Lemma by Ofer Gabber

Thanks to Ofer Gabber
Answers a question posed here
http://math.columbia.edu/~dejong/wordpress/?p=3900

formal-spaces.tex

@@ -2094,6 +2094,127 @@ \section{Completion along a closed subset}
 $$
 between the completions.
 
+\medskip\noindent
+The following lemma is due to Ofer Gabber.
+
+\begin{lemma}
+\label{lemma-completion-countably-indexed}
+\begin{reference}
+Email by Ofer Gabber of September 11, 2014.
+\end{reference}
+Let $S$ be a scheme. Let $X = \Spec(A)$ be an affine scheme over $S$.
+Let $T \subset X$ be a closed subscheme.
+\begin{enumerate}
+\item If the formal completion $X_{/T}$ is countably indexed
+and there exist countably many $f_1, f_2, f_3, \ldots \in A$ such that
+$T = V(f_1, f_2, f_3, \ldots)$, then $X_{/T}$ is adic*.
+\item The conclusion of (1) is wrong if we omit the assumption that
+$T$ can be cut out by countably many functions in $X$.
+\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+The assumption that $X_{/T}$ is countably indexed means that there exists a
+sequence of ideals
+$$
+A \supset J_1 \supset J_2 \supset J_3 \supset \ldots
+$$
+with $V(J_n) = T$ such that every ideal $J \subset A$ with $V(J) = T$
+there exists an $n$ such that $J \supset J_n$.
+
+\medskip\noindent
+To construct an example for (2) let $\omega_1$ be the first uncountable
+ordinal. Let $k$ be a field and let
+$A$ be the $k$-algebra generated by $x_\alpha$, $\alpha \in \omega_1$
+and $y_{\alpha \beta}$ with $\alpha \in \beta \in \omega_1$
+subject to the relations $x_\alpha = y_{\alpha \beta} x_\beta$.
+Let $T = V(x_\alpha)$. Let $J_n = (x_\alpha^n)$.
+If $J \subset A$ is an ideal such that
+$V(J) = T$, then $x_\alpha^{n_\alpha} \in J$ for some $n_\alpha \geq 1$.
+One of the sets $\{\alpha \mid n_\alpha = n\}$ must be unbounded in
+$\omega_1$. Then the relations imply that $J_n \subset J$.
+
+\medskip\noindent
+To see that (2) holds it now suffices to show that $A^\wedge = \lim A/J_n$
+is not a ring complete with respect to a finitely generated ideal.
+For $\gamma \in \omega_1$ let $A_\gamma$ be the quotient of $A$
+by the ideal generated by $x_\alpha$, $\alpha \in \gamma$ and
+$y_{\alpha \beta}$, $\alpha \in \gamma$. As $A/J_1$ is reduced,
+every topologically nilpotent element $f$ of $\lim A/J_n$ is in
+$J_1^\wedge = \lim J_1/J_n$. This means $f$ is an infinite series
+involving only a countable number of generators. Hence $f$ dies in
+$A_\gamma^\wedge = \lim A_\gamma/J_nA_\gamma$ for some $\gamma$.
+If the topology on $A^\wedge$ was $I$-adic for some finitely generated ideal
+$I \subset A^\wedge$, then $I$ would go to zero in some
+$A_\gamma^\wedge$. This would mean that $A_\gamma^\wedge$ is discrete,
+which is not the case as there is a surjective continuous map
+$A_\gamma^\wedge \to k[[t]]$ given by
+$x_\alpha \mapsto t$, $y_{\alpha \beta} \mapsto 1$ for
+$\gamma = \alpha$ or $\gamma \in \alpha$.
+
+\medskip\noindent
+Before we prove (1) we first prove the following: If $I \subset A^\wedge$ is
+a finitely generated ideal whose closure $\bar I$ is open, then $I = \bar I$.
+Since $V(J_n^2) = T$ there exists an $m$ such that $J_n^2 \supset J_m$.
+Thus, we may assume that $J_n^2 \supset J_{n + 1}$ for all $n$ by passing
+to a subsequence. Set $J_n^\wedge = \lim_{k \geq n} J_n/J_k \subset A^\wedge$.
+Since the closure $\bar I = \bigcap (I + J_n^\wedge)$
+(Lemma \ref{lemma-closed}) is open we see that there exists an $m$ such that
+$I + J_n^\wedge \supset J_m^\wedge$ for all $n \geq m$. Fix such an $m$.
+We have
+$$
+J_{n - 1}^\wedge I + J_{n + 1}^\wedge \supset
+J_{n - 1}^\wedge (I + J_{n + 1}^\wedge) \supset
+J_{n - 1}^\wedge J_m^\wedge \supset
+J_n^\wedge
+$$
+for all $n \geq m + 1$. Namely, the first inclusion
+is trivial. The second was shown above. The third as
+$J_{n - 1}J_m \supset J_{n - 1}^2 \supset J_n$,
+hence $J_{n - 1}^\wedge J_m^\wedge \supset J_n^\wedge$.
+Say $I = (g_1, \ldots, g_t)$. Pick $f \in J_{m + 1}^\wedge$.
+Using the displayed inclusions above, valid for all $n \geq m + 1$,
+we can write by induction on $c \geq 0$
+$$
+f = \sum f_{i, c} g_i \mod J_{m + 1+ c}^\wedge
+$$
+with $f_{i, c} \in J_m^\wedge$ and
+$f_{i, c} \equiv f_{i, c - 1} \bmod J_{m + c}^\wedge$.
+It follows that $IJ_m^\wedge \supset J_{m + 1}^\wedge$.
+Combined with $I + J_{m + 1}^\wedge \supset J_m^\wedge$
+we conclude that $I$ is open.
+
+\medskip\noindent
+Proof of (1). Assume $T = V(f_1, f_2, f_3, \ldots)$.
+Let $I_m \subset A^\wedge$ be the ideal generated by $f_1, \ldots, f_m$.
+Case I: For some $m$ the closure of $I_m$ is open. Then $I_m$ is open by
+the result of the previous paragraph. Since in $A^\wedge$ the
+product of open ideals is open, we see that $I_m^k$ is open for all $k$.
+As each element of $I_m$ is topologically nilpotent, we conclude
+that $I_m$ is an ideal of definition which proves that $A^\wedge$
+is adic with a finitely generated ideal of definition, i.e.,
+$X$ is adic*.
+
+\medskip\noindent
+Case II. For all $m$ the closure $\bar I_m$ of $I_m$ is not open.
+Then the topology on $A^\wedge/\bar I_m$ is not discrete. This means
+we can pick $\phi(m) \geq m$ such that
+$$
+\Im(J_{\phi(m)} \to A/(f_1, \ldots, f_m)) \not =
+\Im(J_{\phi(m) + 1} \to A/(f_1, \ldots, f_m))
+$$
+To see this we have used that
+$A^\wedge/(\bar I_m + J_n^\wedge) = A/((f_1, \ldots, f_m) + J_n)$.
+Choose exponents $e_i > 0$ such that $f_i^{e_i} \in J_{\phi(m) + 1}$
+for $0 < m < i$. Let $J = (f_1^{e_1}, f_2^{e_2}, f_3^{e_3}, \ldots)$.
+Then $V(J) = T$. We claim that $J \not \supset J_n$ for all $n$
+which is a contradiction proving Case II does not occur.
+Namely, the image of $J$ in $A/(f_1, \ldots, f_m)$ is contained
+in the image of $J_{\phi(m) + 1}$ which is properly contained in the
+image of $J_m$.
+\end{proof}
+
+
 
 
 \section{Fibre products}