When are arithmetic and geometric monodromy equal? Let $f: Y\to X$ be a finite separable morphism of curves over the finite field $\mathbb{F}_q$. Is there a simple condition under which the arithmetic and geometric monodromy of the covering are equal (i.e. $\mathbb{F}_q$ is algebraically closed in the Galois closure of the extension $\mathbb{F}_q(Y)/\mathbb{F}_q(X)$)?
It is relatively easy to construct examples for which that condition is not satisfied: for example, a Kummer covering of $\mathbb{P}^1$ of degree $n$ coprime to $q-1$. However, I have the vague feeling that for many "natural" coverings I come across, arithmetic and geometric monodromy are indeed equal (although I'm not sure how I'd define "natural" here). Is it possible to give a simple criterion? I'm especially interested in the case when $X$ is an elliptic curve (and if that helps, the characteristic of $\mathbb{F}_q$ is much larger than the degree/genus of everything involved).
 A: I have worked on reasonably similar problems to this, so hopefully what I have to say will be of some use.
In general there is no easy way to do this, as far as I know. There are techniques for proving this kind of statement, and you could phrase an individual technique as a criterion, but it would be unhelpful, because most covers won't satisfy the criterion. This is the sort of statement that is usually true, but can be false for arbitrarily subtle reasons.
That said, here is how a typical proof of this might look:
First, lower bound the geometric monodromy group. The most powerful technique to do this is by evaluating the local monodromy at each ramification point. This is usually much easier to understand then the full monodromy group, and gives you conjugacy classes in the group. You can also use more global facts about the cover - if you can prove it is geometrically irreducible, for instance, then the monodromy must be transitive.
Second, upper bound the arithmetic monodromy group. If there are any automorphisms of your cover defined over $\mathbb F_q$, the arithmetic monodromy must commute with them. If your cover is defined by adding some extra structure to a moduli space, then the arithmetic monodromy group must act on that structure.
If there are any intermediate curves you know exist between $X$ and $Y$, you can use those to both lower bound and upper bound the groups. This might follow from the way they are constructed. Can you compute the discriminant?
Third, use group theory to close the gap between the two groups. The key fact here is that the geometric monodromy group is a normal subgroup of the arithmetic monodromy group, with cyclic quotient. You hope to prove that no pair of groups satisfying all the conditions you set up beforehand can exist. Alternately, prove that one can exist, and try to come up with clever new arguments to rule them out!
This advice might not be too helpful, because it basically says "solve this math problem like you would any other math problem, using math," but I hope that something I said here clicks with something about the "natural" coverings you are defining.
A: I think generically the geometric monodromy group is the symmetric group $S_n$, acting on the $n=[\mathbb F_q(Y):\mathbb F_q(X)]$ conjugates of $\mathbb F_q(Y)$. But the arithmetic monodromy group acts faithfully on the same set, so it can't be bigger.
@JasonStarr: I believe the OP wanted to say that $\mathbb F_q$ is algebraically closed in the Galois closure of $\mathbb F_q(Y)/\mathbb F_q(X)$ if and only if the two groups are equal.
