I was feeling a bit rusty on my field theory, and I was reviewing out of McCarthy's excellent book, Algebraic Extensions of Fields. Out of Chapter 1, I was able to work out everything "left to the reader" or omitted except for one corollary, stated without proof (see here for the page in the book):

Let $K/k$ be a finite normal extension. Then $K$ can be obtained by a purely inseparable extension, followed by a separable extension.

The text immediately preceding this implies that the intermediate field that's going to make this happen is $F=\{a\in K:\sigma(a)=a$ for all $\sigma\in Gal(K/k)\}$, and I understand his argument as to why $F/k$ is purely inseparable (in fact, that's the theorem, Theorem 21, which this is a corollary to). What I don't understand is why $K/F$ is separable; I don't see how we've ruled out it being non-purely inseparable.

Note that I will be making a distinction between non-purely inseparable (inseparable, but not purely inseparable) and not purely inseparable (either separable or non-purely inseparable).

Here are some observations / my general approach:

  • One big thing that seemed promising was Theorem 11 (at the bottom of this page), which is basically the reverse of the corollary I'm having trouble with:

Let $K$ be an arbitrary algebraic extension of $k$. Then $K$ can be obtained by separable extension followed by a purely inseparable extension.

(the separable extension referred to is of course the separable closure of $k$ in $K$). It seems like we want to use Theorem 11 on $K/F$, and argue that there can't be "any more" pure inseparability, but I couldn't figure out a way of doing this.

  • Theorem 21 is actually an "if and only if" (that is, $a\in K$ is purely inseparable over $k$ iff $\sigma(a)=a$ for all $\sigma\in Gal(K/k)$). Because this implies that any $a\in K$ with $a\notin F$ is not purely inseparable over $k$, we have that $F$ is the maximum (not just maximal) purely inseparable extension of $k$ in $K$.

  • If any $a\in K$ were purely inseparable over $F$, by Theorem 8 (see here), there is some $e$ for which $a^{p^e}\in F$. But by the same theorem, since $F/k$ is purely inseparable, there is some $b$ for which $(a^{p^e})^{p^b}=a^{p^{e+b}}\in k$. Thus $a$ would be purely inseparable over $k$ by the converse (Corollary 1 to Theorem 9, see here), and hence be in $F$. Thus, $K$ (and any field between $K$ and $F$, besides $F$ itself) is not purely inseparable over $F$.

So, that's why I don't see how we've ruled out $K/F$ being non-purely inseparable. Sorry about making lots of references to the book - I'm just not sure what previously established results McCarthy intended to be used, and I wanted to point out what I saw as the important ones for people not familiar with the book. I'm sure I'm missing something obvious here. Does anyone see the last bit of the argument?


3 Answers 3


Edit: My mistake, I misread your post. Here's the correct answer.


Proposition 6.11

  • $\begingroup$ No element in K generates a purely inseparable extension of F. That's the issue - I don't see why we couldn't have that an element in K generates a non-purely inseparable extension of F. $\endgroup$ Feb 21, 2010 at 20:58
  • $\begingroup$ It's because if it's normal then you can construct it as a compositum of fields in a nice way using the automorphism groups. $\endgroup$ Feb 21, 2010 at 21:06
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    $\begingroup$ Ah, a very clever argument - and more general than the result McCarthy states! Thanks for your help! $\endgroup$ Feb 21, 2010 at 21:22
  • $\begingroup$ A minor suggestion: Lang's chapters 5,6, 7, and 8 are really some of the best treatments of this subject. Chapter 5 is by far my favorite treatment of algebraic extensoins, and chapter 6 is an excellent treatment of Galois theory based on Artin's famous monograph (although covered with more modern machinery). I suggest you read them rather than McCarthy. $\endgroup$ Feb 21, 2010 at 21:26
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    $\begingroup$ I'll definitely check it out. Though McCarthy is hard to beat on price, I think - I got my copy for $4 :) $\endgroup$ Feb 21, 2010 at 22:08

Dear Zev, as LASER pointed out, Lang's Proposition 6.11 indeed solves your problem. Still, I'd like to add two remarks that might put your problem in perspective.

A) Given a field $K$ and a group of automorphisms $G$ of $K$ we get a fixed field $K^G$ and an extension $K^G \subset K$ . This extension is algebraic if and only if all orbits of $G$ are finite and if this is the case the extension is always separable: each element of $x \in K$ has as minimal polynomial over $K^G$ the product $\prod \limits_{y \in Orb(x)}(T-y)$. [Let me emphasize that i) there is no base field $k$ in this general statement and ii) the group $G$ may very well be infinite with all its orbits finite: every infinite dimensional galois extension gives an example.]

B) The statement is false if you don't assume normality of $k \subset K$ (as I'm sure you guessed!). Here is an example taken from Bourbaki's Algebra V, exercise 3) for §5.

Take a field $F$ of characteristic $p>2$ and define $k=F(x,y)$ where $x,y$ are indeterminates. Let $\theta$ be a zero of the polynomial $P(T)=T^{2p}+xT^p+y$ (in an algebraic closure of $k$, say).Then if we put $K=k(\theta)$, we have our counter-example: no element in $K$ is purely inseparable over $k$, except if it already is in $k$. And yet $K$ is not separable over k since the minimal polynomial of $\theta$ over $k$ is $P(T)$ , which is obviously not separable.

  • $\begingroup$ Is it obvious that no element of $K$ is purely inseparable over $k$? $\endgroup$ Aug 20, 2019 at 15:53

At least in the finite case, I am pretty sure this follows from the fundamental theorem of Galois theory: if $F$ is a field, $G$ a finite group of automorphisms of $F$, and the fixed field of $F$ is $k$, then $f/k$ is Galois with Galois group $G$.

  • $\begingroup$ That's true, but I was trying to do without FTGT because McCarthy doesn't prove it until Chapter 2. $\endgroup$ Feb 21, 2010 at 21:21

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