Is every affine scheme the complement of the closed point $x$ of the spectrum of a local ring $A$?

Question

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 $\operatorname{Spec}(A)$?

Answer 1

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.

Answer 2

$\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).