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For a ring $R$, an $R$-right-module $M$, an $R$-left-module $N$, and an abelian group $P$, one can use the universal property of the tensor product to construct maps $$ M\otimes_R N\to P. $$ It concrete cases, it is often easy to see that the constructed map is surjective by just writing down pre-images.

It seems to be harder to verify that the map is injective, because then one has to consider general sums of elementary tensors in $M\otimes_R N$.

Are there any tricks to avoid this, and to achieve injectivity more elegantly?


More concretely, an example I have in mind is the following: I have a pre-additive category $C$ with finitely many objects. I consider modules over this category (that is, functors from $C$ to Abelian groups; or, equivalently, modules over the category ring of $C$). Now, I consider the "free $C$-right-module $Q_Y$ over an object $Y$ in $C$" which is the hom functor $C(\bullet,Y)$. I want to show that for an arbitrary left-module $M$: $$ Q_Y\otimes_C M\cong M(Y) $$ as abelian groups. Using the module structure of $M$, I can accomplish a natural epimorphism $Q_Y\otimes_C M\to M(Y)$ which should be injective.

Currently I think that the formula $$ M(Y)\to Q_Y\otimes_C M;\quad m\mapsto\mathrm{id}_Y\otimes m $$ gives indeed a (two-sided) inverse. Is this correct?

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Can you give an example? I don't really know what you mean. –  Qiaochu Yuan Oct 28 '10 at 15:28
    
please edit your question to include the example, so that the questyion body contains all the relevant details about what you want to know (and so that we don't have to read the slightly amusingly small font of the comments...) –  Mariano Suárez-Alvarez Oct 28 '10 at 21:27
    
(Isn' t your example application just Yoneda's lemma?) –  Mariano Suárez-Alvarez Oct 28 '10 at 21:29
    
@Mariano: I think you are right. Yoneda's lemma gives $M(Y)\cong\mathrm{Hom}_{C-Mod}(P_Y,M)$, where $P_Y$ is the free left-module over $C$ (the covariant hom-functor). Now I have to understand why $\mathrm{Hom}_{C-Mod}(P_Y,M)\cong Q_Y\otimes_C M$, so that in some sence $Q_Y$ is dual to $P_Y$, right? –  Rasmus Oct 28 '10 at 21:45

1 Answer 1

up vote 2 down vote accepted

If the image $Q$ is easy to find and well understood, then you can try to construct an inverse from $Q\to M\otimes N$. If $Q$ is well understood, then you can define homomorphisms from it (say by defining the images of some generators, and checking that all the relations are sent to $0$ in $M\otimes N$). Then you just check that the composition is the identity (say by sending generators of $Q$ into $M\otimes N$ and then back into $P$, and checking relations to check equality).

This depends heavily on the image being well understood. It works well to prove an abstract tensor product is equal to some concretely understood module. If the image is confusing, then checking that the homomorphism is well defined (or even just defining it) can be hard, and even checking equality of elements in $P$ can be hard in some cases.

There are some other tricks using embeddings and exact sequences, but they also have limited applicability (usually when the entire problem has exact sequences floating around). Feel free to give some more concrete examples for more concrete tricks.

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In the example I describe above, M(Y) can be any (countable) abelian group, so I'm afraid that it is not well understood in your sense. (I realise that we might get a notational clash.) –  Rasmus Oct 28 '10 at 18:24
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Thanks for taking me from "my map is certainly surjective because I can easily find pre-images for everything" to "maybe this procedure could be used as an inverse map"! –  Rasmus Oct 28 '10 at 20:38

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