TITLE: Modules over $k[X,Y]$
QUESTION [8 upvotes]: Over a PID like $k[X]$, all (non-trivial) ideals are free and hence projective. But the ring $k[X,Y]$ is not a PID. Is it possible to describe all ideals of this particular ring which are projective modules?
(What if we restrict to $k$ algebraically closed? ...)

REPLY [5 votes]: One can show that the maximal ideal $M=(x,y)$ is not projective without using Quillen–Suslin, but I'll at least assume tensor products and flat modules (projectives are a special case of flats, flats are the limits of projectives).
In fact, $M$ is not even flat: $M \otimes M \to M^2$ is not injective since $x\otimes y - y \otimes x \mapsto xy-yx = 0$.  One looks at this modulo $M^2$ to see that $M/M^2$ is a nonzero vector space (of dimension 2) over $k$, and so the tensor product is nonzero, but the range is then obviously 0.