pairwise comaximal ideals property

Proposition 1.

Let $R$ be a commutative ring with unity. For every pairwise comaximal ideals $I_{\mathrm{1}}\mathrm{,}{I}_{\mathrm{2}}\mathrm{,}\mathrm{\dots }\mathrm{,}{I}_n$, the following holds:

$$I_1\cap I_2\cap \mathrm{\dots }\cap I_n=I_1{I}_2\mathrm{\dots }{I}_n.$$(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 $n\ge 2$: $J:=I_1\cap I_2\cap \mathrm{\dots }\cap I_n=I_1{I}_2\mathrm{\dots }{I}_n$. Since $I_{n+1}+I_j=R$, for every $j\ne n+1$, 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:

$$I_1{I}_2\mathrm{\dots }{I}_{n+1}=J{I}_{n+1}=J\cap I_{n+1}=I_1\cap I_2\cap \mathrm{\dots }\cap I_n\cap I_{n+1}.$$(3)

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