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Question Let $A$ be a commutative ring and $I_1, I_2, \cdots, I_n$ be ideals of $A$. Any two of the above ideals are relatively prime, i.e., $$I_i+I_j=A, \forall i\ne j.$$ Are $I_i$ and $\bigcap\limits_{j\ne i}$ relatively prime?


This question comes up when I consider the example $\mathbb{Z}$. Suppose that $I_1=2\mathbb{Z}$, $I_2=3\mathbb{Z}$ and $I_3=5\mathbb{Z}$, any two of which are relatively prime. If we intersect $I_2$ and $I_3$, we get $I_2\cap I_3=15\mathbb{Z}$, also relatively prime with $I_1=2\mathbb{Z}$.

As $\mathbb{Z}$ is a typical commutative ring, I guess this is probably true with any commutative ring. But how to prove this? How to use the definition of reletively prime stated above?

Any help would be highly appreciated! Thank you in advance!

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Let $I$, $J$ and $K$ ideals in $A$. Assume that $I+K=A$ and $J+K=A$. The aim is to show that $IJ + K = A$, where $IJ$ is the product of the ideals $I$ and $J$. Indeed: Write $1=a+c=b+d$ with $a\in I$, $b\in J$ and $c,d\in K$, then $$ 1 = 1(1) = (a+c)(b+d) = ab + (ad + cb + cd). $$ Note that $ab\in IJ$ and that $ad+cb+cd\in K$ because $K$ is ideal. Then $1\in IJ+K$ which means that $IJ+K = A$ (recall that an ideal equals the whole ring if and only if it contains an unit of the ring).

Then an induction argument on the number $n$ of ideals shows that if $I_i + I_j = A$ for $i\neq j$, then $$ I_j + I_1 I_2\cdots I_{j-1} I_{j+1}\cdots I_n = A. $$

Finally, note that $$ I_1 I_2\cdots I_{j-1} I_{j+1}\cdots I_n \subseteq \bigcap_{i\neq j} I_i, $$ hence $$ A = I_j + I_1 I_2\cdots I_{j-1} I_{j+1}\cdots I_n \subseteq I_j + \bigcap_{i\neq j} I_i \subseteq A $$ from which your desired result follows.

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  • $\begingroup$ Perfect! Thanks a lot! $\endgroup$ – atlantic0cean Oct 27 '20 at 2:50

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