Lemma 31.4.13. In Situation 31.4.5 if $S$ is affine, then for some $i_0 \in I$ the schemes $S_ i$ for $i \geq i_0$ are affine.

**Proof.**
By Lemma 31.4.12 we may assume that $S_0$ is quasi-affine for some $0 \in I$. Set $R_0 = \Gamma (S_0, \mathcal{O}_{S_0})$. Then $S_0$ is a quasi-compact open of $T_0 = \mathop{\mathrm{Spec}}(R_0)$. Denote $j_0 : S_0 \to T_0$ the corresponding quasi-compact open immersion. For $i \geq 0$ set $\mathcal{A}_ i = f_{i0, *}\mathcal{O}_{S_ i}$. Since $f_{i0}$ is affine we see that $S_ i = \underline{\mathop{\mathrm{Spec}}}_{S_0}(\mathcal{A}_ i)$. Set $T_ i = \underline{\mathop{\mathrm{Spec}}}_{T_0}(j_{0, *}\mathcal{A}_ i)$. Then $T_ i \to T_0$ is affine, hence $T_ i$ is affine. Thus $T_ i$ is the spectrum of

Write $S = \mathop{\mathrm{Spec}}(R)$. We have $R = \mathop{\mathrm{colim}}\nolimits _ i R_ i$ by Lemma 31.4.7. Hence also $S = \mathop{\mathrm{lim}}\nolimits _ i T_ i$. As formation of the relative spectrum commutes with base change, the inverse image of the open $S_0 \subset T_0$ in $T_ i$ is $S_ i$. Let $Z_0 = T_0 \setminus S_0$ and let $Z_ i \subset T_ i$ be the inverse image of $Z_0$. As $S_ i = T_ i \setminus Z_ i$, it suffices to show that $Z_ i$ is empty for some $i$. Assume $Z_ i$ is nonempty for all $i$ to get a contradiction. By Lemma 31.4.8 there exists a point $s$ of $S = \mathop{\mathrm{lim}}\nolimits T_ i$ which maps to a point of $Z_ i$ for every $i$. But $S = \mathop{\mathrm{lim}}\nolimits _ i S_ i$, and hence we arrive at a contradiction by Lemma 31.4.6. $\square$

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