[Theorem] Let $M$ be a finitely generated module over a PID $R$. Then $$M \simeq R^d \oplus R/(r_1) \oplus \cdots \oplus R/(r_n)$$ for some $d \geq 0$ and $r_1| \cdots | r_n$.

If we replace the PID by UFD then the theorem fails. I want to find an example showing this. For proving the theorem, we used the fact that for any free $R$-module of finite rank, its submodule is free as well. This is false if $R$ is not a PID. If $R$ is not a PID, then $R$ contains an ideal that is not a free module. For example, the ideal $(x, y)$ in $K[x, y]$, $K$ is a field, is not a free module.

but I couldn't explicitly show that the theorem fails for $K[x, y]$ with $K$ being a field.

I would appreciate any help!

  • $\begingroup$ For the benefit of many who would like to answer your question, but can't read your mind: what does FTFGMPID stand for? Specifically, the "FT" part isn't readily clear to me at the moment. I assume that "FGM" stands for "finitely generated module," and "PID" is (the common) abbreviation for "principal ideal domain." $\endgroup$ – Cameron Buie Feb 6 at 0:48
  • $\begingroup$ It stands for "Fundamental Theorem for Finitely generated modules over PID", the theorem I provided. $\endgroup$ – Andrew Feb 6 at 1:17
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    $\begingroup$ Hint: $(x,y)$ is torsion-free. $\endgroup$ – Wojowu Feb 6 at 14:57
  • $\begingroup$ And $(x, y)$ is not a free module. Thanks Wojowu. $\endgroup$ – Andrew Feb 8 at 3:29

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