# Kernel of Ring Homomorphism is Subring

## Theorem

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

Then the kernel of $\phi$ is a subring of $R_1$.

## Proof

$\struct {\map \ker \phi, +_1} \le \struct {R_1, +_1}$

where $\le$ denotes subgroup.

Let $x, y \in \map \ker \phi$.

 $\ds \map \phi {x \circ_1 y}$ $=$ $\ds \map \phi x \circ_2 \map \phi y$ Morphism Property $\ds$ $=$ $\ds 0_{R_2} \circ_2 0_{R_2}$ Definition of Kernel $\ds$ $=$ $\ds 0_{R_2}$

Thus $x \circ_1 y \in \map \ker \phi$.

Thus the conditions for Subring Test are fulfilled, and $\map \ker \phi$ is a subring of $R_1$.

$\blacksquare$