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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? ...)

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Every finitely generated projective module over a polynomial ring is free. See the Quillen–Suslin theorem, formely known as Serre's conjecture. –  lhf Oct 29 '11 at 18:22
    
@lhf: well, yes. Do you mean something different from what I said in my answer? –  Georges Elencwajg Oct 29 '11 at 19:17
    
@Georges, no I don't. The question reminded me at once of Serre's conjecture and I just wanted to mention the link to the Quillen-Suslin theorem. –  lhf Oct 29 '11 at 19:27
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@lhf: Thank you for the link to the Quillen-Suslin theorem. –  Georges Elencwajg Oct 29 '11 at 19:35
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2 Answers

up vote 8 down vote accepted

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).

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Ok... Uhm, I didn't expect the big machinery, to be honest :). Is there an easy/elementary way to see that - for example - the non-principal ideal $(X,Y)$ is not projective? –  Evariste Oct 29 '11 at 17:59
    
The ideal $\mathfrak m=(X,Y)$ is not locally free, hence not projective. Indeed $\mathfrak m A_{\mathfrak m}$ is not free as an $A_{\mathfrak m}$- module: this uses no big machinery. –  Georges Elencwajg Oct 29 '11 at 18:15
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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.

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Dear Jack, although $M$ is indeed not flat, I don't understand your arguments. Why is $x\otimes y-y\otimes x$ not zero? And how can you "look at this modulo $M^2$", since $M^2$ is definitely not included in $M \otimes M$ ? –  Georges Elencwajg Oct 29 '11 at 19:30
    
@Georges: If x⊗y−y⊗x = 0 in M⊗M, then it is 0 in every quotient, including the quotient by $M^2 \cdot ( M\otimes M)$, which is isomorphic to $R/M^2 \otimes (M \otimes M) \cong M/M^2 \otimes M/M^2 \cong k^2 \otimes k^2 \cong k^4 \neq 0$. If one traces the isomorphisms, x⊗x = e11, x⊗y = e12, y⊗x = e21, y⊗y = e22, and so of course e12 - e21 ≠ 0. –  Jack Schmidt Oct 29 '11 at 19:45
    
Dear Jack: thank you for your convincing explanations. –  Georges Elencwajg Oct 29 '11 at 19:59
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