Increasing Union of Ideals is Ideal

Theorem
Let $R$ be a ring.

Let $S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$ be ideals of $R$.

Then the increasing union $S$:
 * $\displaystyle S = \bigcup_{i \in \N} S_i$

is an ideal of $R$.

Proof
Let $\displaystyle S = \bigcup_{i \in \N} S_i$.

From Increasing Union of Subrings is Subring, we have that $S$ is a subring of $R$.

Now we need to show that it is an ideal of $R$.

Let $a \in S$.

Then $\exists i \in \N: a \in S_i$.

Let $b \in R$.

Then $a b \in S_i$ and $b a \in S_i$, as $S_i$ is an ideal of $R$.

Thus $a b \in S$ and $b a \in S$.

So $S$ is an ideal of $R$.