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.

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:

*

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


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