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.
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.
$\mathrm{Spec}\,k[x_1,x_2,\ldots]$ 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...
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Commented
Mar 30, 2010 at 6:31
$k[x_1,x_2,\ldots]$ 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 $R\mapsto R^{\oplus \mathbb{N}}$?
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Commented
Mar 30, 2010 at 6:37
Since rings can have infinite Krull dimension, affine schemes can have infinite dimension.
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).
