# Chinese Remainder Theorem in Structure Theorem for f.g. Modules over a PID

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!

• 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 $R^2/(d_1R\times 0)$? Jun 23 at 8:01
• The first one will be $R/d_{1}R$ $\times$ $R/d_{2}R$. Not sure about the second one. Jun 23 at 8:02
• 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, $\mathbb{Z}/(15)=\mathbb{Z}/(3)\times \mathbb{Z}/(5)$. Jun 23 at 8:06
• Would you like to elaborate more on the CRT part? Jun 23 at 8:12