Bryden Cais.  Algebraic Geometry, HW 4.  March 18, 2003

1. Let X = (xy = z^{2} ⊂ 𝔸^{3}) and let D = (x = z = 0) ⊂ X. We show first that D is not Cartier by showing that it is not locally principal at 0. Indeed, let 𝔪 = (x, y, z) be the ideal corresponding to the origin. Now let A = k[x, y, z]/(xy − z^{2}) be the coordinate ring of X and set 𝔭 = (x, z). Then pA_{𝔪} is not principal: indeed, dim_{k}(𝔪/𝔪^{2}) = 3, generated by the images of x, y, z (since X is not smooth at the origin) and the image of 𝔭 in A_{𝔪} is generated by the images of x and z and is therefore not principal. It follows that D is not Cartier.

Now we show that Cl(X) is generated by D. Observe that we have an exact sequence

Z ⟶ Cl(X) ⟶ Cl(X − D)

where Z → Cl(X) is given by 1 ↦ D. Exactness is clear since if Y is a prime divisor of X then Y ∩(X−D) = 0 if and only if Y ∼ D. Now we have X = Spec A as above, and D is given as a subset of X by the single equation x = 0 so that X −D = Spec A_{x} = Spec k[x, y, z, x^{−1}]/(xy −z^{2}) = Spec k[x, z, x^{−1}]. Now k[x, z, x^{−1}] is a UFD so that Cl(X − D) = 0. It follows that we have an exact sequence

Z ⟶ Cl(X) ⟶ 0

which shows that Cl(X) is generated by D. Now we have shown that D is not locally principal (and hence not principal) which in particular shows that Pic(X) = 0.

We finally show that 2D = 0 in Cl(X). Indeed, the ideal of D is 𝔭 = (x, z) so that y is a unit in A_{𝔭}. Since xy = z^{2} and y is a unit, it follows that x ∈ zA_{𝔭} and hence that z generates the ideal of D in A_{𝔭}. Since D = (x = 0) and x ∈ (z)^{2}, we have v_{z}(D) = 2 so that 2D = (x) is principal, as claimed. Thus, we have Cl(X) ≃ Z/2Z.

2. Let X be a smooth surface and P ∈ X a closed point. Let π : X̃ → X be the blowup of the point P and let E = π^{−1}(E). We then have a map π_{∗} : Cl X̃ → Cl X given on prime divisors by π_{∗}(Y) = π(Y) if π(Y) is a divisor (that is, if Y ≠ E) and π_{∗}(Y) = 0 if π(Y) is not a divisor. It is clear that π_{∗} is surjective as for any prime divisor Z ⊂ X we have π^{−1}(Z − P)^{−} a prime divisor of X̃ that maps to Z under P (here we are using the fact that π^{−1} : x−P → X̃ −E is an isomorphism). Now if Y ≠ E is a prime divisor of X̃ that maps via π_{∗} to 0 then π(Y) ∼ (f) for some f ∈ k(X). But we have k(X) = k(X −P) so that, since π^{−1} : X −P → X̃ −E is an isomorphism (in particular a morphism) we obtain a function g ∈ k(X̃ − E) = k(X̃) (since X̃ − E is dense in X̃) with Y = (g) so that Y ∼ 0 as an element of Cl(X̃). Now if Y = E then by definition of π_{∗} we have π_{∗}(Y) = 0. We have thus shown that ker(π_{∗}) is generated by E. It remains to show that the map Z → Cl(X̃) given by 1 ↦ E is injective. Again, suppose that n ↦ 0 for some integer n and for the sake of simplicity, suppose that n > 0. Then there is some f ∈ k(X̃) = k(X̃ − E) with v_{E}(f) = n and v_{Y} (f) = 0 for all other prime divisors Y ⊂ X̃. Since π : X̃ − E → X − P is an isomorphism, this gives a function g = f ∘ π^{−1} ∈ k(X − P) = k(X) vanishing to order n at the point P. But (g = 0) ⊂ X has codimension 1 so that g must vanish on some divisor Z ⊂ X. Then f vanishes on π^{−1}(Z − P)^{−} ≠ E so that (f) ≠ nE. It follows that we have an exact sequence

0 ⟶ Z ⟶ Cl(X̃) ⟶^{π_{∗}} Cl(X) ⟶ 0.

3. Let X ⊂ ℙ^{3} be a smooth cubic surface. Then X contains a line l. Recall that Cl(ℙ^{3}) is generated by any hyperplane H. Without loss of generality, we may suppose that l ⊂ H. We show that the induced map Cl(ℙ^{3}) → Cl(X) given by Y ↦ H ∩ X for any prime divisor Y of ℙ^{3} is not surjective. We claim that for every integer n, n(H ∩ X) is not linearly equivalent to l. Since X is a cubic surface, it is given by a degree 3 homogenous polynomial. Moreover, since we have taken H to contain l, we see that H ∩ X consists of l together with a conic Q. We thus have H ∩ X = l + Q. Suppose that nH ∩ X ∼ l and let H′ be any hyperplane in ℙ^{3} intersecting H in a line. Then n(H ∩ H′ ∩ X) = n(H′ ∩ l + H′ ∩ Q) is a divisor on the irreducible cubic curve X ∩ H′ of degree 3n (since H′ ∩ Q is a divisor of degree 2). On the other hand, nH ∩ H′ ∩ X ∼ l ∩ H′ has degree 1 since l ∩ H′ consists of a single point. This is a contradiction since there is no integer n satisfying 3n = 1, so that l is not in the image of Cl(ℙ^{3}) → Cl(X), as asserted.

5. Let X be a smooth curve of genus 1 over a field k of characteristic not equal to 2. Then there exists a finite map φ : X → ℙ^{1} of degree 2, ramified over 4 points. Let U ⊂ X be some affine open subset mapping to 𝔸_{t}^{1} ⊂ ℙ^{1}. Then we have seen that k(U) = k(X) and that k[U] is the integral closure of k[t] in k(X) (here we use φ to realize k(t) as a subfield of k(X)). Since f is of degree 2, the extension k(X)/k(t) is a degree 2 field extension. Let f ∈ k[U] − k[t]. Then since f ∉ k[t] and k[U] is the integral closure of k[t] in k(X), we have that {1, f}is a basis for k[U] as a k[t] vector space. Thus, there exist α, β ∈ k[t] with f^{2} = αf + β. Since char(k) ≠ 2, it follows that we have (f − α/2)^{2} = β + α^{2}/4 so setting g = f − α/2 we see that g ∉ k[t] and that k(X) = k(t)(g). On the other hand, we clearly have g^{2} ∈ k[t], so write g^{2} = ∏(t − e_{i}) for e_{i} ∈ k̄. Then since U is affine, we see that k[U] = k[t, y]/(y^{2} − ∏(t − e_{i})) and that φ is given by (t, y) ↦ t (that φ takes this form comes from the fact that we have been viewing k(t) as a subfield of k(X) rather than imbedded via ϕ_{∗} in k(X). This is just for notational convenience and clearly does not change anything). Over every point t ∈ 𝔸^{1} there are two points (counted with multiplicity) (t, ±y) satisfying y^{2} − ∏(t − e_{i}) = 0. These points are distinct unless t = e_{i}, over which points the map φ is ramified. Since φ is ramified at exactly 4 points, we see that g^{2} has precisely 4 roots. Thus k(X) = k(t)(g) where g^{2} ∈ k[t] is a polynomial of degree