Bryden Cais.  Algebraic Geometry, HW 5.  April 3, 2003

1. Let X be a smooth variety over k of dimension n and for a closed point P of X, let x_{1}, …, x_{n} generate 𝔪 = 𝔪_{X,P}. Without loss of generality, we may assume that X is affine, say X = Spec B (by working locally). Now we have an isomorphism 𝔪/𝔪^{2} ≃ Ω_{B/k} ⊗_{B} k where x_{i} ↦ dx_{i} ⊗ 1. By Nakayama’s Lemma, the images of the x_{i} freely generate 𝔪/𝔪^{2} so that under the isomorphism above, we see that Ω_{X} is freely generated in a neighborhood of P (since in a neighborhood Spec B of P we have Ω_{X}|U = Ω̃_{B/k}). It follows that ω_{X} is generated by dx_{1} ∧ … ∧ dx_{n} in a neighborhood of P.

Now suppose that ω_{X} is generated by dx_{1} ∧ … ∧ dx_{n} over U and by dx_{1} ∧ … ∧ dx_{n} over V where dx_{i} = ∑_{j} \frac{δx_{i}}{δy_{j}} dy_{j}. Then The matrix L = (\frac{δx_{i}}{δy_{j}}) gives a linear L transformation from ω_{X}|V to Ω_{X}|U so that the transition function g_{UV} is given by ∧^{n}L. By ordinary linear algebra, this is just multiplication on ∧^{n}Ω_{X} by det L, the Jacobian.

3. Let X_{0}, …, X_{n} be homogenous coordinates on ℙ^{n} and let U_{i} = (X_{i} ≠ 0). Then U_{0} has coordinates x_{i} = X_{i}/X_{0} for i ≠ 0 and U_{1} has coordinates y_{i} = X_{i}/X_{1} for i ≠ 1. Observe that x_{i}/x_{1} = y_{i} for i ≠ 0, 1 and y_{0} = 1/x_{1}. We then have

dy_{i} = \frac{1}{x_{1}} dx_{i} − \frac{x_{i}}{x_{1}^{2}} dx_{1}

for i ≠ 0, 1 and

dy_{0} = −\frac{1}{x_{1}^{2}} dx_{1}.

Now let s = dx_{1} ∧ dx_{2} ∧ … ∧ dx_{n} ∈ Γ(U_{0}, ω_{ℙ^{n}}). Then we have by problem (1) that

s ∩ U_{1} = −\frac{1}{x_{1}^{2}} dx_{1} ⋀_{i>1} (\frac{1}{x_{1}} dx_{i} − \frac{x_{i}}{x_{1}^{2}} dx_{1}) = −\frac{1}{x_{1}^{n+1}} dx_{1} ∧ … ∧ dx_{n}.

It follows that the divisor of zeroes and poles of s is the divisor of zeroes and poles of 1/x_{1}^{n+1}, which is −(n + 1)H where H = (x_{1} = 0). But this is exactly the divisor of zeroes and poles of O_{ℙ^{n}}(−n − 1), from which it follows that ω_{ℙ^{n}} ≃ O_{ℙ^{n}}(−n − 1).

4. Let X be a smooth surface and X̃ the blowup of X at P. Recall that we have the exact sequence

0 ⟶ ZE ⟶ Cl X̃ ⟶^{π_{∗}} Cl X ⟶ 0.

We first claim that KX̃ = π^{∗}K_{X} + nE for some integer n. From the exact sequence above, it suffices to see that

π_{∗}KX̃ = π_{∗}π^{∗}K_{X} = K_{X}.

But this follows from the fact that ω_{X̃−E} is the pullback of ω_{X} = ω_{X−P} (since X̃ − E ≃ X − P).

Now we show that n = 1. Let U ≃ 𝔸^{2} be an affine subset of X̃ with P ∈ π(U). We restrict the map π : X̃ → X to U to obtain a map π : 𝔸_{x,y}^{2} → 𝔸_{u,v}^{2} where we realize P as (u, v) = (0, 0) (in other words, we just restrict our attention locally about P and coordinatize an affine neighborhood). Now since π is a blowup, it is given in coordinates on our local patch as (x, y) ↦ (x, xy) = (u, v).

Now by problem (1), we have that ωX̃ is generated over U by dx ∧ dy while ω_{X} is generated over π(U) by du ∧ dv. It follows from our description of π that the pullback of ω_{X} to X̃ is generated over U by dx ∧ d(xy) = xdx ∧ dy. It follows that KX̃ = (x) + π^{∗}K_{X} = E + π^{∗}K_{X} since E = (x = 0) in coordinates. Thus n = 1 and we are done.

5. Let X = ℙ × ℙ and let D ⊂ X be a smooth curve on X in the divisor class (d_{1}, d_{2}). Recall that the canonical divisor K_{X} on X is in the class (−2, −2) (since it is the pullback of the canonical divisor on ℙ^{1} in each component). Moreover, by the adjunction formula we have

K_{D} = (K_{X} + D)|_{D}.

Since Cl(X) ≃ Z ⊕ Z, we see that K_{X} + D is in the divisor class (−2, −2) + (d_{1}, d_{2}) = (d_{1} − 2, d_{2} − 2). We now use this information to determine the degree of K_{D}. It will suffice to determine the degrees of the divisors H_{1}|_{D}, H_{2}|_{D} with H_{1}, H_{2} in the divisor classes (1, 0) and (0, 1) respectively. Since D is cut out by a polynomial F homogenous in X_{0}, X_{1} of degree d_{1} and homogenous in Y_{0}, Y_{1} of degree d_{2}, we find that H_{1}|_{D} has degree d_{2} (since H_{1} is in the class of the hyperplane (X_{0} = 1) ∈ Cl(ℙ^{1}), and without loss of generality we may suppose that this hyperplane intersects D transversally so that F |_{H_{1}} is a polynomial in k[Y_{0}, Y_{1}] of homogenous degree d_{2} and has d_{2} solutions). Similarly, H_{2}|_{D} has degree d_{1}. It follows that K_{D} = (K_{X} + D)|_{D} has degree d_{2}(d_{1} − 2) + d_{1}(d_{2} − 2) = 2(d_{1}d_{2} − d_{2} − d_{1} + 2). Since 2g − 2 = deg K_{D}, where g is the genus of D, it follows that D has genus

g = (d_{1} − 1)(d_{2} − 1).

6. Let X be a compact smooth manifold and let O_{X} be the sheaf of smooth functions on X. Let F be a sheaf of O_{X}-modules on X and U = {U_{i}}_{i∈I} a finite open cover of X. Let ρ_{i} : X → R be a partition of unity subordinate to U. For q ≥ 1, let f ∈ C^{q}(U, F) be a cocycle, i.e. suppose that df = 0. We show that f is in fact a coboundary. Define K ∈ C^{q−1}(U, F)

K_{i_{0}...i_{q−1}} = ∑_{i∈I} ρ_{i}f_{ii_{0}...i_{q−1}}.

Observe that since \overline{Supp ρ_{i}} ⊂ U_{i} we have K_{i_{0}...i_{q−1}} ∈ O_{X}(U_{i_{0}...i_{q−1}}) so indeed K ∈ C^{q−1}(U, F). We have

(dK)_{i_{0}...i_{q}} = ∑_{j=0}^{q}(−1)^{j}K_{i_{0}...î_{j}...i_{q}}

= ∑_{j=0}^{q}(−1)^{j} ∑_{i∈I} ρ_{i}f_{ii_{0}...î_{j}...i_{q}}

= ∑_{i∈I} ρ_{i} ∑_{j=0}^{q}(−1)^{j}f_{ii_{0}...î_{j}...i_{q}}

= ∑_{i∈I} ρ_{i}f_{i_{0}...i_{q}}

since df = 0 by assumption,

= f_{i_{0}...i_{q−1}}

since ρ_{i} is a partition of unity.

Thus we have dK = f so that f is a coboundary. It follows that H^{q}(U, F) = 0 for q ≥ 1.