Definition 111.40.1. Let X be a locally ringed space. An invertible {\mathcal O}_ X-module on X is a sheaf of {\mathcal O}_ X-modules {\mathcal L} such that every point has an open neighbourhood U \subset X such that {\mathcal L}|_ U is isomorphic to {\mathcal O}_ U as {\mathcal O}_ U-module. We say that {\mathcal L} is trivial if it is isomorphic to {\mathcal O}_ X as a {\mathcal O}_ X-module.
111.40 Invertible sheaves
Exercise 111.40.2. General facts.
Show that an invertible {\mathcal O}_ X-module on a scheme X is quasi-coherent.
Suppose X\to Y is a morphism of locally ringed spaces, and {\mathcal L} an invertible {\mathcal O}_ Y-module. Show that f^\ast {\mathcal L} is an invertible {\mathcal O}_ X module.
Exercise 111.40.3. Algebra.
Show that an invertible {\mathcal O}_ X-module on an affine scheme \mathop{\mathrm{Spec}}(A) corresponds to an A-module M which is (i) finite, (ii) projective, (iii) locally free of rank 1, and hence (iv) flat, and (v) finitely presented. (Feel free to quote things from last semesters course; or from algebra books.)
Suppose that A is a domain and that M is a module as in (a). Show that M is isomorphic as an A-module to an ideal I \subset A such that IA_{\mathfrak p} is principal for every prime {\mathfrak p}.
Definition 111.40.4. Let R be a ring. An invertible module M is an R-module M such that \widetilde M is an invertible sheaf on the spectrum of R. We say M is trivial if M \cong R as an R-module.
In other words, M is invertible if and only if it satisfies all of the following conditions: it is flat, of finite presentation, projective, and locally free of rank 1. (Of course it suffices for it to be locally free of rank 1).
Exercise 111.40.5. Simple examples.
Let k be a field. Let A = k[x]. Show that X = \mathop{\mathrm{Spec}}(A) has only trivial invertible {\mathcal O}_ X-modules. In other words, show that every invertible A-module is free of rank 1.
Let A be the ring
A = \{ f\in k[x] \mid f(0) = f(1) \} .Show there exists a nontrivial invertible A-module, unless k = {\mathbf F}_2. (Hint: Think about \mathop{\mathrm{Spec}}(A) as identifying 0 and 1 in {\mathbf A}^1_ k = \mathop{\mathrm{Spec}}(k[x]).)
Same question as in (2) for the ring A = k[x^2, x^3] \subset k[x] (except now k = {\mathbf F}_2 works as well).
Exercise 111.40.6. Higher dimensions.
Prove that every invertible sheaf on two dimensional affine space is trivial. More precisely, let {\mathbf A}^2_ k = \mathop{\mathrm{Spec}}(k[x, y]) where k is a field. Show that every invertible sheaf on {\mathbf A}^2_ k is trivial. (Hint: One way to do this is to consider the corresponding module M, to look at M \otimes _{k[x, y]} k(x)[y], and then use Exercise 111.40.5 (1) to find a generator for this; then you still have to think. Another way to is to use Exercise 111.40.3 and use what we know about ideals of the polynomial ring: primes of height one are generated by an irreducible polynomial; then you still have to think.)
Prove that every invertible sheaf on any open subscheme of two dimensional affine space is trivial. More precisely, let U \subset {\mathbf A}^2_ k be an open subscheme where k is a field. Show that every invertible sheaf on U is trivial. Hint: Show that every invertible sheaf on U extends to one on {\mathbf A}^2_ k. Not easy; but you can find it in [H].
Find an example of a nontrivial invertible sheaf on a punctured cone over a field. More precisely, let k be a field and let C = \mathop{\mathrm{Spec}}(k[x, y, z]/(xy-z^2)). Let U = C \setminus \{ (x, y, z) \} . Find a nontrivial invertible sheaf on U. Hint: It may be easier to compute the group of isomorphism classes of invertible sheaves on U than to just find one. Note that U is covered by the opens \mathop{\mathrm{Spec}}(k[x, y, z, 1/x]/(xy-z^2)) and \mathop{\mathrm{Spec}}(k[x, y, z, 1/y]/(xy-z^2)) which are “easy” to deal with.
Definition 111.40.7. Let X be a locally ringed space. The Picard group of X is the set \mathop{\mathrm{Pic}}\nolimits (X) of isomorphism classes of invertible \mathcal{O}_ X-modules with addition given by tensor product. See Modules, Definition 17.25.9. For a ring R we set \mathop{\mathrm{Pic}}\nolimits (R) = \mathop{\mathrm{Pic}}\nolimits (\mathop{\mathrm{Spec}}(R)).
Exercise 111.40.8. Let R be a ring.
Show that if R is a Noetherian normal domain, then \mathop{\mathrm{Pic}}\nolimits (R) = \mathop{\mathrm{Pic}}\nolimits (R[t]). [Hint: There is a map R[t] \to R, t \mapsto 0 which is a left inverse to the map R \to R[t]. Hence it suffices to show that any invertible R[t]-module M such that M/tM \cong R is free of rank 1. Let K be the fraction field of R. Pick a trivialization K[t] \to M \otimes _{R[t]} K[t] which is possible by Exercise 111.40.5 (1). Adjust it so it agrees with the trivialization of M/tM above. Show that it is in fact a trivialization of M over R[t] (this is where normality comes in).]
Let k be a field. Show that \mathop{\mathrm{Pic}}\nolimits (k[x^2, x^3, t]) \not= \mathop{\mathrm{Pic}}\nolimits (k[x^2, x^3]).
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