# Union of ascending ideals is an ideal

Could you tell me what I'm doing wrong in proving this proposition?

If $I_1 \subset I_2 \subset ... \subset I_n \subset ...$ is an ascending chain of ideals in $R$, then $I := \bigcup _{n \in \mathbb{N}} I_n$ is also an ideal in $R$.

So we need to prove that the group $(I, +)$ is abelian and if $a \in I$, then $ar, ra \in R$ (but I suppose we could assume that $R$ is commutative).

Here are my attempts: $a \in I \Rightarrow \exists n\in \mathbb{N} : a \in I_n \Rightarrow ar, ra \in I_n \Rightarrow ar, ra \in I$. But I think I don't use the fact that the chain is ascending here. Could you point out my mistake and help me correct it?

Thank you a lot!

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What about proving that $(I,+)$ is an abelian group? To me, that seems to be the part where you need that the chain is ascending. –  Svinepels Feb 2 '14 at 12:23
Indeed, thank you ;) –  Bilbo Feb 2 '14 at 12:24

The nested chain is important for showing closure under addition. If we have $a\in I_m$ and $b \in I_n$, how can we guarantee $a+b$ belongs to $I$?