Lemma 6.8.18312218F0503

Proof. We derive the intersection formula from the Artin–Rees lemma, after collecting the facts about Noetherian rings that we need. A commutative ring $A$ is Noetherian if all its ideals are finitely generated. Let $A$ be Noetherian.

(i) Every submodule $N$ of $A^ r$, $r\geq 1$, is finitely generated. For $r=1$ this is the definition. Let $r\geq 2$. The last coordinates of the elements of $N$ form an ideal of $A$; choose $v_1,\ldots ,v_ q\in N$ whose last coordinates generate it, and, by induction on $r$, finitely many generators of the submodule $N\cap (A^{r-1}\times \{ 0\} )$, which is a submodule of $A^{r-1}$. If $v\in N$, subtracting from $v$ a combination of the $v_ i$ with the same last coordinate leaves an element of $N\cap (A^{r-1}\times \{ 0\} )$; hence these finitely many elements generate $N$.

(ii) Every submodule $N$ of a finitely generated $A$-module $M$ is finitely generated. Indeed, let $\pi \colon A^ r\to M$ map the standard basis to generators of $M$. By (i), the submodule $\pi ^{-1}(N)$ of $A^ r$ is finitely generated, and the images of its generators generate $N=\pi (\pi ^{-1}(N))$.

(iii) The polynomial ring $A[t]$ is Noetherian (Hilbert's basis theorem). Let $I\subset A[t]$ be a nonzero ideal. The leading coefficients of the nonzero elements of $I$, together with $0$, form an ideal of $A$: if $f,g\in I$ have degrees $d\leq e$ and leading coefficients $a,b$ with $a+b\neq 0$, then $t^{e-d}f+g\in I$ has leading coefficient $a+b$, and if $c\in A$ with $ca\neq 0$, then $cf\in I$ has leading coefficient $ca$. Choose $p_1,\ldots ,p_ k\in I$ whose leading coefficients generate this ideal, and let $d_0$ be the largest of their degrees. If $f\in I$ has degree $d\geq d_0$, its leading coefficient is $\sum _ ia_ ic_ i$ for some $a_ i\in A$, where $c_ i$ is the leading coefficient of $p_ i$, and $f-\sum _ ia_ it^{d-\deg p_ i}p_ i$ is an element of $I$ of smaller degree. Repeating this, we see that $f$ differs from an element of $I\cap (A+At+\cdots +At^{d_0-1})$ by an element of the ideal generated by $p_1,\ldots ,p_ k$. The set $I\cap (A+At+\cdots +At^{d_0-1})$ is a submodule of the $A$-module $A+At+\cdots +At^{d_0-1}\cong A^{d_0}$, finitely generated by (i). Hence $I$ is generated by $p_1,\ldots ,p_ k$ and finitely many further elements.

(iv) If $I\subset A$ is an ideal, then $A/I$ is Noetherian, since the ideals of $A/I$ are the images of the ideals of $A$ containing $I$. Hence, by (iii) and induction on the number of variables, every quotient of a polynomial ring in finitely many variables over $A$ is Noetherian.

All this applies to $A=R$, which is Noetherian by proposition 9.2.4, since a holomorphic chart at $x$ identifies $R$ with $\mathcal O_ n$.

Next we prove the Artin–Rees lemma. If $\mathfrak m=0$, which happens only in dimension zero, the lemma is trivial; so let $\mathfrak m\neq 0$, and choose generators $\xi _1,\ldots ,\xi _ p$ of $\mathfrak m$. Consider the subring $\mathcal R=\bigoplus _{k\geq 0}\mathfrak m^ kt^ k$ of $R[t]$, consisting of the polynomials whose coefficient of $t^ k$ lies in $\mathfrak m^ k$, where $\mathfrak m^0=R$. Since $\mathfrak m^ k$ is generated by the monomials of degree $k$ in $\xi _1,\ldots ,\xi _ p$, the ring $\mathcal R$ is the image of the homomorphism $R[T_1,\ldots ,T_ p]\to R[t]$, $T_ i\mapsto \xi _ it$, and it is Noetherian by (iii) and (iv).

In the $\mathcal R$-module of polynomials in $t$ with coefficients in $F$, consider the submodule $\bigoplus _{k\geq 0}\mathfrak m^ kFt^ k$. It is generated over $\mathcal R$ by generators of $F$ placed in degree zero, and it contains the submodule $\mathcal E=\bigoplus _{k\geq 0}(E\cap \mathfrak m^ kF)t^ k$, because $\mathfrak m^ j(E\cap \mathfrak m^ kF)\subset E\cap \mathfrak m^{j+k}F$. By (ii), $\mathcal E$ is finitely generated. The homogeneous components of the elements of $\mathcal E$ again belong to $\mathcal E$, so the homogeneous components of finitely many generators generate $\mathcal E$; let these be $e_1t^{d_1},\ldots ,e_ lt^{d_ l}$ with $e_ i\in E\cap \mathfrak m^{d_ i}F$, and let $s=\max _ id_ i$. For $k\geq s$ this gives the Artin–Rees equality

Indeed, let $y\in E\cap \mathfrak m^ kF$, and write $yt^ k=\sum _ i\rho _ ie_ it^{d_ i}$ with $\rho _ i\in \mathcal R$. Comparing the coefficients of $t^ k$, we may replace each $\rho _ i$ by its term of degree $k-d_ i$, which is $a_ it^{k-d_ i}$ with $a_ i\in \mathfrak m^{k-d_ i}$. Thus $y=\sum _ ia_ ie_ i$, where $a_ i\in \mathfrak m^{k-d_ i}=\mathfrak m^{k-s}\mathfrak m^{s-d_ i}$ and $\mathfrak m^{s-d_ i}e_ i\subset E\cap \mathfrak m^ sF$; hence $y\in \mathfrak m^{k-s}(E\cap \mathfrak m^ sF)$. The reverse inclusion is immediate.

Now let $Q$ be a finitely generated $R$-module and $G=\bigcap _ k\mathfrak m^ kQ$. By (ii), $G$ is finitely generated. Let $s$ be the index of the Artin–Rees equality for $G\subset Q$. Since $G\subset \mathfrak m^ kQ$ for all $k$, this equality with $k=s+1$ gives $G=G\cap \mathfrak m^{s+1}Q=\mathfrak m(G\cap \mathfrak m^ sQ)=\mathfrak mG$. If $g_1,\ldots ,g_ r$ generate $G$, write $g_ i=\sum _ ja_{ij}g_ j$ with $a_{ij}\in \mathfrak m$. Multiplying this system of equations by the adjugate of the matrix $(\delta _{ij}-a_{ij})$ gives $\det (\delta _{ij}-a_{ij})\, g_ i=0$ for every $i$. Expanding the determinant, we see that it is congruent to $1$ modulo $\mathfrak m$, hence a unit of the local ring $R$, and therefore $G=0$. Applied to $Q=F/E$, for which $\mathfrak m^ kQ=(E+\mathfrak m^ kF)/E$, this proves the intersection formula.

For the last assertion, let $f_ l\to f$ locally uniformly near $x$, with germs $f_ l\in J$, and work in a holomorphic chart centered at $x$. By section 9.2, a germ belongs to $\mathfrak m^ k$ if and only if its Taylor coefficients of order less than $k$ vanish; hence these coefficients identify $R/\mathfrak m^ k$ with a finite-dimensional vector space. By the integral formula for the coefficients in (5.2.1), the Taylor coefficients of $f_ l$ at $x$ converge to those of $f$, so the images of the $f_ l$ in $R/\mathfrak m^ k$ converge to the image of $f$. The image of $J$ there is a linear subspace of a finite-dimensional vector space, hence closed, so $f\in J+\mathfrak m^ k$ for every $k$. The intersection formula with $F=R$ and $E=J$ gives $f\in J$. $\square$