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]