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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?

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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.

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  • $\begingroup$ Do $[E:F]$ and $[E:F]_{sep}$ coinside in the sense of actual cardinalities? Not that they are just both infinite? $\endgroup$ – Rubertos May 6 '15 at 12:32
  • $\begingroup$ Yes, in my example they are both countably infinite. $\endgroup$ – Zev Chonoles May 6 '15 at 15:20
  • $\begingroup$ How do I prove that they are countably infinite? Would you give me a reference or proof-sketch? Thank you in advance $\endgroup$ – Rubertos May 6 '15 at 17:55
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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!

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