Timeline for Irreducible of finite Krull dimension implies quasi-compact?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 18, 2019 at 8:59 | comment | added | Denis Nardin | @BenWieland I think that should be an answer, it's a better example than mine and it's a pity to leave it in the comments. | |
| Apr 18, 2019 at 0:16 | comment | added | Ben Wieland | A 2d example closer to practical geometry: Let $X_0=\mathbb P^2$ and $p_0$ some point on it. Let $X_{n+1}$ be the blow-up of $X_n$ at $p_n$ and let $p_{n+1}\in D\subset X_{n+1}$ be some point of the exceptional divisor. Let $U_n=X_n-\{p_n\}$. Each is open in the next. So $U=\bigcup U_n$ is a noncompact scheme... The inverse limit of the $X_n$ is probably a compact non-noetherian scheme with open set $U$... The sequence of $p_n$ specifies a 2d valuation ring, necessarily non-noetherian. It has compact spectrum, but the complement of the unique closed point is related to $U$, maybe even 1d. | |
| Apr 18, 2019 at 0:11 | comment | added | Ben Wieland | a separated 1d example: Enumerate the primes: $p_1=2,p_2,...$. Let $R_n$ be the ring of rational numbers whose denominator is not divisible by the first $n$ primes $p_1,...,p_n$. This has spectrum with 1 generic point and $n$ closed points with local rings $\mathbb Z_{(p_i)}$. Each ring is a localization of the next: $R_n=R_{n+1}[p_{n+1}^{-1}]$, so its spectrum is open in the next. Then $X=\bigcup_n \mathrm{Spec}R_n$ is a noncompact scheme with global sections $\mathbb Z$. Clearly not $\mathrm{Spec}\mathbb Z$! | |
| Apr 17, 2019 at 18:31 | history | became hot network question | |||
| Apr 17, 2019 at 17:19 | answer | added | Denis Nardin | timeline score: 10 | |
| Apr 17, 2019 at 17:13 | history | edited | András Bátkai | CC BY-SA 4.0 |
added 23 characters in body; edited tags
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| Apr 17, 2019 at 16:50 | review | First posts | |||
| Apr 17, 2019 at 17:13 | |||||
| Apr 17, 2019 at 16:45 | history | asked | schematic_ftm | CC BY-SA 4.0 |