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Feb 25, 2015 at 19:59 vote accept Will Chen
Feb 24, 2015 at 9:07 answer added Laurent Moret-Bailly timeline score: 7
Feb 24, 2015 at 8:01 comment added Laurent Moret-Bailly More trivially, you can take for $S$ a disjoint union of points $s_n=\mathrm{Spec}(k_n)$ ($n\in\mathbb{N}$ and $k_n$ a field) and $X=\coprod_n X_n$ where $X_n$ is smooth projective of genus $n$ over $k_n$.
Feb 23, 2015 at 22:53 comment added Jason Starr "... what's your counterexample?" Let $A$ be $\mathbb{Q}[x_1,x_2,x_3,\dots]$, let $\mathfrak \subset A$ be $\langle x_1,x_3,\dots \rangle$. Let $S$ be $\text{Spec}(A)\setminus \{ \mathfrak{m} \}$. This is the union of the countably many open subsets $U_n = D(x_1)\cup \dots \cup D(x_n)$. Over each open subset $U_n$, $X\times_S U_n$ will be the blowing up of $\mathbb{P}^1\times U_n$ along a particular closed subscheme. For $U_2$, blowup $s(U_2) = \{[1,0]\}\times U_2$ over $Z(x_1)$. Then, over $U_3$, blowup further the strict transform of $s(U_3)$ over $Z(x_1,x_2)$, etc.
Feb 23, 2015 at 22:29 comment added Will Chen @JasonStarr I wasn't, what's your counterexample?
Feb 23, 2015 at 22:15 comment added Jason Starr Are you assuming that $S$ is quasi-compact? I think you can make a counterexample if $S$ is not quasi-compact.
Feb 23, 2015 at 22:01 history asked Will Chen CC BY-SA 3.0