2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.