Nakayama's lemma

In the special case of a finitely generated module ${\displaystyle M}$ over a local ring ${\displaystyle R}$ with maximal ideal ${\displaystyle {\mathfrak {m}}}$, the quotient ${\displaystyle M/{\mathfrak {m}}M}$ is a vector space over the field ${\displaystyle R/{\mathfrak {m}}}$. Statement 4 then implies that a basis of ${\displaystyle M/{\mathfrak {m}}M}$ lifts to a minimal set of generators of ${\displaystyle M}$. Conversely, every minimal set of generators of ${\displaystyle M}$ 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 ${\displaystyle {\mathcal {M}}}$ be a coherent sheaf of ${\displaystyle {\mathcal {O}}_{X}}$-modules over an arbitrary scheme ${\displaystyle X}$. The stalk of ${\displaystyle {\mathcal {M}}}$ at a point ${\displaystyle p\in X}$, denoted by ${\displaystyle {\mathcal {M}}_{p}}$, is a module over the local ring ${\displaystyle ({\mathcal {O}}_{X,p},{\displaystyle {\mathfrak {m}}_{p}})}$ and the fiber of ${\displaystyle {\mathcal {M}}}$ at ${\displaystyle p}$ is the vector space ${\displaystyle {\mathcal {M}}(p)={\mathcal {M}}_{p}/{\mathfrak {m}}_{p}{\mathcal {M}}_{p}}$. Nakayama's lemma implies that a basis of the fiber ${\displaystyle {\mathcal {M}}(p)}$ lifts to a minimal set of generators of ${\displaystyle {\mathcal {M}}_{p}}$. That is:

• Any basis of the fiber of a coherent sheaf ${\displaystyle {\mathcal {M}}}$ at a point comes from a minimal basis of local sections.

Reformulating this geometrically, if ${\displaystyle {\mathcal {M}}}$ is a locally free ${\displaystyle {\mathcal {O}}_{X}}$-module representing a vector bundle ${\displaystyle E\to X}$, and if we take a basis of the vector bundle at a point in the scheme ${\displaystyle X}$, this basis can be lifted to a basis of sections of the vector bundle in some neighborhood of the point. We can organize this data diagrammatically

${\displaystyle {\begin{matrix}E|_{p}&\to &E|_{U}&\to &E\\\downarrow &&\downarrow &&\downarrow \\p&\to &U&\to &X\end{matrix}}}$

where ${\displaystyle E|_{p}}$ is an n-dimensional vector space, to say a basis in ${\displaystyle E|_{p}}$ (which is a basis of sections of the bundle ${\displaystyle E_{p}\to p}$) can be lifted to a basis of sections ${\displaystyle E|_{U}\to U}$ for some neighborhood ${\displaystyle U}$ of ${\displaystyle p}$.

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

• Let ${\displaystyle R\hookrightarrow S}$ be an integral extension of commutative rings, and ${\displaystyle {\mathfrak {p}}}$ a prime ideal of ${\displaystyle R}$. Then there is a prime ideal ${\displaystyle {\mathfrak {q}}}$ in ${\displaystyle S}$ such that ${\displaystyle {\mathfrak {q}}\cap R={\mathfrak {p}}}$. Moreover, ${\displaystyle {\mathfrak {q}}}$ can be chosen to contain any prime ${\displaystyle {\mathfrak {q}}_{1}}$ of ${\displaystyle S}$ such that ${\displaystyle {\mathfrak {q}}_{1}\cap R\subset {\mathfrak {p}}}$.

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 ${\displaystyle M}$ is a finitely generated ${\displaystyle R}$-module and ${\displaystyle f:M\to M}$ is a surjective endomorphism, then ${\displaystyle f}$ is an isomorphism.[8]

Over a local ring, one can say more about module epimorphisms:[9]

• Suppose that ${\displaystyle R}$ is a local ring with maximal ideal ${\displaystyle {\mathfrak {m}}}$, and ${\displaystyle M,N}$ are finitely generated ${\displaystyle R}$-modules. If ${\displaystyle \phi :M\to N}$ is an ${\displaystyle R}$-linear map such that the quotient ${\displaystyle \phi _{\mathfrak {m}}:M/{\mathfrak {m}}M\to N/{\mathfrak {m}}N}$ is surjective, then ${\displaystyle \phi }$ is surjective.

Nakayama's lemma also has several versions in homological algebra. The above statement about epimorphisms can be used to show:[9]

• Let ${\displaystyle M}$ be a finitely generated module over a local ring. Then ${\displaystyle M}$ is projective if and only if it is free. This can be used to compute the Grothendieck group of any local ring ${\displaystyle R}$ as ${\displaystyle K(R)=\mathbb {Z} }$.

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

More generally, one has[10]

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

${\displaystyle \operatorname {Tor} _{i+1}^{R}(k,M)=0.}$

Here ${\displaystyle k}$ is the residue field of ${\displaystyle R}$ and ${\displaystyle {\text{Tor}}}$ is the tor functor.

Nakayama's lemma is used to prove a version of the inverse function theorem in algebraic geometry:

• Let ${\textstyle f:X\to Y}$ be a projective morphism between quasi-projective varieties. Then ${\textstyle f}$ is an isomorphism if and only if it is a bijection and the differential ${\textstyle df_{p}}$ is injective for all ${\displaystyle p\in X}$.[11]

Let ${\displaystyle X}$ be a finite subset of ${\displaystyle U}$, minimal with respect to the property that it generates ${\displaystyle U}$. Since ${\displaystyle U}$ is non-zero, this set ${\displaystyle X}$ is nonempty. Denote every element of ${\displaystyle X}$ by ${\displaystyle x_{i}}$ for ${\displaystyle i\in \{1,\ldots ,n\}}$. Since ${\displaystyle X}$ generates ${\displaystyle U}$,${\displaystyle \sum _{i=1}^{n}x_{i}R=U}$.

Suppose ${\displaystyle U\cdot \operatorname {J} (R)=U}$, to obtain a contradiction. Then every element ${\displaystyle u\in U}$can be expressed as a finite combination ${\displaystyle u=\sum \limits _{s=1}^{m}u_{s}j_{s}}$ for some ${\displaystyle m\in \mathbb {N} ,\,u_{s}\in U,\,j_{s}\in \operatorname {J} (R),\,s=1,\dots ,m}$.

Each ${\displaystyle u_{s}}$ can be further decomposed as ${\displaystyle u_{s}=\sum \limits _{i=1}^{n}x_{i}r_{i,s}}$ for some ${\displaystyle r_{i,s}\in R}$. Therefore, we have

${\displaystyle u=\sum _{s=1}^{m}\left(\sum _{i=1}^{n}x_{i}r_{i,s}\right)j_{s}=\sum \limits _{i=1}^{n}x_{i}\left(\sum _{s=1}^{m}r_{i,s}j_{s}\right)}$.

Since ${\displaystyle \operatorname {J} (R)}$ is a (two-sided) ideal in ${\displaystyle R}$, we have ${\displaystyle \sum _{s=1}^{m}r_{i,s}j_{s}\in \operatorname {J} (R)}$ for every ${\displaystyle i\in \{1,\dots ,n\}}$, and thus this becomes

${\displaystyle u=\sum _{i=1}^{n}x_{i}k_{i}}$ for some ${\displaystyle k_{i}\in \operatorname {J} (R)}$, ${\displaystyle i=1,\dots ,n}$.

Putting ${\displaystyle u=\sum _{i=1}^{n}x_{i}}$ and applying distributivity, we obtain

${\displaystyle \sum _{i=1}^{n}x_{i}(1-k_{i})=0}$.

Choose some ${\displaystyle j\in \{1,\dots ,n\}}$. If the right ideal ${\displaystyle (1-k_{j})R}$ were proper, then it would be contained in a maximal right ideal ${\displaystyle M\neq R}$ and both ${\displaystyle 1-k_{j}}$ and ${\displaystyle k_{j}}$ would belong to ${\displaystyle M}$, leading to a contradiction (note that ${\displaystyle \operatorname {J} (R)\subseteq M}$ by the definition of the Jacobson radical). Thus ${\displaystyle (1-k_{j})R=R}$ and ${\displaystyle 1-k_{j}}$ has a right inverse ${\displaystyle (1-k_{j})^{-1}}$ in ${\displaystyle R}$. We have

${\displaystyle \sum _{i=1}^{n}x_{i}(1-k_{i})(1-k_{j})^{-1}=0}$.

Therefore,

${\displaystyle \sum _{i\neq j}x_{i}(1-k_{i})(1-k_{j})^{-1}=-x_{j}}$.

Thus ${\displaystyle x_{j}}$ is a linear combination of the elements from ${\displaystyle X\setminus \{x_{j}\}}$. This contradicts the minimality of ${\displaystyle X}$ and establishes the result.[18]