Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
Try to accept solutions given for your questions which will show your appreciations! –  Ehsan M. Kermani Jan 30 '12 at 0:06
1  
Two LaTeX tips: use \times to get a "product" symbol. Use \colon when specifying a map $f\colon X \to Y$. –  Dylan Moreland Jan 30 '12 at 0:16
1  
That will be very hard, dear @Fredrik! –  Georges Elencwajg Jan 30 '12 at 0:29
1  
I'm not picky about accepts, but I don't see any reason to not accept the answer given below. –  Dylan Moreland Jan 30 '12 at 0:37
1  
We should mention that a characterisation of flatness is that $I \otimes_R M \cong IM$ for all ideals $I \lhd R$. –  Paul Slevin Jun 7 '12 at 21:47

1 Answer 1

"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)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.