Let \(Y\) be an irreducible curve on a surface \(X\), and suppose there is a morphism \(f: X \rightarrow X_0\) to a projective variety \(X_0\) of dimension 2, such that \(f(Y)\) is a point \(P\) and \(f^{-1}(P)=Y\). Then show that \(Y^2<0\).
Hint: Let \(|H|\) be a very ample (Cartier) divisor class on \(X_0\), let \(H_0 \in|H|\) be a divisor containing \(P\), and let \(H_1 \in|H|\) be a divisor not containing \(P\). Then consider \(f^* H_0, f^* H_1\) and \(\tilde{H}_0=f^*\left(H_0-P\right)^{-}\).
Solution
1. As literally transcribed, the statement is false if the image of\(f\) is allowed to be one-dimensional.
Let \(C\) be any nonsingular projective curve and take \[ X=C\times\PP^1. \] Let \[ p_1:X\longrightarrow C \] be the first projection. Take a second copy \[ X_0=C\times\PP^1 \] and a point \(q\in\PP^1\), and let \[ i:C\hookrightarrow X_0, \qquad c\longmapsto(c,q) \] be the corresponding closed immersion. Put \[ f=i\circ p_1. \] For any point \(c\in C\), let \[ P=i(c), \qquad Y=\{c\}\times\PP^1. \] Then \[ f(Y)=P, \qquad f^{-1}(P)=Y, \] but two distinct fibres of \(p_1\) are disjoint and linearly equivalent, so \[ Y^2=0. \] Thus the conclusion requires the intended surface-contraction hypothesis \[ \dim f(X)=2. \]
Assume for the rest of the proof that \[ \dim f(X)=2. \] Since \(X\) is projective, \(f(X)\) is closed. Since \(X_0\) is an irreducible surface, this implies \[ f(X)=X_0. \] Hence \(f\) is dominant and generically finite.
2. Let \(H\) be a very ample Cartier divisor on \(X_0\) and put\[ D=f^*H. \]Then\[ \boxed{D^2>0.} \]
Because \(f\) is dominant and generically finite, it has a positive degree \[ d=[K(X):K(X_0)]>0. \] The projection formula for intersection products gives \[ D^2 =(f^*H)^2 =d\,H^2. \] Very ampleness of \(H\) gives \[ H^2>0, \] so \[ D^2>0. \]
3. One has\[ \boxed{D\cdot Y=0.} \]
Choose \[ H_1\in|H| \] with \[ P\notin H_1, \] as in the hint. Since \[ f(Y)=P, \] the divisor \[ f^*H_1 \] is disjoint from \(Y\). Therefore \[ f^*H_1\cdot Y=0. \] Because \[ f^*H_1\sim f^*H=D, \] intersection depends only on numerical equivalence, and hence \[ D\cdot Y=0. \] Equivalently, the projection formula gives \[ f^*H\cdot Y=H\cdot f_*Y=0 \] because \(Y\) is contracted to a point.
4. The numerical class\[ [Y]\in\operatorname{NS}(X)_\RR \]is nonzero.
Let \(A\) be any ample divisor on the nonsingular projective surface \(X\). Since \(Y\) is an irreducible curve, \[ A\cdot Y>0. \] Thus \(Y\) cannot be numerically equivalent to zero.
5. The intersection form is negative definite on\[ D^\perp = \left\{ \alpha\in\operatorname{NS}(X)_\RR: \alpha\cdot D=0 \right\}. \]
The Hodge index theorem T-SRFHODGE says that the intersection form on \[ \operatorname{NS}(X)_\RR \] has signature \[ (1,\rho(X)-1). \] Step 2 gives \[ D^2>0. \] Hence the line \[ \RR D \] already supplies the unique positive direction, so its orthogonal complement \(D^\perp\) is negative definite. Therefore every nonzero \[ \alpha\in D^\perp \] satisfies \[ \alpha^2<0. \]
6. Under the intended hypothesis\[ \dim f(X)=2, \]the contracted curve satisfies\[ \boxed{Y^2<0.} \]
By step 3, \[ [Y]\in D^\perp. \] By step 4 this class is nonzero. Step 5 therefore gives \[ Y^2<0. \] This is exactly the desired conclusion for a genuine contraction to a projective surface, including the birational situation referred to by the retained V.5.7 solution through Exercise V.5.4.
7. The divisors appearing in the printed hint are compatible with thesame calculation.
Choose \[ H_0\in|H|, \qquad P\in H_0, \] and let \(\widetilde H_0\) denote the closure of the pullback away from the contracted point, as in the hint. Since \[ f^{-1}(P)=Y, \] the pullback has the form \[ f^*H_0=\widetilde H_0+mY \] for some integer \[ m>0. \] Intersecting with \(Y\) and using step 3 gives \[ 0 = f^*H_0\cdot Y = \widetilde H_0\cdot Y+mY^2. \] Step 6 says \(Y^2<0\), so necessarily \[ \widetilde H_0\cdot Y=-mY^2>0. \] Thus the two hyperplanes in the hint encode the same geometry: the hyperplane avoiding \(P\) gives zero intersection with \(Y\), while a hyperplane through \(P\) acquires the exceptional component \(mY\).
8. Q.E.D.
Step 1 records the missing hypothesis in the literal transcription. Under the intended two-dimensional-image hypothesis, steps <1>2–<1>5 place the nonzero class of \(Y\) in the negative-definite orthogonal complement of the positive-square divisor \(f^*H\). Step 6 gives \(Y^2<0\), and step 7 reconciles the proof with the printed hint.