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 epimorphism.

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 $$K$$.