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?

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?