# Tuesday October 1st

## Isomorphism Extension Theorem

Suppose we have $F\leq E \leq \overline F$ and $F' \leq E' \leq \overline{F}'$.
Supposing also that we have an isomorphism $\sigma: F \to F'$, we want to extend this to an isomorphism from $E$ to *some* subfield of $\overline{F}'$ over $F'$.

**Theorem:**
Let $E$ be an algebraic extension of $F$ and $\sigma: F \to F'$ be an isomorphism of fields.
Let $\overline{F}'$ be the algebraic closure of $F'$.

Then there exists a $\tau: E \to E'$ where $E' \leq F'$ such that $\tau(f) = \sigma(f)$ for all $f \in F$.

*Proof:*
See Fraleigh.
Uses Zorn's lemma.

$\qed$

**Corollary:**
Let $F$ be a field and $\overline F, \overline F'$ be algebraic closures of $F$.
Then $\overline F \cong \overline F'$.

*Proof:*
Take the identity $F \to F$ and lift it to some $\tau: \overline F \to E = \tau(\overline F)$ inside $\overline F '$.

\begin{center}
\begin{tikzcd}
& \bar F'\ar[d, dash] \\
\bar F\ar[r, "\tau", red]\ar[d, dash] & E = \tau(\bar F)\ar[d, dash] \\
F\ar[r, "\id"] & F
\end{tikzcd}
\end{center}

Then $\tau(\overline F)$ is algebraically closed, and $\overline F' \geq \tau(\overline F)$ is an algebraic extension.
But then $\overline F' = \tau(\overline F)$. $\qed$

**Corollary:**
Let $E \geq F$ be an algebraic extension with $\alpha, \beta \in E$ conjugates.
Then the conjugation isomorphism that sends $\alpha \to \beta$ can be extended to $E$.

*Proof:*

\begin{center}
\begin{tikzcd}
\bar F \ar[d, dash] & \bar F \ar[d, dash, red]  \\
E \ar[r, "\tau", red]\ar[d, dash]  & {\color{red}E }\ar[d, dash]  \\
F(\alpha) \ar[r, "\psi"]\ar[d, dash] & F(\beta)\ar[d, dash] \\
F \ar[r, "\id"] & F \\
\end{tikzcd}
\end{center}



> Note: Any isomorphism needs to send algebraic elements to algebraic elements, and even more strictly, conjugates to conjugates.

Counting the number of isomorphisms:

Let $E \geq F$ be a finite extension. We want to count the number of isomorphisms from $E$ to a subfield of $\overline F$ that leave $F$ fixed.

I.e., how many ways can we fill in the following diagram?

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




Let $G(E/F) \definedas \Gal(E/F)$; this will be a finite group if $[E: F] < \infty$.

**Theorem:**
Let $E \geq F$ with $[E: F] < \infty$ and $\sigma: F \to F'$ be an isomorphism.

Then the number of isomorphisms $\tau: E \to E'$ extending $\sigma$ is *finite*.

*Proof:*
Since $[E: F]$ is finite, we have $F_0 \definedas F(\alpha_1, \alpha_2, \cdots, \alpha_t)$ for some $t\in \NN$.
Let $\tau: F_0 \to E'$ be an isomorphism extending $\sigma$.

Then $\tau(\alpha_i)$ must be a conjugate of $\alpha_i$, of which there are only finitely many since $\deg \min(\alpha_j, F)$ is finite.
So there are at most $\prod_i \deg\min(\alpha_i, F)$ isomorphisms.

*Example:*
$f(x) = x^3 - 2$, which has roots $\sqrt[3] 2, \sqrt[3] 2 \zeta, \sqrt[3] \zeta^2$.

Two other concepts to address:

- Separability (multiple roots)

- Splitting Fields (containing all roots)

**Definition:**
Let
$$
\theset{E: F} \definedas
\abs{ \theset{\sigma: E \to E' \suchthat \sigma \text{ is an isomorphism extending } \id: F \to F}}
,$$

and define this to be the *index*.

**Theorem:**
Suppose $F \leq E \leq K$, then
$$
\theset{K: F} = \theset{K: E} \theset{E: F}.
$$

*Proof:*
Exercise.

$\qed$

*Example:*
$\QQ(\sqrt 2, \sqrt 5)/\QQ$, which is an extension of *degree* 4.
It also turns out that
$$
\theset{\QQ(\sqrt 2, \sqrt 5) : \QQ} = 4.
$$

**Questions:**

1. When does $[E: F] = \theset{E: F}$? (This is always true in characteristic zero.)

2. When is $\theset{E: F} = \abs{\Gal(E/F)}$?

Note that in this example, $\sqrt 5 \mapsto \pm \sqrt 5$ and likewise for $\sqrt 2$, so any isomorphism extending the identity must in fact be an *automorphism*.

We have automorphisms
\begin{align*}
\sigma_1: (\sqrt 2, \sqrt 5) &\mapsto (-\sqrt 2,\sqrt 5) \\
\sigma_2: (\sqrt 2, \sqrt 5) &\mapsto (\sqrt 2, -\sqrt 5)
,\end{align*}

as well as $\id$ and $\sigma_1 \circ \sigma_2$.
Thus $\Gal(E/F) \cong \ZZ_2^2$.

## Separable Extensions

**Goal**:
When is $\theset{E: F} = [E: F]$?
We'll first see what happens for simple extensions.

**Definition:**
Let $f \in F[x]$ and $\alpha$ be a zero of $f$ in $\overline F$.

The maximum $\nu$ such that $(x-\alpha)^\nu \divides f$ is called the *multiplicity* of $f$.

**Theorem:**
Let $f$ be irreducible.

Then all zeros of $f$ in $\overline F$ have the same multiplicity.

*Proof:*
Let $\alpha, \beta$ satisfy $f$, where $f$ is irreducible.
Then consider the following lift:

\begin{center}
\begin{tikzcd}
\bar F\ar[d, dash] & \bar F\ar[d, dash]  \\
F(\alpha) \ar[r, "\psi"]\ar[d, dash]  & F(\beta) \ar[d, dash]  \\
F \ar[r, "\id"] & F \\
\end{tikzcd}
\end{center}


This induces a map
\begin{align*}
F(\alpha)[x] &\mapsvia{\tau} F(\beta)[x] \\
\sum c_i x^i &\mapsto \sum \psi(c_i) x^i
,\end{align*}

so $x\mapsto x$ and $\alpha \mapsto \beta$, so $x\mapsto x$ and $\alpha \mapsto \beta$.

Then $\tau(f(x)) = f(x)$ and
$$
\tau((x-\alpha)^\nu) = (x-\beta)^\nu
.$$

So write $f(x) = (x-\alpha)^\nu h(x)$, then
$$
\tau(f(x)) = \tau((x-\alpha)^\nu) \tau(h(x))
.$$

Since $\tau(f(x)) = f(x)$, we then have
$$
f(x) = (x-\beta)^\nu \tau(h(x))
.$$

So we get $\mathrm{mult}(\alpha) \leq \mathrm{mult}(\beta)$.
But repeating the argument with $\alpha, \beta$ switched yields the reverse inequality, so they are equal.

$\qed$

*Observation:*
If $F(\alpha) \to E'$ extends the identity on $F$, then $E' = F(\beta)$ where $\beta$ is a root of $f \definedas \min(\alpha, F)$.
Thus we have
$$
\theset{F(\alpha): F} = \abs{\theset{\text{distinct roots of } f}}
.$$

Moreover,
$$
[F(\alpha): F] = \theset{F(\alpha) : F} \nu
$$
where $\nu$ is the multiplicity of a root of $\min(\alpha, F)$.

**Theorem:**
Let $E \geq F$, then $\theset{E: F} \divides [E: F]$.