Preimage of Image of Subring under Ring Homomorphism

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

Let $K = \ker \left({\phi}\right)$, and let $J$ be a subring of $R_1$.

Then:


 * $\phi^{-1} \left({\phi \left({J}\right)}\right) = J + K$

Proof

 * Let $x \in \phi^{-1} \left({\phi \left({J}\right)}\right)$.

Then $\phi \left({x}\right) \in \phi \left({J}\right)$.

Thus:
 * $\exists b \in J: \phi \left({x}\right) = \phi \left({b}\right)$

So:


 * $\phi \left({x + \left({-b}\right)}\right) = \phi \left({x}\right) + \left({- \phi \left({b}\right)}\right) = 0_{R_2}$

Thus $x + \left({-b}\right) \in K$.

Now $x = b + \left({x + \left({-b}\right)}\right)$, so $x \in J + K$.

So we have shown that:
 * $\phi^{-1} \left({\phi \left({J}\right)}\right) \subseteq J + K$


 * Now suppose that $x \in J + K$.

Then:
 * $\exists b \in J, a \in K: x = b + a$

So:


 * $\phi \left({x}\right) = \phi \left({b}\right) + \phi \left({a}\right) = \phi \left({b}\right)$

Thus $\phi \left({x}\right) \in \phi \left({J}\right)$.

Therefore $x \in \phi^{-1} \left({\phi \left({J}\right)}\right)$.

So we have shown that:
 * $J + K \subseteq \phi^{-1} \left({\phi \left({J}\right)}\right)$

Hence the result from Equality of Sets.