# Increasing Union of Ideals is Ideal/Sequence

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## Theorem

Let $R$ be a ring.

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

Then the increasing union $S$:

- $\displaystyle S = \bigcup_{i \mathop \in \N} S_i$

is an ideal of $R$.

## Proof

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

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 $R$.

$\blacksquare$

## Sources

- 1970: B. Hartley and T.O. Hawkes:
*Rings, Modules and Linear Algebra*... (previous) ... (next): $\S 2.3$: Some properties of subrings and ideals: Lemma $2.13 \ \text{(ii)}$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $9$: Rings: Exercise $17$