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Showing that if $R$ is local and $M$ an $R$-module, then $M \otimes_R (R/\mathfrak m) \cong M / \mathfrak m M$.

In one of the answers to one of my previous questions the following claim was mentioned:

$R/I \otimes_R M \cong M / IM$

So I tried to prove it. Can you help me finish my proof? Thanks!

We recall that $M \otimes_R -$ is a covariant right-exact functor that it is exact if $M$ is flat and we observe that the following is an exact sequence: $$ 0 \to IM \xrightarrow{i} M \xrightarrow{\pi} M / IM \to 0$$

$R/I$ is an $R$ module since $R/I$ is a subring of $R$ closed with respect to multiplication from $R$ but it is not necessarily flat hence we only get exactness on one side:

$$ (R/I) \otimes_R IM \xrightarrow{id \otimes i} (R/I) \otimes_R M \xrightarrow{id \otimes \pi} (R/I) \otimes_R M / IM \to 0$$

Then $$ \mathrm{Im(id \otimes \pi)} = (R/I) \otimes_R M / IM \cong (R/I) \otimes_R M / \mathrm{Ker} (id \otimes i) $$

so we want to show two things:

(i) that $\mathrm{Ker} (id \otimes i) = \{0\}$. To this end let $r + I \otimes im \in (R/I) \otimes_R IM $ and assume $ id \otimes i(r + I \otimes im ) = r + I \otimes im = 0 + I \otimes 0$. I'm not sure how to proceed from here.

(ii) and that $M /IM \cong (R/I) \otimes_R M / IM $

Can you help me? Thanks.

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marked as duplicate by Gerry Myerson, t.b., sdcvvc, rschwieb, Alex Becker Sep 19 '12 at 22:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Hint: There is also the sequence $0\to I\to R\to R/I\to 0$ – Julian Kuelshammer Jul 27 '12 at 9:29
Also, to prove that something is isomorphic to a tensor product of modules, you can show that it satisfies the universal property of the tensor product. That would be another way of tackling this problem. – Rankeya Jul 27 '12 at 10:04
There was a suggested edit pointing out that $R/I$ is not a subring of $R$. Rather $R/I$ is an $R$-algebra, which is probably what you wanted to say. – t.b. Jul 27 '12 at 11:21
@t.b. Thanks mate, I made a mistake there. I thought $f: r + I \mapsto r$ was an inclusion but it's not even a ring homo because it's not well-defined: $0 = f(0 + I) = f(i + I) = i$. – Rudy the Reindeer Jul 27 '12 at 11:35
up vote 6 down vote accepted

Consider $$ 0 \to I \to R \to R/I \to 0 $$ Apply the right exact functor $-\otimes_R M$, and you get $$ I\otimes_R M \to R\otimes_R M \to (R/I)\otimes_R M \to 0 $$ But $R\otimes_R M$ is canonically identified with $M$ by $a\otimes m \mapsto am$. Then $I\otimes_R M \to R\otimes_R M = M$ is $a\otimes m \mapsto am$ so, by definition, its image is $IM$. So the exactness of the sequence tells you that
$$ M/IM \cong (R/I)\otimes_R M $$

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Note that $M/IM\neq (R/I)\otimes M$: they are not equal as sets! They are just isomorphic as $R$-modules. – KReiser Jul 27 '12 at 9:38
@KReiser Thank you, I'll change $=$ to $\cong$. – Rudy the Reindeer Jul 27 '12 at 10:05

In general, playing around with elements of a tensor product is something that should be avoided if you can help it (defining maps from a tensor product is also usually better done by defining them from the product and then checking that they are in fact $R$-balanced and the using the universal property). A proof which avoids dealing with explicit elements of the tensor product can be done in the following way:

Consider the map $R/I\times M\to M/IM$ given by $(r+I,m)\mapsto rm+IM$. This is well-defined, as the image of any two representatives of $r+I$ differ by an element of $IM$. Since it is $R$-balanced, it descends to a map from the tensor product. Now consider the map $M \to R/I\otimes M$ given by sending $m\mapsto (1+I)\otimes m$. This descends to the quotient $M/IM$ by the 1st isomorphism theorem, and then it is easy to see that these maps are mutual inverses.

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Sorry, I don't understand your proof. What I do understand is that if we can show that $M / MI$ satisfies the universal property of $R/I \otimes M$ then they are isomorphic. So we define a bilinear map $b: R/I \times M \to M / IM$, $(r + I, m) \mapsto rm + IM$. What we need to show now is that for an arbitrary $R$-module $N$ and any bilinear map $b^\prime : R/I \times M \to N$ there exists exactly one linear map $l: M/IM \to N$ such that $l \circ b = b^\prime$. I don't understand where you do this in your second paragraph. – Rudy the Reindeer Jul 27 '12 at 10:54
Sorry, accidentally left out a sentence. The universal property I was referring to was that given an $R$-balanced map from a cartesian product $M\times N$ of right and left $R$-modules (ie one such that $f(ar,b)=f(a,rb)$ and $f(a+b,c)=f(a,c)+f(b,c)$ and $f(a,b+c)=f(a,b)+f(a,c)$) then it descends to a map from their tensor product. – KReiser Jul 27 '12 at 17:38

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