I am doing exercise 4.30 from Eisenbud's Commutative Algebra With A View Towards Algebraic Geometry which I append here:

Exercise 4.30: Suppose $k$ is a Noetherian ring and for every finitely generated $k$ - algebra and maximal ideal $P \subset R$ the $k$ - algebra is finite over $k$. Show that every reduced finitely - generated $k$ - algebra $R$ is such that given any prime ideal $Q \subseteq R$, we have $$Q = \bigcap P$$ where the intersection runs over all primes $P$ of $R$ such that $R/P$ is finite as a module over $k$.

We note that this consequently sows that such an $R$ is a Jacobson ring. The definition of a Jacobson ring given in Eisenbud is that every prime ideal is an intersection of maximal ideals. Necessarily, we need each of these $P$ to contain $Q$ so that $Q \subseteq P$. Now Eisenbud gives a hint for the reverse inclusion:

Hint: If $f \in R$ and $f \notin Q$, 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 $R_f$ and its intersection with $R$.

Now I kinda get the hint of Eisenbud: If we choose a maximal ideal $\mathfrak{m}$ in $R_{f}$, then the "contraction" of that maximal ideal in $R$ is prime. Since we have "inclusions"

$$k \hookrightarrow R/(\mathfrak{m} \cap R) \hookrightarrow R_f/\mathfrak{m}$$

and $R_f/\mathfrak{m}$ is finite as a module over $k$, then necessarily so is $R/(\mathfrak{m} \cap R)$ because $k$ is Noetherian.

My problem is: We are not guaranteed that the ring homomorphism $R \to R_f$ that maps $x$ to $x/1$ is injective. So how can we embed $R$ inside of $R_f$ so that it makes sense to take the intersection of $\mathfrak{m}$ with $R$?


  • $\begingroup$ In the statement, something is missing before "is finite dimensional over $k$". Also I don't known what means finite dimensional over a ring. $\endgroup$
    – user18119
    Commented Mar 6, 2013 at 11:02
  • $\begingroup$ @QiL'8 Sorry, I mistook $k$ for a field, finite dimensional should mean finite as a module over $k$. $\endgroup$
    – user38268
    Commented Mar 6, 2013 at 11:15

1 Answer 1


Why do you need $R\to R_f$ to be an inclusion? The ring homomorphism $R\to R_f$ gives rise to an inclusion $$R/(\mathfrak{m} \cap R) \hookrightarrow R_f/\mathfrak{m}$$ and this is all you need. (Here $\mathfrak{m} \cap R$ denotes the contraction of $\mathfrak m$ to $R$.)

Update. The OP asked why $\mathfrak m\cap R$ contains $Q$. This is also easy to do: take $\mathfrak m$ in $R_f$ a maximal ideal containing $Q_f$.

Remark. I don't think $k$ noetherian is really necessary: we have $$k\to R/(\mathfrak{m} \cap R) \hookrightarrow R_f/\mathfrak{m}$$ and $R_f/\mathfrak{m}$ is finite as a module over $k$. This show that $$R/(\mathfrak{m} \cap R) \hookrightarrow R_f/\mathfrak{m}$$ is finite. Since $R_f/\mathfrak{m}$ is a field, so is $R/(\mathfrak{m} \cap R)$, and therefore $\mathfrak m\cap R$ is maximal. Then $R/(\mathfrak{m} \cap R)$ finite as a module over $k$.

  • $\begingroup$ @BenjaLim As far as I know finite implies integral. But I still don't get it: what's your real problem? (From my point of view all is clear now.) $\endgroup$
    – user26857
    Commented Mar 7, 2013 at 12:20

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