# Thread: Help Proving a Theorem (Ring Isomorphisms)

1. ## Help Proving a Theorem (Ring Isomorphisms)

Can someone help me get started. I start by assuming 1-1 and know that i go the other direction, but can't seem to see how to use 1-1 to show anything...Thanks in advance.

Theorem:
Let $\displaystyle \Phi :R \rightarrow S$ be a ring homomorphism. Show that $\displaystyle \Phi$ is 1-1 iff $\displaystyle ker(\Phi)$ is trivial

2. Originally Posted by ginafara
Can someone help me get started. I start by assuming 1-1 and know that i go the other direction, but can't seem to see how to use 1-1 to show anything...Thanks in advance.

Theorem:
Let $\displaystyle \phi :R \rightarrow S$ be a ring homomorphism. Show that $\displaystyle \phi$ is 1-1 iff $\displaystyle ker(\phi)$ is trivial
1-1 $\displaystyle \Rightarrow ker(\phi)$ is trivial

Since for any ring,$\displaystyle \phi(0) = 0$, 1-1 nature of $\displaystyle \phi$ forces 0 to be the only number of R to map to the 0 of S. Thus $\displaystyle \phi(x) = 0$ only for x=0. Hence $\displaystyle ker(\phi)$ is trivial

$\displaystyle ker(\phi) = \{0\} \Rightarrow \phi$ is 1-1

For any ring,$\displaystyle \phi(0) = 0$. If $\displaystyle \phi(x) = \phi(y)$, then $\displaystyle \phi(x-y) = 0$. But since $\displaystyle ker(\phi) = \{0\}, x=y$