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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
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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
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...it would be of rank 1 –  Mariano Suárez-Alvarez Dec 31 '11 at 16:22
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@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
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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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1 Answer

This is a community wiki answer intended to remove this question from the unanswered queue.


As Chris Eagle hinted in the comments, free ideals in a commutative domain can only be generated by a single element.

If $a,b$ were two elements of a basis of a free ideal in a commutative domain, then $ba+(-a)b=0$ would be a nontrivial $R$-combination of the two, but that is absurd if they are members of an $R$-basis.

So, $(x,y)$, which isn't principal, cannot be a free ideal of $K[x,y]$.

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