# Tensor Product and Flat Module

While doing some exercises about flat modules I got to this particular one:

Let $I$ be an ideal of $A$ (commutative ring with 1), and $M$ an $A$-module. Show that $I \otimes_{A} M \simeq IM$ if $M$ is flat.

The problem I'm having is that I'm not using the hypothesis that $M$ is flat to prove this isomorphism, it seems to me that it is unnecessary.

What I did to prove this was:

Define a function $f: I \times M \rightarrow IM$ that maps $(a,m)\mapsto a \cdot m$. Now this is a bilinear morphism, so by the universal property of the tensor product I know there must exist a unique $A$-module homomorphism $\bar{f}: I \otimes_{A} M \rightarrow IM$ such that: $\bar{f} \circ \pi = f$ in other words $m\otimes a \mapsto a \cdot m$, $\forall a \in A, m \in M$.

$\bar{f}$ is surjective: since every element $x \in IM$ is of the form $\sum a_i \cdot m_i$ so $\bar{f}(\sum a_i \otimes m_i) = x$.

$\bar{f}$ is injective: given $t \in I \otimes_{A} M$, then $t = \sum a_i \otimes m_i$ so $\bar{f}(t) = 0 \implies \sum a_i \cdot m_i = 0$. But since $t = \sum a_i \otimes m_i = \sum (a_i\cdot 1)\otimes m_i = \sum a_i \cdot (1 \otimes m_i) = \sum 1 \otimes (a_i \cdot m_i) = 1 \otimes (\sum a_i \cdot m_i) = 1 \otimes 0 = 0$. So $\ker \bar{f} = \{0\}$.

Am I wrong in some step or is the hypothesis of $M$ being flat really unnecesary here?

rie.

• There is a mistake in the injectivity part: there is no element of the form $1 \otimes m_i$ in $I \otimes M$ unless $1 \in M$. – Fredrik Meyer Jun 5 '16 at 18:08
• Thank you Fredrik, I realized that too. Now I know where I made a mistake. And found a new proof via a short exact sequence – rie Jun 5 '16 at 19:12

Consider the canonical injection $i\colon I\to A$. Tensoring by $M$, it remains an injection \begin{aligned}[t]i\otimes \operatorname{id_M}\colon I\otimes_A M&\to A\otimes_A M (\simeq M),\\\sum\limits_i a_i\otimes m_i&\mapsto \sum\limits_i a_i\otimes m_i=\sum\limits_i 1\otimes a_im_i= 1\otimes \sum\limits_i a_im_i.\end{aligned}

The image is (isomorphic to) IM.