Duke University – Spring 2017 – MATH 627 April 13, 2017
Problem set 3 Olivier Debarre Due Tuesday May 2, 2017
Problem 1. Letkbe a field. We consider two copiesU1 := Spec(k[T1])andU2 := Spec(k[T2])of the affine lineA1k.
a) Compute the Picard groups ofA1kandA1kr{0}(Hint: you may use without proof the fact that ifAis a unique factorization domain, the Picard group ofSpec(A)is trivial).
Proof. SinceA1k = Spec(k[T])andA1kr{0}:= Spec(k[T, T−1])and both ringsk[T]andk[T, T−1] are unique factorization domains, their Picard groups are trivial.
b) Let X be the scheme obtained by glueing U1 and U2 along the open subsets U1 r {0} = Spec(k[T1, T1−1])and U2 r{0} = Spec(k[T2, T2−1])by the isomorphismk[T1, T1−1]→∼ k[T2, T2−1] ofk-algebras sendingT1toT2−1. Which scheme isX?
Asnwer. The schemeX is the projective lineP1k.
c) Compute the Picard group of X (Hint: explain that you may use Leray’s theorem to compute H1(X,OX∗)).
Proof. The scheme X is covered by the affine subsetsU1 andU2. Moreover, Hq(Ui,OX∗) = 0 for q ≥ 2becauseUi has dimension 1, and forq = 1 becausePic(Ui) = Pic(A1k) = 0by question a).
Similarly, Hq(U1 ∩U2,OX∗) = 0forq > 0for the same reasons. We may therefore apply Leray’s theorem to computeH1(X,OX∗)as the cokernel of the map
Γ(U1,OX∗)×Γ(U2,OX∗) −→ Γ(U1 ∩U2,OX∗) (s1, s2) 7−→ s1/s2.
SinceΓ(Ui,OX∗) =k[Ti]∗ =k∗andΓ(U1∩U2,OX∗) =k[T, T−1]∗ =k∗× hTi 'k∗×Z, we obtain Pic(X)'H1(X,OX∗)'Z.
d) Find the global sections of each invertible sheaf onX.
Proof. LetLm be the invertible sheaf onX corresponding to (1, T1m) ∈ k∗ × hT1i. The sections of Lm on X correspond to pairs (P1, P2), where Pi ∈ Γ(Ui,Lm) ' Γ(Ui,OUi) ' k[Ti], with P1(T1)/P2(T1−1) = T1m. It follows thatΓ(X,Lm)is isomorphic to the space of polynomials ink[T] of degree≤m: ifm≥0, the sections are the pairs(P(T1), T2mP(T2−1)), wheredeg(P)≤m.
e) Let Y be the scheme obtained by glueing U1 and U2 as in b), but using now the isomorphism k[T1, T1−1]→∼ k[T2, T2−1] that sends T1 to T2. Compute the Picard group of Y (Hint: proceed as in c)).
Proof. The same proof as in c) givesPic(Y)'H1(Y,OY∗)'Z.
f) Find the global sections of each invertible sheaf onY.
Proof. Let Lm be the invertible sheaf on Y corresponding to (1, T1m) ∈ k∗ × hT1i. The sections of Lm on Y correspond to pairs (P1, P2), where Pi ∈ Γ(Ui,Lm) ' Γ(Ui,OUi) ' k[Ti], with P1(T1)/P2(T1) = T1m. It follows that Γ(Y,Lm) ' k[T] for all m: if m ≥ 0, the sections are the pairs(T1mP(T1), P(T2)); ifm ≤ 0, the sections are the pairs(P(T1), T2−mP(T2)). In all cases, Γ(Y,Lm)is isomorphic tok[T].
g) Prove that there are no ample invertible sheaves onY.
Proof. If m > 0, the global sections of Lm all vanish at the origin on U1; ifm < 0, the global sections of Lm all vanish at the origin onU2. Hence Lm is never generated by global sections if m6= 0. It follows thatLmis not ample for anym(apply the definition of ampleness withF =L1).
Problem 2. Prove that the schemeYn :=Ankr{0}is not an affine scheme for anyn ≥2(Hint:use Leray’s theorem to computeH1(Y2,OY2)).
Proof.The schemeY2is covered by the affine open subsetsU1 :=A1k×k(A1kr{0}) = Spec(k[T1, T2, T2−1]) andU2 := (A1kr{0})×kA1k) = Spec(k[T1, T1−1, T2]). SinceOY2 is a coherent sheaf, we can com- puteH1(Y2,OY2)using Leray’s theorem as the cokernel of the map
Γ(U1,OY∗2)×Γ(U2,OY∗2) −→ Γ(U1∩U2,OY∗2) (P1, P2) 7−→ P1−P2.
We have
Γ(U1,OY∗2) = k[T1, T2, T2−1], Γ(U2,OY∗2) = k[T1, T1−1, T2], Γ(U1∩U2,OY∗2) = k[T1, T1−1, T2, T2−1],
henceH1(Y2,OY2)is an infinite-dimensionalk-vector space with basis(T1mT2n)m,n<0. In particular, Y2 is not affine. SinceY2 is a closed subscheme ofYn for alln ≥ 2and a closed subscheme of an affine scheme is affine,Ynis also not an affine scheme for alln≥2.
Problem 3.LetX be a projective scheme over a field and letL andM be invertible sheaves onX.
a) IfL is generated by global sections andM is very ample, the invertible sheafL ⊗M is very ample (Hint: use a Segre embedding).
Proof. Since L is generated by global sections, there exists a morphism u: X → Pmk such that u∗OPmk(1) = L. SinceM is very ample, there exists a closed embeddingv: X ,→ Pnk such that v∗OPnk(1) = M. The morphism (u, v) :X → Pmk ×Pnkis then also a closed embedding (because its composition with the second projection is) and so is the composition
w: X −−−→(u,v) Pmk ×Pnk−−−→Segre P(m+1)(n+1)−1
k .
Sincew∗OP(m+1)(n+1)−1 k
(1) =L ⊗M, this proves thatL ⊗M is very ample.
b) IfM is ample, the invertible sheaf L ⊗M⊗r is very ample forallsufficiently large integersr (Hint:we proved in class thatL ⊗M⊗ris ample for some integerr >0).
Proof. SinceM is ample, there exists an integer r0 such that L ⊗M⊗r is generated by global sections for all r ≥ r0, and there exists an integer s0 such that M⊗s0 is very ample. For any r≥r0+s0, the invertible sheafL ⊗M⊗r =L ⊗M⊗(r−s0)⊗M⊗s0 is then very ample by a).
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