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.

General Result
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\left({R, +, \circ}\right)$.

Then:
 * $J = J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.

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

Corollary 1
$J$ is contained in every subring of $R$ containing $\displaystyle \bigcup_{k=1}^n {J_k}$.

Corollary 2

 * $J_1 \subseteq J_1 + J_2$
 * $J_2 \subseteq J_1 + J_2$

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 $R$.

Proof of General Result
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\left({R, +, \circ}\right)$.

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.

$P(1)$ is true, as this just says $J_1$ is an ideal of $R$.

Basis for the Induction
$P(2)$ is the case:
 * $J_1 + J_2$ is an ideal of $R$

which has been proved above.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $J_1 + J_2 + \cdots + J_k$ is an ideal of $R$.

Then we need to show:
 * $J_1 + J_2 + \cdots + J_k + J_{k+1}$ is an ideal of $R$.

Induction Step
This is our induction step:

Let $J = J_1 + J_2 + \cdots + J_k$.

From the induction hypothesis, $J$ is an ideal.

From the base case, $J + J_{k+1}$ is an ideal.

That is:
 * $J_1 + J_2 + \cdots + J_k + J_{k+1}$ is an ideal of $R$.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N: J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.

Proof of Corollary 1
Follows directly from the definition of join.

Proof of Corollary 2
We have that $0_R \in J_1 + J_2$ and so $\forall x \in J_1: x + 0_R = x \in J_1 + J_2$.

Similarly for $J_2$.