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



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> 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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Projective embeddings and quasi-compactness
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.


