Nakayama's lemma

Statement

Let R be a commutative ring with identity 1. The following is Nakayama's lemma, as stated in Matsumura (1989):

Statement 1: Let I be an ideal in R, and M a finitely-generated module over R. If IM = M, then there exists an r ∈ R with r ≡ 1 (mod I), such that rM = 0.

This is proven below.

The following corollary is also known as Nakayama's lemma, and it is in this form that it most often appears.

Statement 2: If M is a finitely-generated module over R, J(R) is the Jacobson radical of R, and J(R)M = M, then M = 0.

Proof: r − 1 (with r as above) is in the Jacobson radical so r is invertible.

More generally, one has

Statement 3: If M is a finitely-generated module over R, N is a submodule of M, and M = N + J(R)M, then M = N.

Proof: Apply Statement 2 to M/N.

The following result manifests Nakayama's lemma in terms of generators

Statement 4: If M is a finitely-generated module over R and the images of elements m₁,...,m_n of M in M/J(R)M generate M/J(R)M as an R-module, then m₁,...,m_n also generate M as an R-module.

Proof: Apply Statement 3 to N = Σ_iRm_i.

This conclusion of the last corollary holds without assuming in advance that M is finitely generated, provided that M is assumed to be a complete and separated module with respect to the I-adic topology. Here separatedness means that the I-adic topology satisfies the T₁ separation axiom, and is equivalent to ⋂ k = 1 ∞ I k M = 0.

Local rings

In the special case of a finitely generated module M over a local ring R with maximal ideal m, the quotient M/mM is a vector space over the field R/m. Statement 4 then implies that a basis of M/mM lifts to a minimal set of generators of M. Conversely, every minimal set of generators of M is obtained in this way, and any two such sets of generators are related by an invertible matrix with entries in the ring.

In this form, Nakayama's lemma takes on concrete geometrical significance. Local rings arise in geometry as the germs of functions at a point. Finitely generated modules over local rings arise quite often as germs of sections of vector bundles. Working at the level of germs rather than points, the notion of finite-dimensional vector bundle gives way to that of a coherent sheaf. Informally, Nakayama's lemma says that one can still regard a coherent sheaf as coming from a vector bundle in some sense. More precisely, let F be a coherent sheaf of O_X-modules over an arbitrary scheme X. The stalk of F at a point p ∈ X, denoted by F_p, is a module over the local ring O_p. The fibre of F at p is the vector space F(p) = F_p/m_pF_p where m_p is the maximal ideal of O_p. Nakayama's lemma implies that a basis of the fibre F(p) lifts to a minimal set of generators of F_p. That is:

Any basis of the fiber of a coherent sheaf F at a point comes from a minimal basis of local sections.

Going up and going down

The going up theorem is essentially a corollary of Nakayama's lemma. It asserts:

Let R ⊂ S be an integral extension of commutative rings, and P a prime ideal of R. Then there is a prime ideal Q in S such that Q ∩ R = P. Moreover, Q can be chosen to contain any prime Q₁ of S such that Q₁ ∩ R ⊂ P.

Module epimorphisms

Nakayama's lemma makes precise one sense in which finitely generated modules over a commutative ring are like vector spaces over a field. The following consequence of Nakayama's lemma gives another way in which this is true:

If M is a finitely generated R-module and ƒ : M → M is a surjective endomorphism, then ƒ is an isomorphism.

Over a local ring, one can say more about module epimorphisms:

Suppose that R is a local ring with maximal ideal m, and M, N are finitely generated R-modules. If φ : M → N is an R-linear map such that the quotient φ_m : M/mM → N/mN is surjective, then φ is surjective.

Homological versions

Nakayama's lemma also has several versions in homological algebra. The above statement about epimorphisms can be used to show:

Let M be a finitely generated module over a local ring. Then M is projective if and only if it is free.

A geometrical and global counterpart to this is the Serre–Swan theorem, relating projective modules and coherent sheaves.

More generally, one has

Let R be a local ring and M a finitely generated module over R. Then the projective dimension of M over R is equal to the length of every minimal free resolution of M. Moreover, the projective dimension is equal to the global dimension of M, which is by definition the smallest integer i ≥ 0 such that

Here k is the residue field of R and Tor is the tor functor.

Proof

A standard proof of the Nakayama lemma uses the following technique due to Atiyah & Macdonald (1969).

Let M be an R-module generated by n elements, and φ : M → M an R-linear map. If there is an ideal I of R such that φ(M) ⊂ IM, then there is a monic polynomial

with p_k ∈ I^k, such that p ( φ ) = 0 as an endomorphism of M.

This assertion is precisely a generalized version of the Cayley–Hamilton theorem, and the proof proceeds along the same lines. On the generators x_i of M, one has a relation of the form

φ ( x i ) = ∑ j = 1 n a i j x j

where a_ij ∈ I. Thus

∑ j = 1 n ( φ δ i j − a i j ) x j = 0.

The required result follows by multiplying by the adjugate of the matrix (φδ_ij − a_ij) and invoking Cramer's rule. One finds then det(φδ_ij − a_ij) = 0, so the required polynomial is

p ( t ) = det ( t δ i j − a i j ) .

To prove Nakayama's lemma from the Cayley–Hamilton theorem, assume that IM = M and take φ to be the identity on M. Then define a polynomial p(x) as above. Then

r = p ( 1 ) = 1 + p 1 + p 2 + ⋯ + p n

has the required property.

Noncommutative case

A version of the lemma holds for right modules over non-commutative unitary rings R. The resulting theorem is sometimes known as the Jacobson–Azumaya theorem.

Let J(R) be the Jacobson radical of R. If U is a right module over a ring, R, and I is a right ideal in R, then define U·I to be the set of all (finite) sums of elements of the form u·i, where · is simply the action of R on U. Necessarily, U·I is a submodule of U.

If V is a maximal submodule of U, then U/V is simple. So U·J(R) is necessarily a subset of V, by the definition of J(R) and the fact that U/V is simple. Thus, if U contains at least one (proper) maximal submodule, U·J(R) is a proper submodule of U. However, this need not hold for arbitrary modules U over R, for U need not contain any maximal submodules. Naturally, if U is a Noetherian module, this holds. If R is Noetherian, and U is finitely generated, then U is a Noetherian module over R, and the conclusion is satisfied. Somewhat remarkable is that the weaker assumption, namely that U is finitely generated as an R-module (and no finiteness assumption on R), is sufficient to guarantee the conclusion. This is essentially the statement of Nakayama's lemma.

Precisely, one has:

Nakayama's lemma: Let U be a finitely generated right module over a ring R. If U is a non-zero module, then U·J(R) is a proper submodule of U.

Proof

Let X be a finite subset of U, minimal with respect to the property that it generates U. Since U is non-zero, this set X is nonempty. Denote every element of X by x_i, for i ∈ { 1 , … , n } . Since X generates U,

∑ i = 1 n x i R = U

Suppose, U ⋅ J ( R ) = U , to obtain a contradiction. Since, ∑ i = 1 n x i ∈ U , we conclude,

∑ i = 1 n ( x i r i ) j i = ∑ i = 1 n x i , for r i ∈ R , and j i ∈ J ( R )

By associativity,

∑ i = 1 n x i ( r i j i ) = ∑ i = 1 n x i

Since J(R) is a (two-sided) ideal in R, we have r i j i ∈ J ( R ) for every i, and thus this becomes

∑ i = 1 n x i k i = ∑ i = 1 n x i , for k i ∈ J ( R )

Applying distributivity,

∑ i = 1 n x i ( 1 − k i ) = 0

Since k i ∈ J ( R ) , it is quasiregular and thus, 1 − k i ∈ U ( R ) , for all i, where U(R) denotes the group of units in R. Choose some j and write,

∑ i = 1 n x i ( 1 − k i ) ( 1 − k j ) − 1 = 0

Therefore,

∑ i ≠ j x i ( 1 − k i ) ( 1 − k j ) − 1 = − x j

Thus x_j is a linear combination of the elements of X distinct from x_j. This contradicts the minimality of X and establishes the result.

Graded version

There is also a graded version of Nakayama's lemma. Let R be a ring that is graded by the ordered semi group of non-negative integers, and let R + denote the ideal generated by positively graded elements. Then if M is a graded module over R for which M i = 0 for i sufficiently negative (in particular, if M is finitely generated and R does not contain elements of negative degree) such that R + M = M , then M = 0 . Of particular importance is the case that R is a polynomial ring with the standard grading, and M is a finitely generated module.

The proof is much easier than in the ungraded case: taking i to be the least integer such that M i ≠ 0 , we see that M i does not appear in R + M , so either M ≠ R + M , or such an i does not exist, i.e., M = 0 .