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 $R_1$.