Exercise 3.15 [Atiyah/Macdonald] I have a question regarding a claim in Atiyah, Macdonald. A is a commutative ring with $1$, $F$ is the free $A$-module $A^n$. Assume that $A$ is local with residue field $k = A/\mathfrak m$, and assume we are given a surjective map $\phi: F\to F$ with kernel $N$. Then why is the following true?

Since $F$ is a flat $A$-module, the exact sequence $0\to N \to F\overset\phi\to F\to 0$ gives an exact sequence $0\to k\otimes N \to k\otimes F \overset{1\otimes \phi}\to k\otimes F \to 0$.

I can see that $F$ is a free $A$-module, and that the first sequence is exact. But how does flatness of $F$ tell me something about the second sequence?
Thanks!

A general principle in homological algebra is the following:


Every ses of chain complexes gives rise to a LES in homology.


One can apply this principle to many situations, in our case it can be used to show that every ses of $A$ - modules gives rise to a LES in Tor. The LES in your situation is exactly
$$\ldots \to \text{Tor}_1^A(k, N)  \to \text{Tor}_1^A(k, F) \to \text{Tor}_1^A(k, F)  \to k \otimes_A N \to k \otimes_A F \to k\otimes_A F \to 0.$$
Now we claim that $\text{Tor}_1^A (k,F) = 0$. Indeed because $F$ is free (hence projective) we can always take the tautological projective resolution
$$ \ldots \to 0  \to 0 \to F \to F \to 0 $$
and remove the first $F$, tensor with $k$,  to get the chain complex
$$ \ldots \to 0 \to 0 \to k \otimes_A F  \to 0$$
from which it is clear that the first homology group of this complex is zero, i.e. $\text{Tor}_1^A(k,F) = 0$.
