| Version 3 |
Version 2 |
| \begin{prop} Let $R$ be a commutative ring with unity. For every pairwise comaximal ideals $I_1, I_2, ... , I_n$, the following holds:\begin{equation} |
\begin{prop} Let $R$ be a commutative ring with unity. For every pairwise comaximal ideals $I_1, I_2, ... , I_n$, the following holds:\begin{equation} |
| I_1 \cap I_2 \cap ... \cap I_n = I_1I_2 ... I_n.\end{equation}\end{prop}\begin{proof} We prove by induction on $n$. For $n=2$, $I_1+I_2 = R$ implies:\begin{equation} |
I_1 \cap I_2 \cap ... \cap I_n = I_1I_2 ... I_n.\end{equation}\end{prop}\begin{proof} We prove by induction on $n$. For $n=2$, $I_1+I_2 = R$ implies:\begin{equation} |
| I_1\cap I_2 = R(I_1\cap I_2) = (I_1 + I_2)(I_1 \cap I_2) \subseteq I_1I_2. \end{equation} |
I_1\cap I_2 = R(I_1\cap I_2) = (I_1 + I_2)(I_1 \cap I_2) \subseteq I_1I_2. \end{equation} |
| The converse inclusion is trivial. Assume now that the equality holds for $n \ge 2$: $J:= I_1 \cap I_2 \cap ... \cap I_n = I_1I_2 ... I_n$. Since $ I_n + I_j = R$, for every $j \neq n$, there exist the elements $a_j\in I_j$ and $b_j\in I_n$ such that $a_j + b_j =1$. The product $c:= \prod_{j=1}^n a_j = \prod_{j=1}^n(1-b_j)\in 1 + I_n$. Also $c\in J$, then $1\in J+I_n$ or $J+I_n =R$.\newline |
The converse inclusion is trivial. Assume now that the equality holds for $n \ge 2$: $J:= I_1 \cap I_2 \cap ... \cap I_n = I_1I_2 ... I_n$. Since $ I_n + I_j = R$, for every $j \neq n$, there exist the elements $a_j\in I_j$ and $b_j\in I_n$ such that $a_j + b_j =1$. The product $c:= \prod_{j=1}^n a_j = \prod_{j=1}^n(1-b_j)\in 1 + I_n$. Also $c\in J$, then $1\in J+I_n$ or $J+I_n =R$.\newline |
| Applying the case $2$, the induction step is satisfied:\begin{equation} I_1I_2 ... I_n = J \cap I_n = J\cap I_n = I_1\cap I_2 \cap ... \cap I_n \cap I_{n+1}. \end{equation} \end{proof} |
Applying the case $2$, the induction step is satisfied:\begin{equation} I_1I_2 ... I_n = J \cap I_n = J\cap I_n = I_1\cap I_2 \cap ... \cap I_n \cap I_{n+1}. \end{equation} \end{proof} |
