Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $K$ be imperfect, $K^a$ its algebraic closure and $K^{\rm sep}$ its separable closure. Show $[K^a \colon K]$ and $[K^a\colon K^{\rm sep}]$ are infinite. Is $[K^{\rm sep}\colon K]$ infinite?

Since $K$ is not perfect, I know there is an element $a$ in $K$ that has no $p$th root in $K$, i.e. there is an $a$ in $K$ such that there is a $b$ in $K^a\setminus K$ such that $b^p = a$. Also, $x^{p^n} - a$ is irreducible over $K[x]$ for $a$ in $K\setminus K^p$.

I'm pretty sure that I have to assume these degrees are finite and somehow lead to a contradiction from the fact that $K$ is imperfect, but I'm stuck.

Any help would be greatly appreciated.

share|cite|improve this question
If $[K_a:K]$ were finite, then $p^n> [K_a:K]$ for some natural number $n$, but you already located an irreducible polynomial of degree $p^n$, so... – Jyrki Lahtonen Jun 6 '11 at 13:54
So, would it follow that [K_a:K] has degree p^n, which leads to a contradiction? Also, why is p^n> [K_a:K]? – josh Jun 6 '11 at 15:13
It doesn't have to be, but surely it is, if we select $n$ to be large enough! You did tell that the polynomial you gave is irreducible for any choice of $n$. See also P.L. Clark's answer below. – Jyrki Lahtonen Jun 6 '11 at 15:23
up vote 6 down vote accepted

You have already isolated the key part: since $K$ is imperfect, there exists $a \in K \setminus K^p$, and from this it follows that for all $r > 0$, the polynomial $t^{p^r} - a$ is irreducible (see e.g. Lemma 32 of these notes for corroboration of this fact). Now:

1) You have irreducible polynomials of arbitrary large degree, hence algebraic extensions of arbitrarily large degree, so $[\overline{K}:K]$ must be infinite.

2) Because your irreducible polynomials are purely inseparable, they remain irreducible over the separable closure $K^{\operatorname{sep}}$. (Somewhat more concretely, $a$ does not become a $p$th power in $K^{\operatorname{sep}}$.) So the argument of 1) works with $K$ replaced by $K^{\operatorname{sep}}$.

share|cite|improve this answer
@Josh: And to address the last question, consider the following. It may happen that $K$ is separably closed, but not algebraically closed, or we may have $K=F_p(t)$ form some transcendental element $t$. – Jyrki Lahtonen Jun 6 '11 at 15:24
Thanks Pete and Jyrki! That definitely helped! – josh Jun 6 '11 at 15:28
Would [Ksep : K] be infinite as well? – josh Jun 6 '11 at 15:37
@Josh: not necessarily, no. If you start with any nonperfect field and then take the separable algebraic closure, you will get a field $K = K^{\operatorname{sep}}$ satisfying the hypotheses. – Pete L. Clark Jun 6 '11 at 15:52

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.