Ring Homomorphism Preserves Zero

Theorem
Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism.

Let:
 * $0_{R_1}$ be the zero of $R_1$
 * $0_{R_2}$ be the zero of $R_2$.

Then:
 * $\phi \left({0_{R_1}}\right) = 0_{R_2}$

Proof
By definition, if $\left({R_1, +_1, \circ_1}\right)$ and $\left({R_2, +_2, \circ_2}\right)$ are rings then $\left({R_1, +_1}\right)$ and $\left({R_2, +_2}\right)$ are groups.

Again by definition:
 * the zero of $\left({R_1, +_1, \circ_1}\right)$ is the identity of $\left({R_1, +_1}\right)$
 * the zero of $\left({R_2, +_2, \circ_2}\right)$ is the identity of $\left({R_2, +_2}\right)$.

The result follows from Group Homomorphism Preserves Identity.