Let \(X\) be a normal, projective variety over an algebraically closed field \(k\). Let \(D\) be a linear system of effective Cartier divisors without base points, and assume that \(D\) is not composite with a pencil, which means that if \(f: X \to \PP_k^n\) is the morphism determined by \(D\), then \(\dim f(X) \geq 2\).
Show that every divisor in \(D\) is connected.
Hints: see (10.9.1); use (11.5), (Ex. 5.7) and (7.9).
Solution
Write \(\mathcal D\) for the given base-point-free linear system, and let \[ f:X\longrightarrow\PP_k^n \] be the associated morphism. Put \[ Z=f(X) \] with its reduced induced structure.
1. Every member of \(\mathcal D\) is the inverse image of a hyperplanesection of \(Z\).
The morphism \(f\) is defined by the vector space of sections underlying \(\mathcal D\). Hence \[ \mco_X(D_0)\cong f^*\mco_{\PP^n}(1) \] for the divisor class \(D_0\) represented by the system.
A member \(D\in\mathcal D\) is the zero scheme of a nonzero linear combination of these defining sections. The same linear combination defines a hyperplane \[ H\subset\PP_k^n, \] and scheme-theoretically \[ D=f^{-1}(H)=f^*(H|_Z). \] Since the corresponding section is nonzero, \(Z\) is not contained in \(H\). Thus \(H|_Z\) is an effective Cartier divisor on the integral variety \(Z\).
2. The morphism \(f:X\to Z\) has a factorization\[ X\xrightarrow{h}Y\xrightarrow{g}Z \]such that \(h\) is projective with connected fibres and \(g\) is finitesurjective.
This is the Stein factorization [@Har10a, Corollary III.11.5], applied after replacing the original target by the image \(Z\). By construction, \[ h_*\mco_X=\mco_Y. \] The map \(g\) is surjective because \(f\) is surjective onto \(Z\).
3. The variety \(Y\) is normal, projective, and has dimension at least \(2\).
First, \(Y\) is irreducible because it is the image of the irreducible variety \(X\) under the surjective map \(h\). It is reduced because for every open \(V\subseteq Y\), \[ \Gamma(V,\mco_Y) = \Gamma(h^{-1}(V),\mco_X), \] and \(X\) is reduced.
To prove normality, let \(V\subseteq Y\) be affine and let \[ a\in K(Y) \] be integral over \(\Gamma(V,\mco_Y)\). Pulling back along the dominant map \(h\), the element \(h^*a\in K(X)\) is integral over \[ \Gamma(h^{-1}(V),\mco_X) = \Gamma(V,\mco_Y). \] A normal integral scheme has integrally closed rings of regular functions on all open subsets, since such a ring is the intersection of its normal local rings inside the function field. Hence \[ h^*a\in\Gamma(h^{-1}(V),\mco_X) = \Gamma(V,\mco_Y). \] Since \(h\) is dominant, \(K(Y)\to K(X)\) is injective. The last equality therefore implies \[ a\in\Gamma(V,\mco_Y). \] Thus \(Y\) is normal.
Since \(g\) is finite and \(Z\) is projective, \(Y\) is projective. Moreover a finite surjective morphism preserves dimension, so \[ \dim Y=\dim Z=\dim f(X)\ge2. \]
4. For the hyperplane \(H\) associated to a member \(D\in\mathcal D\), thedivisor\[ E=g^*(H|_Z) \]is an effective ample Cartier divisor on \(Y\).
By step 1, \(H|_Z\) is an effective Cartier divisor and \[ \mco_Z(H|_Z)=\mco_{\PP^n}(1)|_Z \] is ample. Because \(g:Y\to Z\) is finite surjective, pullback preserves ampleness by [@Har10a, Exercise III.5.7(d)]. Therefore \[ \mco_Y(E)=g^*\mco_Z(H|_Z) \] is ample. Pullback of the defining nonzerodivisor of \(H|_Z\) remains nonzero on the integral scheme \(Y\), so \(E\) is an effective Cartier divisor.
5. The support \(|E|\) is connected.
By step 3, \(Y\) is a normal projective variety of dimension at least \(2\). By step 4, \(|E|\) is the support of an effective ample divisor. The Enriques–Severi–Zariski connectedness theorem [@Har10a, Corollary III.7.9] therefore gives \[ \boxed{|E|\text{ is connected}.} \]
6. If \(C\subseteq Y\) is connected, then \(h^{-1}(C)\) is connected.
The projective morphism \(h\) is closed, surjective, and has connected fibres. Suppose, to the contrary, that \[ h^{-1}(C)=A\amalg B \] is a separation into two nonempty subsets closed in \(h^{-1}(C)\). The restriction \[ h^{-1}(C)\longrightarrow C \] is again a closed map: if \(F=F'\cap h^{-1}(C)\) with \(F'\subseteq X\) closed, then \[ h(F)=h(F')\cap C. \] Hence \(h(A)\) and \(h(B)\) are closed in \(C\), and they cover \(C\).
They are disjoint. Indeed, if \[ y\in h(A)\cap h(B), \] then the connected fibre \(h^{-1}(y)\) meets both \(A\) and \(B\). But \[ h^{-1}(y) = (h^{-1}(y)\cap A)\amalg(h^{-1}(y)\cap B) \] would then be a separation of that fibre, a contradiction.
Thus \(C=h(A)\amalg h(B)\) is disconnected, contrary to the hypothesis. Hence \(h^{-1}(C)\) is connected.
7. Every divisor \(D\in\mathcal D\) is connected.
For the hyperplane \(H\) corresponding to \(D\), steps 1 and 2 give \[ D=f^*(H|_Z) = h^*g^*(H|_Z) = h^*E. \] Consequently \[ |D|=h^{-1}(|E|). \] Step 5 says that \(|E|\) is connected, and step 6 says its inverse image under \(h\) is connected. Therefore \(|D|\) is connected, which is exactly to say that the divisor scheme \(D\) is connected.
8. Q.E.D.
The argument applies to every hyperplane, hence to every member of the base-point-free linear system \(\mathcal D\). Thus every divisor in \(\mathcal D\) is connected.