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 $0 \to M \otimes N_1$ ?
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 $\mathbb Q\overset{p}\to \mathbb Q$.