Krull dimension of the holomorphic germ ring

Suppose first that $n=0$. Then ${\mathcal{O}}_{{\mathbb{C}}^0,p}=\mathbb{C}$ is a field by [F1] and [F7]. A nonzero ideal of a field contains a nonzero element, which is a unit, so it is the whole ring; hence $(0)$ is the only prime ideal, there is no strict chain of prime ideals of length $\ge 1$, and $\dim \mathbb{C}=0$ by [F7].

Now let $n\ge 1$, write $\mathcal{O}:={\mathcal{O}}_{n,0}$ and $\mathfrak{m}:={\mathfrak{m}}_{n,0}$ for the origin case; the general-centre case is transported at the end. Choose a representative of $f\in \mathcal{O}$ on a polydisc and let $f(z)={\sum }_{\alpha }{c}_{\alpha }{z}^{\alpha }$ be its expansion with the bound of [F3]. Group the nonzero multi-indices by their first positive coordinate and set $h_i(z):={\sum }_{{\alpha }_i\ge 1,{\alpha }_j=0(j<i)}{c}_{\alpha }{z}^{\alpha -e_i}(1\le i\le n)\mathrm{.}$ The coefficient of $z^{\beta }$ in $h_i$ is $c_{\beta +e_i}$ for the indices with ${\beta }_j=0$ for $j<i$, so it obeys the bound $\mathrm{\mid }{c}_{\beta +e_i}\mathrm{\mid }\le M{r}_i^{-1}{\prod }_k{r}_k^{-{\beta }_k}$; by [F4] each $h_i$ is holomorphic on the polydisc. Every nonzero multi-index has a first positive coordinate, so it contributes to exactly one $h_i$, and the subseries of an absolutely convergent series converge to the corresponding partial sums; hence $f=f(0)+{\sum }_{i=1}^n{z}_i{h}_i$ as germs on the polydisc.

Consequently $f\in \mathfrak{m}$ (that is, $f(0)=0$) if and only if $f$ lies in the ideal $(z_1,\dots ,z_n)$ of $\mathcal{O}$. Conversely each coordinate germ $z_i$ vanishes at $0$, so $(z_1,\dots ,z_n)\subseteq \mathfrak{m}$. Therefore $\mathfrak{m}=(z_1,\dots ,z_n)$, an ideal generated by $n$ elements.

Fix $0\le k\le n$ and define $\phi :{\mathcal{O}}_{n-k,0}\to \mathcal{O}\mathrm{/}(z_1,\dots ,z_k)$ on the germ of $g(z_{k+1},\dots ,z_n)$ as the class of its inclusion in $\mathcal{O}$. This is a well-defined unital ring homomorphism, because addition and multiplication of germs are represented pointwise and the inclusion respects them. It is surjective: expanding any $F\in \mathcal{O}$ as in step 1.2 and grouping the multi-indices with ${\alpha }_1=\cdots ={\alpha }_k=0$ into a germ $G$ of the remaining variables, [F4] makes $G$ holomorphic while the complementary subseries is divisible by one of $z_1,\dots ,z_k$, so $F-G\in (z_1,\dots ,z_k)$ and $\phi (G)=F+(z_1,\dots ,z_k)$.

The map $\phi$ of step 2.2 is injective: if $\phi (g)=0$, then $g\in (z_1,\dots ,z_k)$ as a germ in $\mathcal{O}$, so representatives on a common polydisc satisfy $g(z)={\sum }_{i=1}^k{z}_i{H}_i(z)$ there; setting $z_1=\cdots =z_k=0$ kills the right-hand side, so the representative of $g$, which does not involve the first $k$ variables, vanishes on a polydisc in ${\mathbb{C}}^{n-k}$, which is exactly the zero germ. Hence $\mathcal{O}\mathrm{/}(z_1,\dots ,z_k)\cong {\mathcal{O}}_{n-k,0}$.

By step 3.1 each ideal $(z_1,\dots ,z_k)$ has quotient isomorphic to ${\mathcal{O}}_{n-k,0}$. If $k<n$, then $n-k\ge 1$, so this quotient is an integral domain by [F5]; if $k=n$, it is ${\mathcal{O}}_{0,0}=\mathbb{C}$, a field by [F7] and hence an integral domain. Thus every $(z_1,\dots ,z_k)$ with $0\le k\le n$ is prime by [F6]. In particular $\mathfrak{m}=(z_1,\dots ,z_n)$ is prime, and it is maximal by [F2].

The chain of prime ideals $(0)⊊(z_1)⊊(z_1,z_2)⊊\cdots ⊊(z_1,\dots ,z_n)=\mathfrak{m}$ is strict: $(0)$ is prime because $\mathcal{O}$ is a domain by [F5], each later term is prime by step 4.1, and $z_k\mathrm{\notin }(z_1,\dots ,z_{k-1})$, since otherwise its class in $\mathcal{O}\mathrm{/}(z_1,\dots ,z_{k-1})\cong {\mathcal{O}}_{n-k+1,0}$ would be zero by step 3.1, whereas that class corresponds to the first coordinate germ of ${\mathcal{O}}_{n-k+1,0}$, which is nonzero. This chain has length $n$, so $\dim \mathcal{O}\ge n$.

For the reverse inequality, [F5] makes $\mathcal{O}$ Noetherian, and by step 2.1 the prime ideal $\mathfrak{m}$ is generated by $n$ elements, hence is minimal over that ideal. Under AC, [F8] gives $\mathrm{ht}(\mathfrak{m})\le n$.

Every proper ideal of $\mathcal{O}$ is contained in $\mathfrak{m}$: if $I\subseteq ̸\mathfrak{m}$, then $I$ contains an element outside the maximal ideal, that element is a unit by [F2], and $I=\mathcal{O}$. Hence the primes disjoint from $S=\mathcal{O}\setminus \mathfrak{m}$ are exactly the primes of $\mathcal{O}$. By [F9], contraction is an inclusion-preserving bijection $\mathrm{Spec}({\mathcal{O}}_{\mathfrak{m}})\to \mathrm{Spec}(\mathcal{O})$ whose inverse is the inclusion-preserving prime extension $P\mapsto S^{-1}P$. Thus a strict chain of primes in $\mathcal{O}$ extends to a strict chain of the same length in ${\mathcal{O}}_{\mathfrak{m}}$, so $\dim \mathcal{O}\le \dim {\mathcal{O}}_{\mathfrak{m}}=\mathrm{ht}(\mathfrak{m})\le n$ by step 5.2.

Steps 5.1 and 6.1 give $\dim {\mathcal{O}}_{n,0}=n$ for $n\ge 1$, step 1.1 gives it for $n=0$, and step 2.1 identifies the maximal ideal with $(z_1,\dots ,z_n)$. Transporting along the translation convention replaces each coordinate function $z_i$ by the coordinate difference $z_i-p_i$ and does not change dimensions, so $\dim {\mathcal{O}}_{{\mathbb{C}}^n,p}=n$ and ${\mathfrak{m}}_{{\mathbb{C}}^n,p}=(z_1-p_1,\dots ,z_n-p_n)$ for every $n\ge 1$ and every $p\in {\mathbb{C}}^n$.