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?  
 A: 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.
A: 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.
