MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question
Someone care to explain the close vote? This seems like a perfectly reasonable question to me. – Qiaochu Yuan Mar 30 '10 at 7:54
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
up vote 8 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|cite|improve this answer
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
@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|cite|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|cite|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


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