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Assume it if it´s neccesarly that the ring has an 1 or is commutative ( I´m not sure if it´s needed)

Given a ring $R$ an ideal $I$ of $R$, and a $R$ module $M$ , prove that: $ I \otimes _R M \cong IM $ where $ IM = \left\{ {x \in M:x = \sum\limits_{finite} {i_k m_k \,\,\,i_k \in I\,\,m_k \in M} } \right\} $

This is what I did. First I defined the obvious function $ \varphi\colon I\times M \to \,IM $ which is bilinear, so it defines a R-module-homomorphism $$ \varphi ^ \bullet \colon I \otimes _R M \to IM $$ and satisfies $ \varphi ^ \bullet \left( {i \otimes m} \right) = \varphi \left( {i,m} \right) = im $

I proved that $ \varphi ^ \bullet $ is surjective since, given $ \sum\limits_{finite} {i_k m_k } \in IM $ clearly $ \varphi ^ \bullet \left( {\sum\limits_{finite} {i_k \otimes m_k } } \right) = \sum\limits_{finite} {i_k m_k } $ But the injectivity how can I prove it?

Dylan Moreland
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Susuk
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1 Answers1

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"But the injectivity how can I prove it?"
Dear Susuk, you can't because it is false and it is a good omen for you that you couldn't prove it!
Here is a counterexample:

Let $k$ be any field . Consider $R=k[X]/(X^2)=k[\epsilon]$ and let $I$ be the ideal $I=(\mathbb \epsilon)=k\cdot \epsilon \subset R$ .
Take $M=I$. We have $I\cdot M=I^2=(0)$ and in order to show that your map $ \varphi ^ \bullet :I \otimes _R I \to I^2$ is not injective, it suffices to prove that $I \otimes _R I\neq 0$.
However, since $I$ is killed by $\epsilon$ we have $I \otimes _R I=I \otimes _{R/(\epsilon)} I=I \otimes _k I$ and the latter vector space is one dimensional over the field $k$, hence non-zero.

(Of course if $M$ is flat over $R$, the isomorphism $I \otimes _R M \stackrel {\simeq} {\to}IM $ holds)

Georges Elencwajg
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