Image of Preimage of Subring under Ring Epimorphism

Theorem

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

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

Then:

$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$

Proof

As $\phi$ is an epimorphism, it is a surjection, and so $\Img \phi = R_2$.

So $S_2 \subseteq \Img {R_1}$.

The result then follows from Image of Preimage under Mapping.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 3^\circ$