# Exactness of a short sequence of quotient modules

Suppose R is a commutative ring with 1, I $\subset R$ is an ideal.

We have R-Modules A, B and C with C being flat, as well as a short exact sequence

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

Consider the induced sequence

$0 \rightarrow A/IA \rightarrow B/IB\rightarrow C/IC\rightarrow 0$

How do I prove that this sequence is exact? I have no idea how the flatness of C could come into play, or to be more specific, how I can use the exactness of $C\otimes\_$ (this is the only definition of flatness we have so far).

Any advice in the right direction would be appreciated.

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If $0 \to A \to B \to C \to 0$ is exact, then apply $R/I \otimes_R -$ to get the exact $$\operatorname{Tor}(R/I, C) \to R/I \otimes A \to R/I \otimes B \to R/I \otimes C \to 0.$$ Since $C$ is flat Tor(−,C) = 0, and $R/I \otimes M = M/IM$, you get exactly what you asked for.

Alexander Thumm's answer is basically including a zig-zag lemma to prove Tor is "balanced". In other words, we have by definition that Tor(C,−) = 0, as in, if we tensor with C then there is no "extra term" on the left, but I sneakily used that Tor(−,C) = 0 too, so I can tensor things ending with C and keep exactness.

You might try proving your original question directly for abelian groups. This is a fairly literal translation of "torsion" (as in nx = 0) into the "torsion functor" (as in Tor(−,C)).

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+1 for providing the more general context. – Alexander Thumm Oct 22 '11 at 21:31
Tor (and Ext) will be introduced within the following week. I'll look back to your answer once I know the basics, thanks! – Cedric B Oct 23 '11 at 1:06

Hint: Why is the following diagram exact/commutative?

$$\array{ &&&&&&0& \\ &&&&&&\downarrow& \\ && A \otimes I & \to & B \otimes I & \to & C \otimes I & \to & 0 \\ &&\downarrow&&\downarrow&&\downarrow&& \\ 0 & \to & A \otimes R & \to & B \otimes R & \to & C \otimes R & \to & 0 \\ &&\downarrow&&\downarrow&&\downarrow&& \\ && A \otimes R/I & \to & B \otimes R/I & \to & C \otimes R/I & \to & 0 \\ &&\downarrow&&\downarrow&&\downarrow&& \\ &&0&&0&&0&& \\ }$$

Now try to prove, that the map $A \otimes R/I \to B \otimes R/I$ is a monomorphism.

For completeness of the answer I'll add the diagrammatic yoga, but you should first try to solve this yourself. Here you go:

Let $a'' \in A\otimes R/I$ with $a'' \mapsto 0 \in B\otimes R/I$ and $a \in A \cong A \otimes R$ with $a \mapsto a''$. Now let $b \in B \cong B\otimes R$ be the image of $a$ under the map $A \otimes R \to B \otimes R$. By commutativity of the diagram, we have $b \mapsto 0 \in B \otimes R/I$. By exactness, there is a $b' \in B \otimes I$ with $b' \mapsto b$. Now let $c' \in C \otimes I$ be the image of $b'$ under the map $B\otimes I \to C \otimes I$. By exactness and commutativity we have $c' \mapsto 0$ by $C\otimes I \to C \otimes R$. Again by ecaxtness (the map $C\otimes I \to C \otimes R$ is a monomorphism) we have $c' = 0$, so there is a $a' \in A \otimes I$ with $a' \mapsto b'$ by $A\otimes I \to B \otimes I$. Let $\tilde a \in A \otimes R$ be the image of $a'$ under the map $A\otimes I \to A \otimes R$. By commutativity we have $\tilde a \mapsto b$. By exactness ($A\otimes R \to B \otimes R$ is a monomorphism) we have $\tilde a = a$, so by exacness $a'' = 0$, since $a' \mapsto \tilde a = a \mapsto a''$.

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Thank you very much for the hint, I'll look into it tomorrow and "check" your answer if I succeed. – Cedric B Oct 22 '11 at 18:40

Note that $A/IA \cong A\otimes _R (R/I)$. Now apply the functor $-\otimes R/I$ to your short exact sequence. You get a long exact sequence, because $\otimes$ is not exact, but flatness implies that any tor groups involving $C$ vanish.

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I didn't know that $A/IA \cong A\otimes _R (R/I)$. Thanks, I'll look into your response once I know tor – Cedric B Oct 23 '11 at 1:08