pairwise comaximal ideals property

Proposition 1.

Let $R$ be a commutative ring with unity. For every pairwise comaximal ideals $I_{1},I_{2},...,I_{n}$, the following holds:

 $I_{1}\cap I_{2}\cap...\cap I_{n}=I_{1}I_{2}...I_{n}.$ (1)
Proof.

We prove by induction on $n$. For $n=2$, $I_{1}+I_{2}=R$ implies:

 $I_{1}\cap I_{2}=R(I_{1}\cap I_{2})=(I_{1}+I_{2})(I_{1}\cap I_{2})\subseteq I_{% 1}I_{2}.$ (2)

The converse inclusion is trivial. Assume now that the equality holds for $n\geq 2$: $J:=I_{1}\cap I_{2}\cap...\cap I_{n}=I_{1}I_{2}...I_{n}$. Since $I_{n+1}+I_{j}=R$, for every $j\neq{n+1}$, there exist the elements $a_{j}\in I_{j}$ and $b_{j}\in I_{n+1}$ such that $a_{j}+b_{j}=1$. The product $c:=\prod_{j=1}^{n}a_{j}=\prod_{j=1}^{n}(1-b_{j})\in 1+I_{n+1}$. Also $c\in J$, then $1\in J+I_{n+1}$ or $J+I_{n+1}=R$.
Applying the case $2$, the induction step is satisfied:

 $I_{1}I_{2}...I_{n+1}=JI_{n+1}=J\cap I_{n+1}=I_{1}\cap I_{2}\cap...\cap I_{n}% \cap I_{n+1}.$ (3)

Title pairwise comaximal ideals property PairwiseComaximalIdealsProperty 2013-03-22 16:53:34 2013-03-22 16:53:34 polarbear (3475) polarbear (3475) 9 polarbear (3475) Result msc 16D25