# Constant Mapping to Identity is Homomorphism/Rings

Jump to navigation
Jump to search

## 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 $\map \zeta y = 0_2$ |

The results about image and kernel follow directly by definition.

$\blacksquare$

## Sources

- 1964: Iain T. Adamson:
*Introduction to Field Theory*... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Example $1$ - 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $5$: Rings: $\S 24$. Homomorphisms: Example $42$