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Dec 31, 2014 at 1:32 history edited Question Mark
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Dec 8, 2014 at 18:15 comment added Question Mark Thanks, that resolves the problem. This is probably an overkill, but to finish in the quasi-projective case one may use Thm. 5.1 in Gabber-Liu-Lorenzini "... moving lemma" to reduce to the quasi-affine case. In the quasi-affine case, one may assume that $X$ is the complement of some $V(I)$ in some $\mathrm{Spec}(A)$, choose $f_i \in A$ with $\mathrm{Spec}(A_{f_i}) \subset \mathrm{Spec}(A) \setminus ( \bigcup_{j \neq i} \overline{\{ x_j \}} \cup V(I))$ and then note that $\mathrm{Spec}(A_f)$ with $f = f_1 + \dotsb + f_n$ is a sought affine.
Dec 8, 2014 at 15:00 comment added Jason Starr @QuestionMark: I agree with Moret-Bailly's point. Instead of using Nagata compactification, etc., as I was suggesting, you can just apply Chow's lemma.
Dec 8, 2014 at 8:28 comment added Laurent Moret-Bailly Chow's lemma works directly. There is a proper surjective birational $p:X′\to X$ with $X′$ quasiprojective over $R$. Since $X$ is normal, $p$ is an isomorphism over an open subscheme $U\subset X$ containing each $x_i$. Clearly $U$ is quasiprojective, as a subscheme of $X'$.
Dec 8, 2014 at 5:55 comment added Question Mark Thanks! Could you give more detail? That would be very helpful, since I don't have a lot of experience with the constructions you are using. For instance, doesn't your first sentence imply that $X$ is $R$-proper (how to arrange the map to $Y$ be proper?); to justify the first sentence, should I be using Chow's lemma stacks.math.columbia.edu/tag/088U together with stacks.math.columbia.edu/tag/080A and stacks.math.columbia.edu/tag/0807 ? Doesn't the latter imply that the blowing up in your second sentence is the identity, since the $x_i$ are still of height $1$?
Dec 8, 2014 at 2:00 comment added Jason Starr Okay, then you can replace $X$ by a projective modification that is an isomorphism over each of the $x_i$, and which itself admits a proper birational $R$-morphism to an $R$-projective scheme $Y$. Now replace $Y$ by its blowing up along the closures of the images of the points $x_i$. After a further projective modification of $X$, there is now a proper, birational morphism to $Y$ that is quasi-finite at every point $x_i$. Now replace $X$ by the Stein factorization of this morphism. This is projective. The projective case is straightforward.
Dec 8, 2014 at 1:25 comment added Question Mark I'd be very happy to see any argument that works. For one thing, both Raynaud-Gruson and Nagata compactification have already been used earlier in the book (perhaps nonessentially for the eventual goal, but I didn't care to keep track), so I see no sin in using them again.
Dec 8, 2014 at 1:15 comment added Jason Starr Are we allowed to use Raynaud-Gruson, Nagata compactification, etc., or is the concern that those theorems depend on this result? Probably there is an argument that uses none of those, but my first instinct is to reduce to the case that $X$ is a quasi-projective $R$-scheme.
Dec 8, 2014 at 0:50 history asked Question Mark CC BY-SA 3.0