Prove the increasing union of ideals is an ideal prove that $I_{1} \subseteq I_{2} \subseteq I_{3} \subseteq....$ are ideals of $R$ then $\bigcup_{n =1} I_n$ is an ideal of $R$. I am having a hard time picture this in my mind. Help anyone. 
 A: Let $I_1 \subseteq I_2 \subseteq \cdots$ be an ascending chain of ideals in a ring $R$. Define $I = \bigcup_{n = 1}^\infty I_n$. We first show that $I$ is an additive abelian group:


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*Let $x, y \in I$, then $x \in I_m$, $y \in I_n$ for some $m, n \in \mathbb N$. Since we are in an ascending chain of ideals, either $I_m \subseteq I_n$ or $I_n \subseteq I_m$. Suppose, without loss of generality, that $I_m \subseteq I_n$, then $x, y \in I_n$. Since $I_n$ is an ideal, it is an additive abelian group, and in particular it is closed under addition, that is, $x + y = y + x \in I_n$. Hence $x + y \in I$ and we also get that $I$ is commutative under addition.

*Let $x \in I$, then $x \in I_n$ for some $n \in \mathbb N$. Since $I_n$ is an ideal, it is an additive group, so $-x \in I_n$. Thus $-x \in I$.

*Observe that $0 \in I$ since $0 \in I_1$ and $I_1$ is an ideal and hence additive abelian group, so it has an additive identity: $0$.


Now we show that $I$ absorbs elements from $R$. Since we are working in a commutative ring with identity, we don't have to worry about left and right ideals. Let $d \in R$ and $i \in I$ be arbitrary. Since $i \in I$, then $i \in I_n$ for some $n \in \mathbb N$. Since $I_n$ is an ideal, it has the absorption property, so $id \in I_n$ and hence $id \in I$. Conclude by definition that $I$ is an ideal.
