Preimage of Subring under Ring Homomorphism is Subring

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

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

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

Proof
Let $K = \map \ker \phi$ be the kernel of $R_1$.

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

From Subset Maps to Subset:
 * $\phi^{-1} \sqbrk {\set {0_{R_2} } } \subseteq \phi^{-1} \sqbrk {S_2} = S_1$

But by definition, $K = \phi^{-1} \sqbrk {\set {0_{R_2} } }$

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 $\map \phi r, \map \phi {r'} \in S_2$.

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 $\phi^{-1} \sqbrk {S_2}$ is a subring of $R$ containing $K$.