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Let $D$ be an integral domain, $K$ its field of fractions, and $J_1,...,J_n$ are ideals of $D$ such that $(\sum_{i=1}^{n} J_i)^{-1}=D$. How can we prove that this implies $(\sum_{i=1}^{n} J_i^m)^{-1}=D$ for all $m\ge1$ ?

I know that the converse is true since $\sum_{i=1}^{n}{J_i}^{m}\subseteq\sum_{i=1}^{n}J_i$. I tried to prove it by induction on $m$ but it did not work. This question is part of the proof of the lemma 4.2 in the the paper Some remarks on star-operations by Hedstorm and Houston.

user26857
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Marcos
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  • I have to say that I am a novice in the subject of the stars-operations, I try to understand the result. – Marcos Mar 19 '15 at 18:27

1 Answers1

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Note that $x\in (D:_K \sum_{i=1}^{n} J_i)$ iff $xJ_i\subseteq D$ for all $i=1,\dots,n$.

Let me prove the case $n=2$. I'll show that

$(J_1+J_2)^{-1}=D$ implies $(J_1^{m_1}+J_2^{m_2})^{-1}=D$ for any $m_1,m_2\ge 1$.

First set $m_1=1$. Then pick an $x\in(J_1+J_2^{m_2})^{-1}$, so $xJ_1\subset D$ and $xJ_2^{m_2}\subset D$. Since $(xJ_2)J_2^{m_2-1}\subset D$, and $(xJ_2)J_1\subset D$, by induction we get $xJ_2\subset D$, hence $x\in(J_1+J_2)^{-1}=D$. Now use induction on $m_1$.

user26857
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