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.