The notion of dimension I prefer most is [right global dimension][1], but the question can also be asked for other notions (e.g. weak dimension, injective dimension, Krull dimension). Letting $d$ be whichever dimension you pick, the question then becomes:

> Let $A$ and $B$ be $R$-algebras. When is $d(A\otimes_R B) = d(A)+d(B)$?

This question was addressed in the '50s by [Eilenberg et al][2] and [Auslander][3]. The first proved this result for $R$ commutative and $B$ is a ring of matrices, polynomials, or rational functions. The second proved it for $R$ a field and $A,B,A\otimes B$ semiprimary algebras of finite global dimension. In 1996 [Vladimir Bavula][4] came up with some different sufficient conditions on $A$ and $B$ when $R$ is a field but they seem much more complicated. I can't find anything on the problem since then, so any newer references would be much appreciated. Generally I turn to [Weibel][5] or [Lam][6] for questions like this, but neither book mentions anything. I ask this question out of pure curiosity, so I'm fine with any assumptions on $R$. The field case has been done a lot, so perhaps $R$ could just be a commutative ring or commutative noetherian. Whatever gives a nice answer.

Here's what seems to be known. For $d =$ weak dimension or Krull dimension, $d(A\otimes B) \geq d(A)+d(B)$ and this also holds for left-Noetherian algebras $A$ and $B$ in the case where $d$ is left global dimension. One case where equality fails is $A = B =$ Division ring of $K[x_1,\dots,x_n]$ and $d =$ left global dimension. Then $d(A\otimes_K B) = n$ but $d(A)=d(B)=0$. 

If $K$ is algebraically closed and $A$ and $B$ are finite dimensional $K$-algebras then gl.dim$(A\otimes B)=$ gl.dim$(A)+$ gl.dim$(B)$

Eilenberg et. al give the following as Proposition 10: If $K$ is a field then 

l.gl.dim$(A)+$ weak dim$(B) \leq$ gl.dim$(A\otimes B) \leq$ l.gl.dim$(A) + $ dim$_K(B)$, and

l.proj.dim$(A) +$ weak dim$(B)\leq$ dim$_K(A\otimes B) \leq$ dim$_K(A) +$ dim$_K(B)$.

So this theorem reduces the problem to finding when weak dimension (as a ring) equals dimension as a $K$-algebra. It's not true in general that dim$_K(A\otimes_R B) =$ dim$_K(A)+$ dim$_K(B)$, e.g. if $A$ and $B$ are locally separable algebras over $K$ with $[A:K]=[B:K]=\infty$ then $dim_K(A)=dim_K(B)=dim_K(A\otimes B)=1$ because $A\otimes B$ satisfies the same properties just listed for $A$ and $B$.


  [1]: http://en.wikipedia.org/wiki/Global_dimension
  [2]: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.nmj/1118799929
  [3]: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.nmj/1118799859
  [4]: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.54.5096
  [5]: http://books.google.com/books/about/An_introduction_to_homological_algebra.html?id=flm-dBXfZ_gC
  [6]: http://books.google.com/books/about/Lectures_on_modules_and_rings.html?id=r9VoYbk-8c4C