# Different two definitions for separable extension

Let $$E/F$$ be an algebraic field extension and $$\bar F$$ be an algebraic closure of $$F$$.

Define $$[E:F]_{\text{sep}}$$ as the cardinality of $$\{\sigma\in \operatorname{Mono}(E,\bar F): \sigma \text{ fixes } F\}$$ (it can be proven that this is well-defined)

With this terminology, here are two definitions for separable extension:

Definition 1

If every minimal polynomial $$m$$ of $$\alpha\in E$$ is separable, then $$E/F$$ is called separable.

Definition 2

If $$[E:F]=[E:F]_{\text{sep}}$$, then $$E/F$$ is called separable.

If $$E/F$$ is finite, then these two definitions coincide. However, I'm not really sure why Def. 1 is stronger than Def. 2 for infinite cases. Why Def. 1 is stronger? How do I prove that? And what is the standard one?

Let $p$ be an odd prime, and let $F=\mathbb{F}_p(t)$ and $$E=\mathbb{F}_p(t^{1/p}\cup\{f^{1/2}:f\text{ monic irred in }F[x]\})\subset \overline{F}.$$ There are infinitely many $F$-algebra monomorphisms $E\rightarrow\overline{F}$, since you can choose independently whether to send $f^{1/2}$ to itself or its negative for each monic irreducible $f\in F[x]$. Therefore, we have $$[E:F]=\infty=[E:F]_{\text{sep}}$$ However, the extension $E/F$ isn't separable because the minimal polynomial of $t^{1/p}$ is $$(x-t^{1/p})^p=x^p-t\in F[x]$$ which clearly has repeated roots.
The correct definition for all cases is definition 1, but as you say, they agree when $E/F$ is a finite extension.
• Do $[E:F]$ and $[E:F]_{sep}$ coinside in the sense of actual cardinalities? Not that they are just both infinite? – Rubertos May 6 '15 at 12:32
You can also relate this to the notion of perfect fields. Any field $\mathbb{K}$ of characteristic $p$ with $p$ a prime is said to be perfect if every element is a $p^{th}$ power in $\mathbb{K}$, i.e. $\mathbb{K} = \mathbb{K}^p$. Every field of characteristic $0$ is also called perfect. A result that we know is every irreducible polynomial over a perfect field is separable. If, however, we have a finite field that is not perfect, then as Zev showed, there exists finite inseparable extensions. This same fact cannot carry over to fields of characteristic $0$ since they are perfect by definition. Therefore, in the two definitions above, 1 is stronger for infinite fields because definition 2 is trivially satisfied since the minimal polynomial for $\alpha \in \mathbb{E}$ is by definition irreducible. Hope this helps!