Let $k$ be a field. Are there any useful necessary and sufficient conditions on $k$-algebras $A$ such that $\mathrm{Spec}(A)$ is a projective scheme over $k$? I know that there are very few of these, but I don't think that $k$ or $k\times k$ are the only ones.
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- $\begingroup$ What's the role of $L$ here? $\endgroup$Hoot– Hoot2016-07-11 14:23:11 +00:00Commented Jul 11, 2016 at 14:23
- $\begingroup$ I am not sure about the category of schemes but for the category of varieties the answer is the only affine variety over $k=\bar k$ that is projective is just the one point variety. Projective varieties have only constant regular functions. Affine varieties are defined by their coordinate ring, and the only such coordinate rings are $k[x_1, \dots, x_n]/\mathfrak m$ where $\mathfrak m$ is a maximal ideal corresponding to be some point. $\endgroup$hwong557– hwong5572016-07-11 14:29:14 +00:00Commented Jul 11, 2016 at 14:29
- 5$\begingroup$ Those are just the finite schemes over k. Indeed if Spec(A) is proper over Spec(k), then A is finite over k by Grothendieck's coherence theorem (push forward of coherent under proper morphism is coherent). $\endgroup$Daniele A– Daniele A2016-07-11 14:45:34 +00:00Commented Jul 11, 2016 at 14:45
- $\begingroup$ @Hoot initially I was interested in field extensions, but then I noticed that it is more reasonable to generalize. $\endgroup$Juan Fran– Juan Fran2016-07-12 12:37:53 +00:00Commented Jul 12, 2016 at 12:37
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1 Answer
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2 If $X=\textrm{Spec }A$ is a projective $k$-variety (not necessarily connected), then $A=\Gamma(X,\mathscr O_X)$ is a finite direct sum of copies of $k$, as it occurs for every proper $k$-scheme. This means that the $k$-algebra $A$ is a finite dimensional vector space, i.e. $A$ is an Artin $k$-algebra, hence $$\dim X=\dim_{\textrm{Krull}}A=0.$$
So, as already mentioned in the comments, only finite $k$-schemes can be both affine and projective.
If $A$ is also local and has residue field $k$, then $\textrm{Spec }A$ is called a fat point over $k$.
- $\begingroup$ Thanks! so as I thought not all "points" (e.g. specs of field extensions) are proper, unlike what the "classical" viewpoint might suggest (as in the comment by hwong). It seems like this result isn't quite obvious, is it? I mean you are using a Theorem of Serre on the finiteness of cohomology for proper schemes (or more generally the Grothendieck coherence theorem mentioned by Daniele). $\endgroup$Juan Fran– Juan Fran2016-07-12 12:37:13 +00:00Commented Jul 12, 2016 at 12:37
- $\begingroup$ Yes, it's not trivial but it also does not require the full power of Serre's theorem: you can prove that $\Gamma(X,\mathscr O_X)=k^r$, where $r$ is the number of connected components, without invoking Serre's theorem. $\endgroup$Brenin– Brenin2016-07-13 07:58:17 +00:00Commented Jul 13, 2016 at 7:58