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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!

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  • $\begingroup$ What is $\mathrm{Im}(A')$? It is exactly $d_1R\times d_2 R\times \dots \times d_k R\times 0\times \dots \times 0\subset R^m.$ Now suppose $m=2$, how do you compute the quotient $R^2/(d_1R\times d_2R)$ and $R^2/(d_1R\times 0)$? $\endgroup$
    – Q. Zhang
    Jun 23 at 8:01
  • $\begingroup$ The first one will be $R/d_{1}R$ $\times$ $R/d_{2}R$. Not sure about the second one. $\endgroup$ Jun 23 at 8:02
  • $\begingroup$ The second one is $R/(d_1)\times R$. The general case is similar. About the Chinese Remainder Theorem, it will be useful when you decompose $R/(d)$ into cyclic modules, for example, $\mathbb{Z}/(15)=\mathbb{Z}/(3)\times \mathbb{Z}/(5)$. $\endgroup$
    – Q. Zhang
    Jun 23 at 8:06
  • $\begingroup$ Would you like to elaborate more on the CRT part? $\endgroup$ Jun 23 at 8:12

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