# Exercise 4.30 of Eisenbud: how to connect a maximal ideal in a localization with the original ring's property of being finitely generated?

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.)

• 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. Mar 12 at 0:22
• @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?) Mar 12 at 4:17
• $R_f/m$ 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 Mar 12 at 5:11