# Thread: Quotient Rings and Homomorphisms

1. ## Quotient Rings and Homomorphisms

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.

2. Originally Posted by iwonde
Let $R$ and $S$ be rings. Show that $\pi:R\times S\mapsto R$ given by $(r,s)\mapsto r$ is a surjective homomorphism whose kernel is isomorphic to $S$
Clearly since $0_R\in R,0_S\in S$ we know that $R,S\ne\varnothing$. Obviously every projection is surjective since given any $r\in R$ then for an arbitrary $s\in S$ we have that $(r,s)\in R\times S$ and $(r,s)\mapsto r$.

It is clearly a homomorphism too since $\pi\left((r,s)+(r',s')\right)=\pi\left(\left(r+r', s+s'\right)\right)=r+r'=\pi\left((r,s)\right)+\pi\ left((r',s')\right)$. You can figure out the multiplicative part.

Also, $\ker\pi=\left\{(r,s)r,s)\mapsto 0_R\right\}" alt="\ker\pi=\left\{(r,s)r,s)\mapsto 0_R\right\}" />. Clearly then $\ker \pi=\{0_R\}\times S$. So, simply define $\phi:\ker \pi\mapsto S$ by $(0_R,s)\mapsto s$. This is readily verified to be an isomorphism