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Is it true that: $$M{\otimes}_{A}(A/I) \cong M/IM$$ and $$IM \cong I {\otimes}_AM$$ where $A-$ commutative ring, $M-A$-module, $I \subset A - $ ideal.

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Yes. One reference for this result is Atiyah & MacDonald's commutative algebra textbook (although I believe the first result is an exercise). –  Aaron Jul 12 '11 at 5:15
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There is, however, a natural map $I \otimes_A M \to IM$ which is surjective. This might help you prove the first equation. –  Dylan Moreland Jul 12 '11 at 5:43
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The map Dylan mentions is also injective precisely when $Tor(R/I,M)=0$. So of course this is the only time you get an isomorphism. –  user641 Jul 12 '11 at 6:03
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2 Answers

up vote 2 down vote accepted

There is a proof of your first equation given in page 17 of Osborne's 'Basic Homological Algebra' book (presuming you have defined tensor product in terms of the usual universal property). I'll give an outline here

Define $\phi':M \times A/I \to M/IM$ by $\phi'(m,a+I) = am + IM.$ Check that this is well defined. Suppose there is a bilinear $\psi: M \times A/I \to G$. Using the fact that $A \otimes M \simeq M$ we have the following diagram

image

$\theta$ and $\phi$ come from the above tensor product ($A \otimes M \simeq M$). Convince yourself that $\theta'$ is induced from $\theta$, that the diagram commutes and that $\theta'$ is unique.

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The 1st claim ok. The 2nd not: $A=\mathbf{Z}/4\mathbf{Z}$, $I=2A$, $M=A/2A$. In that case $IM=0$, but because $I$ is isomorphic to $A/2A$ as an $A$-module, we have $I\otimes_AM$ isomorphic to $M$ as well. Basically you are tensoring the short exact sequence $$ 0\rightarrow I\rightarrow A\rightarrow A/I\rightarrow 0 $$ with $M$. The resulting sequence will be exact except possibly at the first spot. Look up flat module for more information. The proof of the first result is also there, because we can identify the image of $I\otimes M\rightarrow A\otimes M\cong M$ with $IM$ (as pointed out by Dylan).

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More generally, $M/A\otimes N/B \cong M\!\otimes\!N/(M\!\otimes\!B\!+\!A\!\otimes\!N)$, right? If $M\!=\!R^m$ and $N\!=\!R^n$, then $M/A=Coker A'$ and $N/B=Coker B'$ where $A',B'$ are matrices with columns the generators of $A,B$ respectively. There holds $Coker A'\otimes Coker B'\cong Coker(I_m\!\otimes\!B'\!+\!A'\!\otimes\!I_n)$, where $\otimes$ is the Kronecker product of matrices, correct? –  Leon Lampret Mar 5 at 1:09
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