The following is a short statement which has far-reaching applications. Since its main applications are for local rings, we will state the result in this context. Throughout this section, is a fixed local ring.
Theorem (Nakayama’s Lemma).
Let M be a finitely generated A-module. Then
In other words, if then , where is the residue field.
Suppose is a minimal set of generators of M. If n = 0, M = 0 and we are done. Otherwise, we have so we can write
But since , we have a unit, so in fact:
and M has a smaller set of generators given by (contradiction). ♦
Let M be a finitely generated module over and . If span it as a vector space over , then generate it as an A-module.
Let be the A-linear map which takes . Let C be the cokernel of f so we get an exact sequence of A-modules: . Since the functor is right-exact (by corollary 2 here), we get an exact:
By the given condition is surjective so . Since C is finitely generated (being a quotient of M), we have C = 0. ♦
Hence, for a finitely generated M over , let and pick such that form a basis over . Then form a minimal generating set for M over A. Furthermore, all minimal generating sets of M have the same cardinality: n.
Prove that if and are minimal generating sets of M, then
for some invertible matrix with entries in A.
Again denotes a local ring. Inspired by our earlier computation of tangent spaces, we define:
The cotangent space of A is given by . More generally, for , the r-th order cotangent space of A is .
Find a local ring which is not a field, satisfying .
[ Hint: take the ring of all differentiable functions ℝ → ℝ and consider the maximal ideal of all functions vanishing at 0. ]
Let us assume that is a finitely generated module; hence so is every . The anomaly in exercise B thus does not occur. From Nakayama’s lemma, we see that a minimal generating set of is obtained from a basis of over .
In the following examples, we will let A be the coordinate ring of a variety V over k (algebraically closed field). For , let so that . We can apply our above reasoning to the local ring and . Note that
since the RHS is already a B-module. Thus we only need to compute:
to find the minimum number of generators of .
Suppose and . Then and has k-basis given by all monomials of degree r. The number of such monomials is the number of solutions to in non-negative integers. By elementary combinatorics, we have
which is polynomial in r of degree n-1.
In the following examples, when A is a quotient of we will write instead of for its image in A.
Let and . Since in A, any monomial of degree r which contains lies in . Thus has basis and
Let and . In , the elements do not have any linear relation modulo , hence we have .
Similarly, in the monomials of degree 2 have no linear relation modulo so we have .
For where , we have in A so a basis is given by where and . Thus . In conclusion
We thus see that in all three case, the function can be represented by a polynomial for sufficiently large r. This is called the Hilbert polynomial of the local ring A; we shall have more to say about this later.
Question to Ponder
What can we say about the degree of the Hilbert polynomial?
F. P. + Flat = F. G. + Projective
Finally, here is an interesting result. Recall that saying a module is finite presented is a stronger condition than saying it is finitely generated. On the other hand, by corollary 1 here, projective modules are flat. It turns out these two pairs of properties complement each other.
An A-module M is finitely presented and flat if and only if it is finitely generated and projective.
Let us keep stock of what we already know; by the above remarks f.p. ⇒ f.g. and flat ⇐ projective. By exercise A.2 here, f.p. ⇐ f.g. + projective. Thus the only remaining case is f.p. + flat ⇒ projective.
By localization, we may assume is local (exercise: write out the detailed argument). Hence it suffices to show the following. ♦
A finitely presented flat module M over a local ring is free.
Pick a minimal generating set over A. Take the surjective map which maps and let K be the kernel of this map. Hence we get an exact sequence
By the exercise after this, tensoring this exact sequence with any A-module N gives an exact sequence. In particular, we set to obtain:
But since ‘s form a minimal generating set of M over A, form a basis over . Thus the map is an isomorphism and we have . By proposition 1 here, since M is finitely presented, K is finitely generated. Thus by Nakayama’s lemma K = 0 and . ♦
It remains to fill in the gap in the proof.
Suppose we have a short exact sequence of A-modules
where M is A-flat. Then tensoring with any A-module N gives a short exact sequence:
[ Hint: let be a surjective map where F is free; let K be its kernel so we get a short exact sequence . Since M and F are flat, we have a diagram where the rows and columns are exact. Do a diagram-chase. ]
For a finitely presented module M, the following are equivalent.
- M is projective.
- M is flat.
- For any maximal ideal , is free over .
We thus see that for such modules, projectivity is a local property and being “locally free” is equivalent to being projective.