This is about problem $5$ in Section IV.$5$ of Hungerford's Algebra book. The question is the following:

If $A'$ is a submodule of the right $R$-module $A$ and $B'$ is a submodule of the left $R$-module $B$, then $$\frac{A}{A'} \otimes _R \frac{B}{B'} \cong \frac{A \otimes _R B}{C}$$ where $C$ is the subgroup of $A \otimes _R B$ generated by all elements $a' \otimes b$ and $a \otimes b'$ with $a \in A$, $a' \in A'$, $b \in B$, $b' \in B'$.

I'm just starting to somehow grasp the tensor product concept and I'm having a lot of trouble when trying to prove that some tensor product is isomorphic to a certain group.

In particular what I feel that's causing me problems is the fact that in the tensor product $A \otimes _R B$ not every element is an elementary tensor of the form $a \otimes b$ so that it is really hard for me sometimes to be able to define an appropriate map to try to produce an explicit isomorphism.

I'm not sure if the universal property could be of help here, since I can try to define a map $$\varphi : \frac{A}{A'} \times \frac{B}{B'} \longrightarrow \frac{A \otimes _R B}{C}$$ by $$\varphi(a + A', b + B') := a \otimes b + C$$

But I've been trying without getting anywhere. Any hints or advice on how to approach this exercise would be very much appreciated. Thank you.

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Tensor product tries to linearize bilinear map. In this case, look at an arbitrary bilinear map from $A/A' \times B/B'$ to an $R$-module. If you can show that every such map can factor through $\phi$ that you constructed, then we are done. Can you lift this bilinear map to $A \times B$ so as to relate back to $A \otimes_R B$? –  Soarer Feb 25 '11 at 4:28
@Soarer I'm sorry but I think I don't fully understand the lifting part of your comment. But since the tensor product is uniquely defined up to isomorphism by the universal property, then I believe that what you mean is that I need to consider a map from $A/A' \times B/B' \longrightarrow M$ where $M$ is an arbitrary abelian group, and if I can factor this map through $\varphi$ as constructed in my question, then that would prove the isomorphism I want by uniqueness of the tensor product? –  Adrián Barquero Feb 25 '11 at 4:40
@Adrian, yes, that's what I mean. But how can you factor through something that involves $A \otimes B$? The cleanest way is to lift your bilinear map $A/A' \times B/B' \to M$ to a bilinear map $A \times B \to M$, via the quotient maps $A \to A/A'$, $B \to B/B'$. Then by definition you can factor through the bilinear map on $A \times B$ through $A \otimes B$. The only thing you need to do now is to check this map on $A \otimes B$ factors through $A \otimes B/C$. –  Soarer Feb 25 '11 at 5:11
You say «I'm not sure if the universal property could be of help here». But you have to keep in mind that the universal property is the only way to define a map on a tensor product! –  Mariano Suárez-Alvarez Feb 25 '11 at 5:11
Adrian: I wrote some course handouts on tensor products which you might find useful: math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf and math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod2.pdf –  KCd Mar 6 '11 at 20:43

Given a bilinear map $A/A' \times B/B' \to M$ to some abelian group $M$, lift it to a bilinear map $A \times B \to M$. It is easy to check that
1. $A \times B \to M$ factors through $A \otimes B/C$ uniquely,
2. $A/A' \times B/B' \to M$ can be factored as $\phi$ and the map $A \otimes B/C \to M$ in 1 uniquely.