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Maps from projective schemes to R #6
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This is false, since for example any infinitesimal disk is a connected projective scheme and admits a non-constant embedding in R. What's true is that such a map cannot assume distinct values. By "compactness", for any map f: X -> R with X projective, the complement of the image of f is open in R, i.e. given by an ideal I in R[X]. If X is non-empty then this ideal is non-zero. Up to not-not we can find a deduplicated set of roots of some non-zero element p of I. Say p has distinct roots a1 ... ak. Now f(x) is not in the open determined by I, so p(f(x)) is not not zero, which means there is a unique i such that f(x) is not not ai. If X is connected then this i is the same for all x : X, which proves the claim. |
I'm confused: Shouldn't X non-empty rather imply something like But other than that, I can follow. So in general one could say: Whenever we have a connected, pointed |
Yes, of course. I think the reason |
Ah. I think the same trick was used for the theorem I didn't remember ;-) |
(Or at least some non-empty open set with empty intersection with the image of f, which is enough) |
So for a projective affine connected pointed type |
I cannot reproduce my last statement, but what we have is that all maps |
Another approach is to look at an embedding |
It seems one can use the above idea to show that We argue as follows. The open |
Very interesting discussion!
Regarding David's hope about finite presentation: This is indeed correct, a reference is Tag 0564 in the Stacks Project; at some point I convinced myself that the proof offered there is constructive (or easily constructivized).
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Nice! I guess we could also phrase it like this: Since only compactness was used above. |
The answer is: yes, those affine schemes are even projective
I remember that @fabianmasato and @mnieper discussed a proof with me, that maps from a connected projective scheme, i.e. a connected closed subset of some$\mathbb{P}^n$ , to the base ring $R$ should be constant. Does anyone remember the details?
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