# Nakayama’s Lemma

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.

**Proof (Lang)**

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). ♦

Corollary 1.Let M be a finitely generated module over and . If span it as a vector space over , then generate it as an A-module.

**Proof**

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*.

**Exercise A**

Prove that if and are minimal generating sets of *M*, then

for some invertible matrix with entries in *A*.

# Cotangent Spaces

Again denotes a local ring. Inspired by our earlier computation of tangent spaces, we define:

Definition.The

cotangent spaceof A is given by . More generally, for , the r-th ordercotangent spaceof A is .

**Exercise B**

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 .

### Example 1

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*.

### Example 2

Let and . Since in *A*, any monomial of degree *r* which contains lies in . Thus has basis and

### Example 3

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

**Note**

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.

Proposition 1.An A-module M is finitely presented and flat if and only if it is finitely generated and projective.

**Note**

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.

**Proof**

By localization, we may assume is local (exercise: write out the detailed argument). Hence it suffices to show the following. ♦

Proposition 2.A finitely presented flat module M over a local ring is free.

**Proof**

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.

**Exercise C**

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. ]

Summary.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.