Complex algebraic boundary points admit normalized curve arcs

Choose an affine open neighbourhood $V_0$ of $z$ in $V$. Then $V_0$ is irreducible of dimension $d=\dim V\ge 1$, and $V_0\cap O\mathrm{\ne }\mathrm{\varnothing }$ because two nonempty open subsets of the irreducible $V$ meet (Irreducible topological spaces and irreducible subsets in the subspace topology). Put $B=V_0\setminus O$; it is closed in $V_0$ and proper.

By [F2] applied to the coordinate ring $A$ of $V_0$, choose a finite morphism $h:V_0\to {\mathbb{A}}^d$ induced by an injection $\mathbb{C}[t_1,\dots ,t_d]\hookrightarrow A$ with $A$ module-finite over the image; the integer $d$ is ${\mathrm{trdeg}}_{\mathbb{C}}\mathrm{Frac}(A)=\dim V_0$ by [F3]. For every irreducible component $W$ of $B$ the image $h(W)$ is closed by [F1], and the induced map $\mathbb{C}[t]\mathrm{/}I(h(W))\hookrightarrow A\mathrm{/}I(W)$ is finite and injective; hence $\mathrm{Frac}(A\mathrm{/}I(W))$ is algebraic over $\mathrm{Frac}(\mathbb{C}[t]\mathrm{/}I(h(W)))$, the two function fields have the same transcendence degree, and $\dim W=\dim h(W)$ by [F3]. A finite union of closed sets has dimension the maximum of the dimensions of its irreducible components, so $\dim h(B)=\dim B<d$, the inequality by [F3] since $B$ is a proper closed subset of the irreducible $V_0$. Thus $h(B)$ is a proper closed subset of ${\mathbb{A}}^d$ and there is a nonzero polynomial $P$ vanishing on it, or $P=1$ if $B=\mathrm{\varnothing }$.

Write $a=h(z)$. The Taylor expansion of a nonzero polynomial $P$ at $a$ has a nonzero homogeneous part, and a nonzero polynomial on ${\mathbb{C}}^d$ is not identically zero (induction on $d$ using that a nonzero one-variable polynomial has finitely many roots), so there is $w\mathrm{\ne }0$ with $P(a+tw)$ not the zero polynomial in $t$; if $B=\mathrm{\varnothing }$ take any $w\mathrm{\ne }0$. The affine line $\mathrm{\ell }=\{a+tw:t\in \mathbb{C}\}$ is therefore not contained in $h(B)$.

Let $g_1,\dots ,g_{d-1}$ be linear forms cutting out $\mathrm{\ell }$ in ${\mathbb{A}}^d$ (no forms occur when $d=1$), and let $C_0$ be an irreducible component of the closed subset $h^{-1}(\mathrm{\ell })\subseteq V_0$ containing $z$; such a component exists because $h(z)=a\in \mathrm{\ell }$. Every component of $h^{-1}(\mathrm{\ell })$ has dimension at least $1$: it is a component of the zero set of the $d-1$ pulled-back forms, so its defining prime is minimal over an ideal generated by $d-1$ elements, has height at most $d-1$ by [F3], and $\dim =\dim A-\mathrm{ht}\ge d-(d-1)=1$; for $d=1$ this reads $\dim V_0=1$. Every component has dimension at most $1$ because it maps finitely into the curve $\mathrm{\ell }$, so its dimension equals that of its closed image by the dimension comparison of step 2.1. Hence every component is a curve, and $C_0$ is one of them.

The curve $C_0$ is not contained in $B$: otherwise $h(C_0)\subseteq h(B)$, while $h(C_0)$ is closed in $\mathrm{\ell }$, irreducible, and of dimension $1=\dim C_0$ by step 4.1, hence equal to $\mathrm{\ell }$, contradicting $\mathrm{\ell }\subseteq ̸h(B)$ from step 3.1. Therefore $C_0\cap O=C_0\setminus B\mathrm{\ne }\mathrm{\varnothing }$. The closure $C:=\overline{C_0}$ in $V$ is an irreducible curve through $z$ meeting $O$, so $C\subseteq ̸V\setminus O$.

View $C_0$ with its reduced induced structure: it is a reduced finite-type $\mathbb{C}$-scheme of pure dimension one, so by [F4] it has a finite normalization $\nu :C_0^{\mathrm{n}\mathrm{u}}\to C_0$, which is surjective by [F1]; choose $z_0\in C_0^{\mathrm{n}\mathrm{u}}$ with $\nu (z_0)=z$. Then $C_0^{\mathrm{n}\mathrm{u}}$ is regular of dimension one and, over $\mathbb{C}$, smooth.

Take an affine chart $\mathrm{Spec}{A}_0$ of $C_0^{\mathrm{n}\mathrm{u}}$ containing $z_0$ and identify it with a closed algebraic set $X=V(I)\subseteq {\mathbb{C}}^M$, $I\subseteq \mathbb{C}[t_1,\dots ,t_M]$. Since ${\mathcal{O}}_{X,z_0}$ is a regular local ring of dimension one, [F6] gives ${\dim }_{\mathbb{C}}{T}_{z_0}X=1$; for a finite generating list $f_1,\dots ,f_r$ of $I$ this tangent space is the kernel of the $r\times M$ Jacobian matrix at $z_0$, so that matrix has rank $M-1$, and some $M-1$ of the $f_i$, say $f_1,\dots ,f_{M-1}$, have an invertible $(M-1)\times (M-1)$ minor at $z_0$. Put $Z=V(f_1,\dots ,f_{M-1})$ and let $J^{\mathrm{\prime }}$ be their Jacobian. Then $\dim {\mathcal{O}}_{Z,z_0}\le \mathrm{edim}{\mathcal{O}}_{Z,z_0}=\dim \ker J^{\mathrm{\prime }}(z_0)=M-(M-1)=1$ by [F6], while $\dim {\mathcal{O}}_{Z,z_0}\ge \dim {\mathcal{O}}_{X,z_0}=1$ because $X\subseteq Z$ induces a surjection of local rings and dimensions do not increase under quotients. Hence $\dim {\mathcal{O}}_{Z,z_0}=1$ and it is regular, therefore a local domain.

The inclusion $X\subseteq Z$ induces a surjection of one-dimensional local rings ${\mathcal{O}}_{Z,z_0}\twoheadrightarrow {\mathcal{O}}_{X,z_0}$. Its kernel is a prime ideal, and since the source is a one-dimensional local domain, a nonzero kernel would be the maximal ideal with a zero-dimensional quotient, contradicting $\dim {\mathcal{O}}_{X,z_0}=1$; so the kernel is zero and ${\mathcal{O}}_{X,z_0}\cong {\mathcal{O}}_{Z,z_0}$. Write $I(X)=(g_1,\dots ,g_k)$ by [F7]. Each $g_i$ maps to zero in ${\mathcal{O}}_{Z,z_0}$, so for each $i$ there is $h_i\mathrm{\notin }{\mathfrak{m}}_{z_0}$ with $h_i{g}_i$ vanishing on $Z$; for $h=h_1\cdots h_k\mathrm{\notin }{\mathfrak{m}}_{z_0}$ all the $g_i$ vanish on $Z\cap \{h\mathrm{\ne }0\}$, and since $X\subseteq Z$ we get $X\cap \{h\mathrm{\ne }0\}=Z\cap \{h\mathrm{\ne }0\}$.

If $M=1$, step 8.1 identifies $X$ with an open neighbourhood of $z_0$ in $Z=\mathbb{C}$, so translation is the required holomorphic chart. For $M>1$, applying [F5] to $Z$ at $z_0$, whose $M-1$ defining functions have rank $M-1$ at $z_0$ by step 7.1, yields a plane domain $A^{\mathrm{\prime }}\subseteq \mathbb{C}$ and a holomorphic map ${\gamma }_0:A^{\mathrm{\prime }}\to {\mathbb{C}}^M$ whose image is a neighbourhood of $z_0$ in $Z$, with ${\gamma }_0$ injective and inverse holomorphic; let $c$ be the value of the free coordinate at $z_0$ and replace ${\gamma }_0$ by $s\mapsto {\gamma }_0(c+s)$ on a disk $\mathrm{\Delta }\subseteq A^{\mathrm{\prime }}-c$ centred at $0$, so that ${\gamma }_0(0)=z_0$. Shrinking $\mathrm{\Delta }$ and using step 8.1, we may assume ${\gamma }_0(\mathrm{\Delta })\subseteq X$. Composing ${\gamma }_0$ with the inclusion $X\subseteq C_0^{\mathrm{n}\mathrm{u}}$ and with $\nu$ gives an arc into $C$; it is nonconstant and holomorphic, because every coordinate function of the affine model $V_0$ pulls back to a regular function on the affine set $X$, is there represented by a polynomial in $t_1,\dots ,t_M$, and substitution of the holomorphic graph coordinates gives a holomorphic function of $s$. The finite normalization of $C$ restricts over the open curve $C_0=C\cap V_0$ to the normalization of $C_0$: both affine descriptions use the same integral closure, localized at the defining open. Thus this parametrizes a point in the normalization of $C$ as stated.

It remains to make the punctured arc avoid $V\setminus O$. Since $C_0\subseteq ̸B$ and $B$ is closed in the irreducible curve $C_0$, the set $C_0\setminus O=C_0\cap B$ has dimension $0$ by [F3] and is therefore finite: a classical variety is Noetherian with finitely many irreducible components, and a zero-dimensional irreducible component is a single point. Because $\nu$ is finite, each of its fibres is finite: on an affine chart the coordinate ring is a finitely generated module over the base with finite-dimensional reduction modulo a maximal ideal. Hence $F:={\nu }^{-1}(C_0\setminus O)$ is finite. The chart ${\gamma }_0$ of step 9.1 is injective, so ${\gamma }_0^{-1}(F)$ is finite with at most $\mathrm{\mid }F\mathrm{\mid }$ elements; replacing the disk by a smaller concentric disk, we may assume no nonzero parameter in $\mathrm{\Delta }$ has ${\gamma }_0(s)\in F$ (if $z\in O$, the parameter $0$ is not in this preimage either). Then for every $s\mathrm{\ne }0$ the point $\nu ({\gamma }_0(s))$ lies in $C_0\setminus \nu (F)=C_0\cap O\subseteq O$. Thus $\gamma :=\nu \circ {\gamma }_0:\mathrm{\Delta }\to V$ is holomorphic with $\gamma (0)=z$ and $\gamma (s)\in O$ for all $s\mathrm{\ne }0$, and the arc was obtained by parametrizing the point $z_0$ over $z$ in the finite normalization $C_0^{\mathrm{n}\mathrm{u}}$.