# 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?

## 1 Answer

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.

• Very nice answer. Just to confirm, the natural map is $\phi: B\otimes C\to A\otimes C$, defined by $b\otimes c\mapsto b\otimes c$ (extend linearly)? Commented Oct 14, 2017 at 16:16
• @yoyostein Exactly: $I$ is generated by the $b\otimes c$, but $I$ is not necessarily isomorphic to $B\otimes C$. Commented Oct 14, 2017 at 16:19
• Also, I can see that the natural map is not injective, but why does that mean $B\otimes C$ is not a submodule of $A\otimes C$? I thought that $B\otimes C$ is a subset of $A\otimes C$ that is also a module (if $R$ is commutative), hence that means it is a submodule? Commented Oct 14, 2017 at 16:21
• Take $R=\Bbb Z$, $A=\Bbb Q$, $B=\Bbb Z$ and $C\cong \Bbb Z/2\Bbb Z$. Can you really see $B\otimes C$ as a submodule of $A\otimes C$ here? Commented Oct 14, 2017 at 16:24
• For a study of when the map $B \otimes C \to A \otimes C$ is injective (or an evaluation of possible obstructions to the map being injective), see homology theory and the $\operatorname{Tor}$ functor. Commented Oct 14, 2017 at 21:25