Constant Mapping to Identity is Homomorphism/Rings

Theorem
Let $\left({R_1, +_1, \circ_1}\right)$ and $\left({R_2, +_2, \circ_2}\right)$ be rings with zeroes $0_1$ and $0_2$ respectively.

Let $\zeta$ be the zero homomorphism from $R_1$ to $R_2$, that is:
 * $\forall x \in R_1: \zeta \left({x}\right) = 0_2$

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

Proof
The additive groups of $\left({R_1, +_1, \circ_1}\right)$ and $\left({R_2, +_2, \circ_2}\right)$ are $\left({R_1, +_1}\right)$ and $\left({R_2, +_2}\right)$ respectively.

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

Thus from the result for Group Homomorphism we have that $\zeta: \left({R_1, +_1}\right) \to \left({R_2, +_2}\right)$ is a group homomorphism:
 * $\zeta \left({x +_1 y}\right) = \zeta\left({x}\right) +_2 \zeta \left({y}\right)$

Then we have:

The results about image and kernel follow directly by definition.