Let \(H\) be a very ample divisor on the surface \(X\), corresponding to a projective embedding \(X \subseteq \PP^N\). If we write the Hilbert polynomial of \(X\) (III, Ex. 5.2) as \[ P(z)=\frac{1}{2} a z^2+b z+c \] show that \(a=H^2\), \(b=\frac{1}{2} H^2+1-\pi\), where \(\pi\) is the genus of a nonsingular curve representing \(H\), and \(c=1+p_a\). Thus the degree of \(X\) in \(\PP^N\), as defined in (I, §7), is just \(H^2\). Show also that if \(C\) is any curve in \(X\), then the degree of \(C\) in \(\PP^N\) is just \(C . H\).
Solution
Let \(K_X\) be a canonical divisor on \(X\).
1. For every integer \(n\),\[ P(n) = \chi\bigl(\OO_X(nH)\bigr) = \chi(\OO_X) +\frac12 n^2H^2 -\frac12 n(H\cdot K_X). \]
Because the embedding is defined by the very ample divisor \(H\), its twisting sheaf is \[ \OO_X(1)\cong\OO_X(H). \] The cohomological description of the Hilbert polynomial in D-COHEULER therefore gives \[ P(n)=\chi\bigl(\OO_X(nH)\bigr). \]
Apply surface Riemann–Roch T-COHRRS to the divisor \(nH\): \[ \begin{aligned} \chi\bigl(\OO_X(nH)\bigr) &= \chi(\OO_X) +\frac12(nH)\cdot(nH-K_X)\\ &= \chi(\OO_X) +\frac12 n^2H^2 -\frac12 n(H\cdot K_X). \end{aligned} \]
2. Comparing coefficients with\[ P(z)=\frac12az^2+bz+c \]gives\[ a=H^2, \qquad b=-\frac12H\cdot K_X, \qquad c=\chi(\OO_X)=1+p_a(X). \]
Step 1 is a polynomial identity in \(n\), so comparison of its quadratic, linear, and constant coefficients gives the first two formulas and \[ c=\chi(\OO_X). \]
Since \(X\) is a surface, the definition of arithmetic genus in D-COHEULER is \[ p_a(X) = (-1)^2\bigl(\chi(\OO_X)-1\bigr) = \chi(\OO_X)-1. \] Thus \[ c=1+p_a(X). \]
3. If \(H_0\in\abs{H}\) is a nonsingular curve of genus \(\pi\), then\[ H\cdot K_X=2\pi-2-H^2. \]
Because \(H\) is very ample, Bertini’s theorem T-BERTINI gives a nonsingular member \[ H_0\in\abs H. \] Its divisor class is \(H\). The adjunction formula T-SRFADJ gives \[ 2\pi-2 = H_0\cdot(H_0+K_X) = H^2+H\cdot K_X. \] Rearranging proves the formula.
4. The linear coefficient is\[ \boxed{ b=\frac12H^2+1-\pi }. \]
Substitute step 3 into the expression for \(b\) from step 2: \[ \begin{aligned} b &=-\frac12(2\pi-2-H^2)\\ &=\frac12H^2+1-\pi. \end{aligned} \]
5. The degree of the embedded surface is\[ \boxed{\deg X=H^2}. \]
For a projective scheme of dimension \(2\), the leading coefficient of its Hilbert polynomial is \[ \frac{\deg X}{2!} \] by D-L6ERW. Step 2 says that the leading coefficient here is \[ \frac12H^2. \] Therefore \[ \deg X=H^2. \]
6. If \(C\subseteq X\) is any curve, then\[ \boxed{\deg_{\PP^N}C=C\cdot H}. \]
The hyperplane bundle of the given embedding restricts to \[ \OO_C(1) \cong \OO_X(H)|_C. \] The degree of the embedded curve \(C\subseteq\PP^N\) is the degree of this hyperplane line bundle. By the definition of the surface intersection pairing D-SRFINT, \[ \deg_C\bigl(\OO_X(H)|_C\bigr)=C\cdot H. \] Hence \[ \deg_{\PP^N}C=C\cdot H. \]
7. Q.E.D.
Steps 2 and 4 give the three coefficients of the Hilbert polynomial, step 5 identifies the degree of \(X\), and step 6 proves the degree formula for every curve \(C\subseteq X\).