# Thursday October 24

## Conjugates

Let $E\geq F$. 
Then $\alpha, \beta \in E$ are **conjugate** iff $\min(\alpha, F) = \min(\beta, F)$.

*Example:*
$\alpha \pm bi \in \CC$.

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

Then $\alpha, \beta$ are conjugates $\iff$ $F(\alpha) \cong F(\beta)$ under the *conjugation map*,
\begin{align*}
\psi: F(\alpha) &\to F(\beta) \\
\sum_{i=1}^{n-1} a_i \alpha^i &\mapsto \sum_{i=1}^{n-1} a_i \beta^i
.\end{align*}

*Proof:*

$\impliedby$:

Suppose that $\psi$ is an isomorphism.
Let $\min(\alpha, F) = p(x) = \sum c_i x^i$ where each $c_i \in F$.
Then
\begin{align*}
0 = \psi(0) = \psi(p(\alpha)) = p(\beta) \implies \min(\beta, F) \divides \min(\alpha, F)
.\end{align*}

Applying the same argument to $q(x) = \min(\beta, F)$ yields $\min(\beta, F) = \min(\alpha, F)$.

$\implies$:

Suppose $\alpha, \beta$ are conjugates.

*Exercise:*
Check that $\psi$ is surjective and
\begin{align*}
\psi(x+y) = \psi(x) + \psi(y) \\
\psi(xy) = \psi(x) \psi(y)
.\end{align*}

Let $z = \sum a_i \alpha^i$.
Supposing that $\psi(z) = 0$, we have $\sum a_i \beta^i = 0$.
By linear independence, this forces $a_i = 0$ for all $i$, and thus $z=0$.
So $\psi$ is injective.

$\qed$

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

1. Any $\phi: F(\alpha) \injects \overline F$ such that $\phi(f) = f$ for all $f\in F$ must map $\alpha$ to a conjugate.

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

*Proof of 1:*

Let $\min(\alpha, F) = p(x) = \sum a_i x^i$.
Note that $0 = \psi(p(\alpha)) = p(\psi(\alpha))$, and since $p$ was irreducible, $p$ must also be the minimal polynomial of $\psi(\alpha)$.
Thus $\psi(\alpha)$ is a conjugate of $\alpha$.

$\qed$

*Proof of 2:*

$F(\alpha)$ is generated by $F$ and $\alpha$, and $\psi$ is completely determined by where it sends $F$ and $\alpha$.
This shows uniquness.

$\qed$

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

*Proof:*
Both $i, -i$ are conjugates and $\min(i, \RR) = \min(-i, \RR) = x^2 + 1 \in \RR[x]$.
We then have a map
\begin{align*}
\psi: \RR[i] &\to \RR[-i] \\
\psi(a+bi) = a + b(-i)
.\end{align*}

So if $f(a+bi) = 0$, then $0 = \psi(f(a+bi)) = f(\psi(a+bi)) = f(a-bi)$.

$\qed$