Artin–Rees lemma

In mathematics, the Artin–Rees lemma is a basic result about modules over a Noetherian ring, along with results such as the Hilbert basis theorem. It was proved in the 1950s in independent works by the mathematicians Emil Artin and David Rees; a special case was known to Oscar Zariski prior to their work.

An intuitive characterization of the lemma involves the notion that a submodule N of a module M over some ring A with specified ideal I holds a priori two topologies: one induced by the topology on M, and the other when considered with the I-adic topology over A. Then Artin-Rees dictates that these topologies actually coincide, at least when A is Noetherian and M finitely-generated.

One consequence of the lemma is the Krull intersection theorem. The result is also used to prove the exactness property of completion. The lemma also plays a key role in the study of ℓ-adic sheaves.

Statement

Let I be an ideal in a Noetherian ring R; let M be a finitely generated R-module and let N a submodule of M. Then there exists an integer k ≥ 1 so that, for n ≥ k,

$I^{n}M\cap N=I^{n-k}(I^{k}M\cap N).$

Proof

The lemma immediately follows from the fact that R is Noetherian once necessary notions and notations are set up.

For any ring R and an ideal I in R, we set $B_{I}R={⨁}_{n=0}^{\mathrm{\infty }}{I}^n$ (B for blow-up.) We say a decreasing sequence of submodules $M=M_0\supset M_1\supset M_2\supset \cdots$ is an I-filtration if $I{M}_n\subset M_{n+1}$; moreover, it is stable if $I{M}_n=M_{n+1}$ for sufficiently large n. If M is given an I-filtration, we set $B_{I}M={⨁}_{n=0}^{\mathrm{\infty }}{M}_n$; it is a graded module over $B_{I}R$.

Now, let M be a R-module with the I-filtration $M_i$ by finitely generated R-modules. We make an observation

$B_{I}M$ is a finitely generated module over $B_{I}R$ if and only if the filtration is I-stable.

Indeed, if the filtration is I-stable, then $B_{I}M$ is generated by the first $k+1$ terms $M_0,\dots ,M_k$ and those terms are finitely generated; thus, $B_{I}M$ is finitely generated. Conversely, if it is finitely generated, say, by some homogeneous elements in ${⨁}_{j=0}^k{M}_j$, then, for $n\ge k$, each f in $M_n$ can be written as $$f=\sum a_j{g}_j,a_j\in I^{n-j}$$ with the generators $g_j$ in $M_j,j\le k$. That is, $f\in I^{n-k}{M}_k$.

We can now prove the lemma, assuming R is Noetherian. Let $M_n=I^{n}M$. Then $M_n$ are an I-stable filtration. Thus, by the observation, $B_{I}M$ is finitely generated over $B_{I}R$. But $B_{I}R\simeq R[It]$ is a Noetherian ring since R is. (The ring $R[It]$ is called the Rees algebra.) Thus, $B_{I}M$ is a Noetherian module and any submodule is finitely generated over $B_{I}R$; in particular, $B_{I}N$ is finitely generated when N is given the induced filtration; i.e., $N_n=M_n\cap N$. Then the induced filtration is I-stable again by the observation.

Krull's intersection theorem

Besides the use in completion of a ring, a typical application of the lemma is the proof of the Krull's intersection theorem, which says: ${\bigcap }_{n=1}^{\mathrm{\infty }}{I}^n=0$ for a proper ideal I in a commutative Noetherian ring that is either a local ring or an integral domain. By the lemma applied to the intersection $N$, we find k such that for $n\ge k$, $$I^n\cap N=I^{n-k}(I^k\cap N).$$ Taking $n=k+1$, this means $I^{k+1}\cap N=I(I^k\cap N)$ or $N=IN$. Thus, if A is local, $N=0$ by Nakayama's lemma. If A is an integral domain, then one uses the determinant trick (that is a variant of the Cayley–Hamilton theorem and yields Nakayama's lemma):

In the setup here, take u to be the identity operator on N; that will yield a nonzero element x in A such that $xN=0$, which implies $N=0$, as $x$ is a nonzerodivisor.

For both a local ring and an integral domain, the "Noetherian" cannot be dropped from the assumption: for the local ring case, see local ring#Commutative case. For the integral domain case, take $A$ to be the ring of algebraic integers (i.e., the integral closure of $\mathbb{Z}$ in $ℂ$). If $𝔭$ is a prime ideal of A, then we have: ${𝔭}^n=𝔭$ for every integer $n>0$. Indeed, if $y\in 𝔭$, then $y={\alpha }^n$ for some complex number $\alpha$. Now, $\alpha$ is integral over $\mathbb{Z}$; thus in $A$ and then in $𝔭$, proving the claim.

References

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External links

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