## 10.19 Nakayama's lemma

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. $\square$

Lemma 10.19.2. Let $A \to B$ be a local homomorphism of local rings. Assume

1. $B$ is finite as an $A$-module,

2. $\mathfrak m_ B$ is a finitely generated ideal,

3. $A \to B$ induces an isomorphism on residue fields, and

4. $\mathfrak m_ A/\mathfrak m_ A^2 \to \mathfrak m_ B/\mathfrak m_ B^2$ is surjective.

Then $A \to B$ is surjective.

Proof. To show that $A \to B$ is surjective, we view it as a map of $A$-modules and apply Lemma 10.19.1 (6). We conclude it suffices to show that $A/\mathfrak m_ A \to B/\mathfrak m_ AB$ is surjective. As $A/\mathfrak m_ A = B/\mathfrak m_ B$ it suffices to show that $\mathfrak m_ AB \to \mathfrak m_ B$ is surjective. View $\mathfrak m_ AB \to \mathfrak m_ B$ as a map of $B$-modules and apply Lemma 10.19.1 (6). We conclude it suffices to see that $\mathfrak m_ AB/\mathfrak m_ A\mathfrak m_ B \to \mathfrak m_ B/\mathfrak m_ B^2$ is surjective. This follows from assumption (4). $\square$

Comment #3850 by Lucy on

In the proof of Nakayama's Lemma, lemma 10.16.2 is referred, but it seems that lemma 10.18.1 is a better choice.

Comment #4236 by Aolong on

In the statement (7) and (8) of Nakayama lemma, it would be better to point out $x_1,\cdots,x_n$ generate $M/IM$ as an $R$-module, since $M/IM$ can be naturally $R/I$-module as well.

Comment #4415 by on

Whether they generate $M/IM$ as an $R$-module or as an $R/I$-module is the same thing.

Comment #4580 by Xavier on

There is a somewhat shorter proof of Lemma 10.19.1(1): applying Lemma 10.15.3 to the map $\mathrm{id}: M \rightarrow M$, we get a polynomial $p$ such that $p(\mathrm{1}) \in \mathrm{1} + I$ and $p(\mathrm{id}) = p(1) \cdot \mathrm{id}$ is the zero map on $M$, thus $f = p(1)$ annihilates $M$ as required.

Comment #4760 by on

@#4580. Yes, this is fine but it isn't really shorter as it uses Cayley-Hamilton.

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