# Please explain the difference between two short exact sequence

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 $$0 \to M \otimes N_1$$ ?

• (***) 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. – Mindlack Oct 29 '20 at 12:31
• 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.) – LetGBeTheGraph Oct 29 '20 at 12:33
• @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? – Masmath Oct 29 '20 at 13:24
• 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 $A/I$. – Mindlack Oct 29 '20 at 17:38

## 1 Answer

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 $$\mathbb Q\overset{p}\to \mathbb Q$$.

• 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 – Masmath Oct 29 '20 at 12:41
• Because the sequence with the $0$ on the left is not an exact sequence, specifically it is not exact at $\mathbb Z/p$ – Maxime Ramzi Oct 29 '20 at 12:44
• ok, thank you very much. I got the point – Masmath Oct 29 '20 at 12:46