# Reconciling two versions of Chinese Remainder Theorem

In an undergraduate number theory class, I was taught the Chinese Remainder Theorem as follows: where $n_1,\cdots,n_k$ are pairwise coprime integers and $\{a_i\}\subseteq \mathbb Z$, the set of equivalences $x\equiv a_i\ \text{mod}(n_i)$ for $i=1,\cdots,k$ has a solution that is unique modulo $\prod_{i=1}^k n_i$.

Now that I am studying Dummitt and Foote's text in graduate school, I get their version of the CRT: Let $A_1,\cdots,A_k$ be ideals of a ring $R$. The map $$R\to R/A_1\times \cdots \times R/A_k$$ defined by $r\mapsto(r+A_1,\cdots,r+A_k)$ is a ring homomorphism with kernel $\cap_{i=1}^k A_i$. If these ideals are pairwise comaximal, then this map is surjective and $\cap_{i=1}^kA_i=A_1A_2\cdots A_k$, so $$R/(A_1A_2\cdots A_k)= R/(\cap_{i=1}^k A_i)\cong R/A_1\times \cdots \times R/A_k$$

I cannot figure out how to reconcile these two versions. More specifically, my question is: Given the latter formulation, how can I show the former formulation to be true?

• $n_i$ and $n_j$ are coprime if and only if the principal ideals generated by them are comaximal, i.e. $(n_i) + (n_j) = \mathbb{Z}$. – Daniel Fischer Dec 1 '17 at 18:49
• @DanielFischer Okay, that lets me get to the following: where $n=\prod_{i=1}^k n_i$, $\mathbb Z/n \cong \mathbb Z/n_1\times\cdots\times \mathbb Z/n_k$. But I still don't see how to pull the first version of the CRT out of that. – Ceph Dec 1 '17 at 18:55
• The system of congruences $x \equiv a_i \pmod{n_i}$ corresponds to the point $(a_1,\dotsc, a_k)$ in $\mathbb{Z}/n_1\times \dotsc \times \mathbb{Z}/n_k$. The system has a solution if and only if the point lies in the image of the canonical map. The kernel of the canonical map is the "unique modulo $n$" part. – Daniel Fischer Dec 1 '17 at 19:01