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.

  • $\begingroup$ What is an example of $Z(A\otimes_RB)\not\cong Z(A)\otimes_RZ(B)$? $\endgroup$ Feb 18, 2021 at 14:54
  • $\begingroup$ @მამუკაჯიბლაძე This is discussed here mathoverflow.net/questions/137584/… . The most upvoted answer is incorrect, but Ben's example seems correct to me. $\endgroup$ Feb 18, 2021 at 16:59
  • $\begingroup$ Sorry, I do not understand that example well enough. Specifically, (1) why would $1\otimes x_1\in S\otimes Z(B)$ imply that $x_1$ is divided by all $n_{1j}$, (2) does "divided" mean "divisible" or something else, and (3) why is it absurd? $\endgroup$ Feb 18, 2021 at 21:36
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    $\begingroup$ @მამუკაჯიბლაძე I simplified the construction so the argument would be more immediate, and now I wonder if this example would not answer my question. $\endgroup$ Feb 20, 2021 at 13:15
  • $\begingroup$ According to the fresh answer to that question by @DavidESpeyer, you indeed found an answer: your algebra is essentially the universal enveloping algebra of a Heisenberg Lie algebra, so it must be central. In more detail, $B$ has a PBW basis spanned by the monomials $y^mz^n$, and on this basis $[y,-]$ acts as $x\frac\partial{\partial z}$ and $[-,z]$ acts as $x\frac\partial{\partial y}$, so any element commuting with both $y$ and $z$ must indeed be in $R$. $\endgroup$ Feb 20, 2021 at 17:56

1 Answer 1


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 $R = Z(A)$.

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.

  • $\begingroup$ Indeed in my mind "central" means that $Z(A)$ is the image of the structural morphism $R\to A$ (maybe it's not standard ?). But in any case it's much nicer to have an example where the centers are isomorphic to $R$! Some claims here are not 100% obvious, but I think everything should be correct, so thanks. $\endgroup$ Feb 21, 2021 at 9:37

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