TITLE: Exactness of a short sequence of quotient modules
QUESTION [11 upvotes]: 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.

REPLY [5 votes]: 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)).

REPLY [3 votes]: 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.