# Thursday September 19th

## Conjugates

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

*Example:*
$a + bi, a-bi$ are conjugate in $\CC/\RR$, since they both have minimal polynomial $x^2 - 2ax + (a^2 + b^2)$ over $\RR$.

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

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

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

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

Let
$$
f \definedas \min (\alpha, F) = \sum c_i x^i
\text{ where }
c_i \in F
,$$
so $f(\alpha) = 0$.

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

so $\beta$ satisfies $f$ as well, and thus $f = \min(\alpha, F) \divides \min(\beta, F)$.

But we can repeat this argument with $f\inv$ and $g(x) \definedas \min(\beta, F)$, and so we get an equality.
Thus $\alpha, \beta$ are conjugates.

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

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

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

Then

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

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

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

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

This isomorphism commutes with $f$, so we in fact have
$$
0 = \psi(f(a+bi)) = f(\psi(a-bi)) = f(a-bi)
.$$

$\qed$

## Fixed Fields and Automorphisms

**Definition:**
Let $F$ be a field and $\psi: F \selfmap$ is an *automorphism* iff $\psi$ is an isomorphism.

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

*Example:*
Let $E = \QQ(\sqrt 2, \sqrt 5) \supseteq \QQ = F$.

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

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

Then
$$
f(c_0 + c_1 \sqrt 2 + c_2 \sqrt 5 + c_3 \sqrt{10}) = c_0 - c_1\sqrt 2 + c_2 \sqrt 5 - c_3\sqrt{10}
.$$

This forces $c_1 = 0, c_3 = 0$, and so $\psi$ fixes $\theset{ c_0 + c_2 \sqrt 5  } = \QQ(\sqrt 5)$.


**Theorem:**
Let $I$ be a set of automorphisms of $E$ and define
$$
E_I = \theset{\alpha \in E \suchthat \sigma(a) = a ~\forall \sigma \in I}
$$

Then $E_I \leq E$ is a subfield.

*Proof:*
Let $a,b \in E_i$.
We need to show $a \pm b, ab, b\neq 0 \implies b\inv \in I$.

We have $\sigma(a\pm b) = \sigma(a) \pm \sigma(b) = a + b \in I$ since $\sigma$ fixes everything in $I$.
Moreover
$$
\sigma(ab) = \sigma(a)\sigma(b) = ab \in I
\quad \text{ and } \quad
\sigma(b\inv) = \sigma(b)\inv = b\inv \in I
.$$

$\qed$

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

**Theorem:**
Let $E$ be a field and $A = \theset{\sigma:E \selfmap \suchthat \sigma\text{ is an automorphism }}$.
Then $A$ is a group under function composition.

**Theorem:**
Let $E/F$ be a field extension, and define
$$
G(E/F) = \theset{\sigma:E\selfmap \suchthat f\in F \implies \sigma(f) = f}
.$$
Then $G(E/F) \leq A$ is a subgroup which contains $F$.

*Proof:*
This contains the identity function.

Now if $\sigma(f) = f$ then $f = \sigma\inv(f)$, and
$$
\sigma, \tau \in G(E/F) \implies (\sigma \circ \tau)(f) = \sigma(\tau(f)) = \sigma(f) = f
.$$

$\qed$

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

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

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

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

\begin{center}
\begin{tikzcd}
\bar F \ar[d, dash] & \bar F' \ar[d, dash, red] \\
E \ar[d, dash]\ar[r, "\tau", red] & {\color{red} E'} \ar[d, dash, red] \\
F \ar[r, "\sigma"] & F
\end{tikzcd}
\end{center}