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.