I have been asked by my Algebra professor, to explicitly determine the Chinese Remainder Theorem in the proof of the Structure Theorem for f.g. Modules over a PID.
Here's what I know:
$\textbf{Theorem : Let R be a Euclidean Domain, and M a finitely generated R-module.}$ $\text{Then,}$ $\exists\text{ } d_{1},d_{2},\cdots,d_{k}\in \textbf{R}, \quad k,r \text{ being non-negative integers such that,}\\$
- $M \cong R/(d_{1}) \oplus \cdots \oplus R/(d_{k}) \oplus R^{r}$
- $d_{1},\cdots,d_{k}$ are non-units, non-zero such that $d_{1}|d_{2}|\cdots|d_{k}$
The proof that I have come across goes something like this:
Let $R$ be a Euclidean Domain and let $M$ be a f.g. $R$-module. Now, $R$ is noetherian $\implies$ $\exists$ a presentation $R^{n}\xrightarrow{\text{$\psi$}} R^{m}\xrightarrow{\text{$\phi$}} M\longrightarrow o$. Let $A$ be the matrix of $\psi$ w.r.t some basis. Take the Smith-normal form $A'$ of $A$. Then, $M\cong R^{m}/im(\psi)\cong R^{m}/im(A')$
By getting rid of unnecessary rows and columns, im($A'$) has the form:
$im(A') \cong d_{1}R \oplus d_{2}R \oplus \cdots \oplus d_{k}R.\\\\ \text{Hence, } M\cong R^{m}/im(A') \cong R/(d_{1})\oplus \cdots \oplus R/(d_{k})\oplus R^{r}$
What I don't understand is the last isomorphism between $R^{m}/im(A')$ and the direct sums. I also have no clue where is the Chinese Remainder Theorem is being used. I detailed help would be appreciated! Thanks!