# The ideal $(x,y)$ is not a free $K[x,y]$-module

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?

-
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

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 $K[x,y]$.