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 $$K = \mathrm{ker} \left({\phi}\right)$$, 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 $$x \in J + K$$.


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