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

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.”

Lemma 10.19.1 (Nakayama's lemma). Let $R$ be a ring with Jacobson radical $\text{rad}(R)$. Let $M$ be an $R$-module. 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 $fM \subset N$ and $M_ f = N_ f$.

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 10.14.5 (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 10.18.1 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. $\square$

Comment #2793 by Kong Bochao on

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

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 $M$.

Comment #3279 by Nicolas on

In the proofs of (3) and (4), I don't see how we can deduce that M/N is finitely generated: after all, N' being finitely generated doesn't imply that IN' is.

Comment #3280 by Nicolas on

Kong is right, claim (4) is definitely false: In his counterexample, you can take $N' = I$ with $I$ satisfying $I^2 = I$ (take for instance $R$ = the valuation ring and $I$ = the valuation ideal of a nondiscrete valued field).

Comment #3281 by Nicolas on

Never mind, now I see it: simply use that $M = N+N'$. My bad. Also, the counterexample is none since $N' = I$ is not f.g.

Comment #3431 by ym on

perhaps add a version with they hypothesis of 3 but with the conclusion $fM \subseteq N$. This is the form used in for example Tag 07CF

Comment #3490 by on

@#3431. What I have done instead is added the conclusion that $fM \subset N$ to part (3). This is actually what the proof shows and the conclusion that $N_f = M_f$ follows immediately from the fact that $N_f \subset M_f$ plus $fM \subset N \Rightarrow (M/N)_f = 0$. See chang here.

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