Irreducible complex varieties are analytically connected

First let $Z$ be irreducible affine with coordinate domain $B$. By [F1] choose $R=\mathbb{C}[z_1,\dots ,z_d]\subseteq B$ with $B$ finite over $R$. Write $K=\mathrm{Frac}(R)$. The localization $(R\setminus \{0\}{)}^{-1}B$ is a finite-dimensional $K$-domain; multiplication by a nonzero element is injective and hence surjective, so this domain is a field, equal to $E=\mathrm{Frac}(B)$. If $d=0$, $B$ is a finite-dimensional complex domain and therefore a field; [F2] forces every algebraic element over $\mathbb{C}$ to lie in $\mathbb{C}$, so $B=\mathbb{C}$ and $Z$ is one point. Otherwise [F2] supplies a primitive element $\eta \in E$ over $K$. Write $\eta =b\mathrm{/}s$ with $b\in B$, $0\mathrm{\ne }s\in R$ and put $\theta =b$; then $K(\theta )=E$, and $\theta$ is integral over $R$ by [F1], using the finite faithful $R[\theta ]$-module $B$ (faithfulness follows by evaluating an annihilator at $1$). By [F1] its monic minimal polynomial is $P(z,T)\in R[T]$, of degree $m=[E:K]\ge 1$. It is irreducible and separable over $K$. For finitely many $R$-module generators of $B$, express each as a $K$-linear combination of $1,\theta ,\dots ,{\theta }^{m-1}$ and clear the finitely many denominators. This gives $0\mathrm{\ne }a\in R$ with $B_a=R_a[\theta ]$. Since $P$ is monic, division by $P$ and linear independence of these powers over $K$ identify this algebra with $R_a[T]\mathrm{/}(P)$.

By separability and [F2], there are polynomials $H,J\in K[T]$ with $HP+J{\mathrm{\partial }}_{T}P=1$. Clear denominators to obtain $H_0,J_0\in R[T]$ and $0\mathrm{\ne }\delta \in R$ with $H_{0}P+J_0{\mathrm{\partial }}_{T}P=\delta$. Put $g=a\delta$ and $V={\mathbb{C}}^d\setminus Z(g)$. This set is nonempty because a nonzero polynomial cannot vanish everywhere on ${\mathbb{C}}^d$ (induction on $d$ using the finiteness of roots of a nonzero one-variable polynomial). It is path connected: for distinct $v,w\in V$, the polynomial $t\mapsto g(v+t(w-v))$ is nonzero at $t=0,1$, hence has only finitely many zeros; the complex parameter plane minus those zeros has a path from $0$ to $1$, obtained from the straight segment by finitely many small circular detours. Its image connects $v$ to $w$ in $V$. The principal open $U\subseteq Z$ where $g$ does not vanish is, by step 1.1's coordinate algebra, the space $\{(z,t):z\in V,P(z,t)=0\}$. All its coordinate identifications and inverses are polynomial quotients, hence analytic homeomorphisms. For each $z\in V$, $P(z,T)$ is monic of degree $m$ and has exactly $m$ distinct complex roots by the Bezout identity and [F2]. The implicit-function theorem [F3] gives disjoint holomorphic graphs near these roots on one common small connected base ball. They exhaust every nearby fibre because there are already $m$ distinct roots of its degree-$m$ polynomial. Thus $U\to V$ is an actual $m$-sheeted holomorphic covering, with no appeal to an unproved normalization or covering comparison.

Suppose $U=U_1⨿U_2$ is a nontrivial analytic clopen partition. On each connected small graph of step 2.1 its membership in $U_1$ or $U_2$ is constant. Consequently the number $m_i$ of roots over $z$ belonging to $U_i$ is locally constant on $V$, hence constant since $V$ is connected. Nonemptiness of both pieces implies $1\le m_i<m$ and $m_1+m_2=m$. Define on $V$ the monic polynomials $Q_i(z,T)$ whose roots are precisely the roots on $U_i$ over $z$. Their coefficients are elementary symmetric functions of those roots; the local holomorphic graphs make them holomorphic, and changing the enumeration changes no symmetric function, so these coefficients glue without any chosen global root functions. Pointwise on $V$ one has $P(z,T)=Q_1(z,T)Q_2(z,T)$.

Write $P(z,T)=T^m+p_1(z)T^{m-1}+\cdots +p_m(z)$. Every root satisfies $\mathrm{\mid }t\mathrm{\mid }\le 1+{\sum }_j\mathrm{\mid }{p}_j(z)\mathrm{\mid }$: if $\mathrm{\mid }t\mathrm{\mid }$ exceeds this quantity, then $\mathrm{\mid }t\mathrm{\mid }>1$ and $\mathrm{\mid }t{\mathrm{\mid }}^m\le {\sum }_j\mathrm{\mid }{p}_j(z)\mathrm{\mid }\mathrm{\mid }t{\mathrm{\mid }}^{m-j}\le ({\sum }_j\mathrm{\mid }{p}_j(z)\mathrm{\mid })\mathrm{\mid }t{\mathrm{\mid }}^{m-1}<\mathrm{\mid }t{\mathrm{\mid }}^m$, a contradiction. The polynomial coefficients therefore give a bound $\mathrm{\mid }t\mathrm{\mid }\le C(1+\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}{)}^D$ for some integer $D\ge 0$ and $C\ge 1$, valid also near $z=0$. Each coefficient of $Q_i$ is a sum of finitely many products of at most $m_i$ such roots; it is locally bounded near $Z(g)$ and has a bound $C^{\mathrm{\prime }}(1+\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}{)}^{D{m}_i}$ on $V$. Apply [F3] to extend each coefficient to all of ${\mathbb{C}}^d$. The same bound holds after extension by continuity: the nonzero polynomial $g$ cannot vanish on an ordinary open ball, since that ball contains a product of disks and successively restricting each variable would force all its coefficients to vanish. Hence $V$ is analytically dense in ${\mathbb{C}}^d$.

An entire function $h$ on ${\mathbb{C}}^d$ satisfying $\mathrm{\mid }h(z)\mathrm{\mid }\le C^{\mathrm{\prime }}(1+\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}{)}^N$ for an integer $N\ge 0$ is a polynomial. Indeed [F4] on the radius-$R$ polydisc centred at zero gives $\mathrm{\mid }{c}_{\alpha }\mathrm{\mid }\le C^{\mathrm{\prime }}(1+R{)}^N\mathrm{/}{R}^{\mathrm{\mid }\alpha \mathrm{\mid }}$; uniqueness makes the coefficients independent of $R$ by comparison on the smaller common polydisc. Letting $R\to \mathrm{\infty }$ forces $c_{\alpha }=0$ for $\mathrm{\mid }\alpha \mathrm{\mid }>N$. Its expansion therefore is the finite polynomial ${\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le N}{c}_{\alpha }{z}^{\alpha }$ on every polydisc, hence on all of ${\mathbb{C}}^d$. Apply this to step 4.1's coefficients: $Q_i\in R[T]$. The equality $P=Q_1{Q}_2$ on $V$ is a polynomial identity, since each coefficient of their difference vanishes on a nonempty ordinary open ball and the preceding one-variable restriction argument makes that coefficient the zero polynomial. Both factors have positive degree smaller than $m$, contradicting the irreducibility of $P$ in $K[T]$. Thus $U$ is analytically connected. This argument factors $P$ itself; it does not infer that one of two polynomial factors is zero from their vanishing on different analytic pieces.

The open $U$ is Zariski nonempty in the irreducible affine $Z$. If $Z$ has positive dimension, [F5] supplies at each $x\in Z$ an arc tending to $x$ with punctured image in $U$, so $U$ is analytically dense in $Z$. The zero-dimensional case was already a singleton in step 1.1. The closure of a connected set is connected: any clopen separation of its closure would restrict to a separation of the set, putting the set in one closed side and hence its closure there. Therefore $Z$ is analytically connected. For a general irreducible $X$, choose a nonempty affine open $Z\subseteq X$. It is irreducible and connected by the affine case; [F5] makes it analytically dense in $X$ when $X$ has positive dimension. If $X$ has dimension zero, its nonempty irreducible affine opens are singletons; irreducibility forces them to meet, so $X$ is one point. Again connectedness passes to the closure, proving the first assertion.

Finally let $A\subseteq X$ be locally closed algebraic, and write $C$ for its Zariski closure. Algebraic closed sets are analytically closed, because their polynomial equations are continuous on affine charts; hence the analytic closure of $A$ lies in $C$. There is a Zariski open $O\subseteq X$ with $A=C\cap O$. For each of the finitely many irreducible components $C_j$ of $C$, the set $A\cap C_j$ is a nonempty open subset: otherwise the dense set $A$ would be contained in the union of the other components, a proper closed subset of $C$. If $\dim C_j>0$, [F5] gives at every point of $C_j$ an arc with punctured image in $A\cap C_j$, proving analytic density there. If $\dim C_j=0$, it is a point by the affine argument and the finite affine-cover property, and nonemptiness of $A\cap C_j$ puts that point in $A$. Thus every $C_j$ lies in the analytic closure, which is therefore exactly $C$. The empty subset has empty closure in both topologies. The final manifold clause only combines this connectedness with supplied holomorphic charts; no chart existence is inferred from irreducibility.