The Stacks project

Email by Ofer Gabber of September 11, 2014.

Lemma 87.14.7. Let $S$ be a scheme. Let $X = \mathop{\mathrm{Spec}}(A)$ be an affine scheme over $S$. Let $T \subset X$ be a closed subscheme.

  1. 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*.

  2. The conclusion of (1) is wrong if we omit the assumption that $T$ can be cut out by countably many functions in $X$.

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 87.4.5. 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 87.4.5) 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 87.4.2) 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 $J_ m$. $\square$

Comments (0)

There are also:

  • 4 comment(s) on Section 87.14: Completion along a closed subset

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0APW. Beware of the difference between the letter 'O' and the digit '0'.