Bryden Cais.  Algebraic Geometry, HW 6.  April 16, 2003

1. Let X = 𝔸_{k}^{1} − {(0, 0)}, with coordinates x, y and consider U_{1}, U_{2} be the affine open cover given by U_{1} = (x ≠ 0) and U_{2} = (y ≠ 0). Let A = k[x, y]. Observe that we have Γ(U_{1}, O_{X}) = A_{x}, Γ(U_{2}, O_{X}) = A_{y}, and Γ(U_{1} ∩ U_{2}, O_{X}) = A_{x,y} (where by A_{x,y} we mean the ring A[x^{−1}, y^{−1}]. We have

C^{p}(U, O_{X}) = \begin{cases}A_{x} × A_{y} & if p = 0 \\ A_{x,y} & if p = 1 \\ 0 & otherwise \end{cases} .

It follows that H^{1}(X, O_{X}) = C^{1}/ im(d : C^{0} → C^{1}), where d is given by d(f_{1}, f_{2}) = f_{1} − f_{2}. Writing f_{1} ∈ A_{x} and f_{2} ∈ A_{y} as f̂_{1}/x^{i} and f̂_{2}/y^{j} respectively, where f̂_{s} ∈ A for s = 1, 2 and f̂_{1}/x ∉ A, f̂_{2}/y ∉ A, we see that f_{1} − f_{2} = F/x^{i}y^{j} where every monomial in F either has degree at least j in y or degree at least i in x. On the other hand, every element of A_{x,y} has the form G/x^{k}y^{l} with G not divisible by x or y. Writing G as a sum of monomials, we see that we can write

\frac{G}{x^{k}y^{l}} = ∑ c_{i,j} \frac{1}{x^{i}y^{j}} + \frac{F}{x^{k}y^{l}}

with c_{i,j} ∈ k and F having every monomial with x degree at least k or y degree at least l. It readily follows that H^{1}(X, O_{X}) is spanned by \{\frac{1}{x^{i}} y^{j} : i, j > 0\} as a k-vector space.

2. Let X be a possibly singular plane curve of degree d so that we have the embedding i : X ↪ ℙ^{2}. Then we have the exact sequence of sheaves on ℙ^{2}

0 ⟶ O_{ℙ^{2}}(−d) ⟶ O_{ℙ^{2}} ⟶ i_{∗}O_{X} ⟶ 0.

This gives rise to a long exact sequence of cohomology. In particular, we have

… → H^{1}(ℙ^{2}, O_{ℙ^{2}}) → H^{1}(ℙ^{2}, i_{∗}O_{X}) → H^{2}(ℙ^{2}, O_{ℙ^{2}}(−d)) → H^{2}(ℙ^{2}, O_{ℙ^{2}}) → ….

From our computations of the cohomology of ℙ^{n} in class, we know that H^{1}(ℙ^{2}, O_{ℙ^{2}}) = 0. Now by Serre Duality, we have

H^{2}(ℙ^{2}, O_{ℙ^{2}}) ≃ H^{0}(X, O_{ℙ^{2}}(−3))^{∗} = 0,

since O_{ℙ^{2}}(−3) has no global sections, so that we have an isomorphism H^{1}(ℙ^{2}, i_{∗}O_{X}) ≃ H^{2}(ℙ^{2}, O_{ℙ^{2}}(−d))^{∗}. Again, by Serre Duality we obtain

H^{2}(ℙ^{2}, O_{ℙ^{2}}(−d)) ≃ H^{0}(X, O_{ℙ^{2}}(d − 3))^{∗}.

Since H^{1}(ℙ^{2}, i_{∗}O_{X}) ≃ H^{1}(X, O_{X}) we have h^{1}(O_{X}) = dim_{k} H^{0}(X, O_{ℙ^{2}}(d − 3)). But this k vector space is spanned by degree d − 3 homogenous polynomials in three variables; it follows that

h^{1}(O_{X}) = \binom{d − 3 + 2}{2} = \frac{1}{2}(d − 1)(d − 2).

Since X is a plane curve, we have dim X = 1 so that H^{i}(X, O_{X}) = 0 for i ≥ 2 whence h^{i}(O_{X}) = 0 for i ≥ 2. Finally, H^{0}(X, O_{X}) = k since X is a projective variety so that h^{0}(O_{X}) = 1.

3. Let X be a curve and D a divisor on X.

(a) Let f ∈ H^{0}(D) = Γ(X, O_{X}(D)). Then ((f) + D) ∩ X ≥ 0, so setting D′ = ((f) + D) ∩ X gives D′ ∼ D with D ≥ 0. This gives a map H^{0}(D) → |D| given by f ↦ ((f) + D) ∩ X. Observe that the image of f̃ coincides with the image of f under this map if and only if f̃ = cf for some c ∈ k. This gives an isomorphism ℙH^{0}(D) ≃ |D|.

(b) Recall that for any f ∈ k(X) we have deg(f) = 0 (since (1 : f) gives a map π : X → ℙ^{1}, we have deg(f) = deg π deg((1 : 0)−(0 : 1)) = 0). It follows that if deg D < 0 then deg(((f)+D)∩X) = deg(f)+deg D = deg D so that D cannot be equivalent to an effective divisor; hence, by part (a), since |D| = 0 we have h^{0}(D) = 0.

(c) Suppose that deg D = 0. Since for any effective divisor D′ we have deg D′ > 0 unless D′ ∼ 0, we see that H^{0}(D) = 0 unless D′ = D + (f) ∼ 0 for some f, i.e. D ∼ 0. In this latter case, there is a unique (up to scalar multiplication) f ∈ k(X)^{×} satisfying D + (f) = 0, namely the function f with divisor −D (since D has degree 0, we know such a function exists). Thus, H^{0}(D) = kf and h^{0}(D) = 1, as required.

(d) We proceed by induction on the degree of D. Suppose that h^{0}(D) ≤ deg D +1 for all divisors D of degree k. Now any divisor of degree k + 1 has the form D + P for some divisor D of degree k and some point P. Let h, g ∈ H^{0}(D + P)/H^{0}(P) (the quotient of k vector spaces makes sense as H^{0}(D) ⊂ H^{0}(D + P)). Then we may assume that v_{P} (h) = v_{P} (g), for if not, then v_{P} (h) < v_{P} (g), say, so that in fact h ∈ H^{0}(D) and h is hence trivial in the quotient. But now since v_{P} (h) = v_{P} (g), there exist a, b ∈ k such that v_{P} (ah − bg) < v_{P} (h) so that ah − bg ∈ H^{0}(D). It follows that g, h are linearly dependent in H^{0}(D + P)/H^{0}(D) so that dim_{k}(H^{0}(D + P)/H^{0}(D)) ≤ 1. This gives h^{0}(D + P) ≤ h^{0}(D) + 1, which is the required induction step. Since the base case holds by part (c) we have shown that

h^{0}(D) ≤ deg D + 1.

Now we have already seen that equality holds if X ≃ ℙ^{1}. For the converse, suppose that equality holds. Let P be any prime divisor of D. We claim that there exists a function f ∈ H^{0}(D) with a single simple pole at P. Indeed, if h, g ∈ H^{0}(D) are linearly independent with v_{P} (h) = v_{p}(g) then there exist a, b ∈ k such that v_{P} (ah − bg) < v_{P} (H) and f = ah − bg is non-constant since h, g are linearly independent. The existence of f now follows by descent. But then the map ϕ : X → ℙ^{1} given by (1 : f) is surjective (since f is nonconstant) and degree 1 (since f has a single simple pole) and is hence an isomorphism.

5. Let X be a curve of genus g ≥ 2 and let G = Aut X have order n. Then G acts on K(X) with fixed field K(X)^{G} = L; let Y be the curve corresponding to L. Then the Galois field extension K(X)/L gives a morphism f : X → Y of degree n. Now let P ∈ X be a ramification point with e_{P} = r. Then for any Q ∈ f^{−1}f(P) we have Q = gP for some g ∈ G (since K(X)/L is Galois with Galois group G) so that the isotropy subgroups of G at P, Q are conjugate and therefore have the same index in G; since this index is exactly e_{P} = e_{Q}, we see that every point in f^{−1}f(P) has the same ramification index. Moreover, from the formula n = ∑_{Q∈f^{−1}f(P)} e_{p}, we see that f^{−1}f(P) consists of n/r points, each having ramification index r. Letting P_{1}, …, P_{s} be a maximal set of ramification points of X over distinct points of Y we obtain, by the Riemann Hurwitz formula

2g − 2 = n(2g(Y) − 2) + ∑_{Q}(e_{Q} − 1)

= n(2g(Y) − 2) + ∑_{i=1}^{s} ∑_{Q∈f^{−1}f(Q)}(e_{P_{i}} − 1)

= n(2g(Y) − 2) + ∑_{i=1}^{s} \frac{n}{r_{i}} (r_{i} − 1).

It follows that

(2g − 2)/n = 2g(Y) − 2 + ∑_{i=1}^{s} (1 − \frac{1}{r_{i}} ) .

We now establish a lower bound for the right hand side. Indeed, the left hand side is positive so that the right hand side must be also. Moreover, we have r_{i} ≥ 2 and g(Y) ≥ 0. Certainly, the minimal value will occur when g(Y) = 0 since this is always an integer. Since r_{i} ≥ 2, we find that 1/2 ≤ (1 − 1/r_{i}) < 1. From this, it is clear that in order to have the right hand side above greater than 0, we must have 2 < s. Moreover, with s = 3 and r_{1} = r_{2} = r_{3} = 6 we get a positive value for the RHS; it follows that s ≤ 3 so that s = 3. We then seek to minimize 1 − 1/r_{1} − 1/r_{2} − 1/r_{3}. We use the greedy algorithm so that r_{1} = 2 and r_{2} = 3 whence 1 − 1/r_{2} − 1/r_{2} = 1/6 so we must take r_{3} = 7. It follows that the minimum value is 1 − 1/2 − 1/3 − 1/7 = 1/42. This gives (2g − 2)/n ≥ \frac{1}{42}, or equivalently,

n ≤ 84(g − 1).

6. Let X be a compact differentiable manifold and let 𝔸 denote the sheaf of smooth C-valued functions on X. Then the exponential sequence

0 ⟶ Z ⟶ 𝔸 ⟶ 𝔸^{×} ⟶ 0.

gives rise to a long exact sequence of cohomology:

… → H^{1}(X, 𝔸) → H^{1}(X, 𝔸^{×}) → H^{2}(X, Z) → H^{2}(X, 𝔸) → ….

Now on Example Sheet 5 we showed using the existence of partitions of unity that H^{i}(X, 𝔸) = 0 for all i > 0. Thus we obtain the isomorphism H^{1}(X, 𝔸^{×}) ≃ H^{2}(X, Z) given by the boundary map c_{1} (the first Chern class). Recall that we have shown that Pic X ≃ H^{1}(X, 𝔸^{×}). It follows that the first Chern class yields an isomorphism

Pic X ≃ H^{2}(X, Z).