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 simple approach that doesn't use $\mathrm{Tor}$ is to show that $0 \to N \to F \to F \to 0$ is split. To do this, use the fact that $F$ is free (or more generally, projective) to find a map $\psi : F \to F$ such that $\phi \circ \psi$ is the identity. Because the tensor product distributes over direct sums, the resulting sequence $0 \to k \otimes N \to k \otimes F \to k \otimes F \to 0$ is split exact.
Note that this proof only works when $F$ is projective, which need not be the case if $F$ is an arbitrary flat module.
