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?
> 
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> 


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.



Non-central tensor product of central algebras
I'm not sure what "central $R$-algebra" means, I have been taking it to mean that the natural map $R \to Z(A)$ is an isomorphism. If it just has to be surjective, the OP has already given a solution. I think the following is an answer to my interpretation where we have to have $R = Z(A)$.




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Let $R$ be the commutative ring $k[u\_1, u\_2]/(u\_1 u\_2)$. For $j=1$, $2$, let
$$A\_j = R\langle x\_j, y\_j \rangle / ( y\_j x\_j - x\_j y\_j - u\_j,\ u\_{3-j} x\_j,\ u\_{3-j} y\_j ).$$
I think, as an $R$-module,
$$A\_j = R \cdot 1 \oplus \bigoplus\_{a+b \geq 1} \left( R/u\_{3-j} R \right) \cdot x\_j^a y\_j^b$$
and that $Z(A\_1) = Z(A\_2) = R$. Thus,
$$A\_1 \otimes\_R A\_2 = $$
$$R \cdot 1 \oplus \bigoplus\_{a+b \geq 1} \left( R/u\_2 R \right) \cdot x\_1^a y\_1^b \oplus \bigoplus\_{a+b \geq 1} \left( R/u\_1 R \right) \cdot x\_2^a y\_2^b \oplus \bigoplus\_{a\_1+b\_1 \geq 1,\ a\_2+b\_2 \geq 1} k \cdot x\_1^{a\_1} y\_1^{b\_1} x\_2^{a\_2} y\_2^{b\_2}.$$
In particular, $Z(A) \otimes\_R Z(B)$ is the first summand, $R \cdot 1$, and so $x\_1 x\_2 \not\in Z(A) \otimes\_R Z(B)$.


Then $x\_1 x\_2$ is central, because it clearly commutes with $x\_1$ and $x\_2$, and we compute that $[x\_1 x\_2, y\_1] = [x\_1, y\_1] x\_2 = u\_1 x\_2 = 0$ and $[x\_1 x\_2, y\_2] = x\_1 [x\_2, y\_2] = x\_1 u\_2 = 0$. More generally, all the monomials $x\_1^{a\_1} y\_1^{b\_1} x\_2^{a\_2} y\_2^{b\_2}$ for $a\_1+b\_1$, $a\_2 + b\_2 \geq 1$ are central in the same way.


