The reason why FTFGMPID fails with UFD instead of PID

[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!

• 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." – Cameron Buie Feb 6 at 0:48
• It stands for "Fundamental Theorem for Finitely generated modules over PID", the theorem I provided. – Andrew Feb 6 at 1:17
• Hint: $(x,y)$ is torsion-free. – Wojowu Feb 6 at 14:57
• And $(x, y)$ is not a free module. Thanks Wojowu. – Andrew Feb 8 at 3:29