# Coprime ideals in commutative unit ring [duplicate]

Possible Duplicate:
Comaximal ideals in a commutative ring

Let $I$ and $J$ are coprime ideals in commutative unit ring $A$. Is it true that $I^m$ and $J^n$ are also coprime for any $m,n\in\mathbb{N}$?

Thanks a lot!

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## marked as duplicate by rschwieb, Sasha, Norbert, Henry T. Horton, Noah SnyderOct 4 '12 at 22:38

@pritam Is coprime ideal a synonim for comaximal ideal? It does not look so. –  Sasha Oct 3 '12 at 16:16
@Sasha: According to Atiyah-Macdonald (page 7), yes they are same –  pritam Oct 3 '12 at 16:22

Although this is a trivial exercise in commutative algebra, I've seen here only solutions based on calculations. Let me try a one-line solution (well, one line and a little):

$I^m+J^n\neq R$ $\Rightarrow$ $\exists P$ prime ideal with $I^m+J^n\subseteq P$ $\Rightarrow$ $I^m\subseteq P$, $J^n\subseteq P$ $\Rightarrow$ $I\subseteq P$, $J\subseteq P$ $\Rightarrow$ $I+J\subseteq P$, contradiction.

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Nice solution ${}{}{}{}{}$ –  Amr May 11 at 9:58

Let's remind ourselves of the definition of coprime: $I,J \subset R$ are coprime iff $I + J = R$. This means, there are $i \in I$ and $j \in J$ such that $i + j = 1$.

We have $1 = 1^{2k} = (i + j)^{2k} = \sum_{\iota =0}^{2k} {2k \choose \iota} i^\iota j^{2k - \iota} \in I^{2k} + \dots + J^{2k} \subset I^k + J^k$ for all $k$.

Choose $k = \max (n,m)$ then $I^k \subset I^n$ and $J^k \subset J^m$ and hence $1 = (i+j)^{2k} \in I^k + J^k \subset I^n + J^m$.

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