Showing that the ideal $(x,y)$ in $k(x,y)$ is not locally free. [duplicate]
Solution 1:
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 $K[x,y]$.