# Thursday October 3rd

When can we guarantee that there is a $\tau: E\selfmap$ lifting the identity?

If $E$ is *separable*, then we have $\abs{ \Gal (E/F) } = \theset{E: F} [E: F]$.

**Fact:**
$\{F(\alpha): F \}$ is equal to number of *distinct* zeros of $\mathrm{min}(\alpha, F)$.

If $F$ is algebraic, then $[F(\alpha): F]$ is the degree of the extension, and $\theset{F(\alpha): F} \divides [F(\alpha): F]$.

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

*Proof:*
If $E \geq F$ is finite, $E = F(\alpha_1, \cdots, \alpha_n)$.

So $\mathrm{min}(\alpha_i, F)$ has $a_j$ as a root, so let $n_j$ be the number of distinct roots, and $v_j$ the respective multiplicities.

Then
$$
[F: F(\alpha_1, \cdots, \alpha_{n-1})] = n_j v_j = v_j \theset{F: F(\alpha_1, \cdots, \alpha_{n-1})}
.$$

So $[E: F] = \prod_j n_j v_j$ and $\theset{E:F} = \prod_j n_j$, and we obtain divisibility.

$\qed$

**Definitions:**

1. An extension $E \geq F$ is **separable** iff $[E:F] = \{E: F\}$

2. An element $\alpha \in E$ is **separable** iff $F(\alpha) \geq F$ is a separable extension.

3. A polynomial $f(x) \in F[x]$ is **separable** iff $f(\alpha) = 0 \implies \alpha$ is separable over $F$.

**Lemma:**

1. $\alpha$ is separable over $F$ iff $\min(\alpha, F)$ has zeros of multiplicity one.

2. Any irreducible polynomial $f(x) \in F[x]$ is separable iff $f(x)$ has zeros of multiplicity one.

*Proof of (1):*
Note that $[F(\alpha): F] = \deg \min(\alpha, F)$, and $\theset{F(\alpha): F}$ is the number of distinct zeros of $\min(\alpha, F)$.

Since all zeros have multiplicity 1, we have $[F(\alpha): F] = \theset{F(\alpha): F}$.

$\qed$

*Proof of (2):*
If $f(x) \in F[x]$ is irreducible and $\alpha\in \overline F$ a root, then $\min(\alpha, F) \divides f(\alpha)$.

But then $f(x) = \ell \min(\alpha, F)$ for some constant $\ell \in F$, since $\min(\alpha, F)$ was monic and only had zeros of multiplicity one.

$\qed$

**Theorem:**
If $K \geq E \geq F$ and $[K:F] < \infty$, then $K$ is separable over $F$ iff $K$ is separable over $E$ and $E$ is separable over $F$.

*Proof:*

\begin{align*}
[K: F]
&= [K:E] [E: F] \\
&= \{K:E\} \{E: F\} \\
&= \{K: F\}
.\end{align*}

**Corollary:**
Let $E \geq F$ be a finite extension.
Then
$$
E \text{ is separable over } F \iff \text{ Every } \alpha \in E \text{ is separable over } F
.$$

*Proof:*

$\implies$:
Suppose $E \geq F$ is separable.

Then $E \geq F(\alpha) \geq F$ implies that $F(\alpha)$ is separable over $F$ and thus $\alpha$ is separable.

$\impliedby$:
Suppose every $\alpha \in E$ is separable over $F$.

Since $E = F(\alpha_1, \cdots, \alpha_n)$, build a tower of extensions over $F$.
For the first step, consider $F(\alpha_1, \alpha_2) \to F(\alpha_1) \to F$.

We know $F(\alpha_1)$ is separable over $F$.
To see that $F(\alpha_1, \alpha_2)$ is separable over $F(\alpha_1)$, consider $\alpha_2$.

$\alpha_2$ is separable over $F \iff \min(\alpha_2, F)$ has roots of multiplicity one.

Then $\min(\alpha_2, F(\alpha_1)) \divides \min(\alpha_2, F)$, so $\min(\alpha_2, F(\alpha))$ has roots of multiplicity one.

Thus $F(\alpha_1, \alpha_2)$ is separable over $F(\alpha_1)$.

$\qed$

## Perfect Fields

**Lemma:**
$f(x) \in F[x]$ has a multiple root $\iff f(x), f'(x)$ have a nontrivial (multiple) common factor.

*Proof*:

$\implies$:
Let $K\geq F$ be an extension field of $F$.

Suppose $f(x), g(x)$ have a common factor in $K[x]$; then $f,g$ also have a common factor in $F[x]$.

If $f, g$ do not have a common factor in $F[x]$, then $\gcd(f, g) = 1$ in $F[x]$, and we can find $p(x), q(x) \in F[x]$ such that $f(x)p(x) + g(x)q(x) = 1$.

But this equation holds in $K[x]$ as well, so $\gcd(f, g) = 1$ in $K[x]$.

We can therefore assume that the roots of $f$ lie in $F$.
Let $\alpha\in F$ be a root of $f$.
Then

\begin{align*}
f(x)  &= (x-\alpha)^m g(x) \\
f'(x) &= m(x-\alpha)^{m-1} g(x) + (x-\alpha)^m g'(x)
.\end{align*}

If $\alpha$ is a multiple root, $m > 2$, and thus $(x-\alpha) \divides f'$.

$\impliedby$:
Suppose $f$ does not have a multiple root.

We can assume all of the roots are in $F$, so we can split $f$ into linear factors.

So

\begin{align*}
f(x) = \prod_{i=1}^n (x-\alpha_i) \\
f'(x) = \sum_{i=1}^n \prod_{j\neq i} (x-\alpha_j)
.\end{align*}

But then $f'(\alpha_k) = \prod{j\neq k} (x - \alpha_j) \neq 0$.
Thus $f, f'$ can not have a common root.

$\qed$

> Moral: we can thus test separability by taking derivatives.

**Definition:**
A field $F$ is *perfect* if every finite extension of $F$ is separable.

**Theorem**:
Every field of characteristic zero is perfect.

*Proof:*
Let $F$ be a field with $\mathrm{char}(F) = 0$, and let $E \geq F$ be a finite extension.

Let $\alpha \in E$, we want to show that $\alpha$ is separable.
Consider $f = \min(\alpha, F)$.
We know that $f$ is irreducible over $F$, and so its only factors are $1, f$.
If $f$ has a multiple root, then $f, f'$ have a common factor in $F[x]$.
By irreducibility, $f \divides f'$, but $\deg f' < \deg f$, which implies that $f'(x) = 0$.
But this forces $f(x) = c$ for some constant $c\in F$, which means $f$ has no roots -- a contradiction.

So $\alpha$ separable for all $\alpha \in E$, so $E$ is separable over $F$, and $F$ is thus perfect.

$\qed$

**Theorem:**
Every finite field is perfect.

*Proof:*
Let $F$ be a finite field with $\mathrm{char} F = p > 0$ and let $E \geq F$ be finite.
Then $E = F(\alpha)$ for some $\alpha\in E$, since $E$ is a simple extension (look at $E^*$?)
So $E$ is separable over $F$ iff $\min(\alpha, F)$ has distinct roots.

So $E\units = E\setminus\theset{0}$, and so $\abs{E} = p^n \implies \abs{E} = p^{n-1}$.
Thus all elements of $E$ satisfy
$$
f(x) \definedas x^{p^n} - x \in \ZZ_p[x]
.$$

So $\min(\alpha, F) \divides f(x)$.
One way to see this is that *every* element of $E$ satisfies $f$, since there are exactly $p^n$ distinct roots.

Another way is to note that
$$
f'(x) = p^nx^{p^n - 1} - 1 = -1 \neq 0
.$$

Since $f(x)$ has no multiple roots, $\min(\alpha, F)$ can not have multiple roots either.

$\qed$

> Note that $[E: F] < \infty \implies F(\alpha_1, \cdots, \alpha_n)$ for some $\alpha_i \in E$ that are algebraic over $F$.

## Primitive Elements

**Theorem (Primitive Element):**
Let $E\geq F$ be a finite extension and separable.

Then there exists an $\alpha \in E$ such that $E = F(\alpha)$.

*Proof:*
See textbook.

**Corollary:**
Every finite extension of a field of characteristic zero is simple.