Algebraic Geometry 3. Exercises sheet 6 Carlos de Vera Piquero, [email protected]

In this sheet we recall some notions about invertible sheaves, the definition of the Picard group of a scheme, and then we study the relation to Cartier divisors. Let X be a scheme, and L be a sheaf of OX -modules. Recall that L is called invertible if the following holds: for every x ∈ X, there exists an open neighborhood U of x and an isomorphism of OU -modules LU 'OU (where we write OU for OX|U ). Given two sheaves of OX -modules F and G, recall also that the association U 7→ HomOU (F|U , G|U ) defines a sheaf, which we denote HomOX (F, G). We put ∨ F := HomOX (F, OX ) (usually referred to as the dual sheaf).

Exercise 1: Let X be a scheme, and L, F be invertible sheaves of OX -modules. Show that L ⊗OX F ∨ ∨ and L are invertible sheaves as well, and that L ⊗OX L 'OX as OX -modules.

0 0 0 0 Given invertible sheaves L, L , F, F of OX -modules, it is immediate that if L'L and F'F 0 0 as OX -modules, then L ⊗OX F'L ⊗OX F . If Pic(X) denotes the set of isomorphism classes (as OX -modules) of invertible sheaves of OX -modules, then the previous exercise gives a group structure on −1 ∨ Pic(X): the identity element is the class cl(OX ) of OX , and cl(L) := cl(L ). This is the so-called Picard group of X.

∗ Exercise 2: Let f : X → Y be a morphism of schemes. If L is a sheaf of OY -modules, define f (L) := −1 ∗ −1 (f L) ⊗f OY OX . Show that if L is invertible then f (L) is an invertible sheaf of OX -modules, and that the association L → f ∗(L) induces a well-defined morphism of groups f ∗ : Pic(Y ) −→ Pic(X).

A fractional ideal over a scheme X is a sub-OX -module of the sheaf KX (cf. Exercises sheet 5). An invertible fractional ideal is a fractional ideal over X which is invertible as OX -module. Exercise 3: Let X be a scheme, and I be a fractional ideal over X. Prove that I is invertible if and only × if for all x ∈ X, there is an open neighborhood U of x and a section f ∈ Γ(U, KX ) such that I|U = OU ·f.

Let D ∈ CaDiv(X) be a Cartier divisor on X, represented by a system {(Ui, fi)}i. Since for every pair × i, j with Ui ∩ Uj 6= ∅ we have fi|Ui∩Uj ∈ fj|Ui∩Uj · Γ(Ui ∩ Uj, OX ), it follows that OUj · fj and OUi · fi define the same sheaf of modules over Ui ∩ Uj. Hence there is a unique fractional ideal I := I(D) over

X such that IUi = OUi · fi for each i. By the previous exercise, I is then an invertible fractional ideal (hence in particular an invertible sheaf of OX -modules, which defines an element in Pic(X)). We define × L(D) := I(−D). Then observe that if D is represented by f ∈ Γ(U, KX ) over an open U, then L(D) is generated by f −1 over U.

Exercise 4: Let X be a scheme. Prove the following assertions. i) The association D 7→ L(D) establishes a one-to-one correspondence between Cartier divisors on X and invertible fractional ideals over X. −1 ii) If D1,D2 ∈ CaDiv(X), then L(D1 − D2) 'L(D1) ⊗ L(D2) . iii) If D1,D2 ∈ CaDiv(X), then D1 ∼ D2 if and only if L(D1) 'L(D2) as invertible sheaves. iv) Deduce that D 7→ L(D) induces an injective morphism CaCl(X) ,→ Pic(X). Hint to prove i). Construct an inverse of D 7→ I(D). Given a fractional ideal I over X, choose an open × covering {Ui}i such that I|Ui = OUi · fi for some fi ∈ Γ(Ui, KX ). Fixing isomorphisms ϕi : OUi 'IUi defined by 1 7→ fi, show that {(Ui, fi)} defines a Cartier divisor D(I) on X and that I 7→ D(I) gives the desired inverse. Hints to prove ii). Consider first the following statements. (a) If I is an invertible sheaf over X, ∨ −1 prove that I is isomorphic (as OX -module) to the invertible fractional ideal I , defined by setting −1 −1 × I|U = OU · f for each open U such that I|U = OU · f with f ∈ Γ(U, KX ). (b) Given two fractional ideals I, J over X, let IJ be the sheaf associated with U 7→ I(U)J (U), where I(U)J (U) ⊂ Γ(U, KX ) denotes the OX (U)-module generated by I(U) and J (U). For each open U, there is a canonical morphism

φU : I(U) ⊗OX (U) J (U) → Γ(U, KX ), and one has I(U)J (U) = Im(φU ). If φ : I ⊗OX J → KX denotes the morphism associated with the collection {φU }U , then IJ = Im(φ), thus in particular one gets a surjective morphism φ : I ⊗OX J → IJ . Prove that if I and J are invertible fractional ideals over X, then φ : I ⊗OX J → IJ is an isomorphism.

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