# Constant Mapping to Identity is Homomorphism/Rings

## Theorem

Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be rings with zeroes $0_1$ and $0_2$ respectively.

Let $\zeta$ be the zero homomorphism from $R_1$ to $R_2$, that is:

$\forall x \in R_1: \map \zeta x = 0_2$

Then $\zeta$ is a ring homomorphism whose image is $\set {0_2}$ and whose kernel is $R_1$.

## Proof

The additive groups of $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ are $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ respectively.

Their identities are $0_1$ and $0_2$ respectively.

Thus from the Constant Mapping to Group Identity is Homomorphism we have that $\zeta: \struct {R_1, +_1} \to \struct {R_2, +_2}$ is a group homomorphism:

$\map \zeta {x +_1 y} = \map \zeta x +_2 \map \zeta y$

Then we have:

 $\displaystyle \map \zeta {x \circ_1 y}$ $=$ $\displaystyle 0_2$ as $x \circ_1 y \in R_1$ $\displaystyle$ $=$ $\displaystyle \map \zeta x \circ_2 \map \zeta y$ as $\map \zeta x = 0_2$ and $\map \zeta y = 0_2$

$\blacksquare$