Preimage of Image of Subring under Ring Homomorphism

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

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

Let $J$ be a subring of $R_1$.

Then:


 * $\phi^{-1} \sqbrk {\phi \sqbrk J} = J + K$

Proof
Let $x \in \phi^{-1} \sqbrk {\phi \sqbrk J}$.

Then:

So we have shown that:
 * $\phi^{-1} \sqbrk {\phi \sqbrk J} \subseteq J + K$

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

Then:

So we have shown that:
 * $J + K \subseteq \phi^{-1} \sqbrk {\phi \sqbrk J}$

Hence the result by definition of set equality.