Let \(X\) be an integral projective scheme of dimension 1 over \(k\), and let \(\tilde{X}\) be its normalization (II, Ex. 3.8). Then there is an exact sequence of sheaves on \(X\), \[ 0 \to \OO_X \to f_* \OO_{\tilde X} \to \sum_{P \in X} \tilde\OO_P/\OO_P \to 0 \] where \(\tilde\OO_P\) is the integral closure of \(\OO_P\). For each \(P \in X\), let \(\delta_P=\operatorname{length}(\tilde\OO_P/\OO_P)\).
Show that \(p_a(X)=p_a(\tilde{X})+\sum_{P \in X} \delta_P\). Hint: Use (III, Ex. 4.1) and (III, Ex. 5.3).
If \(p_a(X)=0\), show that \(X\) is already nonsingular and in fact isomorphic to \(\PP^1\). This strengthens (1.3.5).
* If \(P\) is a node or an ordinary cusp (I, Ex. 5.6, Ex. 5.14), show that \(\delta_P=1\). Hint: Show first that \(\delta_P\) depends only on the analytic isomorphism class of the singularity at \(P\). Then compute \(\delta_P\) for the node and cusp of suitable plane cubic curves. See (V, 3.9.3) for another method.
Solution
Let \[ \nu:\widetilde X\longrightarrow X \] be the normalization and put \[ \mathcal Q = \nu_*\mco_{\widetilde X}/\mco_X. \] Thus \[ \mathcal Q_P = \widetilde\mco_P/\mco_P, \qquad \operatorname{length}_{\mco_P}\mathcal Q_P = \delta_P. \]
1. The sheaf \(\mathcal Q\) is supported at finitely many closed points and\[ \dim_kH^0(X,\mathcal Q) = \sum_{P\in X}\delta_P. \]
The normalization is an isomorphism over the regular locus of the one-dimensional scheme \(X\). Hence \(\mathcal Q\) is supported on the finite singular locus.
At a closed point \(P\), the residue field is \(k\) because \(k\) is algebraically closed. A finite-length \(\mco_P\)-module therefore has \(k\)-dimension equal to its length. Since a sheaf supported on finitely many closed points is the direct sum of its stalk skyscraper sheaves, \[ H^0(X,\mathcal Q) \cong \bigoplus_{P\in X}\mathcal Q_P. \] Taking dimensions gives the formula.
2. There is a short exact sequence\[ 0 \longrightarrow H^0(X,\mathcal Q) \longrightarrow H^1(X,\mco_X) \longrightarrow H^1(\widetilde X,\mco_{\widetilde X}) \longrightarrow 0. \]
Apply cohomology to \[ 0 \longrightarrow \mco_X \longrightarrow \nu_*\mco_{\widetilde X} \longrightarrow \mathcal Q \longrightarrow 0. \] Because \(\nu\) is finite, \[ H^i(X,\nu_*\mco_{\widetilde X}) \cong H^i(\widetilde X,\mco_{\widetilde X}). \] Also \[ H^1(X,\mathcal Q)=0 \] because \(\mathcal Q\) has zero-dimensional support.
Both \(X\) and \(\widetilde X\) are integral and projective, so \[ H^0(X,\mco_X) \cong H^0(\widetilde X,\mco_{\widetilde X}) \cong k. \] The first map is the identity on constants. The long exact sequence therefore reduces to the displayed short exact sequence.
3. One has\[ \boxed{ p_a(X) = p_a(\widetilde X) + \sum_{P\in X}\delta_P. } \]
For an integral projective curve \(C\) over \(k\), \[ p_a(C) = 1-\chi(\mco_C) = h^1(C,\mco_C) \] because \(h^0(C,\mco_C)=1\). Thus step 2 and step 1 give \[ p_a(X) = p_a(\widetilde X) + \sum_{P\in X}\delta_P. \] This proves part (a).
4. If \(p_a(X)=0\), then \(X\) is nonsingular.
By step 3, \[ 0 = p_a(\widetilde X) + \sum_P\delta_P. \] The normalization \(\widetilde X\) is a smooth projective curve, so \(p_a(\widetilde X)\ge0\), and every \(\delta_P\) is a nonnegative integer. Hence \[ p_a(\widetilde X)=0 \qquad\text{and}\qquad \delta_P=0 \] for every \(P\).
Thus \[ \widetilde\mco_P=\mco_P \] for every \(P\), so every local ring of \(X\) is integrally closed. A one-dimensional noetherian integrally closed local domain is a DVR and therefore regular. Hence \(X\) is nonsingular.
5. If \(p_a(X)=0\), then\[ X\cong\PP^1. \]
By step 4, \(X\) is a smooth projective curve. Its genus is \[ g=p_a(X)=0. \] Exercise IV.1.6 gives a finite morphism \[ f:X\longrightarrow\PP^1 \] of degree at most \(1\). Since \(f\) is nonconstant, its degree is positive, so \[ \deg f=1. \] A finite degree-one morphism of integral smooth projective curves is birational, and a finite birational morphism to the normal curve \(\PP^1\) is an isomorphism. This proves part (b).
6. The number \(\delta_P\) depends only on the analytic isomorphism class of the singularity.
Put \[ A=\mco_{X,P}, \qquad B=\widetilde\mco_P. \] The quotient \[ B/A \] has finite length. Completion is exact on finite \(A\)-modules, and a finite-length module is unchanged by completion. Hence \[ \operatorname{length}_A(B/A) = \operatorname{length}_{\widehat A} \left( (B\otimes_A\widehat A)/\widehat A \right). \]
The local ring \(A\) is essentially of finite type over a field, hence excellent. For such a one-dimensional reduced local ring, normalization commutes with completion: \(B\otimes_A\widehat A\) is the normalization of \(\widehat A\) in its total ring of fractions. Therefore \[ \delta_P = \operatorname{length}_{\widehat A} \left( \overline{\widehat A}/\widehat A \right). \] The right-hand side depends only on the isomorphism class of the completed local ring \(\widehat A\). Thus \(\delta_P\) is an analytic invariant.
7. If \(P\) is an ordinary cusp, then\[ \delta_P=1. \]
Up to analytic isomorphism, an ordinary cusp has completed local ring \[ \widehat A \cong k[[t^2,t^3]] \subseteq k[[t]]. \] Its normalization is \(k[[t]]\). Every power \(t^n\) with \(n\ge2\) lies in \(k[[t^2,t^3]]\), so every series in \(k[[t]]\) is congruent modulo \(k[[t^2,t^3]]\) to a unique scalar multiple of \(t\). Hence \[ k[[t]]/k[[t^2,t^3]] \cong k\cdot t \] as a \(k\)-vector space. This quotient has length \(1\). Step 6 therefore gives \[ \delta_P=1. \]
8. If \(P\) is a node, then\[ \delta_P=1. \]
Up to analytic isomorphism, a node has completed local ring \[ \widehat A \cong k[[x,y]]/(xy). \] Its normalization is \[ k[[x]]\times k[[y]], \] with the map \[ k[[x,y]]/(xy) \longrightarrow k[[x]]\times k[[y]] \] sending the class of a power series \(h(x,y)\) to \[ \bigl(h(x,0),h(0,y)\bigr). \] Its image consists exactly of the pairs \[ (a(x),b(y)) \] with the same constant term: \[ a(0)=b(0). \] Therefore the homomorphism \[ k[[x]]\times k[[y]] \longrightarrow k, \qquad (a,b)\longmapsto a(0)-b(0) \] is surjective and has image of \(\widehat A\) as its kernel. Thus \[ \bigl(k[[x]]\times k[[y]]\bigr)/\widehat A \cong k, \] which has length \(1\). Step 6 gives \[ \delta_P=1. \]
9. Q.E.D.
Steps <1>1–<1>3 prove part (a), steps <1>4–<1>5 prove part (b), and steps <1>6–<1>8 prove part (c).