0 .\end{align*}

Since \(F\) is free, there is a generating set \(X = \left\{{x_\alpha}\right\}\) and a map \(\iota:X\hookrightarrow M\) satisfying the 3rd property from (a).

If we construct a map \(f: X\to M\), then the universal property of free modules will give a lift \(\tilde f: F\to M\)

Note \(\left\{{\iota(x_\alpha)}\right\} \subseteq F\) and \(\pi\) is surjective, so choose fibers \(\left\{{y_\alpha}\right\} \subseteq M\) such that \begin{align*}\pi(y_\alpha) = \iota(x_\alpha).\end{align*}

Define a map \begin{align*} f: X&\to M \\ x_\alpha &\mapsto y_\alpha .\end{align*}

By the universal property, this yields a map \(h: F\to M\), commutativity forces \((h\circ \iota)(x_\alpha) = y_\alpha\), i.e. we have a diagram

It remains to check that it’s a section: \begin{align*} f\in F \implies f = \sum_\alpha r_\alpha { \iota(x_\alpha) } \\ \implies (\pi \circ h)(f) &= \pi \qty{h\qty{\sum_\alpha r_\alpha \iota(x_\alpha) }} \\ &= \pi\qty{\sum_\alpha r_\alpha h(\iota(x_\alpha))} \\ &= \pi\qty{ \sum_\alpha r_\alpha y_\alpha} \\ &= \sum_\alpha r_\alpha \pi(y_\alpha) \\ &= \sum_\alpha r_\alpha \iota(x_\alpha) \\ &\mathrel{\vcenter{:}}= f \end{align*}

Checking \((h\circ \pi)(m) = m\): seems to be hard!

Both \(\pi\circ h\) and \(\operatorname{id}_F\) are two maps that agree on the spanning set \(\left\{{\iota(x_\alpha)}\right\}\), so in fact they are

*equal*.

Free implies projective

Universal property of projective modules: for every surjective \(\pi:M\to N\) and every \(f:P\to N\) there exists a unique lift \(\tilde f: P\to M\):

Take the identity map:

- Claim: if \(R\) is a PID and \(M\) is a finitely generated \(R{\hbox{-}}\)module, then \(M \cong M_t \oplus M/M_t\) where \(M_t \leq M\) is the torsion submodule.
- Claim: \(M/M_t\) is torsionfree, and since a f.g. torsion free module over a PID is free, \(M/M_t\) is free.
- Let \(m+M_t \in M/M_t\) and suppose it is torsion, we will show that is must be the zero coset.
- Then there exists an \(r\in R\) such that \(r(m + M_t) = M_t\)
- Then \(rm + M_t = M_t\), so \(rm\in M_t\).
- By definition of \(M_t\), every element is torsion, so there exists some \(s\in R\) such \(s(rm) = 0_M\).
- Then \((sr)m = 0_M\) which forces \(m\in M_t\)
- Then \(m + M_t = M_t\), so \(m+ M_t\) is the zero coset.

- There is a SES
`{=tex} \begin{center} \begin{tikzcd} 0 \ar[r] & M_t \ar[r] & M \ar[r] & M/M_t \ar[r] & 0 \end{tikzcd} \end{center}`

and since \(M/M_t\) is free, by (c) this sequence splits and \(M \cong M \oplus M/M_t\).

Let \(M\) be a finitely generated module over a PID \(R\).

\(M_t\) be the set of torsion elements of \(M\), and show that \(M_t\) is a submodule of \(M\).

Show that \(M/M_t\) is torsion free.

Prove that \(M \cong M_t \oplus F\) where \(F\) is a free module.

Let \(R\) be an integral domain and let \(M\) be a nonzero torsion \(R{\hbox{-}}\)module.

Prove that if \(M\) is finitely generated then the annihilator in \(R\) of \(M\) is nonzero.

Give an example of a non-finitely generated torsion \(R{\hbox{-}}\)module whose annihilator is \((0)\), and justify your answer.

Let \(R = k[x]\) for \(k\) a field and let \(M\) be the \(R{\hbox{-}}\)module given by \begin{align*} M=\frac{k[x]}{(x-1)^{3}} \oplus \frac{k[x]}{\left(x^{2}+1\right)^{2}} \oplus \frac{k[x]}{(x-1)\left(x^{2}+1\right)^{4}} \oplus \frac{k[x]}{(x+2)\left(x^{2}+1\right)^{2}} .\end{align*} Describe the elementary divisors and invariant factors of \(M\).

Let \(I = (2, x)\) be an ideal in \(R = {\mathbb{Z}}[x]\), and show that \(I\) is not a direct sum of nontrivial cyclic \(R{\hbox{-}}\)modules.

Let \(R\) be a PID.

Classify irreducible \(R{\hbox{-}}\)modules up to isomorphism.

Classify indecomposable \(R{\hbox{-}}\)modules up to isomorphism.

Let \(V\) be a finite-dimensional \(k{\hbox{-}}\)vector space and \(T:V\to V\) a non-invertible \(k{\hbox{-}}\)linear map. Show that there exists a \(k{\hbox{-}}\)linear map \(S:V\to V\) with \(T\circ S = 0\) but \(S\circ T\neq 0\).

Let \(A\in M_n({\mathbb{C}})\) with \(A^2 = A\). Show that \(A\) is similar to a diagonal matrix, and exhibit an explicit diagonal matrix similar to \(A\).

Show that the eigenvalues of a Hermitian matrix \(A\) are real and that \(A = PDP^{-1}\) where \(P\) is an invertible matrix with orthogonal columns.

Let \(F\) be a field and let \(V\) and \(W\) be vector spaces over \(F\) .

Make \(V\) and \(W\) into \(F[x]{\hbox{-}}\)modules via linear operators \(T\) on \(V\) and \(S\) on \(W\) by defining \(X \cdot v = T (v)\) for all \(v \in V\) and \(X \cdot w = S(w)\) for all w \(\in\) W .

Denote the resulting \(F[x]{\hbox{-}}\)modules by \(V_T\) and \(W_S\) respectively.

Show that an \(F[x]{\hbox{-}}\)module homomorphism from \(V_T\) to \(W_S\) consists of an \(F{\hbox{-}}\)linear transformation \(R : V \to W\) such that \(RT = SR\).

Show that \(VT \cong WS\) as \(F[x]{\hbox{-}}\)modules \(\iff\) there is an \(F{\hbox{-}}\)linear isomorphism \(P : V \to W\) such that \(T = P^{-1}SP\).

Recall that a module \(M\) is *simple* if \(M \neq 0\) and any proper submodule of \(M\) must be zero. Suppose that \(V\) has dimension 2. Give an example of \(F\), \(T\) with \(V_T\) simple.

Assume \(F\) is algebraically closed. Prove that if \(V\) has dimension 2, then any \(V_T\) is not simple.

Let \(F\) be a field and \(V\) a finite dimensional \(F{\hbox{-}}\)vector space, and let \(A, B: V\to V\) be commuting \(F{\hbox{-}}\)linear maps. Suppose there is a basis \({\mathcal{B}}_1\) with respect to which \(A\) is diagonalizable and a basis \({\mathcal{B}}_2\) with respect to which \(B\) is diagonalizable.

Prove that there is a basis \({\mathcal{B}}_3\) with respect to which \(A\) and \(B\) are both diagonalizable.

Let \(A, B\) be two \(n\times n\) matrices with the property that \(AB = BA\). Suppose that \(A\) and \(B\) are diagonalizable. Prove that \(A\) and \(B\) are *simultaneously* diagonalizable.

Let \(A\) be a square matrix over the complex numbers. Suppose that \(A\) is nonsingular and that \(A^{2019}\) is diagonalizable over \({\mathbb{C}}\).

Show that \(A\) is also diagonalizable over \({\mathbb{C}}\).

\(A\) is diagonalizable iff \(\min_A(x)\) is separable. See further discussion here.

Claim: If \(A \in \operatorname{GL}(m, {\mathbb{F}})\) is invertible and \(A^n/{\mathbb{F}}\) is diagonalizable, then \(A/{\mathbb{F}}\) is diagonalizable.

Let \(A \in \operatorname{GL}(m, {\mathbb{F}})\). Since \(A^n\) is diagonalizable, \(\min_{A^n}(x) \in {\mathbb{F}}[x]\) is separable and thus factors as a product of \(m\) **distinct** linear factors:

\begin{align*} \min_{A^n}(x) = \prod_{i=1}^m (x-\lambda_i), \quad \min_{A^n}(A^n) = 0 \end{align*}

where \(\left\{{\lambda_i}\right\}_{i=1}^m \subset {\mathbb{F}}\) are the **distinct** eigenvalues of \(A^n\).

Moreover \(A\in \operatorname{GL}(m,{\mathbb{F}}) \implies A^n \in \operatorname{GL}(m,{\mathbb{F}})\): \(A\) is invertible \(\iff \det(A) = d \in {\mathbb{F}}^{\times}\), and so \(\det(A^n) = \det(A)^n = d^n \in {\mathbb{F}}^{\times}\) using the fact that the determinant is a ring morphism \(\det: \operatorname{Mat}(m\times m) \to{\mathbb{F}}\) and \({\mathbb{F}}^{\times}\) is closed under multiplication.

So \(A^n\) is invertible, and thus has trivial kernel, and thus zero is not an eigenvalue, so \(\lambda_i \neq 0\) for any \(i\).

Since the \(\lambda_i\) are distinct and nonzero, this implies \(x^k\) is not a factor of \(\mu_{A^n}(x)\) for any \(k\geq 0\). Thus the \(m\) terms in the product correspond to precisely \(m\) **distinct linear** factors.

We can now construct a polynomial that annihilates \(A\), namely \begin{align*} q_A(x) \mathrel{\vcenter{:}}=\min_{A^n}(x^n) = \prod_{i=1}^m (x^n-\lambda_i) \in {\mathbb{F}}[x], \end{align*}

where we can note that \(q_A(A) = \min_{A^n}(A^n) = 0\), and so \(\min_A(x) \divides q_A(x)\) by minimality.

We now claim that \(q_A(x)\) has exactly \(n\cdot m\) distinct linear factors in \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu[x]\), which reduces to showing that no pair \(x^n-\lambda_i, x^n-\lambda_j\) share a root. and that \(x^n-\lambda_i\) does not have multiple roots.

For the first claim, we can factor \begin{align*} x^n - \lambda_i = \prod_{k=1}^n (x - \lambda_i^{1\over n} e^{2\pi i k \over n}) \mathrel{\vcenter{:}}=\prod_{k=1}^n (x-\lambda^{1\over n} \zeta_n^k) ,\end{align*} where we now use the fact that \(i\neq j \implies \lambda_i^{1\over n} \neq \lambda_j^{1\over n}\). Thus no term in the above product appears as a factor in \(x^n - \lambda_j\) for \(j\neq i\).

For the second claim, we can check that \({\frac{\partial }{\partial x}\,}\qty{x^n - \lambda_i} = nx^{n-1}\neq 0\in {\mathbb{F}}\), and \(\gcd(x^n-\lambda_i, nx^{n-1}) = 1\) since the latter term has only the roots \(x=0\) with multiplicity \(n-1\), whereas \(\lambda_i\neq 0 \implies\) zero is not a root of \(x^n-\lambda_i\).

But now since \(q_A(x)\) has exactly distinct linear factors in \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu[x]\) and \(\min_A(x) \divides q_A(x)\), \(\min_A(x) \in {\mathbb{F}}[x]\) can only have distinct linear factors, and \(A\) is thus diagonalizable over \({\mathbb{F}}\).

\(\hfill\blacksquare\)

Let \(k\) be a field and let the group \(G = \operatorname{GL}(m, k) \times\operatorname{GL}(n, k)\) acts on the set of \(m\times n\) matrices \(M_{m, n}(k)\) as follows: \begin{align*} (A, B) \cdot X = AXB^{-1} \end{align*} where \((A, B) \in G\) and \(X\in M_{m, n}(k)\).

State what it means for a group to act on a set. Prove that the above definition yields a group action.

Exhibit with justification a subset \(S\) of \(M_{m, n}(k)\) which contains precisely one element of each orbit under this action.

Let \(G = \operatorname{GL}(3, {\mathbb{Q}}[x])\) be the group of invertible \(3\times 3\) matrices over \({\mathbb{Q}}[x]\). For each \(f\in {\mathbb{Q}}[x]\), let \(S_f\) be the set of \(3\times 3\) matrices \(A\) over \({\mathbb{Q}}[x]\) such that \(\det(A) = c f(x)\) for some nonzero constant \(c\in {\mathbb{Q}}\).

Show that for \((P, Q) \in G\times G\) and \(A\in S_f\), the formula \begin{align*} (P, Q)\cdot A \mathrel{\vcenter{:}}= PAQ^{-1} \end{align*} gives a well defined map \(G\times G \times S_f \to S_f\) and show that this map gives a group action of \(G\times G\) on \(S_f\).

For \(f(x) = x^3(x^2+1)^2\), give one representative from each orbit of the group action in (a), and justify your assertion.

Let \(k\) be a field of characteristic zero and \(A, B \in M_n(k)\) be two square \(n\times n\) matrices over \(k\) such that \(AB - BA = A\). Prove that \(\det A = 0\).

Moreover, when the characteristic of \(k\) is 2, find a counterexample to this statement.

Prove that any nondegenerate matrix \(X\in M_n({\mathbb{R}})\) can be written as \(X = UT\) where \(U\) is orthogonal and \(T\) is upper triangular.

Let \(U\) be an infinite-dimensional vector space over a field \(k\), \(f: U\to U\) a linear map, and \(\left\{{u_1, \cdots, u_m}\right\} \subset U\) vectors such that \(U\) is generated by \(\left\{{u_1, \cdots, u_m, f^d(u_1), \cdots, f^d(u_m)}\right\}\) for some \(d\in {\mathbb{N}}\).

Prove that \(U\) can be written as a direct sum \(U \cong V\oplus W\) such that

- \(V\) has a basis consisting of some vector \(v_1, \cdots v_n, f^d(v_1), \cdots, f^d(v_n)\) for some \(d\in {\mathbb{N}}\), and
- \(W\) is finite-dimensional.

Moreover, prove that for any other decomposition \(U \cong V' \oplus W'\), one has \(W' \cong W\).

Show that two \(3\times 3\) matrices over \({\mathbb{C}}\) are similar \(\iff\) their characteristic polynomials are equal and their minimal polynomials are equal.

Does the conclusion in (a) hold for \(4\times 4\) matrices? Justify your answer with a proof or counterexample.

Let \(F\) be a field and \(T\) an \(n\times n\) matrix with entries in \(F\). Let \(I\) be the ideal consisting of all polynomials \(f\in F[x]\) such that \(f(T) =0\).

Show that the following statements are equivalent about a polynomial \(g\in I\):

\(g\) is irreducible.

If \(k\in F[x]\) is nonzero and of degree strictly less than \(g\), then \(k[T]\) is an invertible matrix.

Let \(V\) be a vector space over a field \(F\) and \(V^\vee\) its dual. A *symmetric bilinear form* \(({\,\cdot\,}, {\,\cdot\,})\) on \(V\) is a map \(V\times V\to F\) satisfying
\begin{align*}
(av_1 + b v_2, w) = a(v_1, w) + b(v_2, w) {\quad \operatorname{and} \quad} (v_1, v_2) = (v_2, v_1)
\end{align*}
for all \(a, b\in F\) and \(v_1, v_2 \in V\). The form is *nondegenerate* if the only element \(w\in V\) satisfying \((v, w) = 0\) for all \(v\in V\) is \(w=0\).

Suppose \(({\,\cdot\,}, {\,\cdot\,})\) is a nondegenerate symmetric bilinear form on \(V\). If \(W\) is a subspace of \(V\), define \begin{align*} W\perp \mathrel{\vcenter{:}}=\left\{{v\in V {~\mathrel{\Big|}~}(v, w) = 0 \text{ for all } w\in W}\right\} .\end{align*}

Show that if \(X, Y\) are subspaces of \(V\) with \(Y\subset X\), then \(X\perp \subseteq Y\perp\).

Define an injective linear map \begin{align*} \psi: Y\perp/X\perp \hookrightarrow(X/Y)^\vee \end{align*} which is an isomorphism if \(V\) is finite dimensional.

Let \(V\) be a finite dimensional vector space over a field (the field is not necessarily algebraically closed).

Let \(\phi : V \to V\) be a linear transformation. Prove that there exists a decomposition of \(V\) as \(V = U \oplus W\) , where \(U\) and \(W\) are \(\phi{\hbox{-}}\)invariant subspaces of \(V\) , \({\left.{{\phi}} \right|_{{U}} }\) is nilpotent, and \({\left.{{\phi}} \right|_{{W}} }\) is nonsingular.

Let \(m(x)\) be the minimal polynomial of \(\phi\). If the polynomial \(f(x) = x\) doesn’t divide \(m\), then \(f\) does not have zero as an eigenvalue, so \(\phi\) is nonsingular and since \(0\) is nilpotent, \(\phi + 0\) works.

Otherwise, write \(\phi(x) = x^m \rho(x)\) where \(\gcd(x, \rho(x)) = 1\).

Then \begin{align*} V \cong \frac{k[x]}{m(x)} \cong \frac{k[x]}{(x^m)} \oplus \frac{k[x]}{(\rho)} \mathrel{\vcenter{:}}= U \oplus W \end{align*} by the Chinese Remainder theorem.

We can now note that \({\left.{{\phi}} \right|_{{U}} }\) is nilpotent because it has characteristic polynomial \(x^m\), and \({\left.{{\phi}} \right|_{{W}} }\) is nonsingular since \(\lambda = 0\) is not an eigenvalue by construction.

Let \(A\) be an \(n \times n\) matrix.

Suppose that \(v\) is a column vector such that the set \(\{v, Av, . . . , A^{n-1} v\}\) is linearly independent. Show that any matrix \(B\) that commutes with \(A\) is a polynomial in \(A\).

Show that there exists a column vector \(v\) such that the set \(\{v, Av, . . . , A^{n-1} v\}\) is linearly independent \(\iff\) the characteristic polynomial of A equals the minimal polynomial of A.

Letting \(\mathbf{v}\) be fixed, since \(\left\{{A^j \mathbf{v}}\right\}\) spans \(V\) we have \begin{align*} B\mathbf{v} = \sum_{j=0}^{n-1}c_j A^j \mathbf{v} .\end{align*}

So let \(p(x) = \sum_{j=0}^{n-1}c_jx^j\). Then consider how \(B\) acts on any basis vector \(A^k \mathbf{v}\).

We have \begin{align*} BA^k \mathbf{v} &= A^k B\mathbf{v} \\ &= A^k p(A) \mathbf{v} \\ &= p(A) A^k \mathbf{v} ,\end{align*}

so \(B = p(A)\) as operators since their actions agree on every basis vector in \(V\).

\(\implies\):

If \(\left\{{A^j \mathbf{v}_k {~\mathrel{\Big|}~}0\leq j \leq n-1}\right\}\) is linearly independent, this means that \(A\) does satisfy any polynomial of degree \(d < n\).

So \(\deg m_A(x) = n\), and since \(m_A(x)\) divides \(\chi_A(x)\) and both are monic degree polynomials of degree \(n\), they must be equal.

\(\impliedby\):

Let \(A\curvearrowright k[x]\) by \(A \curvearrowright p(x) \mathrel{\vcenter{:}}= p(A)\). This induces an invariant factor decomposition \(V =\cong \bigoplus k[x]/(f_i)\). Since the product of the invariant factors is the characteristic polynomial, the largest invariant factor is the minimal polynomial, and these two are equal, there can only be one invariant factor and thus the invariant factor decomposition is \begin{align*} V\cong \frac{k[x]}{(\chi_A(x))} \end{align*} as an isomorphism of \(k[x]{\hbox{-}}\)modules.

So \(V\) is a cyclic \(k[x]\) module, which means that \(V = k[x]\curvearrowright\mathbf{v}\) for some \(\mathbf{v}\in V\) such that \(\operatorname{Ann}(\mathbf{v}) = \chi_A(x)\).

I.e. there is some element \(\mathbf{v}\in V\) whose orbit is all of \(V\).

But then noting that monomials span \(k[x]\), we can write \begin{align*} V &\cong k[x] \curvearrowright\mathbf{v} \\ &\mathrel{\vcenter{:}}=\left\{{f(x) \curvearrowright\mathbf{v} {~\mathrel{\Big|}~}f \in k[x]}\right\} \\ &= {\operatorname{span}}_k \left\{{x^k \curvearrowright\mathbf{v} {~\mathrel{\Big|}~}k \geq 0}\right\} \\ &\mathrel{\vcenter{:}}={\operatorname{span}}_k \left\{{A^k\mathbf{v} {~\mathrel{\Big|}~}k \geq 0}\right\} .\end{align*}

Moreover, we can note that if \(k \geq \deg \chi_A(x)\), then \(A^k\) is a linear combination of \(\left\{{A^j \mathrel{\Big|}0 \leq j \leq n-1}\right\}\), and so \begin{align*} V &\cong {\operatorname{span}}_k \left\{{A^k\mathbf{v} {~\mathrel{\Big|}~}k \geq 0}\right\} \\ &= {\operatorname{span}}_k \left\{{A^k\mathbf{v} {~\mathrel{\Big|}~}1 \leq k \leq n-1}\right\} .\end{align*}

Let \(\{e_1, \cdots, e_n \}\) be a basis of a real vector space \(V\) and let \begin{align*} \Lambda \mathrel{\vcenter{:}}=\left\{{ \sum r_i e_i \mathrel{\Big|}r_i \in {\mathbb{Z}}}\right\} \end{align*}

Let \(\cdot\) be a non-degenerate (\(v \cdot w = 0\) for all \(w \in V \iff v = 0\)) symmetric bilinear form on \(V\) such that the Gram matrix \(M = (e_i \cdot e_j )\) has integer entries.

Define the dual of \(\Lambda\) to be \begin{align*} \Lambda ^\vee\mathrel{\vcenter{:}}=\{v \in V {~\mathrel{\Big|}~}v \cdot x \in {\mathbb{Z}}\text{ for all } x \in \Lambda \} .\end{align*}

Show that \(\Lambda \subset \Lambda ^\vee\).

Prove that \(\det M \neq 0\) and that the rows of \(M^{-1}\) span \(\Lambda^\vee\).

Prove that \(\det M = |\Lambda^\vee/\Lambda|\).

Let \(\mathbf{v} \in \Lambda\), so \(\mathbf{v} = \sum_{i=1}^n r_i \mathbf{e}_i\) where \(r_i \in {\mathbb{Z}}\) for all \(i\).

Then if \(\mathbf{x} = \sum_{j=1}^n s_j \mathbf{e}_j \in \Lambda\) is arbitrary, we have \(s_j \in {\mathbb{Z}}\) for all \(j\) and \begin{align*} {\left\langle {\mathbf{v}},~{\mathbf{x}} \right\rangle} &= {\left\langle {\sum_{i=1}^n r_i \mathbf{e}_i},~{\sum_{j=1}^n s_j \mathbf{e}_j } \right\rangle} \\ &= \sum_{i=1}^n \sum_{j=1}^n r_i s_j {\left\langle {\mathbf{e}_i},~{\mathbf{e}_j } \right\rangle} \in {\mathbb{Z}} \end{align*}

since this is a sum of products of integers (since \({\left\langle {\mathbf{e}_i},~{\mathbf{e}_j} \right\rangle} \in {\mathbb{Z}}\) for each \(i, j\) pair by assumption) so \(\mathbf{v} \in \Lambda^\vee\) by definition.

**\(\det M \neq 0\)**:

Suppose \(\det M = 0\). Then \(\ker M \neq \mathbf{0}\), so let \(\mathbf{v} \in \ker M\) be given by \(\mathbf{v} = \sum_{i=1}^n v_i \mathbf{e}_i \neq \mathbf{0}\).

Note that \begin{align*} M\mathbf{v} = 0 &\implies \left[ \begin{array}{ccc} \mathbf{e}_1 \cdot \mathbf{e}_1 & \mathbf{e}_1 \cdot \mathbf{e}_2 & \cdots \\ \mathbf{e}_2 \cdot \mathbf{e}_1 & \mathbf{e}_2 \cdot \mathbf{e}_2 & \cdots \\ \vdots & \vdots & \ddots \end{array} \right] \left[\begin{array}{c} v_1 \\ v_2 \\ \vdots \end{array}\right] = \mathbf{0} \\ \\ &\implies \sum_{j=1}^n v_j{\left\langle {\mathbf{e}_k},~{\mathbf{e}_j} \right\rangle} = 0 {\quad \operatorname{for each fixed} \quad} k .\end{align*}

We can now note that \({\left\langle {\mathbf{e}_k},~{\mathbf{v}} \right\rangle} = \sum_{j=1}^n v_j {\left\langle {\mathbf{e}_k},~{\mathbf{e}_j} \right\rangle} = 0\) for every \(k\) by the above observation, which forces \(\mathbf{v} = 0\) by non-degeneracy of \({\left\langle {{\,\cdot\,}},~{{\,\cdot\,}} \right\rangle}\), a contradiction.

Write \(M = A^tA\) where \(A\) has the \(\mathbf{e}_i\) as columns. Then \begin{align*} M\mathbf{x} = 0 &\implies A^t A \mathbf{x} = 0 \\ &\implies \mathbf{x}^t A^t A \mathbf{x} = 0 \\ &\implies {\left\lVert {A \mathbf{x}} \right\rVert}^2 = 0 \\ &\implies A\mathbf{x} = 0 \\ &\implies \mathbf{x} = 0 ,\end{align*}

since \(A\) has full rank because the \(\mathbf{e}_i\) are linearly independent.

Let \(A = [\mathbf{e}_1^t, \cdots, \mathbf{e}_n^t]\) be the matrix with \(\mathbf{e}_i\) in the \(i\)th column.

**The rows of \(A^{-1}\) span \(\Lambda^\vee\)**:

Equivalently, the columns of \(A^{-t}\) span \(\Lambda^\vee\).

Let \(B = A^{-t}\) and let \(\mathbf{b}_i\) denote the columns of \(B\), so \(\operatorname{im}B = {\operatorname{span}}{\left\{{\mathbf{b}_i}\right\}}\).

Since \(A \in \operatorname{GL}(n, {\mathbb{Z}})\), \(A^{-1}, A^t, A^{-t} \in \operatorname{GL}(n, {\mathbb{Z}})\) as well.

\begin{align*} \mathbf{v} \in \Lambda^\vee &\implies {\left\langle {\mathbf{e}_i},~{\mathbf{v}} \right\rangle} = z_i \in {\mathbb{Z}}\quad \forall i \\ &\implies A^t \mathbf{v} = \mathbf{z} \mathrel{\vcenter{:}}=[z_1, \cdots, z_n] \in {\mathbb{Z}}^n \\ &\implies \mathbf{v} = A^{-t} \mathbf{z} \mathrel{\vcenter{:}}= B\mathbf{z} \in \operatorname{im}B \\ &\implies \mathbf{v} \in \operatorname{im}B \\ &\implies \Lambda^\vee\subseteq \operatorname{im}B ,\end{align*}

and

\begin{align*} B^t A = (A^{-t})^t A = A^{-1}A = I \\ \implies \mathbf{b}_i \cdot \mathbf{e}_j = \delta_{ij} \in {\mathbb{Z}}\\ \implies \operatorname{im}B \subseteq {\operatorname{span}}~ \Lambda^\vee .\end{align*}

?

Let \(V\) be a finite dimensional vector space over a field \(F\) and let \(T: V\to V\) be a linear operator with characteristic polynomial \(f(x) \in F[x]\).

Show that \(f(x)\) is irreducible in \(F[x] \iff\) there are no proper nonzero subspaces \(W< V\) with \(T(W) \subseteq W\).

If \(f(x)\) is irreducible in \(F[x]\) and the characteristic of \(F\) is 0, show that \(T\) is diagonalizable when we extend the field to its algebraic closure.

Lemma: every \(\mathbf{v}\in V\) is \(T{\hbox{-}}\)cyclic \(\iff \chi_T(x)/{\mathbb{k}}\) is irreducible.

- \(\implies\): Same as argument below.
- \(\impliedby\): Suppose \(f\) is irreducible, then \(f\) is equal to the minimal polynomial of \(T\).

Let \(f\) be the characteristic polynomial of \(T\).

\(\implies\):

- By contrapositive, suppose there is a proper nonzero invariant subspace \(W<V\) with \(T(W) \subseteq W\), we will show the characteristic polynomial \(f \mathrel{\vcenter{:}}=\chi_{V, T}(x)\) is reducible.
- Since \(T(W)\subseteq W\), the restriction \(g\mathrel{\vcenter{:}}=\chi_{V, T}(x) \mathrel{\Big|}_W: W\to W\) is a linear operator on \(W\).

Claim: \(g\) divides \(f\) in \({\mathbb{F}}[x]\) and \(\deg(g) < \deg(f)\).

Matrix-dependent proof

Choose an ordered basis for \(W\), say \({\mathcal{B}}_W \mathrel{\vcenter{:}}=\left\{{\mathbf{w}_1, \cdots, \mathbf{w}_k}\right\}\) where \(k=\dim_F(W)\)

Claim: this can be extended to a basis of \(V\), say \({\mathcal{B}}_V \mathrel{\vcenter{:}}=\left\{{\mathbf{w}_1, \cdots, \mathbf{w}_k, \mathbf{v}_1, \cdots, \mathbf{v}_j}\right\}\) where \(k+j = \dim_F(V)\).

- Note that since \(W<V\) is proper, \(j\geq 1\).

Restrict \(T\) to \(W\) to get \(T_W\), then let \(B = [T_W]_{{\mathcal{B}}_W}\) be the matrix representation of \(T_W\) with respect to \({\mathcal{B}}_W\).

Now consider the matrix representation \([T]_{{\mathcal{B}}_V}\), in block form this is given by \begin{align*} [T]_{{\mathcal{B}}_V} = \begin{bmatrix} B & C \\ 0 & D \end{bmatrix} \end{align*} where we’ve used that \(W<V\) is proper to get the existence of \(C, D\) (there is at least one additional row/column since \(j\geq 1\) in the extended basis.)

Now expand along the first column block to obtain \begin{align*} \chi_{T, V}(x) \mathrel{\vcenter{:}}=\det([T]_{{\mathcal{B}}_V} - xI) = \det(B - xI)\cdot \det(D - xI) \mathrel{\vcenter{:}}=\chi_{T, W}(x) \cdot \det(D-xI) .\end{align*}

Claim: \(\det(D - xI) \in xF[x]\) is nontrivial

The claim follows because this forces \(\deg(\det(D-xI)) \geq 1\) and so \(\chi_{T, W}(x)\) is a proper divisor of \(\chi_{T, V}(x)\).

Thus \(f\) is reducible.

\(\impliedby\)

- Suppose \(f\) is reducible, then we will produce a proper \(T{\hbox{-}}\)invariant subspace.
- Claim: if \(f\) is reducible, there exists a nonzero, noncyclic vector \(\mathbf{v}\).
- Then \({\operatorname{span}}_k\left\{{T^j\mathbf{v}}\right\}_{j=1}^d\) is a \(T{\hbox{-}}\)invariant subspace that is nonzero, and not the entire space since \(\mathbf{v}\) is not cyclic.

Characterization of diagonalizability: \(T\) is diagonalizable over \(F \iff \min_{T, F}\) is squarefree in \(\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu[x]\)?

- Let \(\min_{T, F}(x)\) be the minimal polynomial of \(T\) and \(\chi_{T, F}(x)\) be its characteristic polynomial.
- By Cayley-Hamilton, \(\min_{T, F}(x)\) divides \(\chi_{T, F}\)
- Since \(\chi_{T, F}\) is irreducible, these polynomials are equal.
- Claim: \(T/F\) is diagonalizable \(\iff \min_{T, F}\) splits over \(F\) and is squarefree.
- Replace \(F\) with its algebraic closure, then \(\min_{T, F}\) splits.
- Claim: in characteristic zero, every irreducible polynomial is separable
- Proof: it must be the case that either \(\gcd(f, f') = 1\) or \(f' \equiv 0\), where the second case only happens in characteristic \(p>0\).
- The first case is true because \(\deg f' < \deg f\), and if \(\gcd(f, f') = p\), then \(\deg p < \deg f\) and \(p\divides f\) forces \(p=1\) since \(f\) is irreducible.

- So \(\min_{T, F}\) splits into distinct linear factors
- Thus \(T\) is diagonalizable.

Let \(V\) be a finite-dimensional vector space over a field \(k\) and \(T:V\to V\) a linear transformation.

Provide a definition for the *minimal polynomial* in \(k[x]\) for \(T\).

Define the *characteristic polynomial* for \(T\).

Prove the Cayley-Hamilton theorem: the linear transformation \(T\) satisfies its characteristic polynomial.

Let \(T:V\to V\) be a linear transformation where \(V\) is a finite-dimensional vector space over \({\mathbb{C}}\). Prove the Cayley-Hamilton theorem: if \(p(x)\) is the characteristic polynomial of \(T\), then \(p(T) = 0\). You may use canonical forms.

Consider the following matrix as a linear transformation from \(V\mathrel{\vcenter{:}}={\mathbb{C}}^5\) to itself: \begin{align*} A=\left(\begin{array}{ccccc} -1 & 1 & 0 & 0 & 0 \\ -4 & 3 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 2 \end{array}\right) .\end{align*}

Find the invariant factors of \(A\).

Express \(V\) in terms of a direct sum of indecomposable \({\mathbb{C}}[x]{\hbox{-}}\)modules.

Find the Jordan canonical form of \(A\).

Exhibit the rational canonical form for

- \(A\in M_6({\mathbb{Q}})\) with minimal polynomial \((x-1)(x^2 + 1)^2\).
- \(A\in M_{10}({\mathbb{Q}})\) with minimal polynomial \((x^2+1)^2(x^3 + 1)\).

Exhibit the rational and Jordan canonical forms for the following matrix \(A\in M_4({\mathbb{C}})\): \begin{align*} A=\left(\begin{array}{cccc} 2 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ -2 & -2 & 0 & 1 \\ -2 & 0 & -1 & -2 \end{array}\right) .\end{align*}

Let \(D = {\mathbb{Q}}[x]\) and let \(M\) be a \({\mathbb{Q}}[x]{\hbox{-}}\)module such that \begin{align*} M \cong \frac{\mathbb{Q}[x]}{(x-1)^{3}} \oplus \frac{\mathbb{Q}[x]}{\left(x^{2}+1\right)^{3}} \oplus \frac{\mathbb{Q}[x]}{(x-1)\left(x^{2}+1\right)^{5}} \oplus \frac{\mathbb{Q}[x]}{(x+2)\left(x^{2}+1\right)^{2}} .\end{align*}

Determine the elementary divisors and invariant factors of \(M\).

Let \begin{align*} A=\left[\begin{array}{ccc} 2 & 0 & 0 \\ 4 & 6 & 1 \\ -16 & -16 & -2 \end{array}\right] \in M_{3}(\mathrm{C}) .\end{align*}

Find the Jordan canonical form \(J\) of \(A\).

Find an invertible matrix \(P\) such that \(P^{-1}A P = J\). You should not need to compute \(P^{-1}\).

Write down the minimal polynomial of \(A\).

Let \(p\) be a prime number. Let \(A\) be a \(p \times p\) matrix over a field \(F\) with 1 in all entries except 0 on the main diagonal.

Determine the Jordan canonical form (JCF) of \(A\)

When \(F = {\mathbb{Q}}\),

When \(F = {\mathbb{F}}_p\).

Hint: In both cases, all eigenvalues lie in the ground field. In each case find a matrix \(P\) such that \(P^{-1}AP\) is in JCF.

Work with matrix of all ones instead. Eyeball eigenvectors. Coefficients in minimal polynomial: size of largest Jordan block Dimension of eigenspace: number of Jordan blocks

Let \(A\) be the matrix in the question, and \(B\) be the matrix containing 1’s in every entry.

Noting that \(B = A+I\), we have \begin{align*} &B\mathbf{x} = \lambda \mathbf{x} \\ &\iff (A+I) \mathbf{x} = \lambda \mathbf{x} \\ &\iff A \mathbf{x} = (\lambda - 1) \mathbf{x} ,\end{align*} so we will find the eigenvalues of \(B\) and subtract one from each.

Note that \(B\mathbf{v} = {\left[ {\sum v_i, \sum v_i, \cdots, \sum v_i} \right]}\), i.e. it has the effect of summing all of the entries of \(\mathbf{v}\) and placing that sum in each component.

We proceed by finding \(p\) eigenvectors and eigenvalues, since the JCF and minimal polynomials will involve eigenvalues and the transformation matrix will involve (generalized) eigenvectors.

Claim: each vector of the form \(\mathbf{p}_i \mathrel{\vcenter{:}}=\mathbf{e}_1 - \mathbf{e}_{i+1} = {\left[ {1, 0, 0,\cdots, 0 -1, 0, \cdots, 0 } \right]}\) where \(i\neq j\) is also an eigenvector with eigenvalues \(\lambda_0 = 0\), and this gives \(p-1\) linearly independent vectors spanning the eigenspace \(E_{\lambda_0}\)

- Compute \begin{align*}B \mathbf{p}_i = {\left[ { 1 + 0 + \cdots + 0 + (-1) + 0 + \cdots + 0} \right]} = {\left[ {0, 0, \cdots, 0} \right]}\end{align*}
- So every \(\mathbf{p}_i \in \ker(B)\), so they are eigenvectors with eigenvalue 0.
- Since the first component is fixed and we have \(p-1\) choices for where to place a \(-1\), this yields \(p-1\) possibilities for \(\mathbf{p}_i\)
- These are linearly independent since the \((p-1)\times (p-1)\) matrix \({\left[ { \mathbf{p}_1^t, \cdots, \mathbf{p}_{p-1}^t} \right]}\) satisfies \begin{align*} \det \begin{bmatrix} 1 & 1 & 1 & \cdots & 1\\ -1 & 0 & 0 & \cdots & 0\\ 0 & -1 & 0 & \cdots & 0\\ 0 & 0 & -1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & -1\\ \end{bmatrix} = (1) \cdot \det \begin{bmatrix} -1 & 0 & 0 & \cdots & 0\\ 0 & -1 & 0 & \cdots & 0\\ 0 & 0 & -1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & -1\\ \end{bmatrix} = (-1)^{p-2} \neq 0 .\end{align*}

where the first equality follows from expanding along the first row and noting this is the first minor, and every other minor contains a row of zeros.

Claim: \(\mathbf{v}_1 = {\left[ {1, 1, \cdots, 1} \right]}\) is an eigenvector with eigenvalue \(\lambda_1 = p\).

- Compute \begin{align*}B\mathbf{v} = {\left[ {\sum_{i=1}^p 1, \sum_{i=1}^p 1, \cdots, \sum_{i=1}^p 1} \right]} = {\left[ {p, p, \cdots, p} \right]} = p {\left[ {1, 1, \cdots, 1} \right]} = p\mathbf{v}_1,\end{align*} thus \(\lambda_1 = p\)
- \(\dim E_{\lambda_1} = 1\) since the eigenspaces are orthogonal and \(E_{\lambda_0} \oplus E_{\lambda_1} \leq F^p\) is a subspace, so \(p > \dim(E_{\lambda_0}) + \dim E_{\lambda_1} = p-1 + \dim E_{\lambda_1}\) and it isn’t zero dimensional.

Using that the eigenvalues of \(A\) are \(1+\lambda_i\) for \(\lambda_i\) the above eigenvalues for \(B\), \begin{align*} \operatorname{Spec}(B) \mathrel{\vcenter{:}}=\left\{{(\lambda_i, m_i)}\right\} &= \left\{{(p, 1), (0, p-1)}\right\} \implies \chi_{B}(x) = (x-p)x^{p-1} \\ \implies \operatorname{Spec}(A) &= \left\{{(p-1,1), (-1, p-1) }\right\} \implies \chi_{A}(x) = (x- p+1)(x+1)^{p-1} \\ \end{align*}

Note: we can always read off the

characteristicpolynomial from the spectrum.

The dimensions of eigenspaces are preserved, thus \begin{align*} JCF_{\mathbb{Q}}(A) = J_{p-1}^{1} \oplus (p-1)J_{-1}^1 = \left[\begin{array}{r|r|r|r|r|r} p-1 & 0 & 0 & \cdots & 0 & 0 \\ \hline 0& -1 & 0 & 0 & 0 & 0 \\ \hline 0& 0 & -1 & 0 & 0 & 0 \\ \hline 0& 0 & 0 & \ddots & \ddots & 0 \\ \hline 0& 0 & 0 & \cdots & -1 & 0 \\ \hline 0& 0 & 0 & \cdots & 0 & -1 \\ \end{array}\right] .\end{align*}

The matrix \(P\) such that \(A = PJP^{-1}\) will have columns the bases of the generalized eigenspaces.

In this case, the generalized eigenspaces are the usual eigenspaces, so \begin{align*} P = [\mathbf{v}_1, \mathbf{p}_1, \cdots, \mathbf{p}_{p-1}] = \left[\begin{array}{rrrrrr} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & -1 & 0 & 0 & 0 & 0 \\ 1 & 0 & -1 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 & 0 & 0 \\ 1 & \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 0 & 0 & 0 & 0 & -1 \\ \end{array}\right] .\end{align*}

For \(F = {\mathbb{F}}_p\), all eigenvalues/vectors still lie in \({\mathbb{F}}_p\), but now \(-1 = p-1\), making \((x-(p-1))(x+1)^{p-1} = (x+1)(x+1)^{p-1}\), so \(\chi_{A, {\mathbb{F}}_p}(x) = (x+1)^p\), and the Jordan blocks may merge.

- A computation shows that \((A+I)^2 = pA = 0 \in M_p({\mathbb{F}}_p)\) and \((A+I) \neq 0\), so \(\min_{A, {\mathbb{F}}_p}(x) = (x+1)^2\).
- Thus the largest Jordan block corresponding to \(\lambda = -1\) is of size 2

- Can check that \(\det(A) = \pm 1 \in {\mathbb{F}}_p^{\times}\), so the vectors \(\mathbf{e}_1 - \mathbf{e}_i\) are still linearly independent and thus \(\dim E_{-1} = p-1\)
- So there are \(p-1\) Jordan blocks for \(\lambda = 0\).

Summary: \begin{align*} \min_{A, {\mathbb{F}}_p}(x) &= (x+1)^2 \\ \chi_{A, {\mathbb{F}}_p}(x) &\equiv (x+1)^p \\ \dim E_{-1} &= p-1 .\end{align*}

Thus \begin{align*} JCF_{{\mathbb{F}}_p}(A) = J_{-1}^{2} \oplus (p-2)J_{-1}^1 = \left[\begin{array}{rr|r|r|r|r} -1 & 1 & 0 & \cdots & 0 & 0 \\ 0& -1 & 0 & 0 & 0 & 0 \\ \hline 0& 0 & -1 & 0 & 0 & 0 \\ \hline 0& 0 & 0 & \ddots & \ddots & 0 \\ \hline 0& 0 & 0 & \cdots & -1 & 0 \\ \hline 0& 0 & 0 & \cdots & 0 & -1 \\ \end{array}\right] .\end{align*}

To obtain a basis for \(E_{\lambda = 0}\), first note that the matrix \(P = [\mathbf{v}_1, \mathbf{p}_1, \cdots , \mathbf{p}_{p-1}]\) from part (a) is singular over \({\mathbb{F}}_p\), since \begin{align*} \mathbf{v}_1 + \mathbf{p}_1 + \mathbf{p}_2 + \cdots + \mathbf{p}_{p-2} &= [p-1, 0, 0, \cdots, 0, 1] \\ &= [-1, 0,0,\cdots, 0, 1] \\ &= - \mathbf{p}_{p-1} .\end{align*}

We still have a linearly independent set given by the first \(p-1\) columns of \(P\), so we can extend this to a basis by finding one linearly independent generalized eigenvector.

Solving \((A-I\lambda)\mathbf{x} = \mathbf{v}_1\) is our only option (the others won’t yield solutions). This amounts to solving \(B\mathbf{x} = \mathbf{v}_1\), which imposes the condition \(\sum x_i = 1\), so we can choose \(\mathbf{x} = [1, 0, \cdots, 0]\).

Thus \begin{align*} P = [\mathbf{v}_1, \mathbf{x}, \mathbf{p}_1, \cdots, \mathbf{p}_{p-2}] = \left[\begin{array}{rrrrrr} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & -1 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 & 0 & 0 \\ 1 & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & 0 & 0 & -1\\ 1 & 0 & 0 & 0 & 0 & 0 \\ \end{array}\right] .\end{align*}

Let \begin{align*} A=\left[\begin{array}{lll}{0} & {1} & {-2} \\ {1} & {1} & {-3} \\ {1} & {2} & {-4}\end{array}\right] \in M_{3}(\mathbb{C}) \end{align*}

Find the Jordan canonical form J of A.

Find an invertible matrix \(P\) such that \(P^{-1}AP = J\).

You should not need to compute \(P^{-1}\).

Let \(A\) be an \(n\times n\) matrix with all entries equal to \(0\) except for the \(n-1\) entries just above the diagonal being equal to 2.

What is the Jordan canonical form of \(A\), viewed as a matrix in \(M_n({\mathbb{C}})\)?

Find a nonzero matrix \(P\in M_n({\mathbb{C}})\) such that \(P^{-1}A P\) is in Jordan canonical form.

Let \begin{align*} A=\left(\begin{array}{ccc} -3 & 3 & -2 \\ -7 & 6 & -3 \\ 1 & -1 & 2 \end{array}\right) \in M_{3}(\mathrm{C}) .\end{align*}

Find the Jordan canonical form \(J\) of \(A\).

Find an invertible matrix \(P\) such that \(P^{-1}A P = J\). You do not need to compute \(P^{-1}\).

Let \(F\) be a field and \(n\) a positive integer, and consider \begin{align*} A=\left[\begin{array}{ccc} 1 & \dots & 1 \\ & \ddots & \\ 1 & \dots & 1 \end{array}\right] \in M_{n}(F) .\end{align*}

Show that \(A\) has a Jordan normal form over \(F\) and find it.

Hint: treat the cases \(n\cdot 1 \neq 0\) in \(F\) and \(n\cdot 1 = 0\) in \(F\) separately.

Let \(T\) be a \(5\times 5\) complex matrix with characteristic polynomial \(\chi(x) = (x-3)^5\) and minimal polynomial \(m(x) = (x-3)^2\). Determine all possible Jordan forms of \(T\).

Let \(T: V\to V\) be a linear map from a 5-dimensional \({\mathbb{C}}{\hbox{-}}\)vector space to itself and suppose \(f(T) = 0\) where \(f(x) = x^2 + 2x + 1\).

Show that there does not exist any vector \(v\in V\) such that \(Tv = v\), but there *does* exist a vector \(w\in V\) such that \(T^2 w= w\).

Give all of the possible Jordan canonical forms of \(T\).