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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?

  • 2
    We should mention that a characterisation of flatness is that $I \otimes_R M \cong IM$ for all ideals $I \lhd R$.2012-06-07
  • 0
    Related: http://math.stackexchange.com/questions/4103962017-01-23

1 Answers 1