Picard group of a Affine scheme How do we define a Picard group of an Affine scheme? Is there way to define as for commutative ring? Thanks
 A: The Picard group can be defined for any scheme $X$ as the group of isomorphism classes of invertible sheaves ( $\mathcal{O}_X$-modules that are locally free of rank $1$). Given such an $\mathcal{L}$, its inverse is given by $\mathcal{L}^{\vee} = \mathcal{Hom}(\mathcal{L},\mathcal{O}_X)$ where Hom there is not global Hom but rather sheaf Hom (which is a sheaf). The multiplication in the group is given by the tensor product, and to check $\text{Pic} (X)$ is a  group the only non-trivial calculation is to see that
$$\mathcal{L} \otimes_{\mathcal{O}_X} \mathcal{L}^\vee = \mathcal{O}_X.$$
If you want to check this isomorphism, I suggest you consider the tensor-hom adjunction.
A: Addition to the other answer: If $X = \mathrm{Spec}(A)$, then $\mathrm{Pic}(X)=\mathrm{Pic}(A)$ is simply the group of isomorphism classes of finitely generated projective $A$-modules of rank $1$. It is known that the following are equivalent for some $A$-module $M$:


*

*$M$ is finitely generated projective of rank $1$.

*There are elements $f_1,\dotsc,f_n$ of $A$ which generate the unit ideal such that, for every $i$, $M_{f_i}$ is free of rank $1$ over $A_{f_i}$.

*There is some $A$-module $N$ such that $M \otimes_A N \cong A$.


The group multiplication is given by $[M] \otimes [M'] := [M \otimes_A M']$, the inversion is given by $[M]^{-1} := [M^*]$, where $M^* := \hom_A(M,A)$ is the dual. As a simple example, you can show for a field $k$ that $\mathrm{Pic}(k)$ is trivial. For a nontrivial example, see below.
If $A$ is an integral domain, the Picard group is known as the class group, and we can add
$~~~   4.$ $M$ is isomorphic to an invertible fractional ideal of $A$, i.e. a finitely generated $A$-submodule $I$ of $\mathrm{Q}(A)$ which has an inverse $J$ in the sense that $I \cdot J = A$.
The class group is the quotient group $\{$invertible fractional ideals$\}$ / $\{$principal ideals$\}$, with the simple multiplication $I \cdot I' := \langle a \cdot b : a \in I , b \in I' \rangle$. For example, take $A=\mathbb{Z}[\sqrt{-5}]$ and $I=(2,1+\sqrt{-5})$ and $J = (2,1-\sqrt{-5})$. Then one calculates $I \cdot J = (2)$. Hence, $\frac{1}{2} J$ is inverse to $I$, and $I$ is invertible. But since $I$ not principal, it represents a nontrivial element in $\mathrm{Pic}(\mathbb{Z}[\sqrt{-5}])$ (which turns out to be cyclic of order $2$).
