Proof.
Part (1) is Lemma 87.37.1. If $X$ is McQuillan, then $X = \text{Spf}(A)$ for some weakly admissible topological ring $A$. Then $X_{/T} \to X \to \mathop{\mathrm{Spec}}(A)$ satisfies property (2) of Lemma 87.9.6 and hence $X_{/T}$ is McQuillan, see Definition 87.9.7.
Assume $X$ and $T$ are as in (3). Then $X = \text{Spf}(A)$ where $A$ has a fundamental system $A \supset I_1 \supset I_2 \supset I_3 \supset \ldots $ of weak ideals of definition, see Lemma 87.10.4. By Algebra, Lemma 10.29.1 we can find a finitely generated ideal $\overline{J} = (\overline{f}_1, \ldots , \overline{f}_ r) \subset A/I_1$ such that $T$ is cut out by $\overline{J}$ inside $\mathop{\mathrm{Spec}}(A/I_1) = |X_{red}|$. Choose $f_ i \in A$ lifting $\overline{f}_ i$. If $Z = \mathop{\mathrm{Spec}}(B)$ is an affine scheme and $g : Z \to X$ is a morphism with $g(Z) \subset T$ (set theoretically), then $g^\sharp : A \to B$ factors through $A/I_ n$ for some $n$ and $g^\sharp (f_ i)$ is nilpotent in $B$ for each $i$. Thus $J_{m, n} = (f_1, \ldots , f_ r)^ m + I_ n$ maps to zero in $B$ for some $n, m \geq 1$. It follows that $X_{/T}$ is the formal spectrum of $\mathop{\mathrm{lim}}\nolimits _{n, m} A/J_{m, n}$ and hence countably indexed. This proves (3).
Proof of (4). Here the argument is the same as in (3). However, here we may choose $I_ n = I^ n$ for some finitely generated ideal $I \subset A$. Then it is clear that $X_{/T}$ is the formal spectrum of $\mathop{\mathrm{lim}}\nolimits A/J^ n$ where $J = (f_1, \ldots , f_ r) + I$. Some details omitted.
Proof of (5). In this case $X_{red}$ is the spectrum of a Noetherian ring and hence the assumption that $|X_{red}| \setminus T$ is quasi-compact is satisfied. Thus as in the proof of (4) we see that $X_{/T}$ is the spectrum of $\mathop{\mathrm{lim}}\nolimits A/J^ n$ which is a Noetherian adic topological ring, see Algebra, Lemma 10.97.6.
$\square$
Comments (0)