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Let $R$ be a commutative ring with identity element and let $\operatorname{Spec}(R)$ be the associated affine scheme.

Does for each affine scheme $\operatorname{Spec}(R)$ exist a local ring $A$ with maximal ideal $m$ such that $\operatorname{Spec}(R)\cong \operatorname{Spec}(A)\setminus\{m\}$ as schemes where $\operatorname{Spec}(A)\setminus\{m\}$ carries the unique scheme structure of an open subscheme of $\operatorname{Spec}(A)$?

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    $\begingroup$ What happens when $R$ is a field? when $R$ is a PID? $\endgroup$ – Martin Brandenburg Jul 22 '14 at 7:55
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    $\begingroup$ Here is a try: When $A$ is a DVR, then $\mathrm{Spec}(A) \setminus \{\mathfrak{m}\} = \mathrm{Spec}(Q(A))$. Is every field the ring of fractions of a DVR? Probably not. $\endgroup$ – Martin Brandenburg Jul 22 '14 at 10:48
  • $\begingroup$ Dear @MartinBrandenburg, thank you for the comment. Do you know if every field is the ring of fractions of a DVR? Does somebody know an answer to the question for a homeomorphism of just the topological spaces $\operatorname{Spec}(R)\cong \operatorname{Spec}(A)\setminus\{m\}$ instead of a scheme isomorphism? $\endgroup$ – user8463524 Jul 22 '14 at 11:40
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    $\begingroup$ (1) A finite field cannot be the ring of fractions of a DVR. (2) There's a classification of those topological spaces homeomorphic to the spectrum of a ring, and I think it implies there always exists local $A$ with Spec$(A)\backslash\{m\}$ homeomorphic to Spec$(R)$ (en.wikipedia.org/wiki/Spectral_space) $\endgroup$ – Julian Rosen Jul 22 '14 at 15:45
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    $\begingroup$ An algebraically closed field cannot be the field of fractions of a noetherian ring, because by a theorem of Nagata it would then be the field of fractions of a discrete valuation ring. The latter is impossible, because the value group of a valuation of an algebraically closed field is divisible. $\endgroup$ – Hagen Knaf Jul 23 '14 at 7:37
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No.

Assume that $R$ is a field and $(A,\mathfrak{m})$ is a local ring such that $\mathrm{Spec}(R) \cong \mathrm{Spec}(A) \setminus \{\mathfrak{m}\}$.

Then $A$ has exactly two prime ideals $\mathfrak{p} \subset \mathfrak{m}$ and $R \cong A_{\mathfrak{p}}$. The general fact $Q(A/\mathfrak{p}) \cong A_{\mathfrak{p}} / \mathfrak{p} A_\mathfrak{p}$ implies here that $A/\mathfrak{p}$ embeds into $R$.

Now assume that $R$ is a finite field. Then every subring of $R$ is already a field (since it is integral over the prime field). Hence, $A/\mathfrak{p}$ is a field, i.e. $\mathfrak{p}$ is a maximal ideal - contradiction.

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  • $\begingroup$ PS: It would interesting what happens for $R=\mathbb{Z}$. $\endgroup$ – Martin Brandenburg Jul 22 '14 at 21:34
  • $\begingroup$ For $R$ algebraic over its prime field, it's not true in general that every subring of $R$ is a field (e.g. $R=\mathbb{Q}$, $A=\mathbb{Z}_{(p)}$). The argument still works for $R=\mathbb{F}_p$, though. $\endgroup$ – Julian Rosen Jul 22 '14 at 22:14
  • $\begingroup$ Thank you both. I've fixed the proof once again. (Hopefully now it's correct!) $\endgroup$ – Martin Brandenburg Jul 23 '14 at 8:37
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$\DeclareMathOperator{\Spec}{\operatorname{Spec}}$$\DeclareMathOperator{\codim}{codim}$$\Spec(\mathbb{Z})$ cannot be realized as $\Spec(A) \setminus \{m\}$ for a local ring $(A,m)$ - the argument given here can essentially be adapted verbatim. Set $X := \Spec(A)$, $Y := \{m\}$, $U := X \setminus Y \ne \emptyset$. Notice that $\dim U < \infty \iff \dim X < \infty$ - assume this is so. For any $d > 0$, there is an exact sequence (cf. Hartshorne Ex. III.2.3) $\DeclareMathOperator{\O}{\mathcal{O}}$ $$H^{d-1}(X, \O_X) \to H^{d-1}(U, \O_X \big|_U) \to H^d_Y(Y, \O_X) \to H^d(X, \O_X)$$

Then for $d = \codim Y$, $H^d_Y(Y, \O_X) \ne 0$ by Grothendieck non-vanishing. If $d > 1$, then $H^{d-1}(X, \O_X) = H^d(X, \O_X) = 0$, so $H^{d-1}(U, \O_X \big|_U) \ne 0$, which implies $U$ is not affine (*). This shows that $U$ can be affine only if $d = 1$, in which case $\dim A = 1 \implies \dim U = 0$. Thus the only finite-dimensional rings which can be punctured spectra of local rings are $0$-dimensional.

(*): This is Serre's criterion for affineness: a scheme $X$ is affine iff it is separated, quasi-compact, and $H^i(X, \mathcal{F}) = 0$ for all quasicoherent $\mathcal{F}$, $i > 0$ (see e.g. here).

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    $\begingroup$ 1) In your proof, don't you need some assumptions such as noetherian? Of even of finite type? Otherwise dimension and cohomology don't play well. 2) "a scheme X is affine iff it is separated, quasi-compact, and H^i(X,F)=0". We don't need separated. 3) "Note that any open subset of an affine scheme is separated and quasi-compact. " Not necessarily quasi-compact. In your proof, $\endgroup$ – Martin Brandenburg Jul 23 '14 at 8:28
  • $\begingroup$ @MartinBrandenburg: 1) Finite type is certainly too strong of a hypothesis. I'm not sure if Grothendieck non-vanishing holds without Noetherian hypotheses though - a counterexample would be welcome. 2) An affine scheme is separated, so in any case the statement is correct. 3) I have edited accordingly $\endgroup$ – zcn Jul 23 '14 at 8:49
  • $\begingroup$ Nice answer: +1 $\endgroup$ – Georges Elencwajg Jul 23 '14 at 8:49
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    $\begingroup$ I mean, why should the codimension be finite at all? $\endgroup$ – Martin Brandenburg Jul 23 '14 at 8:53
  • $\begingroup$ @MartinBrandenburg: Do you mean in this particular case (if $U \cong \operatorname{Spec}(\mathbb{Z})$) or in general? $\endgroup$ – zcn Jul 23 '14 at 8:57

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