Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

As we all know, the algebraic closure often has an infinite degree.
Also, this shows the necessary and sufficient condition for a Galois extension to be a finite extension of fields.
However, we may want to characterize the cases of the case where the extension is not Galois, which is actually my question.
And this question is related to this question.

share|improve this question
Geez. The second link is to Pete Clark's notes which are 73 pages long; I'm supposed to go hunting through it to see what specifically you are refering to? Perhaps you might consider quoting it instead of sending the reader on a scavenger hunt through a 73 page long document... –  Arturo Magidin Feb 25 '11 at 16:22
@Arturo Magidin: Sorry, according to the last link, it is in section 12.5, thanks in any case. –  awllower Feb 25 '11 at 16:25
As far as I can see, all the questions asked here are already answered in $\S 12.4$ of my notes. –  Pete L. Clark Feb 25 '11 at 19:31
I am very sorry. –  awllower Feb 26 '11 at 2:55

2 Answers 2

up vote 6 down vote accepted

Let me summarize the results of $\S 12.4$ of my notes. (This involves making explicit some things which were left as "exercises", but I'm okay with that.)

Let $K$ be any field, let $K^{\operatorname{sep}}$ be any separable closure and let $\overline{K}$ be any algebraic closure containing $K^{\operatorname{sep}}$. Then:

1) Suppose first that $K = K^{\operatorname{sep}}$, i.e., $K$ is separably closed. Then either:
1a) $K$ is algebraically closed, or
1b) It isn't, i.e., $K$ has positive characteristic $p$ and there exists $a \in K$ such that the polynomial $t^p - a$ is irreducible. Then $t^{p^n}-a$ is irreducible for all $n$, so $[\overline{K}:K]$ is infinite. (For this, see Lemma 32 in $\S 6.1$.)

2) Suppose that $K$ is not separably closed, so $G = \operatorname{Aut}(K^{\operatorname{sep}}/K)$ is nontrivial. Then by the Artin-Schreier theorem, either
2a) $\# G = 2$, in which case $K$ is formally real and $K(\sqrt{-1})$ is algebraically closed, or
2b) $\# G > 2$, in which case $G$ is infinite, which implies $[\overline{K}:K]$ is infinite.

Note that some of the exercises outline further extensions of the Artin-Schreier Theorem, i.e., if $\overline{K}/K$ is "small" in other ways then $K$ is either real closed or algebraically closed.

share|improve this answer
Thanks very much, @Pete L. Clark. –  awllower Feb 26 '11 at 3:25
I suddenly found that I forgot to accept it as an answer, hope you won't mind, @Pete L. Clark... –  awllower Mar 12 '11 at 3:20

The Artin-Schreier theorem asserts that, if the field $k$ is not algebraically closed and $[\bar{k} : k]$ is finite, then in fact it equals $2$ and $\bar{k} = k[i]$ and $k$ is real closed. So I guess the condition you're looking for is: if and only if $k$ is not algebraically closed and also not real closed.

I am answering the title question, not the body question. The body question is not equivalent to the title question; the answer to it is that the algebraic closure is Galois if and only if it's separable, hence if and only if $k$ is perfect. Perhaps you have only seen the definition of Galois extension for finite extensions. This is not the definition which works in maximal generality. The definition which does work in maximal generality is the following:

Definition: An algebraic extension $k \to L$ is Galois if the fixed field of $\text{Aut}_k(L)$ is $k$.

The separable closure $k_s$ of $k$ is always Galois (in fact it is the maximal Galois extension of $k$), and agrees with the algebraic closure if and only if $k$ is perfect.

share|improve this answer
Thanks very much. –  awllower Feb 25 '11 at 16:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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