Exercise 3.15 [Atiyah/Macdonald]

Solution 1:

From the long exact sequence of Tor the second short exact sequence looks like

$$\text{Tor}_1^A(k,F) \rightarrow k\otimes N \rightarrow k\otimes F \rightarrow k\otimes F \rightarrow 0$$

But $\text{Tor}_1^A(k,F) = 0$ since $F$ is flat. Look at Chapter 2 exercise 24.

Solution 2:

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. $\text{Tor}_1^A(k,F) = 0$.