If $I_1, \dots, I_n \subset R$ are distinct ideals with the property that $I_i + I_j = R$
for any $i \ne j$, show that $I_i + \prod_{j \ne i} I_j = R$ for any $i$ where the product is take over all the integers $1, \dots, n$ except $i$.
My failed attempts : $\prod_{j \ne i} I_j \subset I_j$ for all $j \ne i$ because of definition of ideals, but that does not imply $I_i + \prod_{j \ne i} I_j = R$ because still may not $\prod_{j \ne i} I_j = I_j$. Also, $I_i + I_j = R$ implies that for any $r \in R$, $r=i_1+i_2=i_1+i_3=\dots$ for fixed $i_1 \in I_1$ and some $i_j \in I_j$, though $r-i_1 \in \cap_{j \ne i} I_j$ but that does not lead to any thing!
Source of the exercise : Introduction to Algebraic Geometry by Justin R. Smith
Best Answer
Question: "@hm2020 Thanks! If you would you like you may post your comments as an answer. – TheMagicMountain"
Answer: If $I_i+I_j=(1)$ for $i\neq j$, it follows there are elements (let $i=1$) $(a_j,b_j)\in I_1 \times I_j$ ($j\neq 1$) with $a_j+b_j=1$. You may check that the element
$$\prod_j (a_j+b_j)=1 \in I_1 + \prod_{j\neq 1}I_j$$
hence
$$ I_1 + \prod_{j\neq 1}I_j=(1)$$
is the unit ideal.