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.

  • $\begingroup$ 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

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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.