Let R and S be rings. Show that pi: RxS--> R given by pi(r,s)=r is a surjective homomorphism whose kernel is isomorphic to S.
Clearly since we know that . Obviously every projection is surjective since given any then for an arbitrary we have that and .
It is clearly a homomorphism too since . You can figure out the multiplicative part.
Also, r,s)\mapsto 0_R\right\}" alt="\ker\pi=\left\{(r,s)r,s)\mapsto 0_R\right\}" />. Clearly then . So, simply define by . This is readily verified to be an isomorphism