Preimage of Ideal under Ring Homomorphism is Ideal

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

Let $$S_2$$ be an ideal of $$R_2$$.

Then $$S_1 = \phi^{-1} \left({S_2}\right)$$ is an ideal of $$R_1$$ such that $$\ker \left({\phi}\right) \subseteq S_1$$.

Proof
From Ring Epimorphism Inverse of Subring‎ we have that $$S_1 = \phi^{-1} \left({S_2}\right)$$ is a subring of $$R_1$$ such that $$\ker \left({\phi}\right) \subseteq S_1$$.

We now need to show that $$S_1$$ is an ideal of $$R_1$$.

Let $$s_1 \in S_1, r_1 \in R_1$$.

Then:

$$ $$

Thus:
 * $$r_1 \circ_1 s_1 \in \phi^{-1} \left({S_2}\right)= S_1$$

Similarly for $$s_1 \circ_1 r_1$$.

So $$S_1$$ is an ideal of $$R_1$$.