About the multiplication map $I\otimes M\rightarrow M$

Question

so I am learning about flat modules and I found this criterion for flatness. An $R$-module $M$ is flat iff for every finitely generated ideal $I$, we get that the multiplication map $I\otimes_R M\rightarrow M$ is an injection (Eisenbud Proposition 6.1)

I am confused because say we have any $M$ (an $R$ module), and say that $i\otimes m\mapsto 0$. This means that $im=0$, but wouldn't this imply that $i\otimes m=1\otimes im=1\otimes 0=0$, so the kernel is trivial, and hence the map is injective always?

I am obviously doing something wrong, so I'd appreciate anyone that could clarify things a bit!

Also: Not every element of the tensor product is a pure tensor. The elements are sums of these. — Martin Brandenburg, Oct 13 '13 at 13:24

score 2 · Accepted Answer · answered Oct 13 '13 at 00:54

2

You can't write $i\otimes m=1\otimes im$ unless $1\in I$, i.e., unless $I=R$.

answered Oct 13 '13 at 00:54

Keenan Kidwell

26,020

Thanks. That was very dumb of me. – Daniel Montealegre Oct 13 '13 at 00:59

About the multiplication map $I\otimes M\rightarrow M$

1 Answers1