Sum of Ideals is Ideal

Theorem
Let $J_1$ and $J_2$ be ideals of a ring $\left({R, +, \circ}\right)$.

Then:
 * $J = J_1 + J_2$ is an ideal of $R$

where $J_1 + J_2$ is as defined in subset product.

Proof
By definition, $\left({R, +}\right)$ is an abelian group.

So from Subgroup Product of Abelian Subgroups, we have that:
 * $\left({J, +}\right) = \left({J_1, +}\right) + \left({J_2, +}\right)$

is itself a subgroup of $R$.

Now consider $a \circ b$ where $a \in J, b \in R$.

Then:

Similarly, $b \circ a \in J_1 + J_2$

So by definition $J_1 + J_2$ is an ideal of $R$.