## Chinese Remainder Theorem for Modules, Part 1

Let $R$ be a ring with 1 and let $\{A_i\}_{i \in I}$ be a family of ideals in $R$ indexed by a set $I$ ($I$ need not be finite for the moment.) Let $M$ be a unital left $R$-module. Prove that the mapping $\Phi : M \rightarrow \prod_{i \in I} M/A_iM$ given by $\Phi(m)_i = m + A_iM$ is an $R$-module homomorphism with kernel $\bigcap A_iM$.

[Note: If you’re following along with Dummit & Foote, you may notice that we jumped ahead a bit by assuming that $\prod M/A_iM$ is naturally an $R$-module. It is, and this result could have been proved shortly after the definition of module, so we will let slide the fact that we haven’t yet formally proven that the product of modules is a module.]

First, note that for all $x,y \in M$ and $r \in R$, we have $\Phi(x + r \cdot y)_i = (x+r \cdot y) + A_iM$ $= (x + A_iM) + r \cdot (y + A_iM)$ $= \Phi(x) + r \cdot \Phi(y)$. Thus $\Phi$ is an $R$-module homomorphism.

Now suppose $m \in \mathsf{ker}\ \Phi$. Then for all $i$, we have $0 = \Phi(m)_i = m + A_iM$, and thus $m \in A_iM$. So $m \in \bigcap A_iM$. Conversely, if $m \in \bigcap A_iM$, then $\Phi(m)_i = m + A_iM = 0$ for all $i$. Thus $m \in \mathsf{ker}\ \Phi$.