Hopefully this is not too easy an exercise.
Let $F$ be a field. Let $I \subset \mathbb{N}$ be the set of all positive integers $d$ such that there exists an irreducible polynomial of degree $d$ over $F$. What kind of $I$ can occur?
Of course $1 \in I$, and of course we can have $I = \mathbb{N}$ or $I = \{ 1, 2 \}$ or $I = \{ 1 \}$. The Artin-Schreier theorem implies (I think) that if $I$ is finite, then only the latter two cases occur. So what kind of infinite $I$ can occur?
Edit: For example, correct me if I'm wrong, but we can get $I = \{ 1, p, p^2, … \}$ for any prime $p$. Start with $\mathbb{F}_l, l \neq p$, which has absolute Galois group $\hat{\mathbb{Z}} = \prod_q \mathbb{Z}\_q$ and take the fixed field $K$ of $\prod_{q \neq p} \mathbb{Z}\_q$. Then $G_K = \text{Gal}(\overline{\mathbb{F}_l}/K) = \mathbb{Z}_p$, which (again, correct me if I'm wrong) has the property that its only finite quotients are the groups $\mathbb{Z}/p^n\mathbb{Z}$. Does this work?
Best Answer
Claim 1: The first half of what I said in my comments is correct: for each subset $S$ of the prime numbers, there is a field $F$ having the property that it admits a degree $d$ field extension iff $d$ is divisible only by primes in $S$.
As Qiaochu says, this comes about because of the existence of perfect fields $K$ with absolute Galois group $\hat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$. In particular, take the closed subgroup $H_S = \prod_{p \not \in S} \mathbb{Z}_p$, and let $K_S = \overline{K}^{H_S}$.
We can take $K$ to be any finite field, or $\mathbb{C}((t))$, for instance.
The part of my claim about using $\mathbb{R}((t))$ to get exact $2$-divisibility seems not necessarily correct to me now. The problem is that the absolute Galois group of $\mathbb{R}((t))$ is a nonabelian extension of $\mathbb{Z}/2\mathbb{Z}$ by $\hat{\mathbb{Z}}$: indeed it is the profinite completion of the infinite dihedral group.
As for what I said about the converse: no way, things are definitely more complicated than that! What I had in mind was the "local" observation that by Artin-Schreier, for any prime $p$ the Sylow-$p$-subgroup of the absolute Galois group is either trivial, infinite or has order $2$. (The case of characteristic $p > 0$ has to be given a little separate attention, but I don't think there's a problem there.)
However, the different Sylow p-subgroups will not act independently unless the absolute Galois group is pro-nilpotent. (Note that the profinite dihedral group is not pro-nilpotent.) Thus:
Claim 2: If $K$ has characteristic $0$ and pro-nilpotent Galois group, then the possible orders are exactly those in Claim 1.
Here is an explicit example to show that the converse to Claim 1 is not generally correct: let $F$ be the maximal solvable extension of $\mathbb{Q}$. Then it has no quadratic extensions. However, it certainly does have extensions of even degree, since otherwise -- e.g. by Feit-Thompson! -- the absolute Galois group of $\mathbb{Q}$ would be pro-solvable, which it most certainly is not: many nonabelian simple groups (e.g. $A_n$ for $n \geq 5$) are known to occur as Galois groups over $\mathbb{Q}$.
What I said is the correct answer (with 2 inverted!) to a different question: which supernatural numbers can be the order of the absolute Galois group of a field?