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I'm looking at Serre's paper How to use Finite Fields for Problems Concerning Infinite Fields. 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. And then you can even take the algebraic closure of $\Lambda/\mathfrak{m}$ if you'd like; I suppose it depends on what you need.

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? And then Serre additionally takes the algebraic closure of $\Lambda/\mathfrak{m}$, and I don't see what's gained by this.

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    $\begingroup$ The scheme-theoretic argument is that if $\mathcal X \to \operatorname{Spec} R$ is a nice family (e.g. flat with geometrically integral fibres, although it might work with much weaker hypotheses) where $R$ is a domain of finite type over $\mathbb Z$, then the locus in $\operatorname{Spec} R$ where a map $f \colon \mathcal X \to \mathcal X$ of $R$-schemes is not injective (resp. not surjective) is constructible. Thus, if it contains all closed points, it must be all of $\operatorname{Spec} R$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 23, 2018 at 20:57
  • $\begingroup$ In a more model-theoretic language, the idea is indeed that if there exists a counterexample in characteristic $0$, then there also exists one in large positive characteristic. This is an example of a compactness argument in model theory. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 23, 2018 at 21:00
  • $\begingroup$ @R.vanDobbendeBruyn Thanks! Yeah, I really like the model-theoretic perspective on this proof technique: it hides most of the heavy lifting in the compactness theorem and proving that the theory of algebraically closed fields of a fixed characteristic is complete. I think I'm being a tad paranoid about the details of unpacking those ideas into purely a algebraic-geometry and commutative-algebra language, hence this question :) $\endgroup$
    – Mike Pierce
    Commented May 24, 2018 at 0:44

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