Nakayama lemma

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In mathematics, more specifically modern algebra and commutative algebra, Nakayama's lemma — also known as the Krull–Azumaya theorem[1] — governs the interaction between the Jacobson radical of a ring (typically a commutative ring) and its finitely generated modules. Informally, the lemma immediately gives a precise sense in which finitely generated modules over a commutative ring behave like vector spaces over a field. It is an important tool in algebraic geometry, because it allows local data on algebraic varieties, in the form of modules over local rings, to be studied pointwise as vector spaces over the residue field of the ring.

The lemma is named after the Japanese mathematician Tadashi Nakayama and introduced in its present form in Nakayama (1951), although it was first discovered in the special case of ideals in a commutative ring by Wolfgang Krull and then in general by Goro Azumaya (1951).[2] In the commutative case, the lemma is a simple consequence of a generalized form of the Cayley–Hamilton theorem, an observation made by Michael Atiyah (1969). The special case of the noncommutative version of the lemma for right ideals appears in Nathan Jacobson (1945), and so the noncommutative Nakayama lemma is sometimes known as the Jacobson–Azumaya theorem.[1] The latter has various applications in the theory of Jacobson radicals.[3]

Statement[edit]

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 rR 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.[4]

Statement 2: If M 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 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[5]

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

Proof: Apply Statement 3 to N = ΣiRmi.

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.[6] Here separatedness means that the I-adic topology satisfies the T1 separation axiom, and is equivalent to \textstyle{\bigcap_{k=1}^\infty I^k M = 0.}

Consequences[edit]

Local rings[edit]

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 OX-modules over an arbitrary scheme X. The stalk of F at a point p ∈ X, denoted by Fp, is a module over the local ring Op. The fibre of F at p is the vector space F(p) = Fp/mpFp where mp is the maximal ideal of Op. Nakayama's lemma implies that a basis of the fibre F(p) lifts to a minimal set of generators of Fp. 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[edit]

The going up theorem is essentially a corollary of Nakayama's lemma.[7] 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.

To give geometrical context for this result, integral extensions correspond to proper maps of algebraic varieties. For varieties over the complex field, proper simply means that the inverse image of a compact set in the usual topology is again compact. Going up then implies that the image of an algebraic subvariety under a proper map is again an algebraic subvariety.[8]

Module epimorphisms[edit]

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.[9]

Over a local ring, one can say more about module epimorphisms:[10]

  • 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 φm : M/mM → N/mN is surjective, then φ is surjective.

Homological versions[edit]

Nakayama's lemma also has several versions in homological algebra. The above statement about epimorphisms can be used to show:[10]

  • 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[11]

  • 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
\operatorname{Tor}_{i+1}^R(k,M) = 0.
Here k is the residue field of R and Tor is the tor functor.

Proof[edit]

A standard proof of the Nakayama lemma uses the following technique due to Atiyah & Macdonald (1969).[12]

  • 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
p(x) = x^n + p_1x^{n-1}+\cdots + p_n
with pk ∈ Ik, such that
p(\varphi)=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 xi of M, one has a relation of the form

\varphi(x_i) = \sum_{j=1}^n a_{ij}x_j

where aij ∈ I. Thus

\sum_{j=1}^n\left(\varphi\delta_{ij} - a_{ij}\right)x_j = 0.

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

p(t) = \det(t\delta_{ij}-a_{ij}).

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+\cdots+p_n

has the required property.

Noncommutative case[edit]

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.[13]

Let J(R) be the Jacobson radical of R. If U is a right module over a ring, R, and I is an 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.[14] 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.[15] 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.[16] 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.[17]

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.[17]

Proof[edit]

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 xi, for i\in \{1,\ldots,n\}. Since X generates U,

\sum_{i=1}^n x_i R = U

Suppose, U\cdot J(R) = U, to obtain a contradiction. Since, \sum_{i=1}^n x_i\in U, we conclude,

\sum_{i=1}^n (x_i r_i) j_i = \sum_{i=1}^n x_i, for r_i\in R, and j_i\in J(R)

By associativity,

\sum_{i=1}^n x_i (r_i j_i) = \sum_{i=1}^n x_i

Since J(R) is a (two-sided) ideal in R, we have r_i j_i \in J\left(R\right) for every i, and thus this becomes

\sum_{i=1}^n x_i k_i = \sum_{i=1}^n x_i, for k_i\in J(R)

Applying distributivity,

\sum_{i=1}^n x_i (1 - k_i) = 0

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

\sum_{i=1}^n x_i (1 - k_i) (1 - k_j)^{-1} = 0

Therefore,

\sum_{i\neq j} x_i (1 - k_i) (1 - k_j)^{-1} = -x_j

Thus xj is a linear combination of the elements of X distinct from xj. This contradicts the minimality of X and establishes the result.[18]

Graded version[edit]

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 \ne 0, we see that M_i does not appear in R_+M, so either M \ne R_+M, or such an i does not exist, i.e., M = 0.

See also[edit]

Notes[edit]

  1. ^ a b Nagata 1962, §A.2
  2. ^ Nagata 1962, §A.2; Matsumura 1989, p. 8
  3. ^ Isaacs 1993, Corollary 13.13, p. 184
  4. ^ Eisenbud 1995, Corollary 4.8; Atiyah & Macdonald (1969, Proposition 2.6)
  5. ^ Eisenbud 1995, Corollary 4.8(b)
  6. ^ Eisenbud 1993, Exercise 7.2
  7. ^ Eisenbud 1993, §4.4
  8. ^ Over the complex field, this result is also known as the proper mapping theorem; see Griffiths & Harris (1994, p. 34).
  9. ^ Matsumura 1989, Theorem 2.4
  10. ^ a b Griffiths & Harris 1994, p. 681
  11. ^ Eisenbud 1993, Corollary 19.5
  12. ^ Matsumura 1989, p. 7: "A standard technique applicable to finite A-modules is the 'determinant trick'..." See also the proof contained in Eisenbud (1995, §4.1).
  13. ^ Nagata 1962, §A2
  14. ^ Isaacs 1993, p. 182
  15. ^ Isaacs 1993, p. 183
  16. ^ Isaacs 1993, Theorem 12.19, p. 172
  17. ^ a b Isaacs 1993, Theorem 13.11, p. 183
  18. ^ Isaacs 1993, Theorem 13.11, p. 183; Isaacs 1993, Corollary 13.12, p. 183

References[edit]