Let $X$ be a projective scheme over a ring $R$, and $p : X\to\mathbf{P}^n_R$ a projective embedding.

Does there exist $n$ large enough so that the complement $U\subset \mathbf{P}^n_R$ of $p(X)$ in $\mathbf{P}^n_R$ is quasi-compact?

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Let $X$ be a projective scheme over a ring $R$, and $p : X\to\mathbf{P}^n_R$ a projective embedding.

Does there exist $n$ large enough so that the complement $U\subset \mathbf{P}^n_R$ of $p(X)$ in $\mathbf{P}^n_R$ is quasi-compact?

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Take for $X$ a closed point of $S:=\mathrm{Spec}(R)$. Then $p$ extends to a section $\tilde{p}:S\to \mathbf{P}^n_R$. If $U$ is quasicompact, so is $\tilde{p}^{-1}(U)=S\smallsetminus X$. So there are plenty of counterexamples, e.g. $R=k[t_1,\dots,t_n,\dots]$ and $X=$ the origin.

On the other hand, if $X$ is of finite presentation over $R$, then $p$ is of finite presentation, hence $U$ is (always) quasicompact.