## Nakayama’s Lemma

I promised a minipost on Nakayama when I talked about Flattening Stratifications, and I’ve got a moment now, so I’ll do it quickly. This post is all commutative algebra. So we’ll quickly state Nakyama:

Nakayama’s Lemma: Let $I$ be an ideal contained in the Jacobson radical of a ring $R$ and let $M$ be a finitely generated $R$-module. Then if $IM=M$ we have $M=0$ and if $m_1,\ldots,m_n\in M$ have images in $M/IM$ which generate it as an $R$-module, then $m_1,\ldots,m_n$ generate $M$ as an $R$-module.

First up, the Jacobson radical is just the intersection of all the maximal ideals of a ring. So, for instance, in $k[x]$, the Jacobson radical is zero. Also, this says that we really want to work over a local ring, so for geometry, we want to work at a point and use Nakayama there.

We’ll prove it as a corollary of the following:

Cayley-Hamilton Theorem: Let $M$ be a finitely generated $R$-module, and let $I$ be an ideal of $R$. Let $\phi$ be an endomorphism with $\phi(M)\subset IM$. Then $\phi$ satisfies an equation of the form $\phi^n+a_1\phi^{n-1}+\ldots+a_n=0$ where the $a_i$ are in $I$.

Proof: Let $x_1,\ldots,x_n$ be generators of $M$. Then each $\phi(x_i)\in IM$, so we have $\phi(x_i)=\sum_{j=1}^n a_{ij}x_j$ for $1\leq i\leq n$ and $a_{ij}\in I$. That is, we have $\sum_{j=1}^n (\delta_{ij}\phi-a_{ij})x_i=0$. By multiplying on the left by the adjoint matrix of $\delta_{ij}\phi-a_{ij}$, we get that the determinant of $\delta_{ij}\phi-a_{ij}$ annihilates each $x_i$, and so is zero. Expanding the determinant, we get the desired equation. QED

The first corollary of Cayley-Hamilton is that if $M$ is a finitely generated $R$-module and $I$ is an ideal with $IM=M$, we have $x\equiv 1\mod I$ such that $xM=0$, by taking $\phi$ to be the identity and $x=1+a_1+\ldots+a_n$.

Now we prove Nakayama.

1. We apply the corollary above to get $r\in I$ with $(1-r)M=0$. Since $r$ is in every maximal ideal, $1-r$ is in none of them, so it’s a unit. Thus $M=0$.
2. Now let $N=M/(\sum_i Rm_i)$. We have $N/IN=M/(IM+\sum_iRm_i)=M/M=0$. So $IN=N$. Now the first part says that $N=0$, so $M=\sum_i Rm_i$, and so $M$ is generated by the $m_i$.

That’s it. Nakayama has a nice short proof. Now, here’s a corollary. We define the annihilator of a modules to be $\{r\in R|rM=0\}=\mathrm{ann}(M)$. Now if $M$ and $N$ are two finitely generated modules over $R$, and $M\otimes_R N=0$ then $\mathrm{ann}(M)+\mathrm{ann}(N)=R$. Now, if $R$ is local, this reduces to $M$ or $N$ being zero in the first place.

First, we reduce to the case of a local ring, because if the sum of the annihilators wasn’t $R$, we localize at a prime containing both annihilators, and apply the local result to get a contradiction. Now, we assume that $M$ is nonzero and $P$ is the unique maximal ideal of $R$. Nakayama says that $M/PM\neq 0$ Since this is a $R/P$ vector space, it projects to $R/P$, so there is a surjection $M\to R/P$. Thus $0=M\otimes N$ surjects to $R/P\otimes N=N/PN$. By Nakayama, we have $N=0$.

So this should hint that Nakayama is helpful in general, and especially when dealing with tensor products and flatness, as we have recently. It’s a great tool, and we’ll probably be using it in the future a bit as well.

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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### 13 Responses to Nakayama’s Lemma

1. Mgnbar says:

what is the geometric significance of this result?

2. Charles says:

Of Nakayama’s Lemma? I don’t know a direct geometric analogue of it, but it’s used to prove all sorts of algebraic statements which are more geometric in nature (like the Lying Over theorem)

3. Greg Stevenson says:

Like Charles said it is a good tool, but I can’t really think of a direct geometric interpretation either. But you can get a sheafy version of it which is a bit more geometric (I think so anyway).

If X is a noetherian scheme, Z is a closed subscheme of X and F is a coherent sheaf on X then its sheafy version tells you that if the pullback of F to Z is locally free of rank r then there are open neighbourhoods U of each point of Z in X where F|U is generated by r sections.

4. Mgnbar says:

thanks for the reply, would it be possible to explain how you get this sheafy version,

how do you get a general subscheme, (the ideal in the above lemma has to be in the Jacobson radical)?

where does the locally free condition come in?

5. Greg Stevenson says:

I’ll just outline the proof for you since that should answer both of your other questions (hopefully my attempt at using latex here works).

Since we have a coherent sheaf F on a noetherian scheme X we can extend germs of sections at the stalks of F to open neighbourhoods. Thus it is sufficient to show that $F_{j(z)}$ can be generated by r elements as an $\mathcal{O}_{X,j(z)}$-module for each $z\in Z$, where j is the inclusion of Z and $j^*F$ is locally free of rank r. So let $\mathscr{I}$ be the coherent sheaf of ideals corresponding to Z and observe that $\mathscr{I}_{j(z)}$ is a proper ideal of the local ring $\mathcal{O}_{X,j(z)}$ for each z (hence in the Jacobson radical). Then by assumption
$(j^*F)_z \cong F_{j(z)} \otimes \mathcal{O}_{Z,z} \cong F_{j(z)} / \mathscr{I}_{j(z)}F_{j(z)}$
is free of rank r over $\mathcal{O}_{Z,z}$. The result now follows by Nakayama’s lemma.

6. Mgnbar says:

Thanks Greg, i wasn’t thinking of it on the local ring, this blog is a good place to ask questions, keep up the good work

7. Charles says:

I’d also like to thank Greg. I hadn’t seen the sheafy version of Nakayama before, and I’m certain that it’s going to turn out to be useful to know.

8. Anonymous says:

Here is a restatement of the lemma which doesn’t need noetherianness: Let $F$ be a coherent sheaf on an arbitrary scheme $X$ and let $x$ be an arbitrary point of $X$. Then the fibre of $F$ at $x$, ie, $F(x) = F_x/m_xF_x where$F_x$denotes the stalk of$F$at$x$, is zero if and only if$F_x = 0$or, equivalently, if$F \mid_U = 0$for some open neighbourhood$U$of$x\$. (The converse direction is automatic for any sheaf, it’s the forward direction that is the content of the statement.)

9. Anonymous says:

One can also prove Nakayama using induction. Let M be generated by a_1, … a_n. Suppose IM=M. Then a_1 = i_1 a_1 + … i_n a_n where the i_j are in I. This means (1-i_1)a_1 is a linear combination of the a_2, … a_n, so a_1 isn’t necessary as a generator (1-i_1 is a unit).

Nakayama’s lemma also is useful in algebraic number theory, where the “going-up” theorem is of course important (although you only need it in Noetherian domains).

10. Landau says:

Some typos: ‘generate’ should be ‘generated’ in the first statement of Nakayama. In the statement of Cayley-Hamilton Theorem, A should be R. And in its proof, the first sum should be over j instead of i. In the proof of Nakayama Latex seems to have trouble. And in the second to last paragraph in the end, you say ‘Nayamana’.