Proposition 9.2.4.4B4DFDFC5DBA Every ideal of $\mathcal O_ n$ is finitely generated, and every submodule of $\mathcal O_ n^ r$, $r\geq 1$, is finitely generated. Consequently, every increasing sequence of ideals of $\mathcal O_ n$, or of submodules of a finitely generated $\mathcal O_ n$-module, is eventually constant.
Proposition 9.2.44B4DFDFC5DBA
Proof. We first recall three algebraic facts about a commutative ring $\Lambda $ all of whose ideals are finitely generated. Every submodule $N$ of $\Lambda ^ r$ is finitely generated, by induction on $r$: the projection of $N$ to the last coordinate is an ideal; choose elements of $N$ whose last coordinates generate it, and add generators of $N\cap (\Lambda ^{r-1}\times \{ 0\} )$, which exist by the induction hypothesis. A finitely generated module is a quotient of some $\Lambda ^ r$, and the inverse image of a submodule is a submodule of $\Lambda ^ r$; hence submodules of finitely generated modules are finitely generated. Finally, the union of an increasing sequence of submodules is a submodule; its finitely many generators already lie in one member of the sequence, which is therefore stationary from then on.
It remains to prove that the ideals of $\mathcal O_ n$ are finitely generated, which we do by induction on $n$. The ring $\mathcal O_0=\mathbb C$ is a field. Let $n\geq 1$ and let $I\subset \mathcal O_ n$ be an ideal. If $I=0$ or $I$ contains a unit, it is generated by one element. Otherwise choose a nonzero germ in $I$, and coordinates in which it is distinguished in $w$. By the preparation theorem it is a unit times a Weierstrass polynomial $P$ of some degree $m\geq 1$, and $P\in I$. Divide every $f\in I$ by $P$ using proposition 9.2.2; the remainder $R_ f=f-Pq$ again belongs to $I$. The remainders of degree less than $m$ that lie in $I$ form an $\mathcal O_{n-1}$-submodule of $\mathcal O_{n-1}[w]$, which we identify, through their coefficients, with a submodule of $\mathcal O_{n-1}^ m$. By the induction hypothesis and the first paragraph, it is generated by finitely many remainders $R_1,\ldots ,R_ k$. Then $P,R_1,\ldots ,R_ k$ generate $I$. $\square$
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