Ring Epimorphism Preserves Ideals

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

Let $J$ be an ideal of $R_1$.

Then $\phi \left({J}\right)$ is an ideal of $R_2$.

Proof
$J$ is an ideal of $R_1$, so it is also a subring of $R_1$.

From Ring Homomorphism Preserves Subrings, it follows that $\phi \left({J}\right)$ is a subring of $R_2$.

Now suppose $u \in \phi \left({J}\right)$.

Let $v \in R_2$.

Then $\exists x \in J, y \in R_1$ such that $\phi \left({x}\right) = u, \phi \left({y}\right) = v$.

Thus, by the morphism property:


 * $u \circ_2 v = \phi \left({x}\right) \circ_2 \phi \left({y}\right) = \phi \left({x \circ_1 y}\right)$

So $u \circ_2 v \in \phi \left({J}\right)$ because $x \circ_1 y \in J$.

Similarly $v \circ_2 u \in \phi \left({J}\right)$ also.

The result follows.