# The Picard group of a product of rings.

Reading a book on Theory of Modules, i have found the assertion ${\bf Pic}(A\times B)\cong {\bf Pic}(A)\times {\bf Pic}(B),$ where $A$ and $B$ are commutative rings with unity.

I think that the isomorphism is given by ${\bf Pic}(A)\oplus {\bf Pic}(B)\ni ([E],[M])\mapsto [E\oplus M]\in {\bf Pic}(A\times B).$ The problem is that i have not been able to prove that this map is an epimorphism.

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Following @GeorgesElencwag advice. I changed $A\oplus B$ by $A\times B$ – Hector Pinedo Nov 9 '12 at 21:23

Prelude
First notice that given rings you should never write $A\oplus B$ for rings: in the category of rings you have a product $A\times B$ and a coproduct $A\otimes _\mathbb Z B$ but neither should be written $A\oplus B$.

There is an equivalence of categories $$Mod_{A\times B}\stackrel {\cong}{\to} Mod_A\times Mod_B$$ in which a module $M$ over $A\times B$ is sent to the pair of modules module $(M\otimes_{A\times B} A,M\otimes_{A\times B} B)$.
The quasi-inverse morphism $$Mod_A\times Mod_B \stackrel {\cong}{\to} Mod_{A\times B}$$
sends the pair $(N,P)$ consisting of an $A$-module $N$ and a $B$-module $P$ to the $A\times B$ -module $N\times P$, in which multiplication by scalars is of course given by the formula $(a,b)\cdot(n,p)=(an,bp)$
By restricting this equivalence to invertible modules (=finitely generated projective modules of rank one), you get the required isomorphism $$Pic (A\times B) \stackrel {\cong}{\to} Pic(A)\times Pic(B)$$
Since $Spec(A\times B)=Spec(A) \bigsqcup Spec(B)$, choosing a line bundle on $Spec(A\times B)$ amounts exactly to the independent choice of a line bundle on $Spec(A)$ and a line bundle on $Spec(B)$.