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.


We have $x = x \prod_{i \neq j} (a_{ij} + a_{ji})$. The product has a bunch of terms, namely $2^{\binom{n}{2}}$ terms, which are products of $a$'s. It's a little hard for me to imagine any reasonable way to write all of those terms in a closed form. No doubt someone else will have an idea. For each $i \neq j$ you have to decide whether to multiply by $a_{ij}$ or $a_{ji}$... hmm.

This is a little different from your question of how to write them in closed form, but I think it's pretty easy to see that each of those $2^{\binom{n}{2}}$ terms does lie in the product of the ideals. If this is obvious to you, never mind, but in case it is helpful for you or someone else, here is a sketch.

We consider one of the terms arising on the right hand side. It is a product of $x$ times either $a_{ij}$ or $a_{ji}$ for each $i \neq j$. Write this as $xA$ where $A$ is the product of the $a$'s. For each $i \neq j$, $A$ has a factor in $I_i$ or $I_j$ (or both). So the factors of $A$ are in all but at most one of the ideals; and if there is an ideal missing, the factor of $x$ lies in that ideal.


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.


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