Algebraic Separable Extensions

Example.

Consider $k(\alpha)/k$ where $\alpha$ is algebraic over $k$ with minimal polynomial $f$. 

Then there exists an integer $r$ such that $f(x) = \prod_{i=1}^m(x -\alpha_i)^{r}$ where $\alpha_i$ ranges over all distinct roots of $f$. We call

• $r =[k(\alpha): k]_{i}$ the inseparable degree of $k(\alpha)/k$;
• $m = [k(\alpha): k]_s$ the separable degree of $k(\alpha)/k$;
  Notice that as $rm $ is equal to the degree of $f$, we have $[k(\alpha): k]_i [k(\alpha): k]_s = [k(\alpha): k].$

In general, let $E/k$ be an algebraic field extension. Fix an embedding $\sigma: k \to L$ for some algebraically closed field $L$. We define the separable degree $[E: k]_s$ of $E$ over $k$ to be number of extensions of $\sigma$ to $E/k \to L/k$. It is independent of the choice of $L$ and $\sigma$. 

Comment: It turns out that if $\text{char} k = 0$ then $r = 1$, and if $\text{char} k = p$ then $r$ is a power of $p$.

Separability for Finite Extension

Let $E$ be a finite extension of $k$. We define  the inseparable degree $[E:k]_i$ of $E/k$ to be $[E:k]/[E:k]_s$. We say that $E$ is separable over $k$ if $[E:k]_s = [E:k]$. If $\alpha$ is algebraic over $k$, we say $\alpha$ is separable over $k$ if $k(\alpha)/k$ is (i.e. if the irreducible polynomial of $\alpha$ over $k$ has no multiple roots). A polynomial $f(x) \in k[x]$ is called separable if it has no multiple root.

Theorem. Let $F\supset E \supset k$. Then
$$[F:k]_s = [F:E]_s [E:k]_s.$$
If $F/k$ is finite then so is $[F:k]_s$ and $[F:k]_s \leq [F:k].$

Claim. Moreover if $k \subset E \subset F$ and $\alpha \in F$ is separable over $k$ then it is also separable over $E$.

Theorem. A finite extension $E/k$ is separable iff every element of $E$ is separable over $k$.

Proof. Suppose $E/k$ is separable and $\alpha \in E$. Then consider $E \supset k(\alpha) \supset k$. By tower law, $[E:k] = [E: k]_s = [E:k(\alpha)]_s [k(\alpha):k]_s \leq [E:k(\alpha)] [k(\alpha):k] = [E: k]$ so we must have $[k(\alpha): k]_s = [k(\alpha):k].$

Conversely, suppose every element of $E$ is separable over $k$ and $E= k(\alpha_1, \ldots, \alpha_n).$ Consider the tower $k \subset k(\alpha_1) \subset k(\alpha_1, \alpha_2) \subset \cdots \subset k(\alpha_1, \ldots, \alpha_n) = E$.

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Separability for Algebraic Extension

Let $E/k$ be an algebraic extension. We define $E$ to be separable over $k$ if every finitely generated subextension is separable over $k$.

Theorem.  Suppose $E/k$ is an algebraic extension generated by a (not necessarily finite) family $\{\alpha_i\}$. If each $\alpha_i$ is separable ver $k$ then $E/k$ is separable.

Proof.
Every element $x \in E$ will lie in some $E_x= k(\alpha_{i_1}, \ldots, \alpha_{i_n})$ finitely generated by a finite subset of the $\alpha_i$'s. As each $\alpha_i$ is separable over $k$, so must $E_x$. Now let $F/k$ finitely generated subextension of $E/k$ is of finite degree (as $E/k$ is finite). From the above all of its elements are separable over $k$, so $L/k$ must be separable.

Theorem. Separable extensions form a distinguished class of extensions.

In particular, fixed an algebraic closure $k^a$ of $k$. Then the compositum of all separable subextensions $k^a/k$ is separable over $k$. We call it the separable closure of $k$ and denote it by $k^s$.

A field $k$ is called perfect if $k^p = k$.

Proposition. If $k$ is a perfect field then every algebraic extension of $k$ is separable and perfect.