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 $B_I R = \bigoplus_{n=0}^\infty I^n$ (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 $B_I M = \bigoplus_{n=0}^\infty M_n$ ; 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 $\bigoplus_{j=0}^k M_j$, then, for $$n \ge k$$, each f in $$M_n$$ can be written as $$f = \sum a_{j} g_{j}, \quad a_{j} \in I^{n-j}$$ 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: $\bigcap_{n=1}^\infty I^n = 0$ 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 $$x N = 0$$, 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. For the integral domain case, take $$A$$ to be the ring of algebraic integers (i.e., the integral closure of $$\mathbb{Z}$$ in $$\mathbb{C}$$). If $$\mathfrak p$$ is a prime ideal of A, then we have: $$\mathfrak{p}^n = \mathfrak{p}$$ for every integer $$n > 0$$. Indeed, if $$y \in \mathfrak p$$, then $$y = \alpha^n$$ for some complex number $$\alpha$$. Now, $$\alpha$$ is integral over $$\mathbb{Z}$$; thus in $$A$$ and then in $$\mathfrak{p}$$, proving the claim.