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.

About Charles Siegel

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
This entry was posted in AG From the Beginning, Algebraic Geometry, Big Theorems. Bookmark the permalink.

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

  11. Pingback: Topologies and the Artin-Rees lemma « Delta Epsilons

  12. 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’.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s