# Tuesday October 8th

## Splitting Fields

For $\overline F \geq E \geq F$, we can use the lifting theorem to get a $\tau: E \to E'$.
What conditions guarantee that $E = E'$?

If $E = F(\alpha)$, then $E' = F(\beta)$ for some $\beta$ a conjugate of $\alpha$.
Thus we need $E$ to contain conjugates of all of its elements.

**Definition:**
Let $\theset{f_i(x) \in F[x] \suchthat i\in I}$ be any collection of polynomials.
We way that $E$ is a **splitting field** $\iff E$ is the smallest subfield of $\overline F$ containing all roots of the $f_i$.

*Examples:*

- $\QQ(\sqrt 2, \sqrt 3)$ is a splitting field for $\theset{x^-2, x^2 - 5}$.

- $\CC$ is a splitting field for $\theset{x^2 + 1}$.

- $\QQ(\sqrt[3] 2)$ is *not* a splitting field for any collection of polynomials.

**Theorem:**
Let $F \leq E \leq \overline F$.
Then $E$ is a splitting field over $F$ for some set of polynomials $\iff$ every isomorphism of $E$ fixing $F$ is in fact an automorphism.

*Proof:*

$\implies:$
Let $E$ be a splitting field of $\theset{f_i(x) \suchthat f_i(x) \in F[x], i\in I}$.

Then $E = \generators{\alpha_j \mid j\in J}$ where $\alpha_j$ are the roots of all of the $f_i$.

Suppose $\sigma: E \to E'$ is an isomorphism fixing $F$.
Then consider $\sigma(\alpha_j)$ for some $j \in J$.
We have
$$
\min(\alpha, F) = p(x) = a_0 + a_1 x + \cdots a_{n-1}x^{n-1} + a_n x^n
,$$
and so
$$
p(x) = 0,~~ 0\in F \implies 0 = \sigma(p(\alpha_j)) = \sum_i a_i \sigma(\alpha_j)^i
.$$
Thus $\sigma(\alpha_j)$ is a conjugate, and thus a root of some $f_i(x)$.

$\impliedby:$ Suppose any isomorphism of $E$ leaving $F$ fixed is an automorphism.

Let $g(x)$ be an irreducible polynomial and $\alpha \in E$ a root.

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


Using the lifting theorem, where $F(\alpha \leq E$, we get a map $\tau: E \to E'$ lifting the identity and the conjugation homomorphism.
But this says that $E'$ must contain every conjugate of $\alpha$.

Therefore we can take the collection
$$
S = \theset{g_i(x) \in F[x] \suchthat g_i \text{ irreducible and has a root in } E}
.$$

This defines a splitting field for $\theset{g_j}$, and we're done.

$\qed$

*Examples:*

1. $x^2 + 1 \in \RR[x]$ splits in $\CC$, i.e. $x^2 + 1 = (x+i)(x-i)$.
2. $x^2 - 2 \in \QQ[x]$ splits in $\QQ(\sqrt 2)$.

**Corollary:**
Let $E$ be a splitting field over $F$.
Then every **irreducible** polynomial in $F[x]$ with a root $\alpha \in E$ splits in $E[x]$.

**Corollary:**
The index $\{ E: F \}$ (the number of distinct lifts of the identity).
If $E$ is a splitting field and $\tau:E \to E'$ lifts the identity on $F$, then $E = E'$.
Thus $\{ E : F \}$ is the number of automorphisms, i.e. $\abs{\Gal(E/F)}$.

**Question:**
When is it the case that
$$
[E: F] = \{E: F\} = \abs{\Gal(E/F)}?
$$

- The first equality occurs when $E$ is separable.

- The second equality occurs when $E$ is a splitting field.

> Characteristic zero implies separability

**Definition:**
If $E$ satisfies both of these conditions, it is said to be a **Galois extension**.

Some cases where this holds:

- $E \geq F$ a finite algebraic extension with $E$ characteristic zero.

- $E$ a finite field, since it is a splitting field for $x^{p^n} - x$.

*Example 1:*
$\QQ(\sqrt 2, \sqrt 5)$ is

   1. A degree 4 extension,

   2. The number of automorphisms was 4, and

   3. The Galois group was $\ZZ_2^2$, of size 4.

*Example 2*:
$E$ the splitting field of $x^3 - 3$ over $\QQ$.

This polynomial has roots $\sqrt[3] 3,~ \zeta_3 \sqrt[3] 3,~ \zeta_3^2 \sqrt[3] 3$ where $\zeta_3^3 = 1$.

Then $E = \QQ(\sqrt[3] 3, \zeta_3)$, where
\begin{align*}
\min(\sqrt[3] 3, \QQ) &= x^3 - 3 \\
\min(\zeta_3, \QQ) &= x^2 + x + 1
,\end{align*}

so this is a degree 6 extension.

Since $\ch \QQ = 0$, we have $[E: \QQ] = \{E: \QQ\}$ for free.

We know that any automorphism has to map
\begin{align*}
\sqrt[3] 3 &\mapsto \sqrt[3] 3,~ \sqrt[3] 3 \zeta_3,~ \sqrt[3] 3 \zeta_3^2 \\
\zeta_3 &\mapsto \zeta_3,~ \zeta_3^2
.\end{align*}

You can show this is nonabelian by composing a few of these, thus the Galois group is $S^3$.

*Example 3*
If $[E: F] = 2$, then $E$ is automatically a splitting field.

Since it's a finite extension, it's algebraic, so let $\alpha \in E\setminus F$.

Then $\min(\alpha, F)$ has degree 2, and thus $E = F(\alpha)$ contains all of its roots, making $E$ a splitting field.

## The Galois Correspondence

There are three key players here:
$$
[E: F],\quad  \{E: F\},\quad \Gal(E/F)
.$$

How are they related?

**Definition:**
Let $E \geq F$ be a finite extension.
$E$ is **normal** (or Galois) over $F$ iff $E$ is a separable splitting field over $F$.

*Examples:*

1. $\QQ(\sqrt 2, \sqrt 3)$ is normal over $\QQ$.

2. $\QQ(\sqrt[3] 3)$ is not normal (not a splitting field of any irreducible polynomial in $\QQ[x]$).

3. $\QQ(\sqrt[3] 3, \zeta_3)$ is normal

**Theorem:**
Let $F \leq E \leq K \leq \overline F$, where $K$ is a finite normal extension of $F$.
Then

1. $K$ is a normal extension of $E$ as well,

2. $\Gal(K/E) \leq \Gal(K/F)$.

3. For $\sigma, \tau \in \Gal(K/F)$,
$$
\sigma\mid_E = \tau\mid_E
\iff
\sigma, \tau \text{ are in the same left coset of }~
\frac{\Gal(K/F)}{\Gal(K/E)}
.$$

*Proof of (1):*
Since $K$ is separable over $F$, we have $K$ separable over $E$.

Then $K$ is a splitting field for polynomials in $F[x] \subseteq E[x]$.
Thus $K$ is normal over $E$.

$\qed$

*Proof of (2):*

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

So this follows by definition.

$\qed$

*Proof of (3):*
Let $\sigma, \tau \in \Gal(K/F)$ be in the same left coset.
Then
$$
\tau\inv\sigma \in \Gal(K/E)
,$$
so let $\mu \definedas \tau\inv\sigma$.

Note that $\mu$ fixes $E$ by definition.

So $\sigma = \tau \mu$, and thus
$$
\sigma(e) = \tau(\mu(e)) = \tau(e) \text{ for all } e\in E
.$$

$\qed$

> Note: We don't know if the intermediate field $E$ is actually a *normal* extension of $F$.

> **Standard example:**
$K \geq E \geq F$ where
$$
K = \QQ(\sqrt[3] 3, \zeta_3)\quad E = \QQ(\sqrt[3] 3) \quad F = \QQ
.$$
Then $K \normal E$ and $K\normal F$, since $\Gal(K/F) = S_3$ and $\Gal(K/E) = \ZZ_2$.
But $E \ntrianglelefteq F$, since $\ZZ_2 \ntrianglelefteq S_3$.