TITLE: Comaximal ideals in a commutative ring
QUESTION [19 upvotes]: Let $R$ be a commutative ring and $I_1, \dots, I_n$ pairwise comaximal ideals in $R$, i.e., $I_i + I_j = R$ for $i \neq j$. Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r \in\mathbb N$) also comaximal?

REPLY [6 votes]: A slight variation on other proofs. Suppose $I+J=R$, so $a+b=1$ for some $a\in I, b \in J$. I want to replace $a$ by $a^n$, so I write $a^n + (a-a^n) + b = 1$. If I knew that $a-a^n$ is in $J$, I could group it together with $b$ and get $a^n+b' = 1$, so $I^n+J = R$. But $a-a^n = a(1-a^{n-1})$ is divisible by $1-a = b \in J$, so it is in fact in $J$. From $I^n+J=R$ the general case $I^n+J^m=R$ follows by another application of the same argument.