# Generalized Chinese Remainder Theorem - Proof

Let $R$ be a commutative unitary ring and let $\mathfrak{a_1,a_2,...,a_n}\subset R$ pairwise coprime ideals, meaning that $\mathfrak{a_i}+\mathfrak{a_j}=R$ for $i\ne j$ holds. If $\pi_i\colon R\longrightarrow R/\mathfrak{a_i}$ is the corresponding canonical projection, then the homomorphism $\varphi\colon R\longrightarrow R/\mathfrak{a_1}\times ... \times R/\mathfrak{a_n}, \ x\longmapsto (\pi_1(x),...,\pi_n(x)),$ is surjective and satisfies $\ker \phi = \mathfrak{a_1}\cap...\cap \mathfrak{a_n}$. Thus, it induces an isomorphism $R/\bigcap_{i=1}^n\mathfrak{a_i} \overset{\sim}{\longmapsto} \prod_{i=1}^nR/\mathfrak{a_i}$.

This is the chinese remainder theorem for ideals in ringtheory. I don't understand the first step in the proof of this, that is showing that for $j=1,...,n$ the ideals $\mathfrak{a_j}$ and $\bigcap_{i\ne j}\mathfrak{a_i}$ are coprime.

In the following we fix a $j$. Because of the requirement for the ideals to be pairwise coprime we have for $i\ne j$ two elements $a_i\in \mathfrak{a_j},a_i'\in \mathfrak{a_i}$ with $a_i+a_i'=1$. It then follows that

$1=\prod_{i\ne j}(a_i+a_i')\in\mathfrak{a_j}+\prod_{i\ne j}\mathfrak{a_i}\subset\mathfrak{a_j}+\bigcap_{i\ne j}\mathfrak{a_i} \ \ \ \ \ (*)$

which means that $\mathfrak{a_j}+\bigcap_{i\ne j}\mathfrak{a_i}=R$ holds. I understand the proof from here on. But I do not understand this first part of it. How do i obtain equation $(*)$ and how do i make the statements about where the elements are in?

• Consider the expansion of the product $\prod_{i\neq j}(a_i + a_i ')$. A term of this expansion will either contain an $a_i$, or it will not. If it contains an $a_i$, then since $a_i\in \mathfrak{a_j}$, which is an ideal, this term of the expansion will belong to $\mathfrak{a_j}$. If it does not, then it must be $\prod_{i\neq j}a_i '$, but by definition this belongs to $\prod_{i\neq j}\mathfrak{a_i}$. Then for the inclusion, it's standard that a product of ideals is included in their intersection – Max Jun 7 '17 at 21:10
• Thank you! If you could write a brief answer I can accept it and close the question! – EpsilonDelta Jun 8 '17 at 16:59

As I wrote in the comments, if you expand the product $\prod_{i\neq j} (a_i + a_i')$ you get different terms.
Each of these terms either contains an $a_i$, or it does not. If it does, then all well and good, it belongs to $\mathfrak{a_j}$.
If it doesn't, it must be $\prod_{i\neq j}a_i'$ which, by definition, belongs to $\prod_{i\neq j}\mathfrak{a_i}$, which itself is included in $\displaystyle\bigcap_{i\neq j}\mathfrak{a_i}$.