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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?

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I have only time at the moment to record my guess: let S be a set of prime numbers. Then there exists a field F (of characteristic 0) such that there exists a degree d field extension iff d is divisible only by primes in S. If 2 is not in S, we can build another field such that there exists a degree d field extension iff d or 2d is in S. I suspect the converse is also true. –  Pete L. Clark Mar 1 '10 at 20:40
    
One more remark: all of the sets of degrees referred to above can be realized via algebraic extensions of $\mathbb{R}((t))$. –  Pete L. Clark Mar 1 '10 at 20:44
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@QY: This is a very interesting question. I'm sort of surprised that no one else has tried to answer it. If by tomorrow morning no one else has solved it, I will start thinking seriously about what kind of degree sets one can get by starting with Q and making a "Merkurjev-type construction", i.e., (infinitely) repeatedly applying any of the following basic operations: adjoin to $K$ all roots of irreducible polynomials of degree $d$. –  Pete L. Clark Mar 2 '10 at 3:00
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@Qiaochu: Have you seen the nice paper by J. Shipman, "Improving the fundamental theorem of algebra"? It is in the Math. Intelligencer, 29 (2007), no. 4, 9-14. It is precisely in the direction you want. –  Andres Caicedo Dec 16 '10 at 0:11
    
@Andres: thanks for the link; I think I saw that paper in connection to a different MO problem, but it is certainly relevant here as well. –  Qiaochu Yuan Dec 16 '10 at 0:52

2 Answers 2

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?

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This is a very interesting question, however it seems like a characterization of all subsets of $\mathbb N$ which can be such a degree sequence is very hard. I will point you to the paper "On the degrees of finite extensions of a field" by B. Gordon and E.G. Straus where they study this very problem. Unfortunately, it doesn't seem like the paper is available online so I will include here some of their results. For a field $k$, denote by $S(k)$ the sequence of degrees of irreducible polynomials in $k[x]$, since the only finite $S(k)$ are $\{1\}$ and $\{1,2\}$ by Artin-Schreier we will consider infinite sequences.

  • Let $P$ be any set of prime numbers, denote by $N_P$ the set of integers whose prime divisors are in $P$. There is a field $k$ of any prescribed characteristic with $S(k)=N_P$.
  • For every field $k$, if $a,b\in S(k)$ and $gcd(a,b)=1$ then $ab\in S(k)$. If $K$ is algebraic over $k$ then every element of $S(K)$ divides some element of $S(k)$.
  • For every finite set $A\subset \mathbb{N}$ which doesn't contain $1$, there is a finite $A'\subset \mathbb N$ and a field $k$ so that $S(k)=\mathbb N - A'$ and $A\subset A'$.
  • If $S(k)$ consists only of powers of a prime $p$ then $S(k)=N_p$.
  • If $k$ is a CE field (all finite extensions are cyclic) then $S(k)$ is either equal to some $N_P$ or to all elements in some $N_P$ which are not divisible by $4$.

There is also something you can say from the perspective of computability. If $P$ is $\Sigma_1$ then there is a field $k$ so that $S(k)=N_P$ and you can take $k$ to be computable. See "Sets of degrees of computable fields".

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