The Stacks project

108.70 Affine formal algebraic spaces

Let $K$ be a field and let $(V_ i)_{i \in I}$ be a directed inverse system of nonzero vector spaces over $K$ with surjective transition maps and with $\mathop{\mathrm{lim}}\nolimits V_ i = 0$, see Section 108.3. Let $R_ i = K \oplus V_ i$ as $K$-algebra where $V_ i$ is an ideal of square zero. Then $R_ i$ is an inverse system of $K$-algebras with surjective transition maps with nilpotent kernels and with $\mathop{\mathrm{lim}}\nolimits R_ i = K$. The affine formal algebraic space $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(R_ i)$ is an example of an affine formal algebraic space which is not McQuillan.

Lemma 108.70.1. There exists an affine formal algebraic space which is not McQuillan.

Proof. See discussion above. $\square$

Let $0 \to W_ i \to V_ i \to K \to 0$ be a system of exact sequences as in Section 108.3. Let $A_ i = K[V_ i]/(ww'; w, w' \in W_ i)$. Then there is a compatible system of surjections $A_ i \to K[t]$ with nilpotent kernels and the transition maps $A_ i \to A_ j$ are surjective with nilpotent kernels as well. Recall that $V_ i$ is free over $K$ with basis given by $s \in S_ i$. Then, if the characteristic of $K$ is zero, the degree $d$ part of $A_ i$ is free over $K$ with basis given by $s^ d$, $s \in S_ i$ each of which map to $t^ d$. Hence the inverse system of the degree $d$ parts of the $A_ i$ is isomorphic to the inverse system of the vector spaces $V_ i$. As $\mathop{\mathrm{lim}}\nolimits V_ i = 0$ we conclude that $\mathop{\mathrm{lim}}\nolimits A_ i = K$, at least when the characteristic of $K$ is zero. This gives an example of an affine formal algebraic space whose “regular functions” do not separate points.

Lemma 108.70.2. There exists an affine formal algebraic space $X$ whose regular functions do not separate points, in the following sense: If we write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Formal Spaces, Definition 85.5.1 then $\mathop{\mathrm{lim}}\nolimits \Gamma (X_\lambda , \mathcal{O}_{X_\lambda })$ is a field, but $X_{red}$ has infinitely many points.

Proof. See discussion above. $\square$

Let $K$, $I$, and $(V_ i)$ be as above. Consider systems

\[ \Phi = (\Lambda , J_ i \subset \Lambda , (M_ i) \to (V_ i)) \]

where $\Lambda $ is an augmented $K$-algebra, $J_ i \subset \Lambda $ for $i \in I$ is an ideal of square zero, $(M_ i) \to (V_ i)$ is a map of inverse systems of $K$-vector spaces such that $M_ i \to V_ i$ is surjective for each $i$, such that $M_ i$ has a $\Lambda $-module structure, such that the transition maps $M_ i \to M_ j$, $i > j$ are $\Lambda $-linear, and such that $J_ j M_ i \subset \mathop{\mathrm{Ker}}(M_ i \to M_ j)$ for $i > j$. Claim: There exists a system as above such that $M_ j = M_ i/J_ j M_ i$ for all $i > j$.

If the claim is true, then we obtain a representable morphism

\[ \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathop{\mathrm{Spec}}(\Lambda /J_ i \oplus M_ i) \longrightarrow \text{Spf}(\mathop{\mathrm{lim}}\nolimits \Lambda /J_ i) \]

of affine formal algebraic spaces whose source is not McQuillan but the target is. Here $\Lambda /J_ i \oplus M_ i$ has the usual $\Lambda /J_ i$-algebra structure where $M_ i$ is an ideal of square zero. Representability translates exactly into the condition that $M_ i/J_ jM_ i = M_ j$ for $i > j$. The source of the morphism is not McQuillan as the projections $\mathop{\mathrm{lim}}\nolimits _{i \in I} M_ i \to M_ i$ are not be surjective. This is true because the maps $\mathop{\mathrm{lim}}\nolimits V_ i \to V_ i$ are not surjective and we have the surjection $M_ i \to V_ i$. Some details omitted.

Proof of the claim. First, note that there exists at least one system, namely

\[ \Phi _0 = (K, J_ i = (0), (V_ i) \xrightarrow {\text{id}} (V_ i)) \]

Given a system $\Phi $ we will prove there exists a morphism of systems $\Phi \to \Phi '$ (morphisms of systems defined in the obvious manner) such that $\mathop{\mathrm{Ker}}(M_ i/J_ j M_ i \to M_ j)$ maps to zero in $M'_ i/J'_ j M'_ i$. Once this is done we can do the usual trick of setting $\Phi _ n = (\Phi _{n - 1})'$ inductively for $n \geq 1$ and taking $\Phi = \mathop{\mathrm{colim}}\nolimits \Phi _ n$ to get a system with the desired properties. Details omitted.

Construction of $\Phi '$ given $\Phi $. Consider the set $U$ of triples $u = (i, j, \xi )$ where $i > j$ and $\xi \in \mathop{\mathrm{Ker}}(M_ i \to M_ j)$. We will let $s, t : U \to I$ denote the maps $s(i, j, \xi ) = i$ and $t(i, j, \xi ) = j$. Then we set $\xi _ u \in M_{s(u)}$ the third component of $u$. We take

\[ \Lambda ' = \Lambda [x_ u; u \in U]/(x_ u x_{u'}; u, u' \in U) \]

with augmentation $\Lambda ' \to K$ given by the augmentation of $\Lambda $ and sending $x_ u$ to zero. We take $J'_ k = J_ k \Lambda ' + (x_{u,\ t(u) \geq k})$. We set

\[ M'_ i = M_ i \oplus \bigoplus \nolimits _{s(u) \geq i} K\epsilon _{i, u} \]

As transition maps $M'_ i \to M'_ j$ for $i > j$ we use the given map $M_ i \to M_ j$ and we send $\epsilon _{i, u}$ to $\epsilon _{j, u}$. The map $M'_ i \to V_ i$ induces the given map $M_ i \to V_ i$ and sends $\epsilon _{i, u}$ to zero. Finally, we let $\Lambda '$ act on $M'_ i$ as follows: for $\lambda \in \Lambda $ we act by the $\Lambda $-module structure on $M_ i$ and via the augmentation $\Lambda \to K$ on $\epsilon _{i, u}$. The element $x_ u$ acts as $0$ on $M_ i$ for all $i$. Finally, we define

\[ x_ u \epsilon _{i, u} = \text{image of }\xi _ u\text{ in }M_ i \]

and we set all other products $x_{u'} \epsilon _{i, u}$ equal to zero. The displayed formula makes sense because $s(u) \geq i$ and $\xi _ u \in M_{s(u)}$. The main things the check are $J'_ j M'_ i \subset M'_ i$ maps to zero in $M'_ j$ for $i > j$ and that $\mathop{\mathrm{Ker}}(M_ i \to M_ j)$ maps to zero in $M'_ i/J_ j M'_ i$. The reason for the last fact is that $\xi = x_{(i, j, \xi )} \epsilon _{i, (i, j, \xi )} \in J'_ j M'_ i$ for any $\xi \in \mathop{\mathrm{Ker}}(M_ i \to M_ j)$. We omit the details.

Lemma 108.70.3. There exists a representable morphism $f : X \to Y$ of affine formal algebraic spaces with $Y$ McQuillan, but $X$ not McQuillan.

Proof. See discussion above. $\square$


Comments (2)

Comment #1557 by Matthew Emerton on

In line 3 from bottom, "characteris'' should be "characteristic''.


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