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Let $A=k[x,y]$ and $M$ be a finitely generated graded $A$-module. I want to know if the torsion submodule $T$ of $M$ is a direct summand.

Apparently, Kaplansky, Irving: A characterization of Prufer rings shows that if every finitely generated $A$-module contains its torsion submodule as a direct summand, then $A$ is a Prüfer domain. If I've got it right, then $k[x,y]$ is no Prüfer domain. So there must be a finitely generated module $M$ whose torsion submodule is no summand.

Which?

Bubaya
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1 Answers1

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I would suggest that you try the following $M=Ae_1\oplus Ae_2/(xye_1-x^2e_2, y^2e_1-xye_2)$.

Mohan
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  • Ah, I was thinking in terms of bigraded modules; for these I have also found a module in the meantime. – Bubaya Sep 12 '19 at 15:24