If \(D\) is an ample divisor on the surface \(X\), and \(D^{\prime} \equiv D\), then \(D^{\prime}\) is also ample. Give an example to show, however, that if \(D\) is very ample, \(D^{\prime}\) need not be very ample.
Solution
1. Ampleness of divisors on a smooth projective surface depends only onnumerical equivalence.
Let \(D\) be ample and suppose \[ D'\equiv D. \] Numerical equivalence preserves all intersection numbers, so \[ (D')^2=D^2>0 \] and, for every irreducible curve \(C\subseteq X\), \[ D'\cdot C=D\cdot C>0. \] The two inequalities on the right are exactly the Nakai–Moishezon criterion T-SRFNAKAI. Hence \[ \boxed{D'\text{ is ample}.} \]
2. Let \(C\) be any nonsingular projective curve of genus \(g\geq3\). There isa degree-\(2g\) line bundle \(L\) on \(C\) such that\[ L\not\cong K_C(P+Q) \]for every pair of points \(P,Q\in C\), allowing \(P=Q\).
The variety \(\Pic^{2g}(C)\) is a translate of the Jacobian, hence is irreducible of dimension \(g\) by T-CRVJACFUN. Consider the morphism \[ \Sym^2C\longrightarrow\Pic^{2g}(C), \qquad P+Q\longmapsto K_C(P+Q). \] Its source is projective of dimension two, so its image is closed and has dimension at most two. Since \(g\geq3\), this image is a proper subset of the \(g\)-dimensional variety \(\Pic^{2g}(C)\). Choose \(L\) outside that image.
3. The line bundle \(L\) chosen in step 2 is very ample.
Since \[ \deg L=2g>2g-2, \] Riemann–Roch gives \[ h^0(C,L)=2g+1-g=g+1. \] Let \(Z=P+Q\) be any effective divisor of degree two, including the case \(P=Q\). Riemann–Roch for \(L(-Z)\) gives \[ h^0(C,L(-Z)) = g-1+h^0\qty(C,K_C\tensor L^{-1}(Z)). \] The line bundle \[ K_C\tensor L^{-1}(Z) \] has degree zero. A degree-zero line bundle on an integral projective curve has a nonzero global section only when it is trivial. By the choice of \(L\) in step 2, it is never trivial. Therefore \[ h^0(C,L(-Z))=g-1=h^0(C,L)-2 \] for every length-two effective divisor \(Z\).
Thus the complete linear system of \(L\) separates every pair of points and every tangent direction. The closed-immersion criterion for a complete linear system, T-DIVMAPPN, gives \[ \boxed{L\text{ is very ample}.} \]
4. For any points \(P,Q\in C\), the degree-\(2g\) line bundle\[ L'=K_C(P+Q) \]is not very ample.
Again \(\deg L'=2g\), so Riemann–Roch gives \[ h^0(C,L')=g+1. \] But \[ L'(-P-Q)\cong K_C, \] and therefore \[ h^0(C,L'(-P-Q))=h^0(C,K_C)=g. \] For a very ample line bundle, imposing the length-two subscheme \(P+Q\) would lower \(h^0\) by two, giving \(g-1\). Here it lowers \(h^0\) by only one. Hence the complete linear system fails to separate the corresponding two points or tangent direction, and \[ \boxed{L'\text{ is not very ample}.} \]
5. Put\[ X=C\times\PP^1 \]with projections \(p_1,p_2\), and define\[ M=p_1^*L\tensor p_2^*\OO_{\PP^1}(1), \qquad M'=p_1^*L'\tensor p_2^*\OO_{\PP^1}(1). \]Then\[ M\equiv M' \]numerically on \(X\).
Their quotient is \[ M\tensor(M')^{-1} = p_1^*(L\tensor(L')^{-1}). \] The line bundle \(L\tensor(L')^{-1}\) has degree zero on \(C\).
Let \(\Gamma\subseteq X\) be any irreducible curve and let \(\widetilde\Gamma\to\Gamma\) be its normalization. If the induced map \[ \widetilde\Gamma\longrightarrow C \] is constant, the pullback of \(L\tensor(L')^{-1}\) has degree zero. If it is nonconstant of degree \(e\), then its pullback has degree \[ e\deg\qty(L\tensor(L')^{-1})=0. \] Thus \[ c_1\qty(M\tensor(M')^{-1})\cdot\Gamma=0 \] for every irreducible curve \(\Gamma\). Hence \(M\) and \(M'\) are numerically equivalent.
6. The line bundle \(M\) is very ample on \(X\).
By step 3, \(L\) defines a closed immersion \[ C\hookrightarrow\PP^r. \] The line bundle \(\OO_{\PP^1}(1)\) defines the standard closed immersion of \(\PP^1\). Their product is a closed immersion \[ C\times\PP^1\hookrightarrow\PP^r\times\PP^1. \] Composing with the Segre closed immersion gives an embedding into projective space whose hyperplane bundle pulls back to \[ p_1^*L\tensor p_2^*\OO_{\PP^1}(1)=M. \] Therefore \[ \boxed{M\text{ is very ample}.} \]
7. The numerically equivalent line bundle \(M'\) is not very ample.
4. Hence\[ \boxed{M'\text{ is not very ample}.} \]
8. There are numerically equivalent divisors \(D,D'\) on the surface \(X\) with\(D\) very ample and \(D'\) not very ample.
Because \(X\) is smooth and integral, choose Cartier divisors \(D,D'\) with \[ \OO_X(D)\cong M, \qquad \OO_X(D')\cong M'. \] Step 5 gives \[ D\equiv D'. \] Steps <1>6–<1>7 give respectively that \(D\) is very ample and \(D'\) is not. This is the required counterexample.
9. Q.E.D.
Step 1 proves numerical invariance of ampleness, and steps <1>2–<1>8 give the required failure of numerical invariance for very ampleness.