Jacobson radical of tensor product Suppose $R$ and $S$ are associative rings with unit and that $J(R)$, the Jacobson radical of $R$, is a nil ideal. Clearly if $R$ is commutative then $J(R)\otimes_\mathbb{Z} S$ is a nil ideal. Is this true if $R$ is not commutative?
 A: It's not true in general, but I think all counterexamples are going to be quite complicated. 
Suppose we have a counterexample and $a_1\otimes b_1+\dots+a_n\otimes b_n\in J(R)\otimes_{\mathbb{Z}}S$ is not nilpotent. Let $A$ be the non-unital subring of $R$ generated by $a_1,\dots,a_n$. Then $A$ is a nil ring but not nilpotent.
Conversely, if $A$ is any finitely generated non-nilpotent nil ring (generated by elements $a_1,\dots,a_n$) then if we let $R=\mathbb{Z}\oplus A$ be the unital ring obtained by adjoining a unit to $A$, so that $A=J(R)$, and let $S=\mathbb{Z}\langle x_1,\dots,x_n\rangle$ be the free associative ring on $n$ generators, then
$$a_1\otimes x_1+\dots+ a_n\otimes x_n\in A\otimes_{\mathbb{Z}}S$$
is not nilpotent.
So the question reduces to whether or not there is a finitely generated non-nilpotent nil ring, which there is (although Levitsky's Theorem says that $R$ can't be right or left Noetherian). 
I'm not sure of the history, but in Golod, E.S; Shafarevich, I.R. (1964), "On the class field tower", Izv. Akad. Nauk SSSSR 28: 261–272 there are examples constructed which are algebras over an arbitrary field, but I think it's a lot easier over a countable field, and I suspect examples may have been known before that.
Incidentally, it's a theorem of Smoktunowicz that there are also examples with $S=\mathbb{Z}[x]$, and the Köthe conjecture, a well-known open problem, is equivalent to the assertion that there are no examples with $S=M_2(\mathbb{Z})$
