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