Let $R$ be a subring of $S$, let $N$ be a left $R$-module and let $\iota : N \to S \otimes_RN$ be the $R$-module homomorphism defined by $\iota(n) = 1 \otimes n$. Suppose that $L$ is any left $S$-module and that $\varphi : N \to L$ is an $R$-module homomorphism from $N$ to $L$. Then there is a unique $S$-module homomorphism $\Phi : S \otimes_RN \to L$ such that $\varphi = \Phi \circ \iota$.

My approach so far is:

Define $\Phi : S \otimes_RN \to L$ by $\Phi(s \otimes n) = s\varphi(n)$.

Show $\Phi$ is well defined.

Show $\Phi$ is in fact an $S$-module homomorphism.

Show $\Phi$ is unique.

I am having trouble with showing that $\Phi$ is well-defined. I suppose $s \otimes n = s' \otimes n'$, it follows by property of cosets that $(s, n) - (s', n') \in S \otimes_RN$. We want to show that $\Phi(s \otimes n) = \Phi((s, n) + S\otimes_RN) = s\varphi(n) = s'\varphi(n') = \Phi(s' \otimes n')$. So if I could show that $s \varphi(n) = s'\varphi(n')$ I would be done with showing well-definedness.

Can someone give me a hint on how to do this? Thanks!

  • $\begingroup$ Use the universal property of the tensor product (which should be seen as its definition). Also recall that not every element of the tensor product is a pure tensor (your approach suggests that you assume this to be true). But anyway you don't need this. $\endgroup$ – Martin Brandenburg Jul 14 '13 at 9:35
  • $\begingroup$ The universal property is what I'm trying to show (I think). I understand what you mean about every tensor not necessarily a pure tensor. The tensor product is the set of all finite commutative sums. How do I show well-definedness of the mapping then? $\endgroup$ – Robert Cardona Jul 14 '13 at 9:40

I assume that you use the definition of tensor product that starts with a free module $F$ generated by elements of the form $(s, n)$, then defines the tensor product as $F/A$ where $A$ is the submodule of $F$ generated by bilinearity conditions.

What you want to show is that the map $\overline\Phi: F \to L$ defined by $\overline\Phi((s, n)) = s\phi(n)$ gives the same value for elements in the same coset of $A$ in $F$, i.e., that $\overline\Phi(A) = 0$. This can be verified easily by testing generators of $A$.

  • $\begingroup$ Using the explicit construction of the tensor product is rather clumsy. The universal property of the tensor product is what one needs to work with tensor products. $\endgroup$ – Martin Brandenburg Jul 14 '13 at 9:59
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    $\begingroup$ @MartinBrandenburg I agree totally. I am just under the impression that this is the direction that Robert is following at the moment. $\endgroup$ – Tunococ Jul 14 '13 at 10:01
  • $\begingroup$ Thanks. This is a similar approach to Dummit and Foote. I was trying to avoid looking at it and I tried it my own way, but I guess its wrong :P, thanks! $\endgroup$ – Robert Cardona Jul 14 '13 at 10:14

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