Fix two integers $d$ and $g$. The number of genus $g$ and degree $d$ curves passing through $3d+g-1$ generic points on the complex projective plane is finite and doesn't depend on the choice of points. In this formulation, the problem of counting curves is well understood and there are several different techniques to solve it.

My question is, what happens if you change the field?

For example, in the real case the number of real curves will depend on the choice of points. Welschinger bypassed this by counting curves with certain multiplicities. (As far as I know, it only works when $g=0$).

**Edit:** Let me clarify my rather naive question (which is still very far from being perfect).

Consider an infinite field $k$ and its algebraic closure $\bar k$. Fix a number $d$. For a generic choice of $3d-1$ points in $k \mathbb P^2$ there exist a finite number of rational degree $d$ curves defined over $\bar k$ and the number doesn't depend on the choice of points. However the number of curves defined over $k$ in general will vary.

I'm looking for concrete examples of $k$ for which we can count this curves in some generalized sense and get an invariant. To be more specific, for each generic choice of points we have a bunch of group-theoretic/combinatorial information. We have an action of Galois group $\operatorname{Gal}(\bar k/k)$ on the set of curves passing through the given points, on the set of all double points and moreover on the set of branches at double points. I can imagine many ways of turning this data in to the numerical one.

In particular, Welschinger invariant can be tautologically expressed as the sum of $(-1)^{n(C)}$ where $C$ runs over the set of curves stabilized by $\operatorname{Gal}(\mathbb C/\mathbb R)$ and $n(C)$ is the number of nodes on $C$ invariant under the conjugation, whose branches are permuted by the action.

Are there any other known examples of fields where the similar or more complicated procedure gives an invariant?