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Let $R$ be a ring with identical element and let $I_1, I_2, \dotsc, I_n$ ideals of $R$ such that $I_i+I_j = R$ for all $i \neq j$. How I can to prove that $I_1+\prod_{i=2}^nI_i = R$? I start from double containment. It is easy to see that $I_1+\prod_{i=2}^nI_i \subseteq R$ due to operations with ideals and the definition of ideal but I have not been able to prove the containment $R \subseteq I_1+\prod_{i=2}^nI_i$. Any suggestions?

Juan Herrera
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    I pretend $n = 3$ for simplicity. Then $I_1 + I_2 = R$ means there are $i_1, i_2$ such that $i_1 + i_2 = 1$ (here and later, where each $i_j$ is in the obvious ideal). Similarly, there are $i_1'$ and $i_3$ such that $i_1' + i_3 = 1$. Then $1 = 1 \cdot 1 = (i_1 + i_2)(i_1' + i_3) = (i_1i_1' + i_1 i_3 + i_1' i_2) + i_2i_3$. This shows $1 \in (I_1 + I_2I_3)$, which gives you everything. – davidlowryduda Apr 17 '24 at 05:06
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    @davidlowryduda Why are you posting basically a full answer in the comment section? – Arthur Apr 17 '24 at 05:16
  • See the Lemma in the linked dupe. – Bill Dubuque Apr 17 '24 at 06:31
  • @Arthur Because I didn't want to spend time writing a full solution and instead went to bed. – davidlowryduda Apr 17 '24 at 15:46

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