I am doing Exercise 4.30 in Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry. I will copy the problem statement here:

Exercise 4.30: Suppose that $k$ is a Noetherian ring such that
$*)$ for every finitely generated $k$-algebra $R$ and maximal ideal $P \subset R$ the $k$-algebra $R/P$ is finite over $k$.
Show that for every reduced finitely generated $k$-algebra $R$ and prime ideal $Q \subset R$ we have $Q = \bigcap P$, where the intersection runs over all primes $P$ of $R$ such that $R/P$ is finite over $k$ [error: should be "all primes $P$ of $R$ containing $Q$ such that $R/P$ is finite over $k$"]. (Hint: If $f \in R$, $f \notin Q$, we must find a prime $P$ such that $R/P$ is finite over $k$ and $f \notin P$. Consider a maximal ideal in the $k$-algebra $R[f^{-1}]$ and its intersection with $R$.)

Deduce in particular that Theorem 1.6 [the Nullstellensatz] (for a given field $k$) follows if we prove that $*)$ holds for $k$.

(Square brackets are my annotations; additionally, all rings are commutative and have multiplicative identity.)

Using Zorn's lemma, I found a maximal ideal $\mathfrak{m}$ of $R[f^{-1}]$ containing the image of $Q$ in $R[f^{-1}]$. The preimage of $\mathfrak{m}$ under the natural map from $R$ to $R[f^{-1}]$ is a prime ideal. I don't think it is maximal (I remember that maps $A \to B$ of rings induce maps from the spectrum of $B$ to the spectrum of $A$, but not necessarily maps from the maximal spectrum of $B$ to the maximal spectrum of $A$). This prevents us from directly using condition $*)$. Also, $R[f^{-1}]$ is not necessarily finitely generated, and we also have the issue of the map $R \to R[f^{-1}]$ not being injective.

How do I connect the maximality of $\mathfrak{m}$ in $R[f^{-1}]$ with the finitely-generated property of $R$ in order to use property $*)$?

(I know there is a related question at Exercise 4.30 of Eisenbud, but I don't understand the second-to-last sentence of the answer.)

  • 1
    $\begingroup$ First, $m\cap R$ is a prime ideal which implies $R/(m\cap R)$ is a domain. The map $R/(m\cap R)\hookrightarrow R_f/m$ is finite, so the extension is integral. Eisenbud's corollary 4.17 implies $m\cap R$ is maximal. $\endgroup$
    – user782932
    Mar 12 at 0:22
  • $\begingroup$ @user782932 How do I show that $R/(\mathfrak{m} \cap R) \hookrightarrow R[f^{-1}]/\mathfrak{m}$ is finite? I remember that $R \hookrightarrow R[f^{-1}]$ is not necessarily finite, and I don't think that $R/(\mathfrak{m} \cap R) \hookrightarrow R[f^{-1}]/\mathfrak{m}$ will be either. (Also, why would $R/(\mathfrak{m} \cap R) \hookrightarrow R[f^{-1}]/\mathfrak{m}$ being finite show that it would be an integral extension?) $\endgroup$
    – my2cents
    Mar 12 at 4:17
  • $\begingroup$ $R_f/m$ is finite as a $k$-module, and $R/(m\cap R)$ can be viewed as a $k$-submodule of $R_f/m$. Thus, $R_f/m$ is finite over $R/(m\cap R)$. For another question, see Eisenbud's corollary 4.5 $\endgroup$
    – user782932
    Mar 12 at 5:11


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