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.

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 $M=0$ .
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.

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Mgnbar Says:

July 24, 2008 at 6:09 pm

what is the geometric significance of this result?

Charles Says:

July 24, 2008 at 8:26 pm

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)

Greg Stevenson Says:

July 24, 2008 at 10:06 pm

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.

July 25, 2008 at 11:16 pm

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?

July 26, 2008 at 8:20 am

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.

July 28, 2008 at 9:35 pm

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

July 28, 2008 at 9:51 pm

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.

Anonymous Says:

November 2, 2008 at 1:25 am

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.)

January 1, 2009 at 4:45 pm

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).

Why simple modules are often finite-dimensional II « Delta Epsilons Says:

July 22, 2009 at 8:25 pm

[...] proof uses the Cayley-Hamilton theorem, and is given here. It’s essentially equivalent to other versions of Nakayama’s [...]

Topologies and the Artin-Rees lemma « Delta Epsilons Says:

August 20, 2009 at 10:59 am

[...] at zero for is just itself, i.e. has the indiscrete -adic topology. This means just , so by Nakayama we have [...]

Landau Says:

April 13, 2011 at 5:11 pm

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’.

Charles Siegel Says:

May 6, 2011 at 10:09 am
Thanks, should all be fixed now.

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Nakayama’s Lemma

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