If $A$ is a commutative Noetherian ring, then $D=\lbrace\mathfrak{p}\in \operatorname{Spec}(A): |A/\mathfrak{p}|\leq k\rbrace$ is a finite set.

Question

Let $A$ be a commutative Noetherian ring (with unity) and $k\in \mathbb{N}$. Prove that $D=\lbrace\mathfrak{p}\in \operatorname{Spec}(A): |A/\mathfrak{p}|\leq k\rbrace$ is a finite set.

I'm trying to prove it. We know that $A/\mathfrak{p}$ is a domain. Since $|A/\mathfrak{p}|\leq k$, it follows that $A/\mathfrak{p}$ is a finite domain. Therefore $A/\mathfrak{p}$ is a field. Thus $\mathfrak{p}\in \operatorname{Max}(A)$, i.e., $D\subseteq\operatorname{Max(A)}$...

Answer 1

Assuming $A$ has infinitely many prime ideals of norm $\le k$, there is some $p^d$ such that $A$ has infinitely many maximal ideals of norm $p^d$. Let $$A_0 = A/(p,\{ a^{p^d}-a,a\in A\})$$ It has infinitely many maximal ideals all of norm $p^e, e| d$. By induction for $n=0,1,2,\ldots$,

Take some $a_n\ne 0 \in A_n$ which is in infinitely many maximal ideals and let $A_{n+1}=A_n/(a_n)$.

$A_{n+1}$ has again infinitely many maximal ideals, all of norm $p^e,e|d$.

The sequence $A_n$ of quotient rings implies that $A$ wasn't Noetherian.

If such an $a_n\in A_n$ as above didn't exist then enumerate the maximal ideals $m_j$ of $A_n$, take some $b_n\ne 0\in A_n$, if $b_n=0\bmod m_j$ for infinitely many $j$ then we can take $a_n=b_n$, otherwise $b_n^{p^d-1}-1=0\bmod m_j$ for infinitely many $j$, that $a_n$ doesn't exist means that $b_n^{p^d-1}-1=0$ so $b_n\in A_n^\times$, thus $A_n$ is in fact a field, a contradiction.

Answer 2

The following argument is essentially the same as reuns's answer but I find a more abstract framing to be useful for understanding what's going on. It suffices to show that for each finite field $F$, there are only finitely many homomorphisms $A\to F$ (and thus finitely many prime ideals $P\subset A$ such that $|A/P|=|F|$). Let $S$ be the set of all homomorphisms $A\to F$ and consider the natural map $f:A\to F^S$. Every homomorphism $A\to F$ factors through the image of $f$, so it suffices to show the image of $f$ has only finitely many prime ideals. Thus, it suffices to show that if $B$ is a Noetherian subring of $F^S$ (where $F$ is a finite field and $S$ is some set), then $B$ has only finitely many prime ideals.

To prove this, we will show that every prime ideal of $B$ is maximal, so $B$ is a $0$-dimensional Noetherian ring. Suppose $P\subset B$ is a prime ideal and $x\in B$. Note that $x^{|F|-1}$ is idempotent, since all of its coordinates are $0$ or $1$. So, the image of $x^{|F|-1}$ in the quotient $B/P$ must be $0$ or $1$, and the image of $x$ in $B/P$ must be either $0$ or a unit. Thus $B/P$ is a field and $P$ is maximal.

(Alternatively, the fact that $x^{|F|-1}$ is idempotent and $x^{|F|}=x$ implies that each $x\in B$ generates the same ideal as some idempotent. This then implies that $\operatorname{Spec} B$ can be identified with the Stone space of its Boolean algebra of idempotents, and a Stone space is Noetherian iff it is finite.)