The ring of holomorphic germs is Noetherian

The proof is by induction on $m$. For $m=1$, let $I⊴{\mathcal{O}}_{1,0}$ be a nonzero proper ideal. Choose $f\in I$ of minimal zero order $d$. By [L5], $f=z_1^{d}u$ with $u$ a unit. If $g\in I$, then ${\mathrm{ord}}_0(g)\ge d$, so again by [L5] one has $g=z_1^{d}h$ for some holomorphic germ $h$. Since $u$ is a unit, $z_1^d=u^{-1}f\in I$, so $g=h{z}_1^d\in (f)$. Thus every ideal is principal, hence finitely generated. The zero ideal and whole ring are generated by $0$ and $1$. Therefore [L1] makes ${\mathcal{O}}_{1,0}$ Noetherian.

Assume $m>1$ and that ${\mathcal{O}}_{m-1,0}$ is Noetherian. Let $I⊴{\mathcal{O}}_{m,0}$ be a proper nonzero ideal. Choose nonzero $f\in I$. By [L4], after a complex-linear coordinate change we may assume that $f$ is regular in $z_m$; this replaces $I$ by an isomorphic ideal under a ring automorphism, so finite generation is unaffected. By [L4] and [L5], write $f=uW$ with $u$ a unit and $W$ a Weierstrass polynomial. Since $I$ is an ideal and $u^{-1}$ exists, $W=u^{-1}f$ also lies in $I$.

Let $\pi :{\mathcal{O}}_{m,0}\to {\mathcal{O}}_{m,0}\mathrm{/}(W)$ be the quotient map. By [L3], the quotient is a finite ${\mathcal{O}}_{m-1,0}$-module, so [L2] and the induction hypothesis make it a Noetherian ${\mathcal{O}}_{m-1,0}$-module. Hence the submodule $\pi (I)$ is generated by finitely many classes $\pi (g_1),\dots ,\pi (g_s)$ with $g_i\in I$.

Let $g\in I$. Since $\pi (g)\in \pi (I)$, there are $a_1,\dots ,a_s\in {\mathcal{O}}_{m-1,0}$ with $\pi (g)=a_1\pi (g_1)+\cdots +a_s\pi (g_s)\mathrm{.}$ Thus $g-(a_1{g}_1+\cdots +a_s{g}_s)\in \ker \pi =(W)\subseteq I,$ so $g$ lies in the ideal generated by $W,g_1,\dots ,g_s$. Therefore $I$ is finitely generated. By [L1], ${\mathcal{O}}_{m,0}$ is Noetherian.