Preimage of Ideal under Ring Homomorphism is Ideal

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

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

Then $S_1 = \phi^{-1} \sqbrk {S_2}$ is an ideal of $R_1$ such that $\map \ker \phi \subseteq S_1$.

Proof
From Preimage of Subring under Ring Homomorphism is Subring‎ we have that $S_1 = \phi^{-1} \sqbrk {S_2}$ is a subring of $R_1$ such that $\map \ker \phi \subseteq S_1$.

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:
 * $r_1 \circ_1 s_1 \in \phi^{-1} \sqbrk {S_2} = S_1$

Similarly for $s_1 \circ_1 r_1$.

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