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.

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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.
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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).

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  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’.

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