Sum of Ideals is Ideal

Theorem
Let $J_1$ and $J_2$ be ideals of a ring $\struct{R, +, \circ}$.

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

where $J_1 + J_2$ is as defined in subset product with respect to $\struct{R, +}$.

Proof
By definition, $\struct {R, +}$ is an abelian group.

So from Subset Product of Abelian Subgroups, we have that:
 * $\struct{J, +} = \struct{J_1, +} + \struct{J_2, +}$

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 $R$.