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My question is about exact sequence and tensor product of modules. Consider the following exact sequence of $R$-module for a commutative ring $R$ as in here: \begin{equation} 0 \to N_1 \to N_2 \to N_3 \to 0 \ \ (*)\end{equation} Taking tensor product with $R$-module $M$, we get the exact sequence \begin{equation} M \otimes N_1 \to M \otimes N_2 \to M \otimes N_3 \to 0 \ \ (**)\end{equation} But if $M$ is a flat $R$-module, we get the exact sequence \begin{equation} 0 \to M \otimes N_1 \to M \otimes N_2 \to M \otimes N_3 \to 0 \ \ (***)\end{equation} Why is so ? What is the difference between the exact sequences $(**)$ and $(***)$ ?

What is special about the map $0 \to M \otimes N_1$ ?

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    $\begingroup$ (***) means that the map $M \otimes N_1 \rightarrow M \otimes N_2$ is injective, which isn’t always the case. For instance, if $N_1$ is an ideal $I$ of $A=N_2$ and $M=A/I$, this map is zero. $\endgroup$ – Mindlack Oct 29 '20 at 12:31
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    $\begingroup$ The point is that (**) continues (or can be naturally continued) with more modules and maps, which measure the "obstruction from exactness" that the tensor functor faces. (If M is flat there is no such obstruction.) $\endgroup$ – LetGBeTheGraph Oct 29 '20 at 12:33
  • $\begingroup$ @Mindlack, How is the map in your example is a $zero$ map ? For in your example, we have $A/I \otimes_A I \to A/I \otimes_A A$ i.e., we have the map $A \to A/I \otimes_A A$,if I am correct. How is this map is $zero$ map? $\endgroup$ – Masmath Oct 29 '20 at 13:24
  • $\begingroup$ No, that’s the map $A/I \otimes I \rightarrow A/I \otimes A=A/I$. But if $k \in I$, the image of $1 \otimes k$ under this map is the $1\otimes k \in A/I \otimes_A A$, which is $k \cdot 1_{A/I}=0$ since we’re in $A/I$. $\endgroup$ – Mindlack Oct 29 '20 at 17:38
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The difference is that in the second exact sequence, $M\otimes N_1\to M\otimes N_2$ is injective, whereas in the first one it need not be so.

The best way to understand this type of phenomenon is with an example:

take $0\to \mathbb{Z}\overset{p}\to \mathbb{Z\to Z}/p \to 0$ and tensor it with $\mathbb Z/p$, this gives $\mathbb Z/p\overset{0}\to \mathbb Z/p\to\mathbb Z/p\to 0$, where of course $0$ is not injective.

If you tensor it with a flat $\mathbb Z$-module, for instance $\mathbb Q$, you will get an injection $\mathbb Q\overset{p}\to \mathbb Q$.

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  • $\begingroup$ Thanks, but my point is, why don't we write $ 0 \to \mathbb{Z}/p \to \mathbb{Z}/p \to \mathbb{Z}/p \to 0$ instead of $\mathbb{Z}/p \to \mathbb{Z}/p \to \mathbb{Z}/p \to 0$ ? I am talking the $0 \to \mathbb{Z}/p$ map $\endgroup$ – Masmath Oct 29 '20 at 12:41
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    $\begingroup$ Because the sequence with the $0$ on the left is not an exact sequence, specifically it is not exact at $\mathbb Z/p$ $\endgroup$ – Maxime Ramzi Oct 29 '20 at 12:44
  • $\begingroup$ ok, thank you very much. I got the point $\endgroup$ – Masmath Oct 29 '20 at 12:46

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