# The Stacks Project

## Tag 00DV

Lemma 10.19.1 (Nakayama's lemma). Let $R$ be a ring, let $M$ be an $R$-module, and let $I \subset R$ be an ideal.

1. If $IM = M$ and $M$ is finite, then there exists a $f \in 1 + I$ such that $fM = 0$.
2. If $IM = M$, $M$ is finite, and $I \subset \text{rad}(R)$, then $M = 0$.
3. If $N, N' \subset M$, $M = N + IN'$, and $N'$ is finite, then there exists a $f \in 1 + I$ such that $M_f = N_f$.
4. If $N, N' \subset M$, $M = N + IN'$, $N'$ is finite, and $I \subset \text{rad}(R)$, then $M = N$.
5. If $N \to M$ is a module map, $N/IN \to M/IM$ is surjective, and $M$ is finite, then there exists a $f \in 1 + I$ such that $N_f \to M_f$ is surjective.
6. If $N \to M$ is a module map, $N/IN \to M/IM$ is surjective, $M$ is finite, and $I \subset \text{rad}(R)$, then $N \to M$ is surjective.
7. If $x_1, \ldots, x_n \in M$ generate $M/IM$ and $M$ is finite, then there exists an $f \in 1 + I$ such that $x_1, \ldots, x_n$ generate $M_f$ over $R_f$.
8. If $x_1, \ldots, x_n \in M$ generate $M/IM$, $M$ is finite, and $I \subset \text{rad}(R)$, then $M$ is generated by $x_1, \ldots, x_n$.
9. If $IM = M$, $I$ is nilpotent, then $M = 0$.
10. If $N, N' \subset M$, $M = N + IN'$, and $I$ is nilpotent then $M = N$.
11. If $N \to M$ is a module map, $I$ is nilpotent, and $N/IN \to M/IM$ is surjective, then $N \to M$ is surjective.
12. If $\{x_\alpha\}_{\alpha \in A}$ is a set of elements of $M$ which generate $M/IM$ and $I$ is nilpotent, then $M$ is generated by the $x_\alpha$.

Proof. Proof of (1). Choose generators $y_1, \ldots, y_m$ of $M$ over $R$. For each $i$ we can write $y_i = \sum z_{ij} y_j$ with $z_{ij} \in I$ (since $M = IM$). In other words $\sum_j (\delta_{ij} - z_{ij})y_j = 0$. Let $f$ be the determinant of the $m \times m$ matrix $A = (\delta_{ij} - z_{ij})$. Note that $f \in 1 + I$ (since the matrix $A$ is entrywise congruent to the $m \times m$ identity matrix modulo $I$). By Lemma 10.14.4 (1), there exists an $m \times m$ matrix $B$ such that $BA = f 1_{m \times m}$. Writing out we see that $\sum_{i} b_{hi} a_{ij} = f \delta_{hj}$ for all $h$ and $j$; hence, $\sum_{i, j} b_{hi} a_{ij} y_j = \sum_{j} f \delta_{hj} y_j = f y_h$ for every $h$. In other words, $0 = f y_h$ for every $h$ (since each $i$ satisfies $\sum_j a_{ij} y_j = 0$). This implies that $f$ annihilates $M$.

By Lemma 10.16.2 an element of $1 + \text{rad}(R)$ is invertible element of $R$. Hence we see that (1) implies (2). We obtain (3) by applying (1) to $M/N$ which is finite as $N'$ is finite. We obtain (4) by applying (2) to $M/N$ which is finite as $N'$ is finite. We obtain (5) by applying (3) to $M$ and the submodules $\mathop{\mathrm{Im}}(N \to M)$ and $M$. We obtain (6) by applying (4) to $M$ and the submodules $\mathop{\mathrm{Im}}(N \to M)$ and $M$. We obtain (7) by applying (5) to the map $R^{\oplus n} \to M$, $(a_1, \ldots, a_n) \mapsto a_1x_1 + \ldots + a_nx_n$. We obtain (8) by applying (6) to the map $R^{\oplus n} \to M$, $(a_1, \ldots, a_n) \mapsto a_1x_1 + \ldots + a_nx_n$.

Part (9) holds because if $M = IM$ then $M = I^nM$ for all $n \geq 0$ and $I$ being nilpotent means $I^n = 0$ for some $n \gg 0$. Parts (10), (11), and (12) follow from (9) by the arguments used above. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 3359–3401 (see updates for more information).

\begin{lemma}[Nakayama's lemma]
\label{lemma-NAK}
\begin{reference}
\cite[1.M Lemma (NAK) page 11]{MatCA}
\end{reference}
\begin{history}
We quote from \cite{MatCA}: This simple but
important lemma is due to T.~Nakayama, G.~Azumaya and W.~Krull. Priority
is obscure, and although it is usually called the Lemma of Nakayama, late
Prof.~Nakayama did not like the name.''
\end{history}
Let $R$ be a ring, let $M$ be an $R$-module, and let $I \subset R$
be an ideal.
\begin{enumerate}
\item
\label{item-nakayama}
If $IM = M$ and $M$ is finite, then there exists a $f \in 1 + I$ such that
$fM = 0$.
\item If $IM = M$, $M$ is finite, and $I \subset \text{rad}(R)$, then $M = 0$.
\item If $N, N' \subset M$, $M = N + IN'$, and $N'$ is finite,
then there exists a $f \in 1 + I$ such that $M_f = N_f$.
\item If $N, N' \subset M$, $M = N + IN'$, $N'$ is finite, and
$I \subset \text{rad}(R)$, then $M = N$.
\item If $N \to M$ is a module map, $N/IN \to M/IM$ is
surjective, and $M$ is finite, then there exists a $f \in 1 + I$
such that $N_f \to M_f$ is surjective.
\item If $N \to M$ is a module map, $N/IN \to M/IM$ is
surjective, $M$ is finite, and $I \subset \text{rad}(R)$,
then $N \to M$ is surjective.
\item If $x_1, \ldots, x_n \in M$ generate $M/IM$ and $M$ is finite,
then there exists an $f \in 1 + I$ such that $x_1, \ldots, x_n$
generate $M_f$ over $R_f$.
\item If $x_1, \ldots, x_n \in M$ generate $M/IM$, $M$ is finite, and
$I \subset \text{rad}(R)$, then $M$ is generated by $x_1, \ldots, x_n$.
\item If $IM = M$, $I$ is nilpotent, then $M = 0$.
\item If $N, N' \subset M$, $M = N + IN'$, and $I$ is nilpotent then $M = N$.
\item If $N \to M$ is a module map, $I$ is nilpotent, and $N/IN \to M/IM$
is surjective, then $N \to M$ is surjective.
\item If $\{x_\alpha\}_{\alpha \in A}$ is a set of elements of $M$
which generate $M/IM$ and $I$ is nilpotent, then $M$ is generated
by the $x_\alpha$.
\end{enumerate}
\end{lemma}

\begin{proof}
Proof of (\ref{item-nakayama}). Choose generators $y_1, \ldots, y_m$ of $M$
over $R$. For each $i$ we can write $y_i = \sum z_{ij} y_j$ with
$z_{ij} \in I$ (since $M = IM$).
In other words $\sum_j (\delta_{ij} - z_{ij})y_j = 0$.
Let $f$ be the determinant of the $m \times m$ matrix
$A = (\delta_{ij} - z_{ij})$. Note that $f \in 1 + I$
(since the matrix $A$ is entrywise congruent to the
$m \times m$ identity matrix modulo $I$).
By Lemma \ref{lemma-matrix-left-inverse} (1),
there exists an $m \times m$
matrix $B$ such that $BA = f 1_{m \times m}$. Writing out we see that
$\sum_{i} b_{hi} a_{ij} = f \delta_{hj}$ for all
$h$ and $j$; hence, $\sum_{i, j} b_{hi} a_{ij} y_j = \sum_{j} f \delta_{hj} y_j = f y_h$ for every $h$.
In other words, $0 = f y_h$ for every $h$ (since each
$i$ satisfies $\sum_j a_{ij} y_j = 0$).
This implies that $f$ annihilates $M$.

\medskip\noindent
By Lemma \ref{lemma-Zariski-topology} an element of $1 + \text{rad}(R)$ is
invertible element of $R$. Hence we see that (\ref{item-nakayama}) implies
(2). We obtain (3) by applying (1) to $M/N$ which is finite as $N'$ is finite.
We obtain (4) by applying (2) to $M/N$ which is finite as $N'$ is finite.
We obtain (5) by applying (3) to $M$ and the submodules $\Im(N \to M)$
and $M$. We obtain (6) by applying (4) to $M$ and the submodules
$\Im(N \to M)$ and $M$.
We obtain (7) by applying (5) to the map $R^{\oplus n} \to M$,
$(a_1, \ldots, a_n) \mapsto a_1x_1 + \ldots + a_nx_n$.
We obtain (8) by applying (6) to the map $R^{\oplus n} \to M$,
$(a_1, \ldots, a_n) \mapsto a_1x_1 + \ldots + a_nx_n$.

\medskip\noindent
Part (9) holds because if $M = IM$ then $M = I^nM$ for all $n \geq 0$
and $I$ being nilpotent means $I^n = 0$ for some $n \gg 0$. Parts
(10), (11), and (12) follow from (9) by the arguments used above.
\end{proof}

## References

[MatCA, 1.M Lemma (NAK) page 11]

## Historical remarks

We quote from [MatCA]: ''This simple but important lemma is due to T. Nakayama, G. Azumaya and W. Krull. Priority is obscure, and although it is usually called the Lemma of Nakayama, late Prof. Nakayama did not like the name.''

Comment #2793 by Kong Bochao on September 2, 2017 a 11:54 am UTC

I think M finite is needed, just consider the third version, let $M=N\bigoplus I$ then $M/N\backsimeq I$ unless we further assume $R$ noetherian, $M/N$ is not needed finitely generated.

Comment #2794 by Dario Weißmann on September 3, 2017 a 2:07 pm UTC

No, I think the result is correct as stated. Why your example does not work: what is your finite $N'\subset M$ s.t. $M=N+IN'$? It seems to me you would like it to be $R$, but $R$ is not a submodule of $M$.

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