Ring Homomorphism Preserves Subrings/Proof 3

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 $R_2$.

Proof
Let $S$ be a subring of $R_1$.

Since $S \ne \varnothing$ it follows that $\phi \left({S}\right) \ne \varnothing$.

Let $x, y \in \phi \left({S}\right)$.

Then $\exists s, t \in S: x = \phi \left({s}\right), y = \phi \left({t}\right)$.

So:

As $S$ is a subring of $R_1$, it is closed under $+_1$ and the taking of negatives.

Thus $s +_1 \left({-t}\right) \in S$ and so $x +_2 \left({-y}\right) \in \phi \left({S}\right)$.

Similarly:

As $S$ is a subring of $R_1$, it is closed under $\circ_1$.

Thus $s \circ_1 t \in S$ and so $x \circ_2 y \in \phi \left({S}\right)$.

The result follows from Subring Test.