Sum of Ideals is Ideal/General Result

Theorem

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

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

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

Proof

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

Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

$J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.

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

Basis for the Induction

$\map P 2$ is the case:

$J_1 + J_2$ is an ideal of $R$

which is proved in Sum of Ideals is Ideal.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ 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 $\map P k \implies \map P {k + 1}$ 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 $R$.

$\blacksquare$