Timeline for When spreading out a scheme, does the choice of max ideal matter?
Current License: CC BY-SA 4.0
6 events
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| May 24, 2018 at 2:51 | history | edited | Mike Pierce | CC BY-SA 4.0 |
Added a step that I didn't notice before
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| May 24, 2018 at 0:44 | comment | added | Mike Pierce | @R.vanDobbendeBruyn Thanks! Yeah, I really like the model-theoretic perspective on this proof technique: it hides most of the heavy lifting in the compactness theorem and proving that the theory of algebraically closed fields of a fixed characteristic is complete. I think I'm being a tad paranoid about the details of unpacking those ideas into purely a algebraic-geometry and commutative-algebra language, hence this question :) | |
| May 24, 2018 at 0:35 | history | edited | Mike Pierce | CC BY-SA 4.0 |
k should be algebraically closed. Don't wanna get too crazy
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| May 23, 2018 at 21:00 | comment | added | R. van Dobben de Bruyn | In a more model-theoretic language, the idea is indeed that if there exists a counterexample in characteristic $0$, then there also exists one in large positive characteristic. This is an example of a compactness argument in model theory. | |
| May 23, 2018 at 20:57 | comment | added | R. van Dobben de Bruyn | The scheme-theoretic argument is that if $\mathcal X \to \operatorname{Spec} R$ is a nice family (e.g. flat with geometrically integral fibres, although it might work with much weaker hypotheses) where $R$ is a domain of finite type over $\mathbb Z$, then the locus in $\operatorname{Spec} R$ where a map $f \colon \mathcal X \to \mathcal X$ of $R$-schemes is not injective (resp. not surjective) is constructible. Thus, if it contains all closed points, it must be all of $\operatorname{Spec} R$. | |
| May 23, 2018 at 16:46 | history | asked | Mike Pierce | CC BY-SA 4.0 |