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For what kind of commutative rings $R$, the following property holds?

For any scheme $X$ and any morphism $f:\text{Spec} R \rightarrow X$, there exists a open affine subscheme $U \hookrightarrow X$ such that $f$ factors through $U$.

For example this is true when $R$ is local: choose a $U$ containing the image of the closed point, then $U$ will contain the whole image as it's stable under generalization.

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    A counterexample: Let $R=\mathbb{R}[X,Y,Z]/((X^2+Y^2)Z-1)$. The morphism $\mathrm{Spec} R\to \mathbb{A}_{\mathbb{R}}^2\backslash {0,0}$, $(x,y,z)\mapsto (x,y)$ is surjective, but the latter is not affine. – Phil. Z Nov 18 '17 at 05:00
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    Here's a related question: if one has an $S$-scheme $X$, for what $S$-schemes $P$ is the map $\amalg_i\text{Hom}_S(P,U_i)\rightarrow \text{Hom}_S(P,X)$ surjective where ${U_i}_i$ is an open cover of $X$? If the scheme $P$ has this property then for any morphism $P\rightarrow X$ (as $S$-schemes), and any open cover of $X$ there is a morphism $P\rightarrow U_i\rightarrow X$ that will lift this one. – Eoin Nov 18 '17 at 06:23
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    (cont'd) in this situation, the answer is that this is true if and only if $P$ is local. More generally, one can ask when this is true in the $\tau$-topology and with arbitrary $\tau$-covers. This is also answered for many different topologies in this paper of Gabber and Kelly. – Eoin Nov 18 '17 at 06:25
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    @Eoin Thanks. The condition in the question is a liitle weaker than $(Sch/S, \tau)$-local, so maybe there are counterexamples. –  Nov 18 '17 at 09:53

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I highly doubt that you can extend your example past the case of $R$ local - the presence of even two distinct closed points in $\operatorname{Spec} R$ can cause this property to fail.

Consider the Hironaka example - take an elliptic curve $E$ with a 2-torsion point $t$, then let $P=E\times E/(t) \times E/(t)$ and blow up $(x,x,0)$ and then $(x,0,x)$. Then any 2-point subset consisting of a point in each exceptional divisor is not contained in any affine open. So if I have 2 closed points and I map them in to a set of the form described, your property fails.

KReiser
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