I'm looking at Serre's paper [*How to use Finite Fields for Problems Concerning Infinite Fields*](https://arxiv.org/pdf/0903.0517.pdf). Specifically I'm trying to use the techniques in the proof of Theorem 1.2 to write out the details of the proof of Theorem 3.1, but I'll try to ask a more general question.

Here's how I currently understand *spreading out* a scheme: If you have a scheme $X$ of finite type over an algebraically closed field $k$, you can replace it by a scheme over some finite extension of the prime subring of $k$, either $\mathbf{Z}$ or $\mathbf{Z}/p\mathbf{Z}$. Since $X$ is defined by a finite amount of data (by finitely many polynomials in the case $X$ is a variety) you can simply adjoin this data (the coefficients of the polynomials) to the prime subring of $k$. Call this extension of the prime subring $\Lambda$, and notice that $X$ is still defined over $\Lambda$. You can also adjoin any other finite amount of data from $k$ to the subring too, for example the coefficients of a polynomial map $X \to X$, so it will be preserved when you define $X$ over $\Lambda$. Furthermore you can take a max ideal $\mathfrak{m}$ of $\Lambda$, and $X$ will be defined over the quotient $\Lambda/\mathfrak{m}\,,$ [which will be a finite field](https://mathoverflow.net/q/57515/64073).

In the proof of Theorem 1.2, I'm troubled that Serre says you can choose *any* max ideal $\mathfrak{m}$. Does it really not matter which one? The whole idea behind the proof is that if there is a counterexample over $k$ then it will descend to a counterexample over $\Lambda/\mathfrak{m}$, but don't we have to worry about $\mathfrak{m}$ containing too much of the data that we appended in $\Lambda$? Specifically in the proof of Theorem 1.2 in that paper, what if $\mathfrak{m}$ contains all the coefficients of the $Q_{g,i}$'s that we need to say that the action *doesn't* have a fixed point?