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.