nLab Nakayama's lemma

Idea

Nakayama’s lemma is a family of simple but fundamental results of commutative algebra which are 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 first state and prove a pair of fundamental geometric facts (Prop. ) concerning the relationships between annihilators, supports, and tensor products of modules and then deduce as corollaries two standard formulations (Cor. and Cor. ) of “Nakayama’s lemma” from which various other instances thereof can in turn be derived (as is done in Stacks 00DV).

As motivation, recall the following general fact:

Proposition

(motivation) Given commutative ring $R ,$ natural $n ,$ and $n$-tuple of $R$-modules $\left( M _ i \right) _ { i \in n } ,$

1. for all $i \in n ,$ one has $supp \left( M _ i \right) \subseteq \mathcal{V} \left( Ann \left( M _ i \right) \right) .$

2. $supp \left( \underset{ i \in n }{ \bigotimes } M _ i \right) \subseteq \bigcap _ { i \in n } supp \left( M _ i \right) .$

(Here $\mathcal{V}$ denotes the zero locus,$supp$ denotes the support, and $Ann$ denotes the annihilator. For convenience, we assume the Von Neumann convention so that, e.g., the quantification “$i \in n$” is equivalent to “$i \in \left\{ 0 , \dots , n-1 \right\}$”.)

Proof

As for the first claim, if $i \in n$ and $\mathfrak{p} \in supp \left( M _ i \right) ,$ i.e., if $M _ i \otimes R _ { \mathfrak{p} } \ \,≄\, \ 0 ,$ then the elements of $R \setminus \mathfrak{p}$ act nontrivially on $M _ i \otimes R _ { \mathfrak{p} } ,$ hence nontrivially on $M _ i ,$ hence $Ann \left( M _ i \right) \subseteq \mathfrak{p} .$ As for the second claim, recall that $\left( \underset{ i \in n }{ \bigotimes } M _ i \right) \otimes R _ { \mathfrak{p} } \simeq \underset{ i \in n }{ \bigotimes } \left( M _ i \otimes R _ { \mathfrak{p} } \right) ,$ hence that if $\mathfrak{p} \in supp \left( \underset{ i \in n }{ \bigotimes } M _ i \right) ,$ i.e., if $\left( \underset{ i \in n }{ \bigotimes } M _ i \right) \otimes R _ { \mathfrak{p} } \ \,≄\, \ 0 ,$ then for all $i \in n$ one must have $M _ i \otimes R _ { \mathfrak{p} } \ \,≄\, \ 0 ,$ i.e.,$\mathfrak{p} \in supp \left( M _ i \right) .$

Our first, geometric incarnation of Nakayama’s lemma is just the assertion that if the modules in the statement of Prop. are finitely generated, then the inclusions in the same are equalities.

Proposition

(“geometric Nakayama”) Given commutative ring $R ,$ natural $n ,$ and $n$-tuple of finitely generated $R$-modules $\left( M _ i \right) _ { i \in n } ,$

1. for all $i \in n ,$ one has $supp \left( M _ i \right) = \mathcal{V} \left( Ann \left( M _ i \right) \right) .$

2. $supp \left( \underset{ i \in n }{ \bigotimes } M _ i \right) = \bigcap _ { i \in n } supp \left( M _ i \right) .$

(Here $\mathcal{V}$ denotes the zero locus,$supp$ denotes the support, and $Ann$ denotes the annihilator. For convenience, we assume the Von Neumann convention so that, e.g., the quantification “$i \in n$” is equivalent to “$i \in \left\{ 0 , \dots , n-1 \right\}$”.)

We give an “element-free” proof below. The two key ideas are that finitely-generated $R$-modules are (precisely) those which can be “built up” inductively from cyclic $R$-modules via extensions and that the relevant properties of the former are sufficiently determined by those of the latter, which are standard and/or easily verified.

Proof

Recall the following seven basic facts (essentially two “induction principles”, three “interaction principles”, and two “base cases” of our “induction” up from cyclic modules):

1. The middle term of a short exact sequence is $0$ iff its left and right terms are $0$

2. The middle component of a morphism of short exact sequences is $0$ iff its left and right components are $0 .$

3. Given prime $\mathfrak{p} \in Spec (R) ,$ the functor $N \mapsto N \otimes R _ { \mathfrak{p} }$ is exact.

4. The ($n$-ary) functor of $R$-modules $\left( N _ i \right) _ { i \in n } \mapsto \underset{ i \in n }{ \bigotimes } N _ i$ is right exact in all arguments.

5. Given prime $\mathfrak{p} \in Spec (R)$ and $n$-tuple of $R$-modules $\left( N _ i \right) _ { i \in n } ,$ one has $\left( \underset{ i \in n }{ \bigotimes } N _ i \right) \otimes R _ { \mathfrak{p} } \simeq \underset{ i \in n }{ \bigotimes } \left( N _ i \otimes R _ { \mathfrak{p} } \right) .$

6. Given prime $\mathfrak{p} \in Spec (R)$ and ideal $J \subseteq R ,$ one has $\left( R / J \right) \otimes R _ { \mathfrak{p} } \ \,≄\, \ 0$ iff $\mathfrak{p} \in \mathcal{V} \left( J \right) .$

7. Given $n$-tuple of ideals $\left( J _ i \subseteq R \right) _ { i \in n } ,$ one has $\underset{ i \in n }{ \bigotimes } \left( R / J _ i \right) \simeq R / \left( \sum _ { i \in n } J _ i \right) .$

The argument goes in four parts:

1. Suppose we are given $i \in n .$ By the hypothesis that $M _ i$ is finitely generated there exists natural $d _ i$ such that $M$ is a quotient of $R ^ { d _ i } .$ It follows (by taking the image under said quotient of the standard length-$d _ i$ filtration of $R ^ { d _ i }$) that $M _ i$ admits a length-$d _ i$ filtration with cyclic cokernels; denote this filtration and its associated $d _ i$-tuple of short exact sequences as

   $$0 \ = \ M _ { i , 0 } \ \overset{ \iota _ { i , 0 } }{ \hookrightarrow } \ \dots \ \overset{ \iota _ { i , d _ i - 1 } }{ \hookrightarrow } \ M _ {i , d _ i } \ = \ M _ i$$

   and

   $$\left( \ 0 \ \to \ M_{ i , j } \ \overset{ \iota _ { i , j } }{ \hookrightarrow } \ M_{ i , j + 1 } \ \overset{ \pi _ { i , j } }{ \twoheadrightarrow } \ R / I_{ i , j } \ \to \ 0 \ \right)_{ j \in d _ i }$$

   with $\left( I _ { i , j } \subseteq R \right) _ { j \in d _ i }$ ideals.

2. We now show, given $i \in n ,$ that $\mathcal{V} \left( Ann \left( M _ i \right) \right) = \bigcup _ { j \in d _ i } \mathcal{V} \left( I _ { i , j } \right) .$ Applying recalled fact 2, it follows (inductively) that all $f \in Ann \left( M _ i \right)$ act as $0$ on all the $R / I _ { i , j } ,$ hence that $Ann \left( M _ i \right) \subseteq \bigcap_{ j \in d _ i } I _ { i , j } .$ Conversely, it’s clear for all $j \in d _ i$ that $I _ { i , j } M _ { i , j + 1 } \subseteq M _ { i , j } ,$ from which it follows (reverse inductively) that $\prod _ { j \in d _ i } I _ { i , j } \subseteq Ann \left( M _ i \right) .$ We conclude that $\prod _ { j \in d _ i } I _ { i , j } \subseteq Ann \left( M _ i \right) \subseteq \bigcap_{ j \in d _ i } I _ { i , j } ,$ whence the (sub)claim.

3. We now show, given $i \in n$ and prime $\mathfrak{p} \in Spec (R) ,$ that $M _ i \otimes R _ { \mathfrak{p} } \ \,≄\, \ 0$ iff there exists $j \in d _ i$ such that $\mathfrak{p} \in \mathcal{V} \left( I _ { i , j } \right) .$ Applying recalled fact 3, we have a $d _ i$-tuple of short exact sequences

   $$\left( \ 0 \ \to \ M_{ i , j } \otimes R _ { \mathfrak{p} } \ \overset{ \iota _ { i , j } \otimes 1 _ { R _ { \mathfrak{p} } } }{ \hookrightarrow } \ M_{ i , j + 1 } \otimes R _ { \mathfrak{p} } \ \overset{ \pi _ { i , j } \otimes 1 _ { R _ { \mathfrak{p} } } }{ \twoheadrightarrow } \ R / I_{ i , j } \otimes R _ { \mathfrak{p} } \ \to \ 0 \ \right)_{ j \in d _ i } .$$

   Applying recalled fact 1, it follows (inductively) that $M \ \,≄\, \ 0$ iff there exists $j \in d _ i$ such that $\left( R / I _ { i , j } \right) \otimes R _ { \mathfrak{p} } \ \,≄\, \ 0 .$ Applying recalled fact 6 this holds iff in turn there exists $j \in d _ i$ such that $\mathfrak{p} \in \mathcal{V} \left( I _ { i , j } \right) .$

4. We now show, given a prime $\mathfrak{p} \in Spec (R) ,$ that $\left( \underset{ i \in n }{ \bigotimes } M _ i \right) \otimes R _ { \mathfrak{p} } \ \,≄\, \ 0$ iff for all $i \in n$ there exists $j \in d _ i$ such that $\mathfrak{p} \in \mathcal{V} \left( I _ { i , j } \right) .$ Suppose first that there exists $i _ { nil } \in n$ such that there exists no $j \in d _ { i _ { nil } }$ such that $\mathfrak{p} \in \mathcal{V} \left( I _ { i _ { nil } , j } \right) .$ Applying recalled facts 3 and 6, we obtain a $d _ { i _ { nil } }$-tuple of isomorphisms

   $$\left( \ 0 \ \to \ \ M_{ i _ { nil } , j } \otimes R _ { \mathfrak{p} } \ \overset{ \iota _ { i _ { nil } , j } \otimes 1 _ { R _ { \mathfrak{p} } } }{ \hookrightarrow } \ M_{ i _ { nil } , j + 1 } \otimes R _ { \mathfrak{p} } \ \twoheadrightarrow \ 0 \ \to \ 0 \ \right) _ { j \in d _ { i _ { nil } } } ,$$

   from which it follows (inductively) that the composite

   $$0 \overset{ \iota _ { i _ { nil } , d _ { i _ { nil } } - 1 } \circ \dots \circ \iota _ { i _ { nil } , 0 } }{ \to } M _ { i _ { nil } } \otimes R _ { \mathfrak{p} }$$

   is an isomorphism. Applying recalled fact 5, we conclude by the above that $\left( \underset{ i \in n }{ \bigotimes } M _ i \right) \otimes R _ { \mathfrak{p} } \simeq 0 .$ Suppose now instead that for each $i \in n$ there exists maximal $j _ { i , sup } \in d _ i$ such that $\mathfrak{p} \in \mathcal{V} \left( I _ { i , j _ { i , sup } } \right) .$ Applying recalled facts 3, 4, 6, and 7, we obtain an epimorphism

   $$\left( \underset{ i \in n }{ \bigotimes } M _ { i , j _ { i , sup } } \right) \otimes R _ { \mathfrak{p} } \ \overset{ \left( \underset{ i \in n }{ \bigotimes } \pi _ { i , j _ { i , sup } } \right) \otimes 1 _ { R _ { \mathfrak{p} } } }{ \twoheadrightarrow } \ \left( \underset{ i \in n }{ \bigotimes } \left( R / I _ { i , j _ { i , sup } } \right) \right) \otimes R _ { \mathfrak{p} } \ \ \,≄\, \ \ 0 ,$$

   hence that $\left( \underset{ i \in n }{ \bigotimes } M _ { i , j _ { i , sup } } \right) \otimes R _ { \mathfrak{p} } \ \,≄\, \ 0 .$ Applying recalled facts 3 and 6 (and the maximality of $j _ { i , sup }$), we obtain isomorphisms

   $$\left( \ 0 \ \to \ \ M_{ i , j } \otimes R _ { \mathfrak{p} } \ \overset{ \iota _ { i , j } \otimes 1 _ { R _ { \mathfrak{p} } } }{ \hookrightarrow } \ M_{ i , j + 1 } \otimes R _ { \mathfrak{p} } \ \twoheadrightarrow \ 0 \ \to \ 0 \ \right) _ { \substack{ i \in n \\ j \in \left\{ j _ { i , sup } + 1 , \dots , d _ { i } - 1 \right\} } } ,$$

   from which it follows (inductively) that the tensored composite

   $$\underset{ i \in n }{ \bigotimes } \left( M _ { i , j _ { i , sup } } \otimes R _ { \mathfrak{p} } \right) \overset{ \underset{ i \in n }{ \bigotimes } \left( \left( \iota _ { i , d _ i - 1 } \otimes 1 _ { R _ { \mathfrak{p} } } \right) \circ \dots \circ \left( \iota _ { i , j _ { i , sup } + 1 } \otimes 1 _ { R _ { \mathfrak{p} } } \right) \right) }{ \to } \underset{ i \in n }{ \bigotimes } \left( M _ i \otimes R _ { \mathfrak{p} } \right)$$

   is an isomorphism. Applying recalled fact 5 (twice), we conclude by the above that $\left( \underset{ i \in n }{ \bigotimes } M _ i \right) \otimes R _ { \mathfrak{p} } \ \,≄\, \ 0 .$

Claims 1 and 2 of Prop. are immediate from subarguments 2 and 3 and 2 and 4 above respectively.

Prop. immediately gives the following corollary, which we include for general interest:

Corollary

(Bounds on the annihilator of a tensor product of finitely generated modules) Given commutative ring $R ,$ natural $n ,$ and $n$-tuple of finitely generated $R$-modules $\left( M _ i \right) _ { i \in n } ,$

$$\sum _ { i \in n } Ann \left( M _ i \right) \ \subseteq \ Ann \left( \underset{ i \in n }{ \bigotimes } M _ i \right) \ \subseteq \ rad \left( \sum _ { i \in n } Ann \left( M _ i \right) \right) .$$

(Here $Ann$ denotes the annihilator and $rad$ denotes the radical. For convenience, we assume the Von Neumann convention so that, e.g., the quantification “$i \in n$” is equivalent to “$i \in \left\{ 0 , \dots , n-1 \right\}$”.)

Proof

The left inclusion follows from that each $Ann \left( M _ i \right)$ (manifestly) annihilates $\underset{ i \in n }{ \bigotimes } M _ i .$ The right inclusion follows from that by claims 1 and 2 of Prop.

$$\begin{aligned} \mathcal{V} \left( Ann \left( \underset{ i \in n }{ \bigotimes } M _ i \right) \right) & = supp \left( \underset{ i \in n }{ \bigotimes } M _ i \right) \\ & = \bigcap _ { i \in n } supp \left( M _ i \right) \\ & = \bigcap _ { i \in n } \mathcal{V} \left( Ann \left( M _ i \right) \right) \\ & = \mathcal{V} \left( \sum _ { i \in n } Ann \left( M _ i \right) \right) , \end{aligned}$$

hence $rad \left( Ann \left( \underset{ i \in n }{ \bigotimes } M _ i \right) \right) = rad \left( \sum _ { i \in n } Ann \left( M _ i \right) \right) .$

We now deduce more familiar forms of Nakayama’s lemma can as corollaries of the above. Sometimes, as for instance in the Stacks Project, wherein all 11(!) other formulations of the lemma are straightforwardly reduced to the following, Nakayama’s lemma is stated instead as:

Corollary

(Nakayama, version 1) Given commutative ring $R ,$ finitely generated $R$-module $M ,$ and ideal $I \subseteq R$ such that $I M = M ,$ there exists element $f \in R$ such that $f = 1 \ (mod \ I)$ and $f M = 0 .$

Proof

From that $I M = M$ it follows that $M \otimes \left( R / I \right) \simeq 0 ,$ i.e. that $supp \left( M \otimes \left( R / I \right) \right) = \emptyset ,$ hence by claims 1 and 2 of Prop. that $\mathcal{V} \left( Ann \left( M \right) \right)$ is disjoint from $\mathcal{V} \left( I \right) ,$ i.e. that $Ann \left( M \right)$ and $I$ are comaximal. The Chinese Remainder theorem then constructs the desired $f \in R$ such that $f = 1\ (mod\ I)$ and $f = 0\ \left( mod\ Ann \left( M \right) \right) .$

Alternatively, Nakayama’s lemma is often understood as a means by which information about the fiber of a module at a point can be used to characterize that of its stalk at said point:

Corollary

(Nakayama, version 2) Given local ring $R$ with maximal ideal $\mathfrak{m} \subseteq R$ and finitely generated $R$-module $M ,$ then $M \ \,≄\, \ 0$ iff $M \otimes \left( R / \mathfrak{m} \right) \ \,≄\, \ 0 .$

Proof

Observe that $\mathfrak{m} \in supp \left( R / \mathfrak{m} \right) ,$ hence by claim 2 of Prop. that $\mathfrak{m} \in supp (M)$ iff $\mathfrak{m} \in supp \left( M \otimes \left( R / \mathfrak{m} \right) \right) .$ Conclude by that an $R$-module is (essentially tautologically) $\ \,≄\, \ 0$ iff $\mathfrak{m}$ is in its support.

Examples and consequences

The finiteness hypotheses in the statements of Prop. and Cor. are necessary:

Example

Viewing $\mathbb{Q} / \mathbb{Z}$ as a $\mathbb{Z}$-module,

• $supp \left( \mathbb{Q} / \mathbb{Z} \right) \subsetneq \mathcal{V} \left( Ann \left( \mathbb{Q} / \mathbb{Z} \right) \right) ,$

• $supp \left( \left( \mathbb{Q} / \mathbb{Z} \right) \otimes \left( \mathbb{Q} / \mathbb{Z} \right) \right) \subsetneq supp \left( \mathbb{Q} / \mathbb{Z} \right) \cap supp \left( \mathbb{Q} / \mathbb{Z} \right) ,$

• $Ann \left( \left( \mathbb{Q} / \mathbb{Z} \right) \otimes \left( \mathbb{Q} / \mathbb{Z} \right) \right) \supsetneq rad \left( Ann \left( \mathbb{Q} / \mathbb{Z} \right) + Ann \left( \mathbb{Q} / \mathbb{Z} \right) \right) ,$

contradicting claims 1 and 2 of Prop. and Cor. respectively sans the finiteness hypotheses.

Likewise, the right inclusion in Cor. can be tight even when the left one isn’t:

Example

Let $R = \mathbb{Z} \left[ x _ 0, x _ 1 \right] ,$ $M _ 0 = R / ( x _ 0 ) \oplus R / ( x _ 1 ) ,$ and $M _ 1 = R / ( x _ 0 - x _ 1 ) .$ Then, viewing $M _ 0$ and $M _ 1$ as $R$-modules,

• $Ann \left( M _ 0 \right) = ( x _ 0 x_ 1 ) .$

• $Ann \left( M _ 1 \right) = ( x _ 0 - x _ 1 ) .$

• $Ann \left( M _ 0 \right) + Ann \left( M _ 1 \right) = ( x _ 0 - x _ 1 ) + ( x _ 0 , x_ 1 ) ^ 2 .$

• $rad \left( Ann \left( M _ 0 \right) + Ann \left( M _ 1 \right) \right) = ( x _ 0 , x_ 1 ) .$

• $M _ 0 \otimes M _ 1 \simeq R / ( x _ 0 , x _ 1 ) \oplus R / ( x _ 0 , x _ 1 ) .$

• $Ann \left( M _ 0 \otimes M _ 1 \right) = ( x _ 0 , x_ 1 ) .$

In particular,

$$Ann \left( M _ 0 \right) + Ann \left( M _ 1 \right) \subsetneq Ann \left( M _ 0 \otimes M _ 1 \right) = rad \left( Ann \left( M _ 0 \right) + Ann \left( M _ 1 \right) \right) ,$$

exemplifying the claimed phenomenon.

Cor. has the following two notable consequences:

Example

Suppose $R$ is a local ring with maximal ideal $\mathfrak{m} \subseteq R$ and $f \colon N \to M$ is an $R$-module map. The latter gives rise to an exact sequence

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

Tensoring with $R / \mathfrak{m}$ is a right exact functor, so we have an exact sequence

$$N \otimes (R / \mathfrak{m}) \stackrel{f \otimes 1 _ { R / \mathfrak{m} }}{\to} M \otimes (R / \mathfrak{m}) \stackrel{p \otimes 1 _ { R / \mathfrak{m} }}{\to} (M / N) \otimes (R / \mathfrak{m}) \to 0 .$$

Cor. says that $M / N \simeq 0$ if(f) $(M / N) \otimes (R / \mathfrak{m}) \simeq 0 .$ Equivalently, $f$ is epic if(f) $f \otimes 1 _ { R / \mathfrak{m} }$ is epic. In particular, to check whether a finite set of elements $v_{1}, \ldots, v_{n}$ generates $M ,$ it suffices to check whether the residue classes $v_i \mod \mathfrak{m} M$ generate the vector space $M / \mathfrak{m} M ,$ which is a linear algebra calculation (as $R / \mathfrak{m}$ is a field).

Example

Suppose $R$ is a Noetherian local ring with maximal ideal $\mathfrak{m} \subseteq R .$ A typical example is the stalk at a point $p$ of a Noetherian scheme as locally ringed space, and we will write as if we were in that situation. As $R$ is Noetherian, $\mathfrak{m}$ is finitely generated. Suppose $\mathfrak{m} \otimes (R / \mathfrak{m}) \cong \mathfrak{m} / \mathfrak{m}^2$ – the cotangent space – is a vector space of dimension $n .$ We would like to know whether a collection of functions $f_1, \ldots, f_n$ that vanish at $\mathfrak{m}$ form a local coordinate system.

For this, it suffices to check whether the differentials $d f_1, \ldots, d f_n$ at $\mathfrak{m} ,$ belonging to the cotangent space $\mathfrak{m}/\mathfrak{m}^2 ,$ are linearly independent. (For then they span the cotangent space, and one concludes from Nakayama that the $f_i$ generate $\mathfrak{m}$ as an $R$-module, thereby forming a local coordinate system at $\mathfrak{m} .$) 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 $n$ which is a bijection and has injective derivative at every point is an isomorphism.

References

• Stacks Project: 07RC

• Wikipedia: Nakayama lemma

category: algebra