4.10 Castelnuovo’s Contractibility Criterion
The goal is to prove the following. Eventually we want to do even better.
Let \(X\) be a complex, projective surface and \(E\subset X\) a curve isomorphic to \(\mathbb {P}^1\) with \(E^2 = -1\). Then there is a morphism \(f:X\to Y\) with \(Y\) a complex, projective surface, such that \(f\) is the blow up of a smooth point \(p\in Y\) and \(E\) is the exceptional divisor.
This proof is copied from [ 4 , Theorem V.5.7 ] .
In order to contract \(E\) to a point, we want to find a morphism \(\phi :X\to \mathbb {P}^N\) (for some \(N\)) such that \(f(E)\) is a point but \(\phi \) is one-one elsewhere. Then \(Y\) will just be the image \(\phi (X)\).
Now, such maps \(\phi \) correspond to linear systems on \(X\). So we might as well look for linear systems instead of maps \(\phi \). But we notice that, if \(\phi \) is a morphism as above, and \(\phi ^*\mathcal{O}_{\mathbb {P}^N}(1) = \mathcal{O}_X(D)\), then \(D\cdot E = 0\) because, clearly, \(\phi ^*\mathcal{O}_{\mathbb {P}^N}(1)|_E = \mathcal{O}_E\). So we want to find a divisor \(D\) with \(D\cdot E = 0\), but which doesn’t contract any other curves.
If \(H\) is a very ample divisor on \(X\), then the divisor \(D = H + (H\cdot E) E\) would work. So choose a very ample divisor \(H\) and set \(k = H\cdot E\) and \(D = H + kE\). For reasons which will soon become clear, we choose \(H\) so that \(\operatorname{\mathrm{H}}^1(X, \mathcal{O}_X(H)) = 0\). We’re going to show that \(\mathcal{L} = \mathcal{O}_X(D)\) is base-point free and that the complete linear system \(|\mathcal{L}|\) gives a morphisms \(f:X\to Y\) as in the statement of the theorem.
To prove the basepoint freeness, we use the following.
We have \(\operatorname{\mathrm{H}}^1(X, \mathcal{O}_X(H + iE)) = 0\) for \(0\leq i \leq k\).
We prove it by induction on \(i\) starting with \(i = 0\), where it’s true by hypothesis. If \(0 {\lt} i \leq k\), we have
using the facts that \(E\cong \mathbb {P}^1\), \(E\cdot E = -1\) and \(H\cdot E = k\). For \(i\leq k\), \(\operatorname{\mathrm{H}}^1(\mathbb {P}^1, \mathcal{O}(k-i)) = 0\). So, from the long exact sequence in cohomology, we get a surjection
But then, by induction, the claim is proved.
The linear system \(|\mathcal{L}| = |H+kE|\) is base point free.
Since \(H\) is ample, it’s already clear that \(|D| = |H+kE|\) is bpf away from \(E\). So we just have to prove it’s bpf on \(E\) itself. By construction, we have \(\mathcal{L}|_E = \mathcal{O}_E\). So it suffices to show that the morphism \(\Gamma (X, \mathcal{O}_X(H+kE)) \to \Gamma (E, \mathcal{O}_{\mathbb {P}^1})\) coming from 4.43 is onto. But that follows directly from the long exact sequence in cohomology and the vanishing of \(\operatorname{\mathrm{H}}^1(X,\mathcal{O}_X(H + (k-1)E))\) proved in Claim 4.42.
Set \(N = h^0(X,\mathcal{L}) - 1\). Then \(|\mathcal{L}|\) gives us a morphism \(f_1:X\to \mathbb {P}^N\) with \(f_1(E) = \{ P_1\} \) for \(P\) a closed point. Moreover, if \(Y_1 = f(X)\), then \(f_1\) induces an isomorphism \(f_1:X\setminus E\to Y\setminus \{ P_1\} \).
Since \(H\) is very ample, it’s pretty easy to see that it separates points and tangent vectors away from \(E\) and it also separates points on \(E\) from points away from \(E\). So we get the claim.
Let \(Y\) be the normalization of \(Y\) and let \(f:X\to Y\) be the morphism coming from \(f_1\) using the universal property of normalization. Then again \(f(E) = P\) for a point \(P\in Y\) and \(f:X\setminus E\to Y\setminus \{ P\} \) is an isomorphism.
Kind of obvious.
We have \(f_*\mathcal{O}_X = \mathcal{O}_Y\).
It follows since \(f\) is birational and \(Y\) is normal. (That’s why we wanted \(Y\) to be normal.)
The completion \(\widehat{\mathcal{O}}\) of local ring \(\mathcal{O}_{Y,P}\) at the powers of the \(\mathfrak {m}_{P}\) is isomorphic to the power series ring \(k[[x,y]]\). Consequently \(Y\) is regular.
We’re going to use the theorem on formal functions with \(i = 0\), \(X\) and \(Y\) as in the statement of the theorem, \(y = P\). We have \(f_*\mathcal{O}_X = \mathcal{O}_Y\) by Claim 4.47. So the theorem tells us that \(\widehat{\mathcal{O}} = \varprojlim \operatorname{\mathrm{H}}^0(X_n, \mathcal{O}_{X_n})\) where \(X_n\) is defined by the \(\mathfrak {m}^n\). But if \(\mathcal{I}\) is the ideal of \(E\), we see that the system \(\mathcal{I}^n\) is cofinal in the system defined by \(\mathcal{m}^n\). So we can replace \(Y_n\) with \(V(\mathcal{I}^n\). After doing this we get a short exact sequence
And running the argument we ran before we have that \(\mathcal{I}^n/\mathcal{I}^{n+1} \cong \mathcal{O}_E(n)\). So we get a short exact sequence
Now, for each \(n\), set \(A_n = k[[x,y]]/(x,y)^n\), and set \(B_n = \operatorname{\mathrm{H}}^0(\mathcal{O}_{Y_{n}})\). Clearly \(A_1\cong B_1\) as both are isomorphic to \(k\). On the other hand, taking a basis \(u,v\) of \(\operatorname{\mathrm{H}}^0(E,\mathcal{O}_E(1))\) and sending \(x,y\) to \(u,v\) gives us an isomorphism \(A_2\to B_2\). So assume inductively that we have an isomorphism \(h_n:A_n\to B_n\). Then \(x\) and \(y\) are taken onto generators \(u\) and \(v\). As a vector space we can identify \(\operatorname{\mathrm{H}}^0(\mathcal{O}_E(n))\) with the homogenous polynomials of degree \(n\) in \(x,y\). And this gives the desired isomorphism.