The Stacks project

70.12 Geometrically connected algebraic spaces

If $X$ is a connected algebraic space over a field, then it can happen that $X$ becomes disconnected after extending the ground field. This does not happen for geometrically connected algebraic spaces.

Definition 70.12.1. Let $X$ be an algebraic space over the field $k$. We say $X$ is geometrically connected over $k$ if the base change $X_{k'}$ is connected for every field extension $k'$ of $k$.

By convention a connected topological space is nonempty; hence a fortiori geometrically connected algebraic spaces are nonempty.

Lemma 70.12.2. Let $X$ be an algebraic space over the field $k$. Let $k \subset k'$ be a field extension. Then $X$ is geometrically connected over $k$ if and only if $X_{k'}$ is geometrically connected over $k'$.

Proof. If $X$ is geometrically connected over $k$, then it is clear that $X_{k'}$ is geometrically connected over $k'$. For the converse, note that for any field extension $k \subset k''$ there exists a common field extension $k' \subset k'''$ and $k'' \subset k'''$. As the morphism $X_{k'''} \to X_{k''}$ is surjective (as a base change of a surjective morphism between spectra of fields) we see that the connectedness of $X_{k'''}$ implies the connectedness of $X_{k''}$. Thus if $X_{k'}$ is geometrically connected over $k'$ then $X$ is geometrically connected over $k$. $\square$

Lemma 70.12.3. Let $k$ be a field. Let $X$, $Y$ be algebraic spaces over $k$. Assume $X$ is geometrically connected over $k$. Then the projection morphism

\[ p : X \times _ k Y \longrightarrow Y \]

induces a bijection between connected components.

Proof. Let $y \in |Y|$ be represented by a morphism $\mathop{\mathrm{Spec}}(K) \to Y$ be a morphism where $K$ is a field. The fibre of $|X \times _ k Y| \to |Y|$ over $y$ is the image of $|Y_ K| \to |X \times _ k Y|$ by Properties of Spaces, Lemma 64.4.3. Thus these fibres are connected by our assumption that $Y$ is geometrically connected. By Morphisms of Spaces, Lemma 65.6.6 the map $|p|$ is open. Thus we may apply Topology, Lemma 5.7.6 to conclude. $\square$

Lemma 70.12.4. Let $k \subset k'$ be an extension of fields. Let $X$ be an algebraic space over $k$. Assume $k$ separably algebraically closed. Then the morphism $X_{k'} \to X$ induces a bijection of connected components. In particular, $X$ is geometrically connected over $k$ if and only if $X$ is connected.

Proof. Since $k$ is separably algebraically closed we see that $k'$ is geometrically connected over $k$, see Algebra, Lemma 10.47.4. Hence $Z = \mathop{\mathrm{Spec}}(k')$ is geometrically connected over $k$ by Varieties, Lemma 33.7.5. Since $X_{k'} = Z \times _ k X$ the result is a special case of Lemma 70.12.3. $\square$

Lemma 70.12.5. Let $k$ be a field. Let $X$ be an algebraic space over $k$. Let $\overline{k}$ be a separable algebraic closure of $k$. Then $X$ is geometrically connected if and only if the base change $X_{\overline{k}}$ is connected.

Proof. Assume $X_{\overline{k}}$ is connected. Let $k \subset k'$ be a field extension. There exists a field extension $\overline{k} \subset \overline{k}'$ such that $k'$ embeds into $\overline{k}'$ as an extension of $k$. By Lemma 70.12.4 we see that $X_{\overline{k}'}$ is connected. Since $X_{\overline{k}'} \to X_{k'}$ is surjective we conclude that $X_{k'}$ is connected as desired. $\square$

Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite) Galois extension. For example $\overline{k}$ could be the separable algebraic closure of $k$. For any $\sigma \in \text{Gal}(\overline{k}/k)$ we get a corresponding automorphism $ \mathop{\mathrm{Spec}}(\sigma ) : \mathop{\mathrm{Spec}}(\overline{k}) \longrightarrow \mathop{\mathrm{Spec}}(\overline{k}) $. Note that $\mathop{\mathrm{Spec}}(\sigma ) \circ \mathop{\mathrm{Spec}}(\tau ) = \mathop{\mathrm{Spec}}(\tau \circ \sigma )$. Hence we get an action

\[ \text{Gal}(\overline{k}/k)^{opp} \times \mathop{\mathrm{Spec}}(\overline{k}) \longrightarrow \mathop{\mathrm{Spec}}(\overline{k}) \]

of the opposite group on the scheme $\mathop{\mathrm{Spec}}(\overline{k})$. Let $X$ be an algebraic space over $k$. Since $X_{\overline{k}} = \mathop{\mathrm{Spec}}(\overline{k}) \times _{\mathop{\mathrm{Spec}}(k)} X$ by definition we see that the action above induces a canonical action

70.12.5.1
\begin{equation} \label{spaces-over-fields-equation-galois-action-base-change-kbar} \text{Gal}(\overline{k}/k)^{opp} \times X_{\overline{k}} \longrightarrow X_{\overline{k}}. \end{equation}

Lemma 70.12.6. Let $k$ be a field. Let $X$ be an algebraic space over $k$. Let $\overline{k}$ be a (possibly infinite) Galois extension of $k$. Let $V \subset X_{\overline{k}}$ be a quasi-compact open. Then

  1. there exists a finite subextension $k \subset k' \subset \overline{k}$ and a quasi-compact open $V' \subset X_{k'}$ such that $V = (V')_{\overline{k}}$,

  2. there exists an open subgroup $H \subset \text{Gal}(\overline{k}/k)$ such that $\sigma (V) = V$ for all $\sigma \in H$.

Proof. Choose a scheme $U$ and a surjective ├ętale morphism $U \to X$. Choose a quasi-compact open $W \subset U_{\overline{k}}$ whose image in $X_{\overline{k}}$ is $V$. This is possible because $|U_{\overline{k}}| \to |X_{\overline{k}}|$ is continuous and because $|U_{\overline{k}}|$ has a basis of quasi-compact opens. We can apply Varieties, Lemma 33.7.9 to $W \subset U_{\overline{k}}$ to obtain the lemma. $\square$

Lemma 70.12.7. Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite) Galois extension. Let $X$ be an algebraic space over $k$. Let $\overline{T} \subset |X_{\overline{k}}|$ have the following properties

  1. $\overline{T}$ is a closed subset of $|X_{\overline{k}}|$,

  2. for every $\sigma \in \text{Gal}(\overline{k}/k)$ we have $\sigma (\overline{T}) = \overline{T}$.

Then there exists a closed subset $T \subset |X|$ whose inverse image in $|X_{k'}|$ is $\overline{T}$.

Proof. Let $T \subset |X|$ be the image of $\overline{T}$. Since $|X_{\overline{k}}| \to |X|$ is surjective, the statement means that $T$ is closed and that its inverse image is $\overline{T}$. Choose a scheme $U$ and a surjective ├ętale morphism $U \to X$. By the case of schemes (see Varieties, Lemma 33.7.10) there exists a closed subset $T' \subset |U|$ whose inverse image in $|U_{\overline{k}}|$ is the inverse image of $\overline{T}$. Since $|U_{\overline{k}}| \to |X_{\overline{k}}|$ is surjective, we see that $T'$ is the inverse image of $T$ via $|U| \to |X|$. By our construction of the topology on $|X|$ this means that $T$ is closed. In the same manner one sees that $\overline{T}$ is the inverse image of $T$. $\square$

Lemma 70.12.8. Let $k$ be a field. Let $X$ be an algebraic space over $k$. The following are equivalent

  1. $X$ is geometrically connected,

  2. for every finite separable field extension $k \subset k'$ the algebraic space $X_{k'}$ is connected.

Proof. This proof is identical to the proof of Varieties, Lemma 33.7.11 except that we replace Varieties, Lemma 33.7.7 by Lemma 70.12.5, we replace Varieties, Lemma 33.7.9 by Lemma 70.12.6, and we replace Varieties, Lemma 33.7.10 by Lemma 70.12.7. We urge the reader to read that proof in stead of this one.

It follows immediately from the definition that (1) implies (2). Assume that $X$ is not geometrically connected. Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. By Lemma 70.12.5 it follows that $X_{\overline{k}}$ is disconnected. Say $X_{\overline{k}} = \overline{U} \amalg \overline{V}$ with $\overline{U}$ and $\overline{V}$ open, closed, and nonempty algebraic subspaces of $X_{\overline{k}}$.

Suppose that $W \subset X$ is any quasi-compact open subspace. Then $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are open and closed subspaces of $W_{\overline{k}}$. In particular $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are quasi-compact, and by Lemma 70.12.6 both $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are defined over a finite subextension and invariant under an open subgroup of $\text{Gal}(\overline{k}/k)$. We will use this without further mention in the following.

Pick $W_0 \subset X$ quasi-compact open subspace such that both $W_{0, \overline{k}} \cap \overline{U}$ and $W_{0, \overline{k}} \cap \overline{V}$ are nonempty. Choose a finite subextension $k \subset k' \subset \overline{k}$ and a decomposition $W_{0, k'} = U_0' \amalg V_0'$ into open and closed subsets such that $W_{0, \overline{k}} \cap \overline{U} = (U'_0)_{\overline{k}}$ and $W_{0, \overline{k}} \cap \overline{V} = (V'_0)_{\overline{k}}$. Let $H = \text{Gal}(\overline{k}/k') \subset \text{Gal}(\overline{k}/k)$. In particular $\sigma (W_{0, \overline{k}} \cap \overline{U}) = W_{0, \overline{k}} \cap \overline{U}$ and similarly for $\overline{V}$.

Having chosen $W_0$, $k'$ as above, for every quasi-compact open subspace $W \subset X$ we set

\[ U_ W = \bigcap \nolimits _{\sigma \in H} \sigma (W_{\overline{k}} \cap \overline{U}), \quad V_ W = \bigcup \nolimits _{\sigma \in H} \sigma (W_{\overline{k}} \cap \overline{V}). \]

Now, since $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are fixed by an open subgroup of $\text{Gal}(\overline{k}/k)$ we see that the union and intersection above are finite. Hence $U_ W$ and $V_ W$ are both open and closed subspaces. Also, by construction $W_{\bar k} = U_ W \amalg V_ W$.

We claim that if $W \subset W' \subset X$ are quasi-compact open subspaces, then $W_{\overline{k}} \cap U_{W'} = U_ W$ and $W_{\overline{k}} \cap V_{W'} = V_ W$. Verification omitted. Hence we see that upon defining $U = \bigcup _{W \subset X} U_ W$ and $V = \bigcup _{W \subset X} V_ W$ we obtain $X_{\overline{k}} = U \amalg V$ is a disjoint union of open and closed subsets. It is clear that $V$ is nonempty as it is constructed by taking unions (locally). On the other hand, $U$ is nonempty since it contains $W_0 \cap \overline{U}$ by construction. Finally, $U, V \subset X_{\bar k}$ are closed and $H$-invariant by construction. Hence by Lemma 70.12.7 we have $U = (U')_{\bar k}$, and $V = (V')_{\bar k}$ for some closed $U', V' \subset X_{k'}$. Clearly $X_{k'} = U' \amalg V'$ and we see that $X_{k'}$ is disconnected as desired. $\square$


Comments (2)

Comment #3982 by Matthieu Romagny on

Second sentence: This does not happen for geometrically connected algebraic spaces.


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