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?
The ideal $(x,y)$ is not a free $K[x,y]$-module
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$\begingroup$
commutative-algebra
polynomials
modules
ideals
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0Sorry. I meant the ideal
– 2011-12-31 -
10Hint: any two elements are linearly dependent. So if it were free, it would be one-dimensional. – 2011-12-31
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3...it would be of rank 1 – 2011-12-31
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1@Chris: Very nice! So, in a domain, an ideal is free iff it's principal. (Of course, Mariano is right, as usual...) --- [Related answer](http://math.stackexchange.com/a/55067/660) (generalizing Chris's hint). – 2011-12-31
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0Thank you all very much! – 2011-12-31
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2Dear @Chris: I suggest that you upgrade your comment to an answer. – 2011-12-31
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0Related: https://math.stackexchange.com/questions/2255814 – 2018-11-26