Preimage of Subring under Ring Homomorphism is Subring

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

Let $S_2$ be a subring of $R_2$.

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

Proof
Let $K = \ker \left({\phi}\right)$ be the kernel of $R_1$.

We have that $0_{R_2} \in S_2$ and so $\left\{{0_{R_2}}\right\} \subseteq S_2$.

From Subset Maps to Subset:
 * $\phi^{-1} \left[{\left\{{0_{R_2}}\right\}}\right] \subseteq \phi^{-1} \left[{S_2}\right] = S_1$

But by definition, $K = \phi^{-1} \left[{\left\{{0_{R_2}}\right\}}\right]$

and so $S_1$ is a subset of $R_1$ containing $K$, that is:
 * $K \subseteq S_1 \subseteq R_1$

Now we need to show that $S_1$ is a subring of $R_1$.

Let $r, r' \in S_1$.

Then $\phi \left({r}\right), \phi \left({r'}\right) \in S_2$.

Hence:

So:
 * $r + r' \in S_1$

Then:

So:
 * $-r \in \phi^{-1} S_1$

Finally:

So:
 * $r \circ_1 r' \in S_1$

So from the Subring Test we have that $\phi^{-1} \left[{S_2}\right]$ is a subring of $R$ containing $K$.