Quotient of tensor products I was thinkin about this question. Take a commutative unitary ring $R$ and two left-modules over $R$, let be them $A$, $B$. Now take two left subomdules $I \subset A$, $J \subset B$. In my opinion the following is true. $$\frac{A \otimes_{R}B}{I\otimes_RB+A\otimes_RJ}\cong\frac{A}{I}\otimes_R\frac{B}{J}$$ where the isomorphism is inducted by the natural map from $A\otimes B$ to $\frac{A}{I}\otimes\frac{B}{J}$.
I think that this is "clearly" true when $A$ and $B$ got countable basis and $I$ and $J$ are direct summand in $A$ and $B$. This is the situation which insipred the question. But I think that this could be true in a more general way. I'd like to know something about this question in general case. For example if it is always true, if there is a counterexample but principally if it is true in the case considered.
Thanks in advance
 A: You can prove this using the universal property of tensor products and quotients.  Homomorphisms $f:A\otimes B\to C$ are in natural bijection with bilinear maps $g:A\times B\to C$.  Such a homomorphism $f$ vanishes on $I\otimes B+A\otimes J$ iff $g(a,b)=0$ whenever either $a\in I$ or $b\in J$ (since $f$ and $g$ are related by $f(a\otimes b)=g(a,b)$).  Now I claim that such a $g$ actually descends to a well-defined map $A/I\times B/J\to C$.  That is, I claim that if $a'-a\in I$ and $b'-b\in J$ then $g(a,b)=g(a',b')$.  To prove this, just note that $g(a,b)=g(a,b)+g(a'-a,b)=g(a',b)$ (since $g(a'-a,b)=0$ and $g$ is bilinear) and then similarly $g(a',b)=g(a',b)+g(a',b'-b)=g(a',b')$.  So $g$ descends to a map $A/I\times B/J\to C$ which is easily seen to be bilinear.  Conversely, any $g$ which does descend to such a map must satisfy $g(a,b)=0$ whenever $a\in I$ or $b\in J$.
Thus, homomorphisms $A\otimes B\to C$ which vanish on $I\otimes B+A\otimes J$ are in natural bijection with bilinear maps $A/I\times B/J\to C$.  This gives an isomorphism $A\otimes B/(I\otimes B+A\otimes J)\to A/I\otimes B/J$ and you can easily chase the definitions to see this map is induced by the canonical map $A\otimes B\to A/I\otimes B/J$.
