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

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

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

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