Separable Extensions and Perfect Fields

Definition.

Let $F$ be a field and $\alpha \in \overline{F}$. We say that $\alpha$ is separable over $F$ if its minimal polynomial $m_{\alpha, F}(x)$ is separable.
Let $E/F$ be an algebraic extension. We say that $E/F$ is a separable extension if every $\alpha \in E$ is separable over $F$.

Theorem (Towers of separable extensions).
Let $F \subset E \subset K$ be a tower of algebraic extensions. Then $K/F$ is separable if and only if both $E/F$ and $K/E$ are separable.

Proof.
Suppose $K/F$ is separable. Then every $\alpha \in E \subset K$ is separable over $F$, so $E/F$ is separable. For any $\alpha \in K$, the minimal polynomial $m_{\alpha, E}(x)$ divides $m_{\alpha, F}(x)$ in $E[x]$. Since $m_{\alpha, F}(x)$ is separable by assumption, so is $m_{\alpha, E}(x)$. Thus, $K/E$ is separable.

The converse direction will be proven after we define the separable closure.

Definition/Lemma (Perfect fields).
A field $F$ is called perfect if it satisfies any (and hence all) of the following equivalent conditions:

The algebraic closure $\overline{F}/F$ is a separable extension.
Every algebraic extension $E/F$ is separable.
Every polynomial $p(x) \in F[x]$ is separable.

Proof.
$1 \Rightarrow 2$: Follows directly from the tower theorem above.
$2 \Rightarrow 3$: Let $p(x) \in F[x]$ be a polynomial, and let $p_i(x)$ be an irreducible factor of $p(x)$. Let $\alpha \in \overline{F}$ be a root of $p_i(x)$. Then $F(\alpha)/F$ is algebraic and hence separable. So $m_{\alpha, F}(x)$ is separable, and since $p_i(x)$ is irreducible, $p_i(x)$ is a scalar multiple of $m_{\alpha, F}(x)$ and thus also separable. Therefore, all irreducible factors of $p(x)$ are separable, so $p(x)$ is separable.
$3 \Rightarrow 1$: Let $\alpha \in \overline{F}$. Then $m_{\alpha, F}(x)$ is separable by assumption, so $\alpha$ is separable over $F$. Hence, $\overline{F}/F$ is separable.

Theorem (Characterization of perfect fields).
A field $F$ is perfect if and only if one of the following holds:

$\operatorname{char} F = 0$;
$\operatorname{char} F = p > 0$ and $F^p = F$, where $F^p := { a^p : a \in F }$.

Proof.
In characteristic zero, all polynomials are separable (see Separable Polynomials), so $F$ is perfect.

Suppose $\operatorname{char} F = p > 0$ and $F^p = F$. Then every $a \in F$ is a $p$-th power: $a = b^p$. By induction, for all $k \geq 0$, we have $a = c^{p^k}$ for some $c \in F$.

Let $f(x) \in F[x]$ be irreducible. By an earlier corollary, there exists an irreducible separable $g(x) \in F[x]$ and $k \geq 0$ such that
\[f(x) = g\left(x^{p^k}\right).\]
Since $F^p = F$, each coefficient $c_i$ of $f(x)$ satisfies $c_i = d_i^{p^k}$ for some $d_i \in F$, and we have:
\[\begin{align*} f(x) &= c_0 + c_1 x^{p^k} + \cdots + c_n x^{n p^k} \\ &= (d_0 + d_1 x + \cdots + d_n x^n)^{p^k} \end{align*}\]
Since $f(x)$ is irreducible, we must have $p^k = 1$, so $f(x) = g(x)$ is separable. Hence, $F$ is perfect.

Conversely, if $F^p \ne F$, then there exists $a \in F \setminus F^p$. Let $\alpha \in \overline{F}$ satisfy $\alpha^p = a$. Then $m_{\alpha, F}(x) \mid x^p - a = (x - \alpha)^p$, and since $\alpha \notin F$, we have $\deg m_{\alpha, F}(x) > 1$, so $m_{\alpha, F}(x)$ has repeated roots. Hence, $\alpha$ is not separable over $F$, and $F$ is not perfect.

Corollary (Finite fields are perfect).
Every finite field is perfect.

Proof.
The Frobenius map $\sigma_p : \mathbb{F}_q \to \mathbb{F}_q$ given by $x \mapsto x^p$ is bijective (since the field is finite). Therefore, $\mathbb{F}_q^p = \mathbb{F}_q$, and $\mathbb{F}_q$ is perfect by the theorem above.

Definition/Lemma (Separable closure).
Let $E/F$ be an algebraic extension. The separable closure of $F$ in $E$ is defined as

\[S = \{ \alpha \in E : \alpha \text{ is separable over } F \}.\]

Then $S$ is a subfield of $E$, and $S/F$ is a separable extension.

Given a fixed algebraic closure $\overline{F}$, we define the separable closure of $F$ as

\[F^{\text{sep}} = \{ \alpha \in \overline{F} : \alpha \text{ is separable over } F \}.\]

Then $F^{\text{sep}}/F$ is separable, and the separable closure of $F$ in $E$ is $F^{\text{sep}} \cap E$.

Proof.
Let $\alpha, \beta \in E$ be separable over $F$. Then their minimal polynomials $m_{\alpha, F}(x)$ and $m_{\beta, F}(x)$ are separable. The splitting field of their product is a Galois extension (We will prove this later in the characterization of finite Galois extension), hence normal and separable. Therefore, $\alpha \pm \beta$, $\alpha \beta^{\pm 1}$ are also separable over $F$. Thus, $S$ is a subfield of $E$.

Lemma.
Let $F$ be a field of characteristic $p > 0$, and let $E/F$ be an algebraic extension. For any $\alpha \in E$, there exists $k \ge 0$ such that $\alpha^{p^k}$ is separable over $F$.

Proof.
Let $m_{\alpha, F}(x)$ be the minimal polynomial of $\alpha$. By Separable Polynomials, there exists an irreducible separable $g(x) \in F[x]$ and $k \ge 0$ such that
\[m_{\alpha, F}(x) = g(x^{p^k}).\]
Then $g(x)$ is the minimal polynomial of $\alpha^{p^k}$, which is separable. Thus, $\alpha^{p^k}$ is separable over $F$.

Now, we complete the proof of the tower theorem:

Theorem (Tower of separable extensions, part 2).
Let $F \subset E \subset K$ be a tower of algebraic extensions, with $E/F$ and $K/E$ both separable. Then $K/F$ is separable.

Proof.
If $\operatorname{char} F = 0$, then all algebraic extensions are separable, and there’s nothing to prove. Assume $\operatorname{char} F = p > 0$. Let $S$ be the separable closure of $F$ in $K$. Since $E/F$ is separable, $E \subset S$.

Let $\alpha \in K$. By the lemma above, there exists $k \ge 0$ such that $\alpha^{p^k} \in S$. Assume $k$ is minimal with this property. Then,
\[m_{\alpha, S}(x) \mid x^{p^k} - \alpha^{p^k} = (x - \alpha)^{p^k}.\]
Since $k$ is minimal, $m_{\alpha, S}(x) = (x - \alpha)^{p^k}$.

But $E \subset S$, so $m_{\alpha, S}(x) \mid m_{\alpha, E}(x)$. Since $K/E$ is separable, $m_{\alpha, E}(x)$ must be separable. This forces $p^k = 1$, or else $m_{\alpha, E}(x)$ will have repeated roots. Therefore, we have $\alpha \in S$. We conclude that every $\alpha \in K$ is separable over $F$, and $K/F$ is separable.