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.

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