Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is infinite (say complex) projective space a scheme? More generally, can schemes have infinite cardinal dimension? It seems that infinite dimensional projective space is not a manifold, since it is not locally Euclidean for any R^n.

Related question. If inifinite projective space is a scheme, then take a nonclosed point. Taking the closure of this nonclosed point, can we get infinite dimensional subschemes? Sorry for I'm quite foreign to schemes.

share|improve this question
1  
Someone care to explain the close vote? This seems like a perfectly reasonable question to me. –  Qiaochu Yuan Mar 30 '10 at 7:54
2  
perhaps first we should clarify what the infinite affine space is (see the other comments). $Spec k[x_1,x_2,...]$ represents the functor $R \mapsto R^\mathbb{N}$, whereas the functor $R \mapsto R^{(\mathbb{N})}$ cannot represented by a scheme (this was discussed on MO), but is rather and ind-scheme. also, you can take the colimit of the finite affine spaces $\mathbb{A}^n$ and obtain a locally ringed space (but which, again, is not a scheme, since it can be shown that every quasicompact open subset is empty). –  Martin Brandenburg Mar 30 '10 at 11:57

3 Answers 3

up vote 7 down vote accepted

Starting with the affine case, if you try to define infinite dimensional affine space as Spec of k{x1,x2,...], then you realise that this is not a vector space of countable dimension, but something much larger. If you want a vector space over k of countable dimension, then this will not be a scheme, but instead will be an ind-scheme. A similar description should hold in the projective case.

Edit: Regarding why I am saying that Spec(k[x1,x2,...]) is too big: A (k-)point of Spec(k[x1,x2,...]) is an infinite sequence a1,a2,... of elements of k. If I wanted a vector space of countable dimension, then I should be asking for sequences a1,a2,... of elements of k, only finitely many of which are non-zero. This latter space is the inductive limit of affine n-space as n tends to infinity.

share|improve this answer
1  
Can you elaborate on your answer a bit? Why is it clear that $\mathrm{Spec}\,k[x_1,x_2,\ldots]$ is "too big" to be what one would intuitively seek in an affine space of countable dimension? Here is one thought I have. Reasoning by analogy with the f.d. case, where the affine space corresponding to a f.d. $k$-v.s. $V$ is $\mathrm{Spec}\,\mathrm{Sym}^\cdot V^\vee$, then $\mathbb{A}^\infty$ should be $\mathrm{Spec}\, R$ where $R=\mathrm{Sym}^\cdot V^\vee$ where $V$ is a $k$-vector space with a countable basis. But $V^\vee$ is huge... –  Sam Lichtenstein Mar 30 '10 at 6:31
    
... so this would seem to suggest to me that in fact $k[x_1,x_2,\ldots]$ is actually too small to be functions on $\mathbb{A}^\infty$. I'm not sure whether this does or does not accord with your assertion that $\mathrm{Spec}\,k[x_1,x_2,\ldots]$is "bigger" than $\mathbb{A}^\infty$ ought to be. On the other hand, $\mathbb{A}^n$ represents the functor $\mathbb{A}^n(R)=R^n$. It seems to me $k[x_1,x_2,\ldots]$ represents the functor $R\mapsto R^\mathbb{N}$ (countable direct product). Perhaps $\mathbb{A}^\infty$ should represent $R\mapsto R^{\oplus \mathbb{N}}$? –  Sam Lichtenstein Mar 30 '10 at 6:37
    
Anyhow, enough of my rambling. But can you say what are the desiderata for $\mathbb{A}^\infty$ you had in mind, and why the naive guess for what scheme this should be is wrong? –  Sam Lichtenstein Mar 30 '10 at 6:39
    
If $k[x_1,\cdots]$ is defined as finite polynomials then it does have a countable vector space basis. Are you thinking of formal polynomials in some way? –  Thomas Kragh Mar 30 '10 at 9:01
1  
@TK: A ring homomorphism $k[x_1,\dots]\to k$ can map each variable to anything, with no constraints. As a vector space, then, the set of such maps does not have a countable basis. –  Mariano Suárez-Alvarez Mar 30 '10 at 15:43

Since rings can have infinite Krull dimension, affine schemes can have infinite dimension.

share|improve this answer

You can define $Proj S$ for any graded ring $S$ and this is certainly a scheme; this is in Hartshorne II.2. Infinite projective space is $Proj S$ where $S = k[x_0, x_1, ....]$ and $k$ is the base field.

Regarding your second question, if you take any homogeneous element $f \in S$, then the vanishing of this should define a closed subscheme of codimension 1 (in particular still infinite dimensional).

Maybe I should say $Proj S$ is the algebraic analogue of infinite projective space. As a topological space it is very different from $\mathbb{C}^{\infty} - 0$/scaling. But this is even true in the finite dimensional case (Zariski topology is not the same as topology considered as a real or complex manifold).

share|improve this answer
    
I'd say your $\mathrm{Proj}\, S$ is a candidate algebraic analogue of infinite dimensional projective space, but it's far from obvious (to me) that it "deserves" to be $\mathbb{P}^\infty$. For example, one of the most interesting things about $\mathbb{P}^n$ is that maps to it correspond to a choice of line bundle $L$ and $n+1$ generating global sections of $L$. I'll bet your $\mathrm{Proj}\, S$ does NOT enjoy the analogous property. (Proving interesting things about Proj of a graded ring generally requires assuming the ring is finitely generated over the degree 0 piece.) –  Sam Lichtenstein Mar 30 '10 at 6:54
    
I agree with you $Proj\ S$ is probably not as nice as you might hope it to be; I don't know for sure and I don't know if there's anything better. I have suspicion this is addressed somewhere in EGA, but to be honest when it comes to algebraic geometry, I don't mind restricting to finite dimensional things. –  solbap Mar 30 '10 at 8:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.