Let $X$ be the underlying space of a scheme.
- If $X$ is irreducible of finite Krull dimension, is it necessarily quasi-compact?
- Is it necessarily Noetherian?
- What if we assume not only that Krull dimension is finite but also that it is 1?
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1$\begingroup$ 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 $\mathrm{Spec}\mathbb Z$! $\endgroup$– Ben WielandCommented Apr 18, 2019 at 0:11
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1$\begingroup$ 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. $\endgroup$– Ben WielandCommented Apr 18, 2019 at 0:16
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$\begingroup$ @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. $\endgroup$– Denis NardinCommented Apr 18, 2019 at 8:59
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The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $\mathbb{A}^1$ along the open subsets $\mathbb{G}_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.
answered Apr 17, 2019 at 17:19
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1$\begingroup$ Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)? $\endgroup$– user137767Commented Apr 17, 2019 at 17:44
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1$\begingroup$ @StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it. $\endgroup$ Commented Apr 17, 2019 at 18:13