Simplifying Quotient of Tensor Products Consider
$$(A\otimes C)/(B\otimes C)$$
where $B$ is a submodule of $A$. ($A,B,C$ are $R$-modules).
Is it true that $$(A\otimes C)/(B\otimes C)\cong(A/B)\otimes C$$?
Thanks. If no, are there any easy counter-examples?

This is sort of true.
There is a natural map from $B\otimes C$ to $A\otimes C$, but in general
it is not injective, so that we cannot think of
$B\otimes C$ as a submodule of $A\otimes C$. But the image $I$ of
this map is a submodule of $A\otimes C$, and
$(A\otimes C)/I$ is isomorphic to $(A/B)\otimes C$.
What is going on is that the tensor product is right exact. We have
an exact sequence
$$0\to B\to A\to A/B\to 0$$
and when we tensor with $C$ we get that
$$B\otimes C\to A\otimes C\to (A/B)\otimes C\to 0$$
is exact.
