tensor an exact sequence

Question

Let $R$ be a ring, $M$ an $R$-module, and $I$ an ideal of $R$. Then given the exact sequence: $$0\rightarrow I\rightarrow R\rightarrow R/I\rightarrow0$$

when is $M$ flat. By some theorem, $M$ is flat if and only if $I\otimes M\xrightarrow{f\otimes 1} R\otimes M$ is injective.

But how can I proof that $I\otimes M\xrightarrow{f\otimes 1}R\otimes M$ is injective?

I should make it clear: My question is

Given the exact sequence: $$0\rightarrow I\rightarrow R\rightarrow R/I\rightarrow0$$
and an $R$-module $M$, how to proof the sequence: $$0\rightarrow I\otimes M\rightarrow R\otimes M\rightarrow R/I\otimes M\rightarrow0$$ is exact?

One way to prove it is to prove $I\otimes M\xrightarrow{f\otimes 1} R\otimes M$ is injective.But how to prove this map is injective?

$\underline{edit 2:}$

More straight forwardly, given $R$ (a ring), $I$ (an $R$-ideal), an exact sequence $$0\rightarrow I\rightarrow R\rightarrow R/I\rightarrow0$$ and a module $M$, prove that the sequence $$0\rightarrow I\otimes M\rightarrow R\otimes M\rightarrow R/I\otimes M\rightarrow0$$ is exact.

Answer 1

No, the theorem is that $M$ is flat if and only if $I\otimes M \to R\otimes M$ is injective for every ideal $I\subset R$.

For the direction you are asking about: if $M$ is flat, then tensoring by it preserves exact sequences (by definition). $0\to I\to R$ is an exact sequence, so $0\to I\otimes M \to R\otimes M$ is an exact sequence. In other words, $I\otimes M \to R\otimes M$ is injective.

Answer 2

Your question is unfortunately still not clear. There are several ways to define flatness, and one of them is to say $M$ is a flat $R$-module if and only if for every ideal $I\subset R$, $M\otimes_R I \rightarrow M\otimes_R R$ is injective. The only time you would have to prove $M\otimes_R I \rightarrow M\otimes_R R$ is injective is if you define flatness differently, and want to show that this other definition implies the above. How you would do this depends of course on how you defined flatness! But in fact using every definition of flatness this implication is the easy one. It's the reverse implication that's hard.

To give an example, another common way to define flatness is to say $M$ is flat if whenever some exact sequence of $R$-modules

$$ 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0,$$

is given, then it remains exact if you tensor everything by $M$. In other words, $M$ is flat if for all such $A,B,C$, the sequence

$$ 0 \rightarrow M\otimes_R A \rightarrow M\otimes_R B \rightarrow M \otimes_R C\rightarrow 0$$ is also exact.

Now if this were your definition of flatness, then what you want to prove is easy. We know that for any ideal $I \subset R$, the sequence

$$0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0,$$

is almost by definition exact. If $M$ is flat, then it must be true that

$$0 \rightarrow M\otimes_R I \rightarrow M\otimes_R R \rightarrow M\otimes_R (R/I) \rightarrow 0,$$

is also exact. In particular $M\otimes_R I \rightarrow M\otimes_R R$ must be injective. The hard part is to prove that for a general $M$ if this is true for any ideal $I\subset R$, then $M$ is flat.