# Tuesday September 17th

## Finite Fields and Roots of Polynomials

*Recall from last time:*

Let $\FF$ be a finite field.
Then $\FF\units = \FF\setminus\theset{0}$ is *cyclic* (this requires some proof).

Let $f \in \FF[x]$ with $f(\alpha) = 0$.
Then $\alpha$ is a *multiple root* if $f'(\alpha) = 0$.

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

has $p^n$ distinct roots.

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

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

$\qed$

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

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

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

We also have
$$
(\alpha\beta)^{p^n} = \alpha^{p^n}\beta^{p^n} - \alpha\beta
\text{ and }
\alpha^{-p^n} = \alpha\inv
.$$

So $K$ is a field and $\abs K = p^n$.

$\qed$

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

*Proof:*
Let $F$ be a finite field, so $\abs F = p^r$.
By the previous lemma, there exists a $K$ such that $\ZZ_p \subseteq k \subseteq \overline F$.

$K$ is defined as
$$
K \definedas \theset{\alpha \in F \mid \alpha^{p^n} - \alpha = 0}
.$$

We also have
$$
F  = \theset{\alpha \in \overline F \suchthat \alpha^{p^n} - \alpha = 0}.$$

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

Then
$$
\alpha^{p^{rn}} = \alpha^{p^r p^{r(n-1)}} = (\alpha^{p^r})^{p^{r(n-1)}} = \alpha^{p^{r(n-1)}}
,$$

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

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

$\qed$

## Simple Extensions

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

denote the evaluation map.

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

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

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

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

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

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

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

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

*Proof:*

*Existence:*
We have
$$
F(\alpha) = \theset{\sum_{i=1}^r \beta_i \alpha^i \suchthat \beta_i \in F}
,$$
so all elements look like polynomials in $\alpha$.

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

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

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

$\qed$

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

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

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

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

Moreover, any $\beta \in F(\alpha)$, is *also* algebraic over $F$,and $\deg\min(\beta, F) \divides \deg \min(\alpha, F)$.

*Proof of first part:*
Exercise.

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

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

$\qed$

## Automorphisms and Galois Theory

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

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

*Examples:*

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

- $\alpha = a+bi \in \CC$ has conjugate $\bar \alpha = a-bi$, and
$$
\min(\alpha, \RR) = \min(\bar \alpha, \RR) = x^2 - 2ax + (a^2 + b^2)
.$$