Show that the union of a chain of ideals is an ideal.

Question

Here is my proof:

Let $I=I_1\cup\ I_2\cup\ I_3 \cup\ ..... \cup\ I_n$, $a\in I$ and $r\in R$.

Then $a\in I_i$ for some $i$ varying from $1$ to $n$. Since $I_i$ is an ideal of $R$, we have $ar\in I_i$.

Since $I_i\subset I$, $ar\in I$.

Thus, implying $I=I_1\cup\ I_2\cup\ I_3 \cup\ ..... \cup\ I_n$ is an ideal of $R$.

Is it correct?

Answer 1

Let

$$a,b\in\bigcup_{i=1}^\infty I_i\implies a\in I_{i_1}\,,\,\,b\in I_{i_2}\implies a,b\in I_j\;,\;\text{with}\;j=\max\{i_1,i_2\}$$

and since $\;I_j\;$ is an abelian group then $\;a\pm b\in I_j\;$ and etc., this together with what you already did proves what you want.

Answer 2

Let $a, b \in I$. Then there exists an $n$ such that $a, b \in I_{n}$. Hence $a - b \in I_{n}$, and so $a - b \in I$. Let $r \in R$. Then $ra, ar \in I_{n}$, so $ra, ar \in I$, and hence $I$ is an ideal.