# Thursday September 5th

## Rings

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

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

*Examples:*
$R = \ZZ, \QQ, \RR, \CC$, or the ring of $n\times n$ matrices, or $\ZZ_n$.

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

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

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

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

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

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

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

*Example:*
In $\ZZ_n$, an element $a$ is a zero divisor iff $\gcd(a, n) \neq 1$.

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

**Theorem:**
Every field is an integral domain.

*Proof:*
Let $R$ be a field. If $ab=0$ and $a\neq 0$, then $a\inv$ exists and so $b=0$.

$\qed$

**Theorem:**
Any finite integral domain is a field.

*Proof:*

> Idea: Similar to the pigeonhole principle.

Let $D = \theset{0, 1, a_1, \cdots, a_n}$ be an integral domain.
Let $a_j \neq 0, 1$ be arbitrary, and consider $a_j D = \theset{a_j x \mid x\in D\setminus\theset{0}}$.

Then $a_j D = D\setminus\theset{0}$ as sets.
But
$$
a_j D = \theset{a_j, a_j a_1, a_j a_2, \cdots, a_j a_n}.
$$

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

$\qed$

## Field Extensions

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

**Definition**:
We can consider
$$
\aut(E/F) \definedas \theset{\sigma: E \selfmap \mid f\in F \implies \sigma(f) = f},
$$
i.e. the field automorphisms of $E$ that fix $F$.

*Examples of field extensions:*
$\CC \to \RR \to \QQ$.

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


\begin{center}
\begin{tikzcd}
 & F(x, y) & \\
F(x)\ar[ur] & & F(y)\ar[ul] \\
& F\ar[ul] \ar[ur] \ar[uu] &
\end{tikzcd}
\end{center}

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

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

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

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

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

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

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

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

$\qed$

*Example:*
$\RR[x] / (x^2 + 1)$ over $\RR$ is isomorphic to $\CC$ as a field.

## Algebraic and Transcendental Elements

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

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

*Examples:*

- $\sqrt 2 \in \RR \from \QQ$ is algebraic, since it satisfies $x^2 - 2$.

- $\sqrt{-1} \in \CC \from \QQ$ is algebraic, since it satisfies $x^2 + 1$.

- $\pi, e \in \RR \from \QQ$ are transcendental

    > This takes some work to show.

An *algebraic number* $\alpha \in \CC$ is an element that is algebraic over $\QQ$.

*Fact:*
The set of algebraic numbers forms a field.

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

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

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

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

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

## Minimal Polynomials

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

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

2. $p$ is irreducible.

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

*Proof:*

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

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

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

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

$\qed$

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

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