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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