# Free modules and Polynomial rings

Given a field $K,$ we have the polynomial ring $K[x,y]$ in $2$ variables, which is also a left module (over itself). How can we prove that the ideal $(x,y)$ is not a free module?

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Sorry. I meant the ideal <x,y> – Ben Berger Dec 31 '11 at 16:05
Hint: any two elements are linearly dependent. So if it were free, it would be one-dimensional. – Chris Eagle Dec 31 '11 at 16:14
...it would be of rank 1 – Mariano Suárez-Alvarez Dec 31 '11 at 16:22
@Chris: Very nice! So, in a domain, an ideal is free iff it's principal. (Of course, Mariano is right, as usual...) --- Related answer (generalizing Chris's hint). – Pierre-Yves Gaillard Dec 31 '11 at 16:37
Dear @Chris: I suggest that you upgrade your comment to an answer. – Pierre-Yves Gaillard Dec 31 '11 at 22:45
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