I am searching for two infinite dimensional algebras such that the center of their tensor product is bigger than the tensor product of their centers. Who knows of such examples? Thanks a lot.
The following result tells us we can't work over $k$algebras; as suggested in the comments, this doesn't preclude an example where the algebra is over a commutative ring instead.
Result: Let $A$, $B$ be associative $k$algebras. Then $Z(A \otimes_k B) = Z(A) \otimes_k Z(B)$.
Proof: Let $z = \sum_{i=1}^n a_i \otimes b_i$ be an element of $Z(A \otimes_k B)$, and assume wlog that the $b_i$ are $k$linearly independent. Since $z$ is central, it must commute with all elements of the form $a \otimes 1$, $a \in A$. Therefore \begin{equation*}0 = z(a\otimes 1)  (a\otimes 1)z = \sum_{i =1}^n (a_ia  aa_i) \otimes b_i\end{equation*} and this holds iff $a_i \in Z(A)$ for all $i$, since $a \in A$ was arbitrary and the $b_i$ are linearly independent.
We can assume that the $a_i$ are linearly independent in $Z(A)$. Since $z$ must also commute with all elements of the form $1\otimes b$, $b \in B$, we get that $b_i \in Z(B)$ for all $i$ as well. Thus $Z(A\otimes_kB) \subseteq Z(A) \otimes_k Z(B)$, and the reverse inclusion is clear.
Let's try to use the fact this fails over $R$algebras to construct an example. There are probably simpler ones, but here's what I've come up with.
Let $R = k[x]$, the polynomial ring in one variable. Define two Ore extensions:
\begin{equation*} A = k[x^{\pm1},u_1][u_2;\alpha], \quad B = k[x,y,t_1][t_2;\beta] \end{equation*} where $\alpha: x \mapsto x, u_1 \mapsto qu_1$, $\beta: x\mapsto x, y\mapsto y, t_1 \mapsto qt_1$ and $q \in k^{\times}$ is not a root of unity. In other words, $A$ and $B$ are both nearly polynomial or Laurent polynomial, but we've enforced the relations $u_2u_1 = qu_1u_2$ and $t_2t_1 = qt_1t_2$. $Z(A) = k[x^{\pm1}]$, $Z(B) = k[x,y]$ and we're viewing both of them as algebras over $R = k[x]$.
Now define $z = u_1 \otimes y  u_1x^{1} \otimes xy$, which is not in $Z(A) \otimes_R Z(B)$ since neither $u_1$ nor $u_1x^{1}$ are in $Z(A)$. However,
\begin{eqnarray*} z(a\otimes b)  (a\otimes b)z &=& u_1a \otimes yb  u_1x^{1}a \otimes xyb  au_1\otimes by + au_1x^{1}\otimes xyb \\ &=& u_1a \otimes yb  u_1a \otimes yb  au_1 \otimes by + au_1 \otimes yb \\ &=& 0 \end{eqnarray*} for all $a \in A$, $b \in B$, using the centrality of $x^{1}$ in $A$ and the fact that the tensor product is over $k[x]$. Therefore $z \in Z(A \otimes_RB)$.
(I suppose you could just take $B = k[x,y]$ or even $B = k[x]$ if you prefer, we only really need one ring to be noncommutative for this to work.)

5$\begingroup$ I think that the key point is that you assume that the algebra is over a field instead of a commutative ring. $\endgroup$ – Maximiliano Valle Jul 24 '13 at 11:08

$\begingroup$ Good point; I'm so used to "algebra" meaning $k$algebra I completely forgot about $R$algebras. I've updated my post with an $R$algebra example that hopefully works. $\endgroup$ – eithil Jul 25 '13 at 11:02

1$\begingroup$ Isn't $z=0$? The tensor product is over $k[x]$, so $u_1 x^{1}\otimes xy$ is just $u_1 \otimes y$. I got doubtful because your Argument would in fact work for $B=k[x,y]$, and I'm sure that can't be. $\endgroup$ – Ben Apr 17 '14 at 11:17

$\begingroup$ I think @Ben is correct, and this example does not work. Usually I wouldn't write a comment just to say I agree with someone, but given the number of upvotes for this answer it can't hurt to attract some attention to its incorrectness. $\endgroup$ – Captain Lama Feb 7 at 14:14
It seems to me that a simple example is as follows: Let $R = \mathbb{Z}$, let $A$ be the noncommutative $\mathbb{Z}$algebra on two generators $x$, $y$, obeying $xy=yx+2$ and let $B = \mathbb{Z}/2 \mathbb{Z}$.
Note that $A$ is basically the Weyl algebra and its center is $\mathbb{Z}$. Here is a more detailed argument: Let $W = \mathbb{Q}\langle x,y \rangle / (xyyx2)$; $W$ is isomorphic to the Weyl algebra. There is an obvious map $A \to W$. We claim that it is injective. Proof: By a PBW like argument, $x^i y^j$ spans $A$ over $\mathbb{Z}$, and these monomials are linearly independent in $W$. $\square$ So $x^i y^j$ is a $\mathbb{Z}$basis of $A$, and we can compute in it in the usual way to see that $\mathbb{Z} = Z(A)$.
So $Z(A) \otimes_R Z(B)$ is $\mathbb{F}_2$, but $A \otimes_R B$ is $\mathbb{F}_2[x,y]$.

$\begingroup$ Is it obvious that the argument about reducing the center to scalars works without invertibility of $[x,y]$? $\endgroup$ – მამუკა ჯიბლაძე Feb 20 at 16:34

1$\begingroup$ @მამუკაჯიბლაძე You can decide whether or not it is obvious, I added the details. $\endgroup$ – David E Speyer Feb 20 at 17:14

$\begingroup$ Thanks! I actually meant to ask about the part that you call "compute in the usual way" but this is also straightforward: $[x,]$ acts on the PBW basis as $2\frac\partial{\partial y}$ and $[,y]$ as $2\frac\partial{\partial x}$... $\endgroup$ – მამუკა ჯიბლაძე Feb 20 at 17:50
I needed an example myself with one of the algebras being commutative (and I'm sure you can't take $k[x,y]$ in the other answer). In case someone is still interested in this question, here it goes.
We let $A = \mathbb{Z}[e_{ij},n_{ij}]/(e_{ij}n_{ij})$ for all positive integers with $i<j$ and define $S$ to be the commutative (flat) $A$algebra $A[n_{ij}^{1}]$ (the localization at all $n_{ij}$, $i<j$). Also let $B$ be the quotient of $A\langle x_{i}\rangle_{i\geq 1}$ by the two sided ideal generated by $[x_i,x_j]e_{ij}$ for all $i<j$. Now I claim that $S\otimes_AZ(B)\not=Z(S\otimes_AB)$. In fact, since localising at all the $n_{ij}$ kills all the $e_{ij}$, the $A$algebra $S\otimes_AB$ is commutative, hence $Z(S\otimes_AB) = S\otimes_AB$. However, $1\otimes x_1\in S\otimes_AB$ can't be an element of $S\otimes_AZ(B)$, because this would require $x_1$ to be divided by each $n_{1j}$, $j\geq 2$, and this is absurd.
This example involves $B$ being not finitely generated, but as long as we require $S$ to be flat, any counterexample has to be of this kind, because if $S$ is flat and $B$ is finitely generated as $A$algebra (or without this finiteness assumption on $B$ if $S$ acts sufficiently nicely on $B$), then $S\otimes_A Z(B) = Z(S\otimes_A B)$.