Modules over $k[X,Y]$ 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? ...)
 A: 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.
A: If $k$ is a field the ideals $I\subset A=k[X,Y]$ which are projective modules over $A$ are exactly the principal ideals.
Indeed every projective module over $A$ is free and a free  ideal in a domain can only be of dimension $0$ or $1$, because two non-zero elements $f,g$ are always  linearly dependent: $g.f+ (-f).g=0$ !  
So the free (= projective) ideals in $A$  are $I=(0)$ with $dim_A(0)=0 $ and $I=fA, \; f\neq 0$ with $dim_A (fA)=1$ (where $\lbrace f \rbrace $ is a basis of $I$).
Since I'm in an aggressive mood [:-)], I'll use a sledge-hammer to justify my claim above that $I$ is free if it is projective: the  Quillen-Suslin theorem, which indeed says that any finitely generated projective module over a polynomial ring $k[X_1,X_2,...,X_n]$ ($ k$ a field) is free.
Edit
 I'm in a more peaceful mood now, and I'll give a more elementary argument that projective implies free.
If a non-zero  ideal in a domain is projective, it must be projective of rank one (this is proved in Bourbaki, Algèbre Commutative,chap.II, §5, Théorème 4).
And now you just use that projective modules of rank one in a UFD are free (in geometric language: the Picard group of a UFD is trivial. You can find that in this book by Görtz-Wedhorn, on page 309).
