Email by Ofer Gabber of September 11, 2014.

**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 $J \supset J_ n$.

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 $J_ n \subset J$.

To see that (2) holds it now suffices to show that $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/J_ n$ 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 $\mathop{\mathrm{lim}}\nolimits A/J_ n$ is in $J_1^\wedge = \mathop{\mathrm{lim}}\nolimits J_1/J_ n$. This means $f$ is an infinite series involving only a countable number of generators. Hence $f$ dies in $A_\gamma ^\wedge = \mathop{\mathrm{lim}}\nolimits A_\gamma /J_ nA_\gamma $ for some $\gamma $. Note that $A^\wedge \to A_\gamma ^\wedge $ is continuous and open by Lemma 84.4.5. 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 and open (by Lemma 84.4.5) 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 $.

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 = \mathop{\mathrm{lim}}\nolimits _{k \geq n} J_ n/J_ k \subset A^\wedge $. Since the closure $\bar I = \bigcap (I + J_ n^\wedge )$ (Lemma 84.4.2) 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

\[ J_{n - 1}^\wedge I + J_{n + 1}^\wedge \supset J_{n - 1}^\wedge (I + J_{n + 1}^\wedge ) \supset J_{n - 1}^\wedge J_ m^\wedge \]

for all $n \geq m + 1$. Namely, the first inclusion is trivial and the second was shown above. Because $J_{n - 1}J_ m \supset J_{n - 1}^2 \supset J_ n$ these inclusions show that the image of $J_ n$ in $A^\wedge $ is contained in the ideal $J_{n - 1}^\wedge I + J_{n + 1}^\wedge $. Because this ideal is open we conclude that

\[ J_{n - 1}^\wedge I + J_{n + 1}^\wedge \supset J_ n^\wedge . \]

Say $I = (g_1, \ldots , g_ t)$. Pick $f \in J_{m + 1}^\wedge $. Using the last displayed inclusion, 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.

Proof of (1). Assume $T = V(f_1, f_2, f_3, \ldots )$. Let $I_ m \subset A^\wedge $ be the ideal generated by $f_1, \ldots , f_ m$. We distinguish two cases.

Case I: For some $m$ the closure of $I_ m$ is open. Then $I_ m$ is open by the result of the previous paragraph. For any $n$ we have $(J_ n)^2 \supset J_{n+1}$ by design, so the closure of $(J_ n^\wedge )^2$ contains $J_{n+1}^\wedge $ and thus is open. Taking $n$ large, it follows that the closure of the product of any two open ideals in $A^\wedge $ is open. Let us prove $I_ m^ k$ is open for $k \ge 1$ by induction on $k$. The case $k = 1$ is our hypothesis on $m$ in Case I. For $k > 1$, suppose $I_ m^{k - 1}$ is open. Then $I_ m^ k = I_ m^{k - 1} \cdot I_ m$ is the product of two open ideals and hence has open closure. But then since $I_ m^ k$ is finitely generated it follows that $I_ m^ k$ is open by the previous paragraph (applied to $I = I_ m^ k$), so we can continue the induction on $k$. 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., $X_{/T}$ is adic*.

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

\[ \mathop{\mathrm{Im}}(J_{\phi (m)} \to A/(f_1, \ldots , f_ m)) \not= \mathop{\mathrm{Im}}(J_{\phi (m) + 1} \to A/(f_1, \ldots , f_ m)) \]

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 $J_ m$.
$\square$

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