Today’s goal is to partially finish the proof of the generic freeness lemma; the more general case, with finitely generated algebras, will have to wait for a later time though.

Recall that our goal was the following:

Theorem 1 Let ${A}$ be a Noetherian integral domain, ${M}$ a finitely generated ${A}$-module. Then there there exists ${f \in A - \{0\}}$ with ${M_f}$ a free ${A_f}$-module.

The argument proceeds using dévissage. By the last post, we can find a filtration $\displaystyle 0 = M_0 \subset M_1 \subset \dots \subset M_n = M,$

with ${M_{i+1}/M_i}$ isomorphic to ${A/\mathfrak{p}_i}$ for prime ideals ${\mathfrak{p}_i}$. Now, consider the nonzero prime ideals ${\mathfrak{p}'_j}$ that occur in the above filtration. Since ${A}$ is a domain, we have $\displaystyle \prod \mathfrak{p}'_j \neq 0,$

and we may choose ${f \neq 0}$ with ${f \in \prod \mathfrak{p}_j'}$. Then when we localize at ${f}$, there is still a filtration $\displaystyle 0 = (M_0)_f \subset (M_1)_f \subset \dots \subset (M_n)_f = M_f,$

such that ${(M_{i+1})_f/(M_i)_f = (A/\mathfrak{p}_i)_f}$. (Essentially, this uses the fact that localization is an exact functor.) By the definition of localization and the choice of ${f}$, this is zero when ${\mathfrak{p}_i \neq 0}$; this is ${A_f}$ when ${\mathfrak{p}_i=0}$.

So we have a filtration of ${M_f}$ by ${A_f}$-modules: $\displaystyle 0 = N_0 \subset N_1 \subset \dots \subset N_m = M_f$

such that each successive quotient is a free module of rank 1.

The proof will be completed by the following lemma:

Lemma 2 Suppose ${F', F''}$ are free modules over a ring ${A}$ and ${M}$ is an ${A}$-module. If there is an exact sequence $\displaystyle 0 \rightarrow F' \rightarrow M \rightarrow F'' \rightarrow 0,$

then ${M}$ is free.

Proof: The exact sequence splits. Indeed, we can lift a basis of ${F''}$ to elements of ${M}$ by surjectivity; then define a map ${F'' \rightarrow M}$ from that lifting, which gives a section of ${M \rightarrow F''}$. Thus the sequence splits. $\Box$

Now, by induction, we can show that the ${N_k}$‘s are free over ${A_f}$. Hence ${N_m = M_f}$ is free.

[For a better proof, see the comments below.  -AM, 8/17]