# The degree of the algebraic closure over the separable closure of an imperfect field

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.

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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

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 $K^{\operatorname{sep}}$.
@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
@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