Ring Homomorphism Preserves Subrings

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

If $S$ is a subring of $R_1$, then $\phi \left({S}\right)$ is a subring of $R_2$.

Corollary
The image of a ring homomorphism $\phi: R_1 \to R_2$ is a subring of $R_2$.

Proof 1

 * Since $S \ne \varnothing$, $\phi \left({S}\right) \ne \varnothing$.


 * From Group Homomorphism Preserves Subgroups, $\left({\phi \left({S}\right), +_2}\right)$ is a subgroup of $\left({R_2, +_2}\right)$.


 * From Homomorphism Preserves Subsemigroups, $\left({\phi \left({S}\right), \circ_2}\right)$ is a subsemigroup of $\left({R_2, \circ_2}\right)$.

Thus, as $\left({R_2, +_2}\right)$ is a group and $\left({R_2, \circ_2}\right)$ is a semigroup, the result follows.

Proof 2
From Morphism Property Preserves Closure, $\phi \left({R_1}\right)$ is a closed algebraic structure.

From Epimorphism Preserves Rings, $\phi \left({S}\right)$ is a ring.

Hence the result, from the definition of subring.

Proof of Corollary
Follows from the fact that from Null Ring and Ring Itself Subrings $R_1$ is a subring of itself.