Possible degrees of separable irreducible polynomials Let $K$ be a field, and suppose that $[K^{sep} : K] = \infty$. Can we find, for any prime number $p$ and any $k \geq 0$, a separable irreducible polynomial $P$ such that $p^k$ divides the degree of $P$? If not, what are some examples?
 A: Not always. Let $p, q$ be prime numbers, and let $K$ be the pro-$q$ closure of $GF(p)$, i.e. the union of all $GF(p^{q^n})$. Then $K^{sep}/K = \overline K/K$ is infinite (of Galois group $\prod_{\ell \neq q} \mathbb Z_\ell$). But $K$ has no finite extension of degree divisible by $q$. Indeed, because $Gal(\overline K / K)$ is abelian, $K$ would then have an extension of degree $q$.
A: EDIT . Given a field $K$ and a prime number $p$, the maximal pro-$p$-extension $K(p)$ of $K$ is the compositum of all the finite galois $p$-extensions of $K$ (i.e. of degree equal to a power of $p$). By construction $K(p)/K$ is galois, and its Galois group $G_K (p)$ is a pro-$p$-group, i.e. the projective limit of its finite quotients. Your question amounts to whether $[K(p):K]$ is infinite if $[K^{sep}:K]$ is infinite. Actually, $[K(p):K]$ can be finite or infinite. Examples :
1) Finite degree.  The extension $K(p)/K$  is maximal in the following sense: if $F$ is a finite $p$-extension of $K(p)$, so are all its conjugates inside $K^{sep}/K$, hence the normal closure of $F/k$ is a pro-$p$-extension contained in $K(p)$, and $F=K(p)$. Putting $L=K(p)$, we have $L(p)=L$. Note that the example given by @punctured dusk is of the same type.
2) Infinite degree. (i) If $char K=p$, then $G_K (p)$ is a free pro-$p$-group on $dim_{\mathbf F_p}K^+/P(K^+)$, where $P$ is the Artin-Schreier operator $P(x)=x^p-x$. 
(ii) In the following, a "local field" will be an algebraic extension of the $p$-adic field $\mathbf Q_p$ or of the field of formal power series $\mathbf F_p [[X]]$. Denote by $\chi(K)$ the characteristic of the residual field of $K$, and put $\delta(K)=1$ (resp. $0$) if $K$ contains (resp. does not contain) a primitive $p$-th root of unity. Then : (a) If $char(K)= p$ ,  $G_K (p)$ is a free pro-$p$-group on a countable number of generators ; (b) If $char(K)=0$, $G_K (p)$ is a  pro-$p$-group on $2$ (resp. $[K:\mathbf Q_p]+1$) generators if $\chi(K)\neq p$ (resp. $= p$), with one relation ; (iii)  If $\chi(K)\neq p$ and $\delta(K)=0$, $G_K (p)$ is pro-$p$-free on one generator. In all cases, $K(p)$ contains the maximal unramified pro-$p$-extension  of $K$, whose Galois group over $K$  is isomorphic to the additive group of the ring $\mathbf Z_p$ of $p$-adic integers, hence $G_K (p)$ is infinite. For all this, see e.g. H. Koch, "Galois theory of $p$-extensions", chap. 10.
(iii) The case of number fields is much more complicated. But it will be enough to consider $K_S(p)^{ab}$, the maximal abelian pro-$p$-extension of $K$ which is uramified outside a finite set of primes of $K$ which contains the archimedean primes and the primes above $p$. Class field theory shows that Gal$(K_S(p)^{ab})/K$ is a $\mathbf Z_p$-module of rank $\ge 1+c$, where $2c$ is the number of complex embeddings of $K$. It follows that $G_K (p)$ is infinite. 
