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 $\displaystyle 0_R\in R,0_S\in S$ we know that $\displaystyle R,S\ne\varnothing$. Obviously every projection is surjective since given any $\displaystyle r\in R$ then for an arbitrary $\displaystyle s\in S$ we have that $\displaystyle (r,s)\in R\times S$ and $\displaystyle (r,s)\mapsto r$.
It is clearly a homomorphism too since $\displaystyle \pi\left((r,s)+(r',s')\right)=\pi\left(\left(r+r', s+s'\right)\right)=r+r'=\pi\left((r,s)\right)+\pi\ left((r',s')\right)$. You can figure out the multiplicative part.
Also, $\displaystyle \ker\pi=\left\{(r,s)r,s)\mapsto 0_R\right\}$. Clearly then $\displaystyle \ker \pi=\{0_R\}\times S$. So, simply define $\displaystyle \phi:\ker \pi\mapsto S$ by $\displaystyle (0_R,s)\mapsto s$. This is readily verified to be an isomorphism