Commutative Algebra 34

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, (A, \mathfrak m) is a fixed local ring.

Theorem (Nakayama’s Lemma).

Let M be a finitely generated A-module. Then

\mathfrak m M = M \implies M = 0.

In other words, if M\ne 0 then k(\mathfrak m) \otimes M \cong M/\mathfrak m M \ne 0, where k(\mathfrak m) is the residue field.

Proof (Lang)

Suppose m_1, \ldots, m_n \in M is a minimal set of generators of M. If n = 0, M = 0 and we are done. Otherwise, we have m_1 \in M = \mathfrak m M so we can write

m_1 = a_1 m_1 + \ldots + a_n m_n, \quad a_1, \ldots, a_n \in \mathfrak m.

But since a_1 \in \mathfrak m, we have 1-a_1 \in A a unit, so in fact:

m_1 = (1 - a_1)^{-1} (a_2 m_2 + \ldots + a_n m_n)

and M has a smaller set of generators given by m_2, \ldots, m_n (contradiction). ♦

Corollary 1.

Let M be a finitely generated module over (A, \mathfrak m) and m_1, \ldots, m_n \in M. If \overline m_1, \ldots, \overline m_n \in M/\mathfrak m M span it as a vector space over A/\mathfrak m, then m_1, \ldots, m_n \in M generate it as an A-module.

Proof

Let f:A^n \to M be the A-linear map which takes e_i \mapsto m_i. Let C be the cokernel of f so we get an exact sequence of A-modules: A^n \stackrel f \to M \to C \to 0. Since the functor N \mapsto N/\mathfrak m N is right-exact (by corollary 2 here), we get an exact:

(A/\mathfrak m)^n \stackrel{\overline f}\longrightarrow M/\mathfrak m M \longrightarrow C/\mathfrak m C \longrightarrow 0

By the given condition \overline f is surjective so C = \mathfrak m C. Since C is finitely generated (being a quotient of M), we have C = 0. ♦

Hence, for a finitely generated M over (A, \mathfrak m), let n = \dim_{A/\mathfrak m} M/\mathfrak m M and pick m_1, \ldots, m_n such that \overline m_i \in M/\mathfrak m M form a basis over A/\mathfrak m. Then m_1, \ldots, m_n 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 \{m_1, \ldots, m_n\} and \{m_1', \ldots, m_n'\} are minimal generating sets of M, then

\begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{pmatrix} \begin{pmatrix} m_1 \\ m_2 \\ \vdots \\ m_n\end{pmatrix} = \begin{pmatrix} m_1 \\ m_2 \\ \vdots \\m_n\end{pmatrix}

for some invertible matrix (a_{ij}) with entries in A.

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Cotangent Spaces

Again (A, \mathfrak m) denotes a local ring. Inspired by our earlier computation of tangent spaces, we define:

Definition.

The cotangent space of A is given by \mathfrak m / \mathfrak m^2. More generally, for r\ge 1, the r-th order cotangent space of A is \mathfrak m^r / \mathfrak m^{r+1}.

Exercise B

Find a local ring (A, \mathfrak m) which is not a field, satisfying \mathfrak m = \mathfrak m^2.

[ 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 \mathfrak m is a finitely generated module; hence so is every \mathfrak m^r. The anomaly in exercise B thus does not occur. From Nakayama’s lemma, we see that a minimal generating set of \mathfrak m^r is obtained from a basis of \mathfrak m^r/ \mathfrak m^{r+1} over A/\mathfrak m.

In the following examples, we will let A be the coordinate ring of a variety V over k (algebraically closed field). For P\in V, let \mathfrak m = \mathfrak m_P so that A/\mathfrak m \cong k. We can apply our above reasoning to the local ring B := A_{\mathfrak m} and \mathfrak n = \mathfrak m B. Note that

\mathfrak n^r /\mathfrak n^{r+1} \cong \mathfrak m^r / \mathfrak m^{r+1}

since the RHS is already a B-module. Thus we only need to compute:

P(r) = \dim_{k} \mathfrak m^r / \mathfrak m^{r+1}

to find the minimum number of generators of \mathfrak n^r.

Example 1

Suppose A = k[X_1, \ldots, X_n] and \mathfrak m = (X_1, \ldots, X_n). Then A / \mathfrak m = k and \mathfrak m^r / \mathfrak m^{r+1} has k-basis given by all monomials of degree r. The number of such monomials is the number of solutions to a_1 + \ldots + a_n = r in non-negative integers. By elementary combinatorics, we have

P(r) = {r+n-1 \choose n-1},

which is polynomial in r of degree n-1.

In the following examples, when A is a quotient of k[X_1, \ldots, X_n] we will write X_i instead of \overline X_i for its image in A.

Example 2

Let A = k[X, Y]/(Y^2 - X^3 + X) and \mathfrak m =(X, Y). Since X = X^3 - Y^2 \in \mathfrak m^2 in A, any monomial of degree r which contains X lies in \mathfrak m^{r+1}. Thus \mathfrak m^r / \mathfrak m^{r+1} has basis Y^r and

P(r) = \dim_k (\mathfrak m^r / \mathfrak m^{r+1}) = 1.

Example 3

Let A = k[X, Y, Z]/(Z^3 - X^3 - Y^3) and \mathfrak m = (X, Y, Z). In k[X, Y, Z], the elements X, Y, Z do not have any linear relation modulo (X, Y, Z)^2 + (Z^3 - X^3 - Y^3) = (X, Y, Z)^2, hence we have \dim_k \mathfrak m/\mathfrak m^2 = 3.

Similarly, in k[X, Y, Z] the monomials of degree 2 have no linear relation modulo (X, Y, Z)^3 + (Z^3 - X^3 - Y^3) = (X, Y, Z)^3 so we have \dim_k \mathfrak m^2 / \mathfrak m^3 = 6.

For \mathfrak m^r /\mathfrak m^{r+1} where r\ge 3, we have Z^3 = X^3 + Y^3 in A so a basis is given by X^a Y^b Z^c where a+b+c = r and c=0,1,2. Thus P(r) = 3r. In conclusion

P(r) = \begin{cases} 1, \quad &\text{if } r = 0,\\ 3r, &\text{if } r\ge 1.\end{cases}

Note

We thus see that in all three case, the function P(r) 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?

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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 (A, \mathfrak m) 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 (A, \mathfrak m) is free.

Proof

Pick a minimal generating set m_1, \ldots, m_n \in M over A. Take the surjective map A^n \to M which maps e_i \mapsto m_i and let K be the kernel of this map. Hence we get an exact sequence

0 \longrightarrow K \stackrel \subseteq \longrightarrow A^n \longrightarrow M \longrightarrow 0.

By the exercise after this, tensoring this exact sequence with any A-module N gives an exact sequence. In particular, we set N = A/\mathfrak m to obtain:

0 \longrightarrow K/\mathfrak m K \longrightarrow (A/\mathfrak m)^n \longrightarrow M/\mathfrak m M \longrightarrow  0.

But since m_i‘s form a minimal generating set of M over A, \overline m_i \in M/\mathfrak m M form a basis over A/\mathfrak m. Thus the map (A/\mathfrak m)^n \to M/\mathfrak m M is an isomorphism and we have K = \mathfrak m K. By proposition 1 here, since M is finitely presented, K is finitely generated. Thus by Nakayama’s lemma K = 0 and M \cong A^n. ♦

It remains to fill in the gap in the proof.

Exercise C

Suppose we have a short exact sequence of A-modules

0 \longrightarrow M_1 \longrightarrow M_2 \longrightarrow M \longrightarrow 0

where M is A-flat. Then tensoring with any A-module N gives a short exact sequence:

0 \longrightarrow M_1 \otimes_A N \longrightarrow M_2 \otimes_A N \longrightarrow M \otimes_A N \longrightarrow 0.

[ Hint: let F\to N be a surjective map where F is free; let K be its kernel so we get a short exact sequence 0\to K \to F \to N \to 0. 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 \mathfrak m\subset A, M_{\mathfrak m} is free over A_{\mathfrak m}.

We thus see that for such modules, projectivity is a local property and being “locally free” is equivalent to being projective.

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