## 109.40 Invertible sheaves

Definition 109.40.1. 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.

Exercise 109.40.2. General facts.

1. Show that an invertible ${\mathcal O}_ X$-module on a scheme $X$ is quasi-coherent.

2. 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 109.40.3. Algebra.

1. 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.)

2. 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 109.40.4. 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 109.40.5. Simple examples.

1. 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.

2. 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 ${\mathbf A}^1_ k = \mathop{\mathrm{Spec}}(k[x])$.)

3. 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 109.40.6. Higher dimensions.

1. 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 109.40.5 (1) to find a generator for this; then you still have to think. Another way to is to use Exercise 109.40.3 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.)

2. 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].

3. 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 109.40.7. 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 17.23.9. For a ring $R$ we set $\mathop{\mathrm{Pic}}\nolimits (R) = \mathop{\mathrm{Pic}}\nolimits (\mathop{\mathrm{Spec}}(R))$.

Exercise 109.40.8. Let $R$ be a ring.

1. 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 109.40.5 (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).]

2. 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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