nLab Nakayama's lemma

Contents

Contents

Idea

Nakayama’s lemma is a simple but fundamental result of commutative algebra frequently used to lift information from the fiber of a sheaf over a point (as for example a coherent sheaf over a scheme) to give information on the stalk at that point.

Statement

We give a general version of the Nakayama’s lemma (Prop. ) and then two special cases that are often considered (Prop. and Prop. ).

Proposition

(Nakayama’s lemma)
Let RR be a commutative ring and MM be a finitely generated R R -module. Then:

  1. supp(M)=𝒱(Ann(M))\text{supp}(M) = \mathcal{V}(Ann(M)).

  2. If moreover IRI\subseteq R is an ideal with supp(M)V(I)supp(M)\cap V(I) inhabited, then M RR/IM\otimes_{R} R/I is nonzero.

Recall here that supp(M)={𝔭Spec(R) s.t. M RR 𝔭 is nonzero}supp(M) = \{\mathfrak{p}\in Spec(R)\ \text{ s.t. }\ M\otimes_{R} R_{\mathfrak{p}}\ \text{ is nonzero}\}, Ann(M)={fR s.t. fM=0}Ann(M) = \{f \in R\ \text{ s.t. }\ f M = 0\}, and 𝒱(I)={𝔭Spec(R) s.t. 𝔭I}\mathcal{V}(I) = \{\mathfrak{p} \in Spec(R)\ \text{ s.t. }\ \mathfrak{p} \supseteq I\}. The first claim is not usually considered part of Nakayama’s lemma but will be useful here and can be proved with a closely related argument to that which we will use for the main result.)

The following proof uses relatively little pre-existing commutative algebraic theory (i.e., just some basic facts about 𝒱\mathcal{V}, nonzeroness in short exact sequences, localization, tensoring, and localizing/tensoring cyclic modules)

Proof

The argument proceeds in the following four steps:

  1. As MM is finitely generated, it admits a filtration with cyclic cokernels. To be explicit, if the nn-tuple of elements (m kM) k=0,,n1(m_{k}\in M)_{k=0,\dots,n-1} generates MM, then it begets an nn-tuple of short exact sequences

    (0M kM k+1R/I k0) k=0,,n1,\left(\ 0 \to M_{k} \to M_{k+1} \to R/I_{k} \to 0\ \right)_{k=0,\dots,n-1},

    where M kM_{k} is the submodule of MM spanned by (m kM) k=0,,k1(m_{k}\in M)_{k=0,\dots,k-1} and in particular M 0=0M_{0}=0 and M n=MM_{n}=M. (Here the maps M kM k+1M_{k} \to M_{k+1} are the evident inclusions; as the cokernel is generated by the image of m km_{k}, it is in particular isomorphic to R/I kR/I_{k} for some ideal I kRI_{k}\subseteq R.) The basic idea is that the process of extension by which MM is built up from these R/I kR/I_{k}s interacts sufficiently nicely with the operations of localization/taking fibers that the properties of the latter determine those of the former.

  2. For instance, it is clear that any fAnn(M)f \in Ann(M) must descend to 00 on all the R/I kR/I_{k}s; on the other hand, for all k=0,,n1k=0,\dots,n-1 we have that I kM k+1M kI_{k}M_{k+1} \subseteq M_{k}. We conclude that k=0 n1I kAnn(M) k=0 n1I k\prod_{k=0}^{n-1} I_{k} \subseteq Ann(M) \subseteq \bigcap_{k=0}^{n-1} I_{k}, whence 𝒱(Ann(M))= k=0 n1𝒱(I k)\mathcal{V}(Ann(M)) = \bigcup_{k=0}^{n-1} \mathcal{V}(I_{k}).

  3. Let us now show that supp(M)= k=0 n1𝒱(I k)supp(M) = \bigcup_{k=0}^{n-1} \mathcal{V}(I_{k}); this will complete the proof of our first claim. Recall that for any 𝔭Spec(R)\mathfrak{p}\in Spec(R) the functor RR 𝔭-\otimes_{R} R_{\mathfrak{p}} is exact; we therefore in any case obtain an nn-tuple of short exact sequences

    (0M k RR 𝔭M k+1 RR 𝔭R/I k RR 𝔭0) k=0,,n1.\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to R/I_{k} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=0,\dots,n-1}.

    If 𝔭 k=0,,n1𝒱(I k)\mathfrak{p} \notin \bigcup_{k=0,\dots,n-1} \mathcal{V}(I_{k}), then each R/I k RR 𝔭R/I_{k} \otimes_{R} R_{\mathfrak{p}} vanishes, so the above SESs collapse to isomorphisms

    (0M k RR 𝔭M k+1 RR 𝔭0) k=0,,n1\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=0,\dots,n-1}

    and thus

    M RR 𝔭M n RR 𝔭M 0 RR 𝔭0.M \otimes_{R} R_{\mathfrak{p}} \simeq M_{n} \otimes_{R} R_{\mathfrak{p}} \simeq M_{0} \otimes_{R} R_{\mathfrak{p}} \simeq 0.

    Conversely, if 𝔭 k=0,,n1𝒱(I k)\mathfrak{p} \notin \bigcup_{k=0,\dots,n-1} \mathcal{V}(I_{k}), then there is a maximal index k maxk_{max} such that 𝔭I k max\mathfrak{p} \supseteq I_{k_{max}} and in particular an epimorphism

    M k max+1 RR 𝔭R/I k max RR 𝔭0,M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}} \to R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}} \to 0,

    with R/I k max RR 𝔭R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}} nonzero. But there are also, as before, isomorphisms

    (0M k RR 𝔭M k+1 RR 𝔭0) k=k max+1,,n1\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=k_{max}+1,\dots,n-1}

    whence

    M RR 𝔭M n RR 𝔭M k max+1 RR 𝔭M \otimes_{R} R_{\mathfrak{p}} \simeq M_{n} \otimes_{R} R_{\mathfrak{p}} \simeq M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}}

    is epic onto a nonzero module and thus itself nonzero.

  4. The proof of the second claim is in the same vein, requiring just a little bit more delicacy. Suppose that 𝔭\mathfrak{p} inhabits supp(M)V(I)\text{supp}(M)\cap V(I); by the above this means that 𝔭I\mathfrak{p}\supseteq I and that there exists some index kk for which 𝔭I k\mathfrak{p}\supseteq I_{k}. Again consider the maximal index k maxk_{max} for which 𝔭I k max\mathfrak{p} \supseteq I_{k_{max}}. As before, there is an epimorphism

    M k max+1 RR 𝔭R/I k max RR 𝔭0M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}} \to R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}} \to 0

    and isomorphisms

    (0M k RR 𝔭M k+1 RR 𝔭0) k=k max+1,,n1\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=k_{max}+1,\dots,n-1}

    By the right-exactness of the functor RR/𝔭-\otimes_{R} R/\mathfrak{p}, we obtain from the former an epimorphism

    M k max+1 RR 𝔭/𝔭R/I k max RR 𝔭/𝔭0M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}} /\mathfrak{p} \to R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \to 0

    (implicitly using that ( RR 𝔭) RR/𝔭 RR 𝔭/𝔭(- \otimes_{R} R_{\mathfrak{p}})\otimes_{R} R/\mathfrak{p}\ \simeq\ - \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p}). By its functoriality, there are likewise isomorphisms

    (0M k RR 𝔭/𝔭M k+1 RR 𝔭/𝔭0) k=k max+1,,n1\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \to 0\ \right)_{k=k_{max}+1,\dots,n-1}

    Now,

    R/I k max RR 𝔭/𝔭R 𝔭/𝔭R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq R_{\mathfrak{p}}/\mathfrak{p}

    is nonzero (here we use that 𝔭I k max\mathfrak{p} \supseteq I_{k_{max}}). It follows that

    M RR 𝔭/𝔭M n RR 𝔭/𝔭M k max+1 RR 𝔭/𝔭M \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq M_{n} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p}

    is epic onto a nonzero module and thus itself nonzero. But

    M RR 𝔭/𝔭(M RR/I) RR 𝔭/𝔭M \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq (M \otimes_{R} R/I) \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p}

    (here we use that 𝔭I\mathfrak{p} \supseteq I), so the latter must itself be nonzero, as claimed.

It is now straightforward to recover more familiar forms of Nakayama’s lemma as a corollary of the above.

Proposition

(special version 1)
Let RR be a local ring with maximal ideal 𝔪R\mathfrak{m}\subseteq R and MM be a finitely generated RR-module. Then if MM is nonzero, so too is M RR/𝔪M \otimes_R R/\mathfrak{m}.

Proof

Indeed, 𝔪\mathfrak{m} is practically tautologically in (and so 𝒱(𝔪)\mathcal{V}(\mathfrak{m}) practically tautologically intersects) the support of any nonzero RR-module; the conditions of Prop. , and we conclude that M RR/𝔪M \otimes_R R/\mathfrak{m} is nonzero as advertised.

Sometimes (for instance in the Stacks Project), Nakayama’s lemma is stated instead as follows:

Proposition

(special version 2)
Let RR be a commutative ring, MM be a finitely generated RR-module, and IRI\subseteq R be an ideal with IM=MI M = M. Then there exists fRf\in R such that f=1(modI)f = 1\ (mod\ I) and fM=0f M=0.

Proof

Of course, IM=MM RR/I0I M = M\ \iff\ M \otimes_{R} R/I \simeq 0; it follows from Prop. that II is comaximal with Ann(M)Ann(M). The Chinese Remainder theorem then asserts the existence of fRf\in R such that f=1(modI)f = 1\ (mod\ I) and f=0(modAnn(M))f = 0\ (mod\ Ann(M)); the desired ff.

Examples and consequences

Example

Suppose f:NMf \colon N \to M is an RR-module map, giving rise to an exact sequence

NfMpM/N0.N \stackrel{f}{\to} M \stackrel{p}{\to} M/N \to 0.

Tensoring with kk is a right exact functor, so we have an exact sequence

k RNk Rfk RMk RM/N0.k \otimes_R N \stackrel{k \otimes_R f}{\to} k \otimes_R M \to k \otimes_R M/N \to 0.

Nakayama’s lemma says that if k RM/N0k \otimes_R M/N \cong 0, then M/N0M/N \cong 0. Equivalently, that if k Rfk \otimes_R f is epic, then ff is epic. In particular, to check whether a finite set of elements v 1,,v nv_1, \ldots, v_n generates MM, it suffices to check whether the residue classes v imod𝔪Mv_i \mod \mathfrak{m}M generate the vector space M/𝔪MM/\mathfrak{m}M, which is a linear algebra calculation.

Example

Suppose OO is a Noetherian local ring. A typical example is the stalk at a point pp of a Noetherian scheme as locally ringed space, and we will write as if we were in that situation. Being Noetherian, its maximal ideal 𝔪\mathfrak{m} is finitely generated. Suppose k O𝔪𝔪/𝔪 2k \otimes_O \mathfrak{m} \cong \mathfrak{m}/\mathfrak{m}^2 – the cotangent space – is a vector space of dimension nn. We would like to know whether a collection of functions f 1,,f nf_1, \ldots, f_n that vanish at pp form a local coordinate system.

For this, it suffices to check whether the differentials df 1,,df nd f_1, \ldots, d f_n at pp, belonging to the cotangent space 𝔪/𝔪 2\mathfrak{m}/\mathfrak{m}^2, are linearly independent. (For then they span the cotangent space, and one concludes from Nakayama that the f if_i generate 𝔪\mathfrak{m} as an OO-module, thereby forming a local coordinate system at pp.) In this way, Nakayama’s lemma operates as a kind of “inverse function theorem”.

To cement this further, the following statement is offered in Harris as a corollary of Nakayama’s lemma (corollary 14.10, page 179):

Proposition

(Inverse Function Theorem) A map between complex projective varieties of dimension nn which is a bijection and has injective derivative at every point is an isomorphism.

Last revised on December 12, 2023 at 21:33:44. See the history of this page for a list of all contributions to it.