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I'm trying to prove that $M/IM\simeq A/I\otimes_A M$. I defined $\varphi:M\to A/I\otimes_A M$ with $\varphi(m)=[1]\otimes m$. I've already proven that $\varphi$ is $A$-linear and also surjective.

But I'm having trouble to prove that $\ker(\varphi)=IM$. The inclusion $IM\subset \ker(\varphi)$ is clear. On the other hand, if $m\in \ker(\varphi)$, then $[1]\otimes m=0$. If $m=0$, then $m\in IM$, and we are done. On the other hand, if $m\neq 0$, then I must prove that there are $i\in I$, $n\in M$ such that $m=in$. I really don't know how to prove that.

What is the trick?

rmdmc89
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    As studiosus said, in this type of problems is better to find an explicit inverse instead of proving bijectivity. Generally is easy to show the surjectivity and from there you can construct the inverse map. I recommend you to check K. Conrad's blur on tensorial products. – Xam Mar 29 '17 at 20:12

2 Answers2

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Here is another (more advanced) way to prove it:

Consider the exact sequence of $A$-modules

$I \to A \to A/I \to 0$

Tensoring with $M$, we get an exact sequence: $I \otimes M \to A \otimes M \to A/I \otimes M \to 0$

Now $A \otimes M$ is naturally isomorphic to $M$ via $a \otimes m \to am$ and it is easy to see that if we take this isomorphism as an identification, the image of $I \otimes M$ in $M$ is $IM$. Thus by exactness, we have $M/IM \cong A/I \otimes M$.

Lukas Heger
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  • I \bigotimes M is not naturally isomorphic to IM unless M is flat. – Jishu Das Jun 04 '19 at 10:40
  • https://math.stackexchange.com/questions/298195/tensor-product-of-a-module-with-an-ideal-is-isomorphic-to-their-standard-product – Jishu Das Jun 04 '19 at 10:41
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    @JishuDas this is not a problem: the image of $I \otimes M$ under the isomorphism $A \otimes M \cong M$ composed with the morphism $I \otimes M \to A \otimes M$ is $IM$ (I have not claimed that $I \otimes M$ is isomorphic to $IM$) – Lukas Heger Jun 04 '19 at 16:39
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I think it is more convenient to find an inverse map, rather than showing injectivity and surjectivity. As you showed, $IM\subset\text{Ker}(\varphi)$, which implies that $\varphi$ induces a map $$\overline{\varphi}:M/IM\rightarrow A/I\otimes_{A}M:[m]\mapsto[1]\otimes m.$$ To make an inverse map, we can consider $$\psi:A/I\times M\rightarrow M/IM:([a],m)\mapsto[am],$$ which induces $$\overline{\psi}:A/I\otimes_{A}M\rightarrow M/IM:[a]\otimes m\mapsto[am].$$ You can check that $\overline{\varphi}$ and $\overline{\psi}$ are inverses of each other.

studiosus
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  • How to show that $\overline{\psi}$ is well defined? – Victor Mar 21 '23 at 15:37
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    @Victor It follows from the universal property of the tensor product (since the map $\psi:A/I\times M\to M/IM$ is bilinear) – Jon Oct 19 '23 at 12:42