I am trying to prove that the product of $n$ pairwise coprime ideals is their intersection. I can easily do the proof for $n=2$, and I see how to generalize the argument but there are details I don't know how to handle. Here's my attempt:
Let $R$ be a commutative ring, $I_1,\dots,I_n$ be a collection of ideals such that $I_i+I_j=R$ for all $i\neq j$. Then $\cap_i I_i\subseteq\prod_i I_i$.
For all $i\neq j$ I can find elements $a_{ij}\in I_i$, $a_{ji}\in I_j$ such that $a_{ij}+a_{ji}=1$. Thus if $x\in \cap_i I_i$, it can be written as \begin{equation}
x=\prod_{i\neq j} (a_{ij}+a_{ji}) x. \end{equation}
My problem is: how to write the above product as a sum of products of elements of each ideal. I can see it works, but the product gets super-messy already with $n=3$, so I don't see how to write it in close form.
Best Answer
Try induction. Suppose that $\mathfrak{b}=\prod_{i=1}^{n-1}I_i=\cap_{i=1}^{n-1}I_i$.
Then, since $I_i+I_n=(1)$ for all $i \leq n-1$ by assumption, we have that$x_i+y_i=1$ for some $x_i,y_i$ in $I_i$ and $I_n$ respectively.
But then $\prod_{i=1}^{n-1}x_i=\prod_{i=1}^{n-1}(1-y_i) \equiv 1 (\mod I_n)$
But then $I_n+\mathfrak{b}=(1)$, from which you can apply the same argument for the $n=2$ case, since $I_n$ is coprime to the whole product.
Sorry I couldn't help with the closed form. As an aside, this argument is from Proposition 1.10 in Atiyah-Macdonald, which I just went back and found.