(page 1) Chapter 13

numerical equivalence regular very ample

Good Curves

Our immediate task is to clarify the notion of a “good curve” on a surface X, so that we can choose a numerical equivalence class ξ satisfying Condition I of the previous chapter.

Start by fixing an embedding of X into projective space, and let O(1) be the corresponding very ample invertible sheaf. We can think of at least four properties that might single out an invertible sheaf L that is represented by a good curve.

(I) L is 0-regular, so that H^{i}(X, L(n)) = 0 for all i > 0 and n + i ≥ 0.

(II) L is generated by its global sections. Equivalently, for every closed point

x ∈ X there exists a curve D ⊂ X such that O_{X}(D) ≈ L and x ∉ Supp(D). (III) L is very ample. (IV) There is a curve D ⊂ X such that O_{X}(D) ≈ L and D has no multiple

components.

Lemma 13.1. If L has any of the properties (I)-(IV), then L(n) has the same property for all n > 0.

Proof: For (I), we already proved this. For (II), we start with a sheaf L such that the natural map

O_{X} ⊗ Γ(L) → L

is surjective. Now consider the composite map

O_{X} ⊗ Γ(O_{X}(1)) ⊗ Γ(L) → O_{X}(1) ⊗ Γ(L) → L(1).

The first map is surjective since O(1) is generated by global sections, and the second map is surjective because it is a twist of the earlier surjection. Thus the composite is surjective. Since the composite factors through the natural map

O_{X} ⊗ Γ(L(1)) → L(1),

we see that L(1) is also generated by global sections.

For (III), it is enough to show that if L is very ample and if M is generated by global sections, then L ⊗ M is also very ample. To prove this claim, let ϕ : X → P^{n} be the closed immersion corresponding to the very ample sheaf L and let

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2 Chapter 13: Good Curves

Segre embedding Bertini’s Theorem Castelnuovo’s Regularity Theorem Bertini’s Theorem Serre’s Finiteness Theorem Riemann-Roch Theorem for Curves Serre Duality

ψ : X → P^{m} be the morphism defined by the sheaf M. Composing with the Segre embedding gives a map

X ⟶^{(ϕ,ψ)}P^{n} × P^{m}⟶^{σ}P^{n+m}.

This map is a closed immersion, and

(σ ∘ (ϕ, ψ))^{∗}(O(1)) = ϕ^{∗}(O(1)) ⊗ ψ^{∗}(O(1)) = L ⊗ M.

For (IV), we use a weak form of Bertini’s Theorem: If L is a very ample invertible sheaf on X, then there exists a nonsingular irreducible curve C ⊂ X such that O_{X}(C) ≈ L. If L has property (IV), we can represent L by a divisor D = ∑_{i=1}^{n} D_{i}, where the D_{i} are distinct irreducible curves. Now it is possioble that some of those curves represent (positive) multiples of O(1). Collect all such factors together into a single piece E = ∑_{i=1}^{n_{0}} D_{i}. By Bertini’s Theorem, we can represent the very ample sheaf O_{X}(E + 1) by a single irreducible curve C, and thus we can represent L(1) by the reduced divisor C + ∑_{i=n_{o}+1}^{n} D_{i}.

Theorem 13.2. There exists an integer m depending only on the embedding of X into projective space such that if an invertible sheaf L has any of the properties (I)-(IV), then L(m) has all four properties.

Proof: Castelnuovo’s Regularity Theorem shows that if L has property (I), then L also has property (II). In the proof of the previous result, we showed that if L has property (II), then L(1) has property (III). The form of Bertini’s Theorem used in the previous result also shows that if L has property (III), then it also has property (IV). The only hard part, therefore, is to cycle back around from (IV) to (I).

Write L = O_{X}(D) where D is a curve with no multiple components. Write L_{D} = L ⊗ O_{D}. Twist the exact sequence

0 → O_{X}(D) → O_{X} → O_{D} → 0

by L(n) to obtain the sequence

(13.3)  0 → O_{X}(n) → L(n) → L_{D}(n) → 0.

By Serre’s Finiteness Theorem, there is an integer n_{0} such that H^{i}(X, O_{X}(n)) = 0 for all i > 0 and n ≥ n_{0}. It now follows from (13.3) that

H^{2}(L(n)) = 0 and H^{1}(L(n)) = H^{1}(L_{D}(n)).

Apply the Riemann-Roch Theorem for Curves on D to compute this final cohomology group. Let ω_{X} be the canonical sheaf on X. By the genus formula, we can write the canonical sheaf on D as ω_{D} = (ω_{X} ⊗ O_{X}(D)) ⊗ O_{D}. By Serre Duality, we have

dim(H^{1}(L_{D}(n))) = dim(H^{0}(L_{D}^{−1}(−n) ⊗ ω_{D})).

However,

L_{D}^{−1}(−n) ⊗ ω_{D} = L^{−1} ⊗ O_{X}(−n) ⊗ ω_{X} ⊗ O_{X}(D) ⊗ O_{D} = ω_{X}(−n) ⊗ O_{D}.

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Now there is an integer n_{1} such that H^{i}(ω_{X}^{−1}(n)) = 0 for all i > 0 and all n ≥ n_{1}. So, ω_{X}^{−1}(n) is 0-regular provided that n ≥ n_{1} +2. By the first part of the argument, we see that ω_{X}^{−1}(n) is very ample if n ≥ n_{1} + 3. Restricting to D, it remains very ample. It follows that H^{0}(ω_{X}(−n)⊗O_{D}) = 0 if n ≥ n_{1} +3. Thus, L(n) is 0-regular provided that

n ≥ max(n_{0} + 2, n_{1} + 4).

Lemma 13.4. There is a constant c_{1} depending only on X and O_{X}(1) such that for all invertible sheaves L, if deg(L) ≥ c_{1}, then H^{2}(L) = 0.

Proof: Let H be a nonsingular irreducible curve on X representing the sheaf O(1). For all integers k one has the standard sequence

0 → L(k − 1) → L(k) → L_{H}(k) → 0.

If deg(L) = deg_{H}(L_{H}) > 2p_{a}(H) − 2, then by the usual corollary of the Riemann-Roch Theorem on the curve H, one has H^{1}(L_{H}) = 0. Necessarily, one also has H^{1}(L_{H}(k)) = 0 for all k > 0. From our standard sequence, one concludes that

H^{2}(L) = H^{2}(L(1)) = H^{2}(L(2)) = · · · .

Since the cohomology group H^{2}(L(k)) vanishes for k >> 0, the result follows.

Corollary 13.5. With c_{1} as above, if deg(L) ≥ c_{1} and χ(L) > 0, then there exists an effective curve D representing L.

Proof: We have

χ(L) = h^{0}(L) − h^{1}(L) + h^{2}(L) = h^{0}(L) − h^{1}(L).

So, the assumptions force h^{0}(L) > 0.

Corollary 13.6. With c_{1} as above, let h = deg(O_{X}(1)). If an invertible sheaf L satisifes deg(L) ≥ c_{1} + h and if h^{1}(L) = 0, then L(2) satisfies all the properties (I)-(IV).

Proof: We have

deg(L(−1)) = deg(L) − h ≥ c_{1} + h − h = c_{1}.

So, by the previous corollary, H^{2}(L(−1)) = 0. Thus, L(1) is 0-regular. Therefore, L(2) is very ample.

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Theorem 13.7. There exists a constant c_{2} and a positive constant ϵ, depending only on X and O_{X}(1), with the following property. If L is any invertible sheaf on X satisfying the conditions

(i) deg(L) ≥ c_{2}, and

(ii) χ(L) ≥ (1 − ϵ)\frac{(deg L)^{2}}{2(O(1).O(1))},

then L is 0-regular and very ample.

Proof: Here’s the plan. We are actually going to show that L(−n) is 0-regular and very ample for some positive integer n. It then follows from our earlier work that L, being a positive twist of a good sheaf, has the same properties. So, let c_{1} be the constant of Lem. 13.4, and let m be the constant of Thm. 13.2. Define

h = (O(1).O(1)).

Step 1. We will first explain how to choose an integer k such that deg(L(−k)) ≥ c_{1}. We have

deg(L(−k)) = deg(L) − k deg(O(1)) = deg(L) − kh.

Since we’d like to factor the term deg(L) out of this entire expression, let’s write k in the form

(13.8)  k = (1 − u)\frac{deg L}{h}

for some constant u. Now we have

deg(L(−k)) = deg(L) − \frac{(1 − u) deg(L)}{h}h = u deg(L)

Using condition (i), we can get deg(L(−k)) ≥ c_{1} provided

(13.9)  u ≥ 0 and c_{2} ≥ \frac{c_{1}}{u} .

Step 2. Next, we will explain how to ensure that χ(L(−k)) > 0 for the same value of k. To do this, let H be a nonsingular irreducible curve that represents O(1). Write

p = χ(O_{X}) − χ(O_{X}(−1)) = χ(O_{H}).

Recall that we have an exact sequence

0 → L(j − 1) → L(j) → L_{H}(j) → 0

for all integers j. Inductively, we find that

χ(L(−k)) = χ(L) − ∑_{i=0}^{k−1} χ(L_{H}(−i)).

Applying the Riemann-Roch Theorem for Curves to all these sheaves on H, we get Riemann-Roch Theorem for Curves

χ(L(−k)) = χ(L) − kχ(O_{H}) − ∑_{i=0}^{k−1} deg_{H}(L_{H}(−i))

= χ(L) − kp − ∑_{i=0}^{k−1}(deg(L) − ih)

= χ(L) − kp − k deg(L) + \frac{k(k − 1)}{2}h

= χ(L) − \frac{k}{2} (2 deg(L) + 2p + h) + \frac{k^{2}h}{2}.

Now substitute in the known value for k = (1 − u) deg L/h to get

χ(L(−k)) = χ(L) − \frac{(1 − u)(2 deg(L) + 2p + h)}{2h} deg(L) + \frac{(1 − u)^{2}}{2h} (deg(L))^{2}

= χ(L) − \frac{(1 − u)(2p + h)}{2h} deg(L) − \frac{(1 − u^{2})}{2h}(deg(L))^{2}.

Using the bound on χ(L) given by condition (ii), we have

χ(L(−k)) ≥ \frac{(u^{2} − ϵ)}{2h}(deg(L))^{2} − \frac{(1 − u)(2p + h)}{2h} deg(L)

= \frac{deg(L)(u^{2} − ϵ)}{2h} (deg(L) − \frac{(1 − u)(2p + h)}{(u^{2} − ϵ)}) .

If u^{2} > ϵ, then we can replace the second occurrence of deg(L) with c_{2}. Thus, we see that χ(L(−k)) > 0 provided

(13.10)  u^{2} > ϵ and c_{2} ≥ \frac{(1 − u)(2p + h)}{(u^{2} − ϵ)}.

Step 3. In this step, we will use the fact that for any effective curve D of degree d, the sheaf O_{X}(−D + d) is generated by global sections. The proof of this result will be deferred to the end of the chapter.

Now it follows from Cor. 13.5 that L(−k) = O_{X}(D) for some effective curve D. Letting d = deg(D), we see in particular that O_{X}(−D + d) = O_{X}(E) for some effective curve E. The degree of E is

deg(E) = − deg(D) + dh = d(h − 1).

Applying the same result a second time, we see that the sheaf O_{X}(−E + d(h − 1)) is generated by global sections. However,

O_{X}(−E + d(h − 1)) = O_{X}(D − d + d(h − 1))
= O_{X}(D) ⊗ O_{X}(d(h − 2))
= L(−k) ⊗ O_{X}(d(h − 2))
= L(−k + d(h − 2)).

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It follows from Thm. 13.2 that L(−k + d(h − 2) + m) is 0-regular and very ample. Step 4. The theorem will be proven if we can determine how to ensure that the quantity m − k + d(h − 2) is negative. Since

d = deg(D) = deg(L) − kh,

we have

m − k + d(h − 2) = m − k + (deg(L) − kh)(h − 2)

= m + (h − 2) deg(L) − k(1 + h(h − 2))

= m + (h − 2) deg(L) − \frac{(1 − u) deg(L)}{h}(1 + h(h − 2))

= m + \frac{deg(L)}{h}(h(h − 2) − (1 − u)(1 + h(h − 2)))

= m + \frac{deg(L)}{h}(−1 + u(h − 1)^{2})

= −\frac{1 − u(h − 1)^{2}}{h} (deg(L) − \frac{mh}{1 − u(h − 1)^{2}} ) .

Now suppose we can arrange things so that

1 − u(h − 1)^{2} ≥ 0.

Then we have

m − k + d(h − 2) ≤ −\frac{1 − u(h − 1)^{2}}{h} (c_{2} − \frac{mh}{1 − u(h − 1)^{2}} ) .

Thus, the entire expression is negative if we can satisfy the conditions

(13.11)  1 − u(h − 1)^{2} ≥ 0 and c_{2} ≥ \frac{mh}{1 − u(h − 1)^{2}} .

Each of our three steps is conditional on satisfying certain constraints; namely, the equations and inequalities listed in (13.9), (13.10), and (13.11). To finish the proof, we nned to explain how to satisfy all these constraints. But this is easy. We start by choosing any number u such that

0 < u ≤ \frac{1}{(h − 1)^{2}} ,

and use that u to define k by formula (13.8). Next, we define

ϵ = \frac{u^{2}}{2}

and

c_{2} = max (\frac{c_{1}}{u} , \frac{2(1 − u)(2p + h)}{u^{2}}, \frac{mh}{1 − u(h − 1)^{2}} ) .

Now all the necessary conditions are met, and the theorem follows.

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Corollary 13.12. Let h, ω ∈ Num(X) be the numerical equivalence classes of the invertible sheaves O(1) and of ω_{X}. There exist constants c_{2} and ϵ such that if an element λ ∈ Num(X) satisfies

(i) λ.h ≥ c_{2}

(ii) λ.(λ − ω) ≥ (1 − ϵ) [\frac{(λ.h)^{2}}{h.h} ]

then every L ∈ P ic(X) representing = lambda is 0-regular and very ample.

Proof: The first condition is a direct transaltion of the result of the theorem, since by definition one has

deg(λ) = λ.h.

For the second condition, we use the Riemann-Roch Theorem for Surfaces, which tells us that

χ(λ) = \frac{1}{2}(λ.(λ − ω) + χ(O_{X})

≥ (1 − ϵ) [\frac{(λ.h)^{2}}{2h.h} ] + χ(O_{X})

≥ (1 − ϵ) [\frac{(deg(L))^{2}}{2h.h} ] + χ(O_{X}).

We get the result by increasing ϵ slightly, as necessary to remove the constant term χ(O_{X}).

The final result in this chapter is the proof of the result that was used in Step 3 of the preceding theorem.

Theorem 13.13. Let D ⊂ X be a curve of degree d. Then O_{X}(−D + d) is generated by global sections.

Proof: let X → P^{n} be the closed immersion defined by the very ample sheaf O_{X}(1). Let L ⊂ P^{n} be a linear subspace of dimension n−3, and let π : P^{n}−L → P^{2} be the projection with center L. (More concretely, we can write L = H_{1} ∩ H_{2} ∩ H_{3} as an intersection of three hyperplanes, and let h_{i} ∈ H^{0}(P^{n}, O(1)) be a linear form defining H_{i}. Then the three sections h_{1}, h_{2}, h_{3} have no common zero in P^{n} − L, so they define a morphism π.) Choose L so that L ∩ X = ∅. Then the restriction of π defines a morphism ϕ : X → P^{2}. The theorem will follow from the following claims:

(i) ϕ is finite and flat.

(ii) ϕ^{∗}(ϕ_{∗}(D)) = D + E, where E is an effective divisor.

(iii) For all closed points x ∈ X, there exists a linear subspace L of dimension n − 3 such that the divisor E produced by the previous construction does not pass through x.

We need (i) in order to know that the push forward of D is a divisor on P^{2}. In fact, it follows that the push forward is π_{∗}(D) = dH, where H is a line in P^{2}. So,

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the pullback satisfies

π^{∗}π_{∗}(O_{X}(D)) = π^{∗}O(d) = O_{X}(d).

Thus, the second claim tells us that

O_{X}(−D + d) = O_{X}(E).

Now the theorem follows from the third claim and the definition of “generated by global sections”.

To verify claim (i), we observe first that π is an affine morphism. Thus, its restriction ϕ is also an affine morphism. Now ϕ factors as the composite

X⟶^{(ι,ϕ)}P^{n} × P^{t}⟶^{p_{2}}P^{2}.

Since the first map is an isomorphism onto a closed subscheme, we can identify the direct image ϕ_{∗}(O_{X}) with the direct image (p_{2})_{∗}(O_{X}). Since the projection is proper, the latter sheaf is coherent. Therefore, ϕ is a finite map. That ϕ is flat follows from the following result in commutative algebra (which we shall not prove, but let you look up in Nagata @@ @@):

Lemma 13.14. Let A be a regular local ring of dimension n, and let B be an A-algebra that is finitely generated as an A-module. If all the localizations of B with respect to maximal ideals are Cohen-Macaulay of dimension n, then B is a free A-module.

To verify claim (ii), we must realize that the direct image of a Cartier divisor under a finite, flat morphism is computed by taking the norms of the local defining equations. The effectivity follows from the fact that if one has a finite flat map of algebras A → B, then for all β ∈ B there exists β′ ∈ B such that

N_{B/A}(β) = ββ′.

This claim, in turn, is a consequnce of the Cayley-Hamilton Theorem, applied to the matrix T_{β} that represents the A-linear map B → B defined by multiplication by β.

Finally. let’s verify claim (iii). Let x ∈ X be closed, and let y = ϕ(x) ∈ P^{2}. Let O = O_{P^{2},y} be the local ring at y in P^{2}, and let R be the stalk of ϕ_{∗}(O_{X}) at y. Let g ∈ R be a local equation of D at the points ϕ^{−1}(y), and let M ⊂ R be the maximal ideal such that R_{M} = O_{X,x}. Passing to the completions of the local rings, we find that

R̂ = R ⊗_{O} Ô = R̂_{M} ⊕ ⨁_{othermax} R̂_{M′}.

The image of N(g) in Ô is the product of the norms of g from all the summands. However, we want g and N(g) to differ by a unit. So, we need two properties to hold.

(a) g is a unit at all other localizations R_{M′} of R.

(b) Ô ≈ R̂_{M}.

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The first condition says that Supp(D) does not contain any points x′ ≠ x such that ϕ(X′) = ϕ(x). We can reinterpret this condition in terms of the linear subspace L. If we write L_{x} for the n − 2-dimensional linear subspace spanned by L and x, then x must be the only intersection point of L_{x} with Supp(D).

Now the second condition says that R_{M} is unramified over O, or equivalently, that the map from the Zariski tangent space of X at x to the Zariski tangent space of P^{2} at y is an isomorphism. Looking at the full Zariski tangent space T of P^{n} at x, we see that it contains both the tangent space T_{L_{x}} of L_{x}, which has dimension n − 2, and the tangent space T_{X} of X, which has dimension 2. Moreover, the fill projection π identifies the quotient T/T_{L_{x}} with the tangent space of P^{2} at y. So, the second condition simply asserts that T_{L_{x}} should intersect T_{X} transversely at x.

To complete the proof, we need to show how, given x, to choose an L such that L_{x} satisfies the reinterpreted versions of conditions (a) and (b). We simply let P denote the 2-dimensional linear space through x with tangent space T_{X} at x. (In other words, P is the tangent plane to X at x, viewed as a subspace of P^{n}.) Now choose a linear form h_{1} such that h_{1}(x) = 0 but h_{1}(η) ≠ 0 if η is a generic point of P or of Supp(D). Let H_{1} be the hyperplane defined by h_{1} = 0. Now choose a second linear form h_{2} such that h_{2}(x) = 0 but h_{2}(η) ≠ 0 if η is any of

• a generic point of P ∩ H_{1},

• a generic point of X ∩ H_{1}, or

• any point of Supp(D) ∩ H − {x}. Let H_{2} be the corresponding hyperplane. The n − 2-dimensional linear space L_{x} = H_{1} ∩ H_{2} satisfies (a) and (b) by construction. Moreover, L_{x} ∩ X is a finte set. So, we can take L to be any linear subspace of L_{x} ∩ X that misses the finite set L_{x} ∩ X.

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