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\begin{document}

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\section*{Nakayama&#39;s lemma}

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{statement_and_consequences}{Statement and consequences}\dotfill \pageref*{statement_and_consequences} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{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.

\hypertarget{statement_and_consequences}{}\subsection*{{Statement and consequences}}\label{statement_and_consequences}

Nakayama's lemma is frequently stated in a general but slightly unilluminating form. We begin with an easier and more intuitive form. In this article, all rings are assumed to be commutative.

\begin{prop}
\label{}\hypertarget{}{}
Let $R$ be a [[local ring]], with maximal [[ideal]] $\mathfrak{m}$ and residue [[field]] $k = R/\mathfrak{m}$. Let $M$ be a finitely generated $R$-[[module]]. Then $M \cong 0$ if and only if $k \otimes_R M \cong 0$.

\end{prop}
Here is a sample application. Suppose $f \colon N \to M$ is an $R$-module map, giving rise to an exact sequence

\begin{displaymath}
N \stackrel{f}{\to} M \stackrel{p}{\to} M/N \to 0.
\end{displaymath}
Tensoring with $k$ is a right exact functor, so we have an exact sequence

\begin{displaymath}
k \otimes_R N \stackrel{k \otimes_R f}{\to} k \otimes_R M \to k \otimes_R M/N \to 0.
\end{displaymath}
Nakayama's lemma says that if $k \otimes_R M/N \cong 0$, then $M/N \cong 0$. Equivalently, that if $k \otimes_R f$ is epic, then $f$ 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.

\begin{remark}
\label{}\hypertarget{}{}
Suppose $O$ is a [[Noetherian ring|Noetherian]] local ring. 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. Being Noetherian, its maximal ideal $\mathfrak{m}$ is finitely generated. Suppose $k \otimes_O \mathfrak{m} \cong \mathfrak{m}/\mathfrak{m}^2$ -- 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 $p$ form a local coordinate system.

For this, it suffices to check whether the differentials $d f_1, \ldots, d f_n$ at $p$, 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 $O$-module, thereby forming a local coordinate system at $p$.) In this way, Nakayama's lemma operates as a kind of ``inverse function theorem''.

\end{remark}
To cement this further, the following statement is offered in \hyperlink{Harris}{Harris} as a corollary of Nakayama's lemma (corollary 14.10, page 179):

\begin{prop}
\label{}\hypertarget{}{}
\textbf{(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.

\end{prop}
We turn now to a general statement of Nakayama's lemma.

(To be continued)

[[!redirects Nakayama lemma]]



\end{document}