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Motivation:

Say you have a noetherian scheme of finite type $X$ over $\mathbb Z$. Then to every closed point $x$ corresponds a norm, which is the number of elements in the residue field (which we know to be finite)

$$ N(x) = |k(x)|. $$

So, this number will be $p^n$ for some prime $p$ and integer $n$. We want to prove that the number

$$ a(p,k) = \{x \text{ closed point }\mid N(x) = p^k \} $$

is finite.

Given this, we reduced the problem to the following:

Let $X = \textrm{spec }\mathbb{F}_p[T_1, \dots, T_n] = \textrm{spec }R$ defined over $\mathbb F_p$. The closed points now correspond to maximal ideals of $R$, and therefore, we want to prove that for each $k$ there are finitely many ideals $\mathfrak m$ such that $R/\mathfrak{m} = p^k$.

If $n=1$ this is easy to determine (since $R$ will be a PID, we can count irreducible polymonials of given degree). For a generic $n$ we have no clue how to proceed.

One strategy may be to use the Nullstellensatz on $\overline{\mathbb{F}_p}$ and hope we can get something back by using the Frobenius morphism.

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Thiago
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  • hmmm I'll try to read it later. Still interested in a direct proof of the second part of this. Is there a purely "commutative algebra" answer to the second question? I feel there should. – Thiago Aug 27 '19 at 00:30
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    Sure - any such ideal $\mathfrak{m}$ is exactly the ideal generated by the polynomials $m_i(T_i)$ for all $1\leq i\leq n$ where $m_i$ is the minimal polynomial of the image of $T_i$ in $\Bbb F_{p^k}$ over $\Bbb F_p$. There are only finitely many such polynomials, and that does it. – KReiser Aug 27 '19 at 01:04

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