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 be an ideal contained in the Jacobson radical of a ring and let be a finitely generated -module. Then if we have and if have images in which generate it as an -module, then generate as an -module.

First up, the Jacobson radical is just the intersection of all the maximal ideals of a ring. So, for instance, in , 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 be a finitely generated -module, and let be an ideal of . Let be an endomorphism with . Then satisfies an equation of the form where the are in .

*Proof:* Let be generators of . Then each , so we have for and . That is, we have . By multiplying on the left by the adjoint matrix of , we get that the determinant of annihilates each , and so is zero. Expanding the determinant, we get the desired equation. QED

The first corollary of Cayley-Hamilton is that if is a finitely generated -module and is an ideal with , we have such that , by taking to be the identity and .

Now we prove Nakayama.

- We apply the corollary above to get with . Since is in every maximal ideal, is in none of them, so it’s a unit. Thus .
- Now let . We have . So . Now the first part says that , so , and so is generated by the .

That’s it. Nakayama has a nice short proof. Now, here’s a corollary. We define the annihilator of a modules to be . Now if and are two finitely generated modules over , and then . Now, if is local, this reduces to or 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 , we localize at a prime containing both annihilators, and apply the local result to get a contradiction. Now, we assume that is nonzero and is the unique maximal ideal of . Nakayama says that Since this is a vector space, it projects to , so there is a surjection . Thus surjects to . By Nakayama, we have .

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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July 24, 2008 at 6:09 pm

what is the geometric significance of this result?

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)

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 can be generated by r elements as an -module for each , where j is the inclusion of Z and is locally free of rank r. So let be the coherent sheaf of ideals corresponding to Z and observe that is a proper ideal of the local ring for each z (hence in the Jacobson radical). Then by assumption

is free of rank r over . 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.

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).

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 [...]

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 [...]

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’.

May 6, 2011 at 10:09 am

Thanks, should all be fixed now.