This is an export of https://math.stackexchange.com/questions/4016545/non-central-tensor-product-of-central-algebras which despite a bounty has sadly attracted no answer.

I repeat the question here: for (unital associative) algebras over a field $K$, it is easy to show that $Z(A\otimes_K B)=Z(A)\otimes_K Z(B)$. In particular, a tensor product of central algebras is central. But over a general commutative ring $R$, it is no longer true that $Z(A\otimes_R B)=Z(A)\otimes_R Z(B)$; for instance you can have $A\otimes_R B$ commutative but not generated by $Z(A)$ and $Z(B)$.

**EDIT:** An example of that is to take $k$ any commutative ring, $R=k[x]$, $A=k$ seen as an $R$-algebra through $A\simeq R/(x)$, and $B=R\langle y,z\rangle/([y,z]=x)$. Then $A\otimes_R B\simeq k[y,z]$ since $x$ is sent to $0$, but $y$ and $z$ are not in $Z(B)$.

Is it also false that a tensor product of central algebras is central?

I strongly suspect that there will be counter-examples, but I cannot write one down, so if someone can give a reference or a sketch of construction, that would be great.

**EDIT:** Now I wonder if the example above works; precisely, is $B$ central over $R$? It seems like it should be, but working in non-commutative quotient rings is always a bit tricky so I'm not sure.