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Let $k$ be any field. Fix an algebraic closure $\overline{k}$ and let $k_s$ denote the separable closure of $k$ within $\overline{k}$. Define an étale $k$-scheme as one with discrete underlying space such that for every $x\in X$, the local ring $\mathcal{O}_{X,x}$ is a field, which is a finite separable extension of $k$.

  1. Is $X$ affine?
  2. $X$ is a disjoint union of spectrums of finite separable extensions of $k$;
  3. $X(k_s) \cong X(\overline{k})$.

I think that 1. follows from 2. for if $X=\sqcup_i Spec(K_i)$ with $K_i$ a finite separable extension of $k$, then $X=Spec (\prod_i K_i)$ (by extending Exercise II.2.19 in Hartshorne to the possibly infinite case) and $X$ is affine. For 3. I only know that $k_s\subseteq \overline{k}$ implies that the canonical map $X(k_s)\to X(\overline{k})$ is injective since $X$ is separated. However I'm having no luck showing the other "inclusion."

Thank you for your help.

SamiNami
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    Doesn't 2. implies 3. as well? As for any separable exxtension $l$ of $k$ one has that a $k$-morphism $l\to \overline{k}$ lands in $k_s$? Also, 1. is false unelss $X$ is also quasi-compact -- the infinite disjoint union of affine schemes is not affine. – Alex Youcis Feb 28 '22 at 15:35
  • @AlexYoucis Thank you for your comment. If I understand correctly, the statement "disjoint union of specrta is the spectrum of the product" does not generalize to infinite products. For 2, is it correct to think of X as always being a disjoint union of the spectra of its local rings? – SamiNami Feb 28 '22 at 16:26
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    Yes, that is correct. – Alex Youcis Feb 28 '22 at 16:35
  • @AlexYoucis Is the statement "X is a disjoint union of the spectra of its local rings" true for any scheme X? I have never worked out such a result. – SamiNami Feb 28 '22 at 20:02
  • No, that is certainly not true in general. It is true for schemes of dimension zero, which $X$ must be here by your hypothesis on the local rings: take an affine open neighborhood of $x\in X$ and apply [04MG](https://stacks.math.columbia.edu/tag/04MG) for instance. – KReiser Mar 01 '22 at 00:58
  • Dear @KReiser, the link you provide shows that X is zero-dimensional (which can already be seen knowing that dim X is the supremum of the dimensions of the local rings), but not how the zero-dimensionality implies the ability of writing X as a disjoint union of the spectra of its local rings. Sorry if I'm missing something obvious here. – SamiNami Mar 01 '22 at 01:45
  • The link shows that a zero-dimension scheme is discrete. This enables you to write it as the disjoint union of the spectra of its local rings. – KReiser Mar 01 '22 at 01:57
  • @KReiser 1. Isn't the spectrum of a countable product of copies of a field an example of a zero-dimensional scheme which is not discrete? 2. If X is discrete, then how is each x in X identified with the spectrum of the local ring of X at x? – SamiNami Mar 01 '22 at 02:16
  • 1. No. Why do you think this? 2. As $X$ is discrete, $\{x\}$ is open, so $(\{x\},\mathcal{O}_X|_{\{x\}})$ is a scheme. But that restriction is just the sheaf taking the value $\mathcal{O}_{X,x}$ on $\{x\}$. Further, every 1-point scheme is affine, so there you go. – KReiser Mar 01 '22 at 02:28
  • @KReiser Thank you for the explanation! For 1. please see point (d) in Georges' comment here https://math.stackexchange.com/questions/474873/why-is-every-noetherian-zero-dimensional-scheme-finite-discrete – SamiNami Mar 01 '22 at 02:33
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    Oops, forgot about that counterexample, thanks for the reminder. A zero-dimensional scheme with essentially any sort of finiteness hypothesis will satisfy that conclusion, though - see [0CKV](https://stacks.math.columbia.edu/tag/0CKV) for instance. And let me just point out that you assumed your underlying space was discrete in the problem statement too :) – KReiser Mar 01 '22 at 02:48

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