Constant Mapping to Identity is Homomorphism/Rings

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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