# Kernel of Ring Homomorphism is Ideal

## Theorem

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.

The kernel of $\phi$ is an ideal of $R_1$.

## Proof

By Kernel of Ring Homomorphism is Subring, $\map \ker \phi$ is a subring of $R_1$.

Let $s \in \map \ker \phi$, so $\map \phi s = 0_{R_2}$.

Suppose $x \in R_1$. Then:

 $\ds \map \phi {x \circ_1 s}$ $=$ $\ds \map \phi x \circ_2 \map \phi s$ Definition of Morphism Property $\ds$ $=$ $\ds \map \phi x \circ_2 0_{R_2}$ as $s \in \map \ker \phi$ $\ds$ $=$ $\ds 0_{R_2}$ Properties of $0_{R_2}$

and similarly for $\map \phi {s \circ_1 x}$.

The result follows.

$\blacksquare$