## Chinese Remainder Theorem for Modules, Part 2

Let $R$ be a commutative ring with 1 and let $\{A_i\}_{i =1}^n$ be a finite family of ideals of $R$. Suppose further that $A_i + A_j = R$ if $i \neq j$. (I.e. the ideals $A_i$ are pairwise comaximal.) Let $M$ be a unital left $R$-module. Prove that the $R$-module homomorphism $\Phi : M \rightarrow \prod_{i=1}^n M/A_iM$ discussed here is surjective and has kernel $(\prod_{i=1}^n A_i)M$, where $\prod_{i=1}^n A_i$ denotes the ideal product in $R$ rather than the direct product of sets. Deduce that $M/(\prod_{i=1}^n A_i)M \cong_R \prod_{i=1}^n M/A_iM$.

We will prove this result by induction on $n$. For the base case $n = 2$, we have $\Phi(m) = (m+A_1M, m+A_2M)$. First we show that $\Phi$ is surjective. Let $(m_1 + A_1M, m_2 + A_2M)$ be in $M/A_1M \times M/A_2M$. Since $A_1$ and $A_2$ are comaximal and $R$ has a 1, there exist $a_1 \in A_1$ and $a_2 \in A_2$ such that $a_1+a_2 = 1$. Note then that $\Phi(a_1m_2 + a_2m_1) = (a_1m_2 + a_2m_1 A_1M, a_1m_2 + a_2m_1 + A_2M)$ $= (a_2m_1 + A_1M, a_1m_2 + A_2M)$ $= ((1-a_1)m_1 + A_1M, (1-a_2)m_2 + A_2M)$ $= (m_1 - a_1m_1 + A_1M, m_2 - a_2m_2 + A_2M)$ $= (m_1 + A_1M, m_2 + A_2M)$. Thus $\Phi$ is surjective. Next we show that $\mathsf{ker}\ \Phi = (A_1A_2)M$. We showed in Part 1 that $\mathsf{ker}\ \Phi = A_1M \cap A_2M$. Clearly $(A_1A_2)M \subseteq A_1M \cap A_2M$. Now suppose $m \in A_1M \cap A_2M$. Recall that $a_1+a_2 = 1$ for some $a_1 \in A_1$ and $a_2 \in A_2$; now $m = 1 \cdot m$ $= (a_1+a_2) \cdot m$ $= a_1 \cdot m + a_2 \cdot m$. Since $m \in A_2$, $a_1 \cdot m \in (A_1A_2)M$. Since $m \in A_1$, $a_2 \cdot m \in (A_2A_1)M = (A_1A_2)M$. Thus $m \in (A_1A_2)M$ as desired. By the First Isomorphism Theorem for modules, the induced mapping $\Psi : M/(A_1A_2)M \rightarrow M/A_1M \times M/A_2M$ given by $m+(A_1A_2)M \mapsto (m+A_1M, m+A_2M)$ is an $R$-module isomorphism.

Now for the inductive step, suppose the result holds for $2$ and $n$. Let $\{A_i\}_{i=1}^{n+1}$ be a family of pairwise comaximal ideals in $R$ and let $M$ be a unital left $R$-module. We claim that $A_1$ and $\prod_{i=2}^{n+1}$ are comaximal; to see this, note that $1 = a_{1,i} + a_i$ for some $a_{1,i} \in A_i$ and $a_i \in A_i$ for all $2 \leq i \leq n+1$. Then $1 = \prod_{i=2}^{n+1}(a_{1,i} + a_i) \in A_1 + \prod_{i=2}^{n+1} A_i$, so that $A_1 + \prod_{i=2}^{n+1} A_i = R$. Now $M/(\prod_{i=1}^{n+1} A_i)M \cong_R M/A_1M \times M/(\prod_{i=2}^{n+1} A_i)M$ $\cong_R M/A_1M \times \prod_{i=2}^{n+1} M/A_iM$ $\cong_R \prod_{i=1}^{n+1} M/A_iM$.