# Thursday October 10th


## Computation of Automorphisms

Setup:

- $F \leq E \leq K \leq \overline F$

- $[K: F] < \infty$

- $K$ is a normal extension of $F$

**Facts:**

- $\Gal(K/E) = \theset{\sigma \in \Gal(K/F) \suchthat \sigma(e) = e ~\forall e\in E}$.

- $\sigma, \tau \in \Gal(K/F)$ and $\restrictionof{\sigma}{E} = \restrictionof{\tau}{E} \iff \sigma, \tau$ are in the same left coset of $\Gal(K/F) / \Gal(K/E)$.

*Example*:
$K = \QQ(\sqrt 2, \sqrt 5)$.

Then $\Gal(K/\QQ) \cong \ZZ_2^2$, given by the following automorphisms:

\begin{align*}
\id:     \sqrt 2 &\mapsto \sqrt 2,  \quad& \sqrt 5 &\mapsto \sqrt 5 \\
\rho_1:  \sqrt 2 &\mapsto \sqrt 2,  \quad& \sqrt 5 &\mapsto -\sqrt 5 \\
\rho_2:  \sqrt 2 &\mapsto -\sqrt 2, \quad& \sqrt 5 &\mapsto \sqrt 5 \\
\rho_1 \circ \rho_2:  \sqrt 2 &\mapsto -\sqrt 2, \quad& \sqrt 5 &\mapsto -\sqrt 5
.\end{align*}

We then get the following subgroup/subfield correspondence:

```{=latex}
\begin{center}
\begin{tikzcd}[column sep=small]
&  & \ZZ_2^2                                           &  &                                                  &  &                           &  & {\QQ(\sqrt 2, \sqrt 5)} \arrow[lldd] \arrow[dd] \arrow[rrdd] &  &                             \\
&  &                                                   &  &                                                  &  &                           &  &                                                              &  &                             \\
{\theset{\id, \rho_1}} \arrow[rruu] &  & {\theset{\id, \rho_2}} \arrow[uu]                 &  & {\theset{\id, \rho_1 \circ \rho_2}} \arrow[lluu] &  & \QQ(\sqrt 2) \arrow[rrdd] &  & \QQ(\sqrt 5) \arrow[dd]                                      &  & \QQ(\sqrt{10}) \arrow[lldd] \\
&  &                                                   &  &                                                  &  &                           &  &                                                              &  &                             \\
&  & \theset{\id} \arrow[lluu] \arrow[uu] \arrow[rruu] &  &                                                  &  &                           &  & \RR                                                          &  &
\end{tikzcd}

\end{center}
```

## Fundamental Theorem of Galois Theory

Recall that $\definedas \Gal(K/E)$.

**Theorem (Fundamental Theorem of Galois Theory):**

Let $\mathcal D$ be the collection of subgroups of $\Gal(K/F)$ and $\mathcal C$ be the collection of subfields $E$ such that $F \leq E \leq K$.

Define a map
\begin{align*}
\lambda: \mathcal C &\to \mathcal D \\
\lambda(E) &\definedas \theset{\sigma \in \Gal(K/F) \mid \sigma(e) = e ~\forall e\in E}
.\end{align*}

Then $\lambda$ is a bijective map, and

1. $\lambda(E) = \Gal(K/E)$

2. $E = K_{\lambda(E)}$

3. If $H \leq \Gal(K/F)$ then
$$
\lambda(K_H) = H
$$

4. $[K: E] = \abs{\lambda(E)}$ and
$$
[E: F] = [\Gal(K/F): \lambda(E)]
$$

5. $E$ is normal over $F \iff \lambda(E) \normal \Gal(K/F)$, and in this case
$$
\Gal(E/F) \cong \Gal(K/F) / \Gal(K/E)
.$$

6. $\lambda$ is order-reversing, i.e.
$$
E_1 \leq E_2 \implies \lambda(E_2) \leq \lambda(E_1)
.$$

*Proof of 1:*
Proved earlier.

$\qed$

*Proof of 2:*
We know that $E \leq L_{\Gal(K/E)}$.
Let $\alpha \in K\setminus E$; we want to show that $\alpha$ is not fixed by all automorphisms in $\Gal(K/E)$.

We build the following tower:

```{=latex}
\begin{center}
\begin{tikzcd}
K \arrow[rr, "\tau'", dotted]          &  & K                  \\
E(\alpha) \arrow[rr, "\tau"] \arrow[u] &  & E(\beta) \arrow[u] \\
E \arrow[rr, "\id"] \arrow[u]          &  & E \arrow[u]        \\
F \arrow[rr, "\id"] \arrow[u]          &  & F \arrow[u]
\end{tikzcd}
\end{center}
```

This uses the isomorphism extension theorem, and the fact that $K$ is normal over $F$.

If $\beta\neq \alpha$, then $\beta$ must be a conjugate of $\alpha$, so $\tau'(\alpha) \neq \alpha$ while $\tau' \in \Gal(K/E)$.

$\qed$

> **Claim:**
> $\lambda$ is injective.
>
> *Proof:*
> Suppose $\lambda(E_1) = \lambda(E_2)$.
> Then by (2), $E_1 = K_{\lambda(E_1)} = K_{\lambda(E_2)} = E_2$.
> $\qed$

*Proof of 3:*
We want to show that if $H\leq \Gal(K/F)$ then $\lambda(K_H)  = H$.

We know $H \leq \lambda(K_H) = \Gal(K/K_H) \leq \Gal(K/F)$, so suppose $H \lneq \lambda(K_H)$.

Since $K$ is a finite, separable extension, $K = K_H(\alpha)$ for some $\alpha \in K$.

Let
\begin{align*}
n = [K: K_H] = {K: K_H} = \abs{\Gal(K/K_H)}
.\end{align*}

Since $H \lneq \lambda(K_H)$, we have $\abs{H} < n$.
So denote $H = \theset{\sigma, \sigma_2, \cdots}$ and let define
\begin{align*}
f(x) = \prod_i (x - \sigma_i(\alpha))
.\end{align*}

We then have

- $\deg f = \abs{H}$

- The coefficients of $f$ are symmetric polynomials in the $\sigma_i(\alpha)$ and are fixed under any $\sigma\in H$

- $f(x) \in K_H(\alpha)[x]$

- $f(\alpha) = 0$ since $\sigma_i(\alpha) = \alpha$ for every $i$.

This is a contradiction, so we must have
$$
[K_H: K] = n = \deg \min(\alpha, K_H) \leq \deg f = \abs{H}
.$$

$\qed$

> Assuming (3), $\lambda$ is surjective, so suppose $H < \Gal(K/F)$.
> Then $\lambda(K_H) = H \implies \lambda$ is surjective.

*Proof of 4:*
\begin{align*}
\abs{\lambda(E)} &= \abs{\Gal(K/E)} =_{\text{splitting field}} [K: E]  \\
[E: F] &=_{\text{separable}} \{E: F\} =_{\text{previous part}} [\Gal(K/F): \lambda(E)]
.\end{align*}

*Proof of 5:*

We have $F\leq E \leq K$ and $E$ is separable over $F$, so $E$ is normal over $F \iff E$ is a splitting field over $F$.

That is, every extension $E'/E$ maps $K$ to itself, since $K$ is normal.
```{=latex}
\begin{center}
\begin{tikzcd}
K                            &  & K            \\
E \arrow[u]                  &  & E' \arrow[u] \\
F \arrow[rr, "id"] \arrow[u] &  & F \arrow[u]
\end{tikzcd}
\end{center}
```

So $E$ is normal over $F \iff$ for all $\sigma \in \Gal(K/F), \sigma(\alpha) \in E$ for all $\alpha \in E$.

By a previous property, $E = K_{\Gal(K/E)}$, and so
\begin{align*}
\sigma(\alpha) \in E
&\iff \tau(\sigma(\alpha)) = \sigma(\alpha)
&\quad \forall \tau \in \Gal(K/E) \\
&\iff (\sigma\inv \tau \sigma) (\alpha) = \alpha
S&\quad \forall \tau \in \Gal(K/E) \\
&\iff \sigma\inv \tau\sigma \in \Gal(K/E) \\
&\iff \Gal(K/E) \normal \Gal(K/F)
.\end{align*}

Now assume $E$ is a normal extension of $F$, and let
\begin{align*}
\phi: \Gal(K/F) &\to \Gal(E/F) \\
\sigma &\mapsto \restrictionof{\sigma}{E}
.\end{align*}

Then $\phi$ is well-defined precisely because $E$ is normal over $F$, and we can apply the extension theorem:

```{=latex}
\begin{center}
\begin{tikzcd}
K                              &  & K           \\
E \arrow[u] \arrow[rr, "\tau"] &  & E \arrow[u] \\
F \arrow[rr, "\id"] \arrow[u]  &  & F \arrow[u]
\end{tikzcd}

\end{center}
```

$\phi$ is surjective by the extension theorem, and $\phi$ is a homomorphism, so consider $\ker \phi$.

Let $\phi(\sigma) = \restrictionof{\sigma}{E} = \id$.
Then $\phi$ fixes elements of $E \iff \sigma \in \Gal(K/E)$, and thus $\ker \phi = \Gal(K/E)$.

$\qed$

*Proof of 6:*
\begin{align*}
E_1 \leq E_2 \iff &\Gal(K/E_2) \leq &\Gal(K/E_1) \\
& \shortparallel &\shortparallel \\
&\lambda(E_2) \leq &\lambda(E_1)
.\end{align*}

> Example:
> $K = \QQ(\sqrt[3] 2, \zeta_3)$.
> Then $\min(\zeta, \QQ) = x^2 + x + 1$ and $\Gal(K/\QQ) = S_3$.
> There is a subgroup of order 2, $E = \Gal(K/\QQ(\sqrt[3] 2)) \leq \Gal(K/\QQ)$, but $E$ doesn't correspond to a normal extension of $F$, so this subgroup is not normal.
> On the other hand, $\Gal(\QQ(\zeta_3), \QQ) \normal \Gal(K/ \QQ)$.