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Let $A$ be a commutative ring with $1$ and let $M,N$ be $A$-modules. Let $I$ be an ideal of $A$.

How do we show the following isomorphism?

$(M \otimes_{A} N)/(I \cdot (M \otimes_{A} N)) \cong M/(IM) \otimes_{A/I} N/(IN)$

What confuses me is how to pass from tensoring with respect $A$ to tensoring with respect $A/I$.

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By the way, it is a general fact that if $M$ is an $A$-module and $B$ an $A$-algebra, the functor $M \mapsto M \otimes_A B$ (which I'll abbreviate to $M \mapsto M_B$) is a monoidal functor; that is, $(M \otimes_A N)_B = M_B \otimes_B N_B$ (up to natural and nice isomorphisms). This is the case where $B = A/I$. –  Akhil Mathew Mar 24 '11 at 7:18

1 Answer 1

up vote 1 down vote accepted

There is one sensible way to define maps $$\phi:(M \otimes_{A} N)/(I \cdot (M \otimes_{A} N)) \to M/(IM) \otimes_{A/I} N/(IN)$$ and $$\psi: M/(IM) \otimes_{A/I} N/(IN) \to (M \otimes_{A} N)/(I \cdot (M \otimes_{A} N)).$$ Indeed, you should check that maps exist such that $$\phi(\overline{m\otimes n})=\overline m\otimes\overline n$$ and $$\psi(\overline m\otimes\overline n)=\overline{m\otimes n}$$ for all $m\in M$ and all $n\in N$; here the overlines mean coset with respect to the relevant submodule.

Once you have done that, show that $\phi\circ\psi$ and $\psi\circ\phi$ are both identities.

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Thanks! –  user6495 Mar 28 '11 at 1:08

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