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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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 $K[x,y]$.