# Subrings of Jacobson rings

Suppose $$A\subset B$$ is an inclusion of commutative rings with $$B$$ Jacobson.
If $$B$$ is finitely generated as an algebra over $$A$$ does it follow that $$A$$ is Jacobson?
If $$B$$ is finitely generated as a module over $$A$$ does it follow that $$A$$ is Jacobson?

For the module-finite (i.e. second) question:

Yes, it is true: The point is that $$A \subseteq B$$ is integral, and the claim is true for all integral extensions.

When $$A \subseteq B$$ is integral, every prime $$\mathfrak{p} \subseteq A$$ comes as $$\mathfrak{P}\cap A$$ for a prime $$\mathfrak{P} \subseteq B$$ ("lying over theorem"), and morever, if $$\mathfrak{P}$$ is maximal then so is $$\mathfrak{p}$$ (e.g. by "going up theorem").

So given a prime $$\mathfrak{p}$$ of $$A$$, choose such lift $$\mathfrak{P} \subseteq B$$. By $$B$$ being Jacobson one has $$\mathfrak{P}=\bigcap_{\mathfrak{P}\subseteq{\mathfrak{M}}\subseteq_{\mathrm{max}}B}\mathfrak{M}$$. Intersecting this with $$A$$ yields $$\mathfrak{p}=\bigcap_{\mathfrak{P}\subseteq{\mathfrak{M}}\subseteq_{\mathrm{max}}B}(\mathfrak{M}\cap A)$$ where all the ideals $$\mathfrak{M}\cap A$$ are maximal in $$A$$. Thus, $$A$$ is Jacobson.

• Dear Pavel, maybe it is worth emphasizing from the beginning that your elegant argument proves the result for an arbitrary integral overring $B$ of $A$, not only for finite ones. At my first superficial reading of your post I thought that integrality was only used as a step in the proof, whereas it suffices for the whole proof to go through. Anyway, bravo! (And +1, of course.) – Georges Elencwajg Feb 23 at 9:01
• @GeorgesElencwajg Dear Georges, thank you, I have edited the answer as you proposed. – Pavel Čoupek Feb 23 at 18:07
• Well done Pavel: I hope your answer will get the many more upvotes it deserves – Georges Elencwajg Feb 23 at 18:25

A counterexample for the first question is any DVR $$R$$. Clearly, $$R$$ is not Jacobson. But if $$\pi$$ is the uniformizer, then $$Q(R) = R[\frac{1}{\pi}]$$ is a finitely generated $$R$$-algebra and a field, hence Jacobson.

• I would not be surprised if the second statement is true, though. – Martin Brandenburg Feb 22 at 21:03