# Thursday October 17th

## Example Galois Group Computation

*Example:*
What is the Galois group of $x^4-2$ over $\QQ$?

First step: find the roots.
We can find directly that there are 4 roots given by
$$
\theset{\pm \sqrt[4] 2, \pm i \sqrt[4] 2} \definedas \theset{r_i}
.$$

The splitting field will then be $\QQ(\sqrt[4] 2, i)$, which is separable because we are in characteristic zero.
So this is a normal extension.

We can find some automorphisms:
$$
\sqrt[4] 2 \mapsto r_i, \quad i \mapsto \pm i
.$$

So $\abs G = 8$, and we can see that $G$ can't be abelian because this would require every subgroup to be abelian and thus normal, which would force every intermediate extension to be normal.

But the intermediate extension $\QQ(\sqrt[4] 2)/\QQ$ is not a normal extension since it's not a splitting field.

So the group must be $D_4$.

$\qed$

## Insolubility of the Quintic

### Symmetric Functions

Let $F$ be a field, and let
$$
F(y_1, \cdots , y_n) = \theset{\frac{f(y_1, \cdots, y_n)}{g(y_1, \cdots, y_n)} \suchthat f, g \in F[y_1, \cdots, y_n]}
$$
be the set of *rational* functions over $F$.

Then $S_n \actson F(y_1, \cdots, y_n)$ by permuting the $y_i$, i.e.

\begin{align*}
\sigma \left(\frac{
f(y_1, \cdots, y_n)
}{
g(y_1, \cdots, y_n)
}\right) = \frac{
f(\sigma(y_1), \cdots, \sigma(y_n))
}{
g(\sigma(y_1), \cdots, \sigma(y_n))
}
.\end{align*}

**Definition:**
A function $f \in F(\alpha_1, \cdots, \alpha_n)$ is **symmetric** $\iff$ under this action, $\sigma\actson f = f$ for all $\sigma \in S_n$.

*Examples:*

1. $f(y_1, \cdots, y_n) = \prod y_i$
2. $f(y_1, \cdots, y_n) = \sum y_i$.

### Elementary Symmetric Functions

Consider $f(x) \in F(y_1, \cdots, y_n)[x]$ given by $\prod (x-y_i)$.
Then $\sigma f = f$, so $f$ is a symmetric function.
Moreover, all coefficients are fixed by $S_n$.
So the coefficients themselves are symmetric functions.

Concretely, we have

| Coefficient | Term |
| --- | --- |
| 1           | $(-1)^n$
| $x^{n-1}$   | $-y_1 - y_2 - \cdots - y_n$
| $x^{n-2}$   | $y_1y_2 + y_1y_3 + \cdots + y_2y_3 + \cdots$

The coefficient of $x^{n-i}$ is referred to as the *$i$th elementary symmetric function*.

Consider an intermediate extension $E$ given by joining all of the elementary symmetric functions:

![Image](figures/2019-10-17-09:56.png)\

Let $K$ denote the base field with *all* symmetric functions adjoined; then $K$ is an intermediate extension, and we have the following results:

**Theorem**:

1. $E \leq K$ is a field extension.

2. $E \leq F(y_1, \cdots, y_n)$ is a finite, normal extension since it is the splitting field of $f(x) = \prod (x-y_i)$, which is separable.

We thus have
$$
[F(y_1, \cdots, y_n): E] \leq n! < \infty
.$$

*Proof:*

We'll show that in fact $E = K$, so all symmetric functions are generated by the elementary symmetric functions.

By definition of symmetric functions, $K$ is exactly the fixed field $F(y_1, \cdots, y_n)_{S_n}$, and $\abs S_n = n!$.

So we have

\begin{align*}
n! &= \abs{ \Gal(F(y_1, \cdots, y_n / K))} \\
& \leq \{F(y_1, \cdots, y_n) : K\} \\
& \leq [F(y_1, \cdots, y_n): K]
.\end{align*}

But now we have
$$
n! \leq [F(y_1, \cdots, y_n):K] \leq [F(y_1, \cdots, y_n) : E] \leq n!
$$
which forces $K=E$.

$\qed$

**Theorem**:

1. Every symmetric function can be written as a combination of sums, products, and possibly quotients of elementary symmetric functions.

2. $F(y_1, \cdots, y_n)$ is a finite normal extension of $F(s_1, \cdots, s_n)$ of degree $n!$.

3. $\Gal(F(y_1, \cdots, y_n) / F(s_1, \cdots, s_n)) \cong S_n$.

We know that every group $G \injects S_n$ by Cayley's theorem.
So there exists an intermediate extension
$$
F(s_1, \cdots, s_n) \leq L \leq F(y_1, \cdots, y_n)
$$
such that $G = \Gal(F(y_1, \cdots, y_n) / L)$.

> Open question: which groups can be realized as Galois groups over $\QQ$?
Old/classic question, possibly some results in the other direction (i.e. characterizations of which groups *can't* be realized as such Galois groups).

### Extensions by Radicals

Let $p(x) = \sum a_i x^i \in \QQ[x]$ be a polynomial of degree $n$.
Can we find a formula for the roots as a function of the coefficients, possibly involving radicals?

- For $n = 1$ this is clear

- For $n=2$ we have the quadratic formula.

- For $n = 3$, there is a formula by work of Cardano.

- For $n = 4$, this is true by work of Ferrari.

- For $n \geq 5$, there can **not** be a general equation.

**Definition:**
Let $K \geq F$ be a field extension.
Then $K$ is an **extension of $F$ by radicals** (or a **radical extension**) $\iff$ $K = \alpha_1, \cdots, \alpha_n$ for some $\alpha_i$ such that

1. Each $\alpha_i^{m_i} \in F$ for some $m_i > 0$.

2. For each $i$, $\alpha_i^{\ell_i} \in F(\alpha_1, \cdots, \alpha_{i-1})$ for some $\ell_i < m_i$ (?).

**Definition:**
A polynomial $f(x) \in F[x]$ is **solvable by radicals** over $F$ $\iff$ the splitting field of $f$ is contained in some radical extension.

*Example:*
Over $\QQ$, the polynomials $x^5-1$ and $x^3-2$ are solvable by radicals.

Recall that $G$ is *solvable* if there exists a normal series
$$
1 \normal H_1 \normal H_2 \cdots \normal H_n \normal G
\text{ such that } H_n/H_{n-1} \text{ is abelian } \forall n
.$$


### The Splitting Field of $x^n-a$ is Solvable

**Lemma**:
Let $\ch F = 0$ and $a\in F$.
If $K$ is the splitting field of $p(x) = x^n-a$, then $\Gal(K/F)$ is a solvable group.

*Example:*
Let $p(x) = x^4-2 / \QQ$, which had Galois group $D_4$.

*Proof:*
Suppose that $F$ contains all $n$th roots of unity, $\theset{1, \zeta, \zeta^2, \cdots, \zeta^[n-1]}$ where $\zeta$ is a primitive $n$th root of unity.
If $\beta$ is any root of $p(x)$, then $\zeta^i\beta$ is also a root for any $1\leq i \leq n-1$.
This in fact yields $n$ distinct roots, and is thus all of the them.
Since the splitting field $K$ is of the form $F(\beta)$, then if $\sigma \in \Gal(K/F)$, then $\sigma(\beta) = \zeta^i \beta$ for some $i$.
Then if $\tau \in \Gal(K/F)$ is any other automorphism, then $\tau(\beta) = \zeta^k \beta$ and thus (exercise) the Galois group is abelian and thus solvable.

Suppose instead that $F$ does not contain all $n$th roots of unity.
So let $F' = F(\zeta)$, so $F \leq F(\zeta) = F' \leq K$.
Then $F \leq F(\zeta)$ is a splitting field (of $x^n-1$) and separable since we are in characteristic zero and this is a finite extension.
Thus this is a normal extension.

We thus have $\Gal(K/F) / \Gal(K/F(\zeta)) \cong \Gal(F(\zeta)/ F)$.
We know that $\Gal(F(\zeta)/ F)$ is abelian since this is a cyclotomic extension, and so is $\Gal(K/F(\zeta))$.
We thus obtain a normal series
$$
1 \normal \Gal(K/F(\zeta)) \normal \Gal(K/F)
$$
Thus we have a solvable group.

$\qed$