Ring Epimorphism Preserves Ideals

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 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.