Timeline for Rigidity of the category of schemes
Current License: CC BY-SA 3.0
13 events
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| S Mar 10, 2017 at 6:17 | history | suggested | Ali Caglayan | CC BY-SA 3.0 |
changed mathrm to operatorname
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| Mar 10, 2017 at 5:54 | review | Suggested edits | |||
| S Mar 10, 2017 at 6:17 | |||||
| Feb 23, 2012 at 11:01 | comment | added | Martin Brandenburg | Okay. Anyway it is the most profound answer so far. Reduced schemes are categorical and strong specializations are categorical. If we already knew that dimension is categorical, then the same would be true for "irreducible of dimension $\leq d$" for fixed $d$. | |
| Feb 23, 2012 at 10:29 | history | edited | Laurent Moret-Bailly | CC BY-SA 3.0 |
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| Feb 23, 2012 at 10:19 | comment | added | Laurent Moret-Bailly | Martin, about "strong" vs "immediate" specialization I am afraid you are right again! I'm editing my answer. | |
| Feb 20, 2012 at 12:26 | comment | added | Martin Brandenburg | What about the following: [$X$ is irreducible of dimension $\leq d$ iff there is some point $y$ such that every point $x$ can be reached by $\leq d$ many strong specializations, but not by $>d$ many proper strong specializations.] It is easy to prove $\Rightarrow$, but for $\Leftarrow$ it is unclear whether $X$ has finite dimension at all. The basic problem is that it is unclear if we can compose a specialization into finitely many intermediate (and thus strong) specializations. | |
| Feb 20, 2012 at 7:46 | comment | added | Martin Brandenburg | [Lemma: $x$ is an intermediate specialization of $y$ if and only if $x$ is a strong specialization of $y$ and there is no proper finite chain of strong specializations $x,x_1,...,x_n,y$.] But I can only prove this lemma for noetherian schemes. | |
| Feb 20, 2012 at 7:41 | comment | added | Martin Brandenburg | Let $x$ be an intermediate strong specialization of $y$. Choose some affine neighborhood $\mathrm{Spec}(A)$ of $x$, then $y,x$ correspond to prime ideals $\mathfrak{p} \subseteq \mathfrak{q} \subseteq A$. There is no prime ideal $\mathfrak{r}$ such that $\mathfrak{p} \subseteq \mathfrak{r} \subseteq \mathfrak{q}$ is an intermediate chain, because otherwise two two maps $A \to (A/\mathfrak{p})_\mathfrak{q}$, $(A/\mathfrak{r})_\mathfrak{q}$ contradict the asumption that $\mathfrak{p} \subseteq \mathfrak{q}$ is immediate strong. But what about longer chains? Perhaps the followig adjustment works: | |
| Feb 20, 2012 at 7:36 | comment | added | Martin Brandenburg | Thanks for providing the details. I've deleted the irrelevant comments above. a) In the proof, "only one monomorphism" / "infinitely many monomorphisms" are meant to be "up to isomorphism". b) Let's say that $x$ is a strong specialization of $y$ if this is witnessed by a two-point scheme as above. Then it is easy to verify that every strong specialization is a specialization and that every immediate specialization is an immediate strong specialization. Now you claim that also the converse holds, but I have difficulties with the proof: | |
| Feb 19, 2012 at 13:54 | history | edited | Laurent Moret-Bailly | CC BY-SA 3.0 |
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| Feb 19, 2012 at 13:32 | comment | added | Laurent Moret-Bailly | Martin, you are right about specialization, but somethig can be saved. I'll edit my answer. | |
| Feb 18, 2012 at 18:23 | history | edited | Laurent Moret-Bailly | CC BY-SA 3.0 |
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| Feb 18, 2012 at 15:22 | history | answered | Laurent Moret-Bailly | CC BY-SA 3.0 |