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If $I$ and $J$ are ideals in a ring $R$ with 1 which are co-maximal, i.e $I+J = R$, show that $I^m$ and $J^n$ are co-maximal for all $m,n$ in $\mathbb{N}$

Work done:

Should I proceed using Zorn's Lemma?

By taking $$K=I^m+J^m \neq R$$ Let $$S=\{I,J \in R \mid I^m+J^n \neq R \}$$ be the required partially order set ordered by inclusion then $$I+J \subseteq I^n+J \subseteq I^n+J^m \subseteq \ldots $$ Then, I think this chain doesn't have an upper bound...as $I^m+J^n \neq R$ for all $ m,n \in \mathbb N$. Am I right??

Thanks for any help!!

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If $I+J=R$, there exist $i\in I$, $j\in J$ with $1 = i+j$. Then using the binomial theorem \begin{align*} 1 &= 1^{m+n-1} = (i+j)^{m+n-1}\\ &= \sum_{k=0}^{m+n-1}\binom {m+n-1}k i^{m+n-k-1}j^k\\ &= \underbrace{\left(\sum_{k=0}^{n-1}\binom {m+n-1}k i^{n-1-k}j^k\right)\cdot i^m}_{\in I^m} + \underbrace{\left(\sum_{k=n}^{m+n-1}\binom {m+n-1}k i^{m+n-k-1}j^{k-n}\right)\cdot j^n}_{\in J^n}\\ &\in I^m+J^n. \end{align*} This shows $I^m + J^n = R$.

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$I+J=R \Rightarrow \sqrt {I^n}+\sqrt {J^m}=R \Rightarrow \sqrt{\sqrt {I^n}+\sqrt {J^m}}=R \Rightarrow \sqrt{ {I^n}+ {J^m}}=R \Rightarrow I^n+J^m=R$

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  • $\begingroup$ I think the third implication needs an explanation. $\endgroup$ – user26857 Jun 10 '16 at 14:40

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