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I was asked to solve this simple question in one of my assignments:

If $I_1, \ldots, I_n$ are ideals of $A$ ($A$ is commutative ringe with unity) such that $I_i + I_j = A$ for all $i \ne j$, show that $$ I_1 + (I_2 \cdots I_n) = A. $$

My spontaneous answer was that $I_1+I_2 \subseteq I_1 + (I_2 \cdots I_n) $, then $1\in I_1+I_2 \implies 1\in I_1 + (I_2 \cdots I_n) \implies I_1 + (I_2 \cdots I_n)=A$. However, my teacher uses induction as some other sources I looked into do. I understand their proof but I can´t see why my proof is being naive, if it is. Might I be missing something trivial when stating that $I_1+I_2 \subseteq I_1 + (I_2 \cdots I_n) $?

Thank you guys in advance!

Alexandre Zagara
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    Actually, we have $I_2\cdots I_n\subseteq I_2$. – Bowei Tang Jun 21 '25 at 10:05
  • Why do you believe $,I_1+I_2 \subseteq I_1 + (I_2 \cdots I_n)? $ Generally, For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Jun 21 '25 at 10:15
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    To help remember the correct inclusion direction, note that - just like for principal ideals - it is true for any ideals that divides $\Rightarrow$ contains, so $, I\mid IJ ,\Rightarrow, I\supseteq IJ,$ (but unlike the principal case the converse only holds in very special (Prufer) domains, e.g. Dedekind domains, e.g. number rings). $\ \ $ – Bill Dubuque Jun 21 '25 at 10:51
  • @BillDubuque My question was not of a solution-verification type since I suspected my solution to be too naive. I did not explained the steps you are referring too since it was trivial for me and it turned out to be trivially wrong! Thank you very much for your explanation, it made me realize where my mistake was and completely solved my doubts. – Alexandre Zagara Jun 21 '25 at 11:09

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