Sum and product of comaximal ideals

Question

Let $R$ be a commutative ring with unity. If $R=I_{i}+I_{j}$, for all $i\ne j$, where $I_1,I_2,...,I_n$ are ideals of $R$, I want to show that $$R=I_{n}+I_{1}I_{2}\cdots I_{n-1}.$$

I started off showing for $n=3$, which is easy to show. I assumed that it is true for $k$. Then $R=I_k+I_1I_2..I_{k-1}$. I need to show it for $k+1$.

I am kind of stuck here.

THanks for the help!!

Answer 1

By hypotheses $R=I_{i}+I_{j}$, for all $i\ne j$ whence $$1 = a_j + b_j \ \ \ \ \ j = 1, \ldots n-1 $$ with $a_j \in I_n $ for all $j$ , and $b_j \in I_j $.

Thus $$1 = \prod_{j=1}^{n-1} (a_j + b_j) = b_1 b_2 \cdots b_{n-1} + r $$ with $r \in I_n $ and $b_1 b_2 \cdots b_{n-1} \in I_1I_2..I_{n-1} $. This implies $$R=I_{n}+I_{1}I_{2}...I_{n-1}$$

Answer 2

If $R\ne I_{n}+I_{1}I_{2}\cdots I_{n-1}$, then there is a maximal ideal $M$ containing $I_{n}+I_{1}I_{2}\cdots I_{n-1}$, so $I_n\subset M$ and $I_{1}I_{2}\cdots I_{n-1}\subset M$. From $I_{1}I_{2}\cdots I_{n-1}\subset M$ there is $I_j\subset M$ with $1\le j\le n-1$, so $R=I_n+I_j\subset M$, a contradiction.