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.
 A: 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 $A\oplus B$. 
The answer to your question
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  $$Pic (A\times B)   \stackrel {\cong}{\to}   Pic(A)\times Pic(B)  $$   
Scheme-theoretic interpretation
All this is crystal-clear geometrically:
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 $Spec(B)$.
