Let $f: R \rightarrow S$ be a ring homomorphism of Noetherian rings, $M$ finite $R$-module and $N$ finite $S$-module which is flat over $R$. Let $J$ be an arbitrary ideal of $S$. I want to prove that there exists a canonical isomorphism $(M\otimes N)/J (M \otimes N) \cong M \otimes \left(N/JN \right)$. I have come up with a proof, but i am not sure it is as rigorous/elegant as possible, so i would appreciate such an argument.
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The tensor product is right exact. – Martin Brandenburg Dec 17 '13 at 20:52
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@MartinBrandenburg: I am aware of that, please be more specific. – Manos Dec 17 '13 at 20:53
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You have too many hypotheses on your rings and modules. All you need is a homomorphism $f: R \rightarrow S$ of commutative rings, an $R$-module $M$, an $S$-module $N$, and an $S$-ideal $J$. No flatness or finiteness assumptions are needed anywhere.
Then one has the following isomorphisms: $$ \frac{(M \otimes_R N)}{J(M \otimes_R N)} \cong (M \otimes_R N) \otimes_S (S/J) \cong M \otimes_R (N \otimes_S S/J) \cong M \otimes_R (N/JN). $$
These follow from associativity of tensor product, along with the fact that for any $S$-module $L$, we have $L \otimes_S (S/J) \cong L/JL$.
Actually, even the above argument makes too many assumptions. You can put the same isomorphisms in a non-commutative context if you are careful about which modules and ideals have which left- and right-actions by the rings in question.
neilme
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