## 33.7 Geometrically connected schemes

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

Definition 33.7.1. Let $X$ be a scheme over the field $k$. We say $X$ is geometrically connected over $k$ if the scheme $X_{k'}$ is connected for every field extension $k'$ of $k$.

By convention a connected topological space is nonempty; hence a fortiori geometrically connected schemes are nonempty. Here is an example of a variety which is not geometrically connected.

Example 33.7.2. Let $k = \mathbf{Q}$. The scheme $X = \mathop{\mathrm{Spec}}(\mathbf{Q}(i))$ is a variety over $\mathop{\mathrm{Spec}}(\mathbf{Q})$. But the base change $X_{\mathbf{C}}$ is the spectrum of $\mathbf{C} \otimes _{\mathbf{Q}} \mathbf{Q}(i) \cong \mathbf{C} \times \mathbf{C}$ which is the disjoint union of two copies of $\mathop{\mathrm{Spec}}(\mathbf{C})$. So in fact, this is an example of a non-geometrically connected variety.

Lemma 33.7.3. Let $X$ be a scheme 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 33.7.4. Let $k$ be a field. Let $X$, $Y$ be schemes 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. The scheme theoretic fibres of $p$ are connected, since they are base changes of the geometrically connected scheme $X$ by field extensions. Moreover the scheme theoretic fibres are homeomorphic to the set theoretic fibres, see Schemes, Lemma 26.18.5. By Morphisms, Lemma 29.23.4 the map $p$ is open. Thus we may apply Topology, Lemma 5.7.6 to conclude. $\square$

Lemma 33.7.5. Let $k$ be a field. Let $A$ be a $k$-algebra. Then $X = \mathop{\mathrm{Spec}}(A)$ is geometrically connected over $k$ if and only if $A$ is geometrically connected over $k$ (see Algebra, Definition 10.48.3).

Proof. Immediate from the definitions. $\square$

Lemma 33.7.6. Let $k \subset k'$ be an extension of fields. Let $X$ be a scheme 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 10.48.4. Hence $Z = \mathop{\mathrm{Spec}}(k')$ is geometrically connected over $k$ by Lemma 33.7.5 above. Since $X_{k'} = Z \times _ k X$ the result is a special case of Lemma 33.7.4. $\square$

Lemma 33.7.7. Let $k$ be a field. Let $X$ be a scheme 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 33.7.6 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. $\square$

Lemma 33.7.8. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $A$ be a $k$-algebra. Let $V \subset X_ A$ be a quasi-compact open. Then there exists a finitely generated $k$-subalgebra $A' \subset A$ and a quasi-compact open $V' \subset X_{A'}$ such that $V = V'_ A$.

Proof. We remark that if $X$ is also quasi-separated this follows from Limits, Lemma 32.4.11. Let $U_1, \ldots , U_ n$ be finitely many affine opens of $X$ such that $V \subset \bigcup U_{i, A}$. Say $U_ i = \mathop{\mathrm{Spec}}(R_ i)$. Since $V$ is quasi-compact we can find finitely many $f_{ij} \in R_ i \otimes _ k A$, $j = 1, \ldots , n_ i$ such that $V = \bigcup _ i \bigcup _{j = 1, \ldots , n_ i} D(f_{ij})$ where $D(f_{ij}) \subset U_{i, A}$ is the corresponding standard open. (We do not claim that $V \cap U_{i, A}$ is the union of the $D(f_{ij})$, $j = 1, \ldots , n_ i$.) It is clear that we can find a finitely generated $k$-subalgebra $A' \subset A$ such that $f_{ij}$ is the image of some $f_{ij}' \in R_ i \otimes _ k A'$. Set $V' = \bigcup D(f_{ij}')$ which is a quasi-compact open of $X_{A'}$. Denote $\pi : X_ A \to X_{A'}$ the canonical morphism. We have $\pi (V) \subset V'$ as $\pi (D(f_{ij})) \subset D(f_{ij}')$. If $x \in X_ A$ with $\pi (x) \in V'$, then $\pi (x) \in D(f_{ij}')$ for some $i, j$ and we see that $x \in D(f_{ij})$ as $f_{ij}'$ maps to $f_{ij}$. Thus we see that $V = \pi ^{-1}(V')$ 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 a scheme 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

33.7.8.1
$$\label{varieties-equation-galois-action-base-change-kbar} \text{Gal}(\overline{k}/k)^{opp} \times X_{\overline{k}} \longrightarrow X_{\overline{k}}.$$

Lemma 33.7.9. Let $k$ be a field. Let $X$ be a scheme 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. By Lemma 33.7.8 there exists a finite subextension $k \subset k' \subset \overline{k}$ and an open $V' \subset X_{k'}$ which pulls back to $V$. This proves (1). Since $\text{Gal}(\overline{k}/k')$ is open in $\text{Gal}(\overline{k}/k)$ part (2) is clear as well. $\square$

Lemma 33.7.10. Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite) Galois extension. Let $X$ be a scheme 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 $X_{\overline{k}}$ is $\overline{T}$.

Proof. This lemma immediately reduces to the case where $X = \mathop{\mathrm{Spec}}(A)$ is affine. In this case, let $\overline{I} \subset A \otimes _ k \overline{k}$ be the radical ideal corresponding to $\overline{T}$. Assumption (2) implies that $\sigma (\overline{I}) = \overline{I}$ for all $\sigma \in \text{Gal}(\overline{k}/k)$. Pick $x \in \overline{I}$. There exists a finite Galois extension $k \subset k'$ contained in $\overline{k}$ such that $x \in A \otimes _ k k'$. Set $G = \text{Gal}(k'/k)$. Set

$P(T) = \prod \nolimits _{\sigma \in G} (T - \sigma (x)) \in (A \otimes _ k k')[T]$

It is clear that $P(T)$ is monic and is actually an element of $(A \otimes _ k k')^ G[T] = A[T]$ (by basic Galois theory). Moreover, if we write $P(T) = T^ d + a_1T^{d - 1} + \ldots + a_0$ the we see that $a_ i \in I := A \cap \overline{I}$. By Algebra, Lemma 10.38.5 we see that $x$ is contained in the radical of $I(A \otimes _ k \overline{k})$. Hence $\overline{I}$ is the radical of $I(A \otimes _ k \overline{k})$ and setting $T = V(I)$ is a solution. $\square$

Lemma 33.7.11. Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent

1. $X$ is geometrically connected,

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

Proof. 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 33.7.7 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.

Suppose that $W \subset X$ is any quasi-compact open. Then $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are open and closed in $W_{\overline{k}}$. In particular $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are quasi-compact, and by Lemma 33.7.9 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 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 $W \subset X$ 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. Also, by construction $W_{\bar k} = U_ W \amalg V_ W$.

We claim that if $W \subset W' \subset X$ are quasi-compact open, 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 33.7.10 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. $\square$

Lemma 33.7.12. Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite) Galois extension. Let $f : T \to X$ be a morphism of schemes over $k$. Assume $T_{\overline{k}}$ connected and $X_{\overline{k}}$ disconnected. Then $X$ is disconnected.

Proof. Write $X_{\overline{k}} = \overline{U} \amalg \overline{V}$ with $\overline{U}$ and $\overline{V}$ open and closed. Denote $\overline{f} : T_{\overline{k}} \to X_{\overline{k}}$ the base change of $f$. Since $T_{\overline{k}}$ is connected we see that $T_{\overline{k}}$ is contained in either $\overline{f}^{-1}(\overline{U})$ or $\overline{f}^{-1}(\overline{V})$. Say $T_{\overline{k}} \subset \overline{f}^{-1}(\overline{U})$.

Fix a quasi-compact open $W \subset X$. There exists a finite Galois subextension $k \subset k' \subset \overline{k}$ such that $\overline{U} \cap W_{\overline{k}}$ and $\overline{V} \cap W_{\overline{k}}$ come from quasi-compact opens $U', V' \subset W_{k'}$. Then also $W_{k'} = U' \amalg V'$. Consider

$U'' = \bigcap \nolimits _{\sigma \in \text{Gal}(k'/k)} \sigma (U'), \quad V'' = \bigcup \nolimits _{\sigma \in \text{Gal}(k'/k)} \sigma (V').$

These are Galois invariant, open and closed, and $W_{k'} = U'' \amalg V''$. By Lemma 33.7.10 we get open and closed subsets $U_ W, V_ W \subset W$ such that $U'' = (U_ W)_{k'}$, $V'' = (V_ W)_{k'}$ and $W = U_ W \amalg V_ W$.

We claim that if $W \subset W' \subset X$ are quasi-compact open, then $W \cap U_{W'} = U_ W$ and $W \cap V_{W'} = V_ W$. 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 $X = U \amalg V$. 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 $f(T)$ by construction. $\square$

Lemma 33.7.13. Let $k$ be a field. Let $T \to X$ be a morphism of schemes over $k$. Assume $T$ is geometrically connected and $X$ connected. Then $X$ is geometrically connected.

Proof. This is a reformulation of Lemma 33.7.12. $\square$

Lemma 33.7.14. Let $k$ be a field. Let $X$ be a scheme over $k$. Assume $X$ is connected and has a point $x$ such that $k$ is algebraically closed in $\kappa (x)$. Then $X$ is geometrically connected. In particular, if $X$ has a $k$-rational point and $X$ is connected, then $X$ is geometrically connected.

Proof. Set $T = \mathop{\mathrm{Spec}}(\kappa (x))$. Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. The assumption on $k \subset \kappa (x)$ implies that $T_{\overline{k}}$ is irreducible, see Algebra, Lemma 10.47.8. Hence by Lemma 33.7.13 we see that $X_{\overline{k}}$ is connected. By Lemma 33.7.7 we conclude that $X$ is geometrically connected. $\square$

Lemma 33.7.15. Let $k \subset K$ be an extension of fields. Let $X$ be a scheme over $k$. For every connected component $T$ of $X$ the inverse image $T_ K \subset X_ K$ is a union of connected components of $X_ K$.

Proof. This is a purely topological statement. Denote $p : X_ K \to X$ the projection morphism. Let $T \subset X$ be a connected component of $X$. Let $t \in T_ K = p^{-1}(T)$. Let $C \subset X_ K$ be a connected component containing $t$. Then $p(C)$ is a connected subset of $X$ which meets $T$, hence $p(C) \subset T$. Hence $C \subset T_ K$. $\square$

The following lemma will be superseded by the stronger Lemma 33.7.17 below.

Lemma 33.7.16. Let $k \subset K$ be a finite extension of fields and let $X$ be a scheme over $k$. Denote by $p : X_ K \to X$ the projection morphism. For every connected component $T$ of $X_ K$ the image $p(T)$ is a connected component of $X$.

Proof. The image $p(T)$ is contained in some connected component $X'$ of $X$. Consider $X'$ as a closed subscheme of $X$ in any way. Then $T$ is also a connected component of $X'_ K = p^{-1}(X')$ and we may therefore assume that $X$ is connected. The morphism $p$ is open (Morphisms, Lemma 29.23.4), closed (Morphisms, Lemma 29.44.7) and the fibers of $p$ are finite sets (Morphisms, Lemma 29.44.10). Thus we may apply Topology, Lemma 5.7.7 to conclude. $\square$

Lemma 33.7.17 (Gabber). Let $k \subset K$ be an extension of fields. Let $X$ be a scheme over $k$. Denote $p : X_ K \to X$ the projection morphism. Let $\overline{T} \subset X_ K$ be a connected component. Then $p(\overline{T})$ is a connected component of $X$.

Proof. When $k \subset K$ is finite this is Lemma 33.7.16. In general the proof is more difficult.

Let $T \subset X$ be the connected component of $X$ containing the image of $\overline{T}$. We may replace $X$ by $T$ (with the induced reduced subscheme structure). Thus we may assume $X$ is connected. Let $A = H^0(X, \mathcal{O}_ X)$. Let $L \subset A$ be the maximal weakly étale $k$-subalgebra, see More on Algebra, Lemma 15.104.2. Since $A$ does not have any nontrivial idempotents we see that $L$ is a field and a separable algebraic extension of $k$ by More on Algebra, Lemma 15.104.1. Observe that $L$ is also the maximal weakly étale $L$-subalgebra of $A$ (because any weakly étale $L$-algebra is weakly étale over $k$ by More on Algebra, Lemma 15.103.9). By Schemes, Lemma 26.6.4 we obtain a factorization $X \to \mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(k)$ of the structure morphism.

Let $L'/L$ be a finite separable extension. By Cohomology of Schemes, Lemma 30.5.3 we have

$A \otimes _ L L' = H^0(X \times _{\mathop{\mathrm{Spec}}(L)} \mathop{\mathrm{Spec}}(L'), \mathcal{O}_{X \times _{\mathop{\mathrm{Spec}}(L)} \mathop{\mathrm{Spec}}(L')})$

The maximal weakly étale $L'$-subalgebra of $A \otimes _ L L'$ is $L \otimes _ L L' = L'$ by More on Algebra, Lemma 15.104.4. In particular $A \otimes _ L L'$ does not have nontrivial idempotents (such an idempotent would generate a weakly étale subalgebra) and we conclude that $X \times _{\mathop{\mathrm{Spec}}(L)} \mathop{\mathrm{Spec}}(L')$ is connected. By Lemma 33.7.11 we conclude that $X$ is geometrically connected over $L$.

Let's give $\overline{T}$ the reduced induced scheme structure and consider the composition

$\overline{T} \xrightarrow {i} X_ K = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(K) \xrightarrow {\pi } \mathop{\mathrm{Spec}}(L \otimes _ k K)$

The image is contained in a connected component of $\mathop{\mathrm{Spec}}(L \otimes _ k K)$. Since $K \to L \otimes _ k K$ is integral we see that the connected components of $\mathop{\mathrm{Spec}}(L \otimes _ k K)$ are points and all points are closed, see Algebra, Lemma 10.36.19. Thus we get a quotient field $L \otimes _ k K \to E$ such that $\overline{T}$ maps into $\mathop{\mathrm{Spec}}(E) \subset \mathop{\mathrm{Spec}}(L \otimes _ k K)$. Hence $i(\overline{T}) \subset \pi ^{-1}(\mathop{\mathrm{Spec}}(E))$. But

$\pi ^{-1}(\mathop{\mathrm{Spec}}(E)) = (X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(K)) \times _{\mathop{\mathrm{Spec}}(L \otimes _ k K)} \mathop{\mathrm{Spec}}(E) = X \times _{\mathop{\mathrm{Spec}}(L)} \mathop{\mathrm{Spec}}(E)$

which is connected because $X$ is geometrically connected over $L$. Then we get the equality $\overline{T} = X \times _{\mathop{\mathrm{Spec}}(L)} \mathop{\mathrm{Spec}}(E)$ (set theoretically) and we conclude that $\overline{T} \to X$ is surjective as desired. $\square$

Let $X$ be a scheme. We denote $\pi _0(X)$ the set of connected components of $X$.

Lemma 33.7.18. Let $k$ be a field, with separable algebraic closure $\overline{k}$. Let $X$ be a scheme over $k$. There is an action

$\text{Gal}(\overline{k}/k)^{opp} \times \pi _0(X_{\overline{k}}) \longrightarrow \pi _0(X_{\overline{k}})$

with the following properties:

1. An element $\overline{T} \in \pi _0(X_{\overline{k}})$ is fixed by the action if and only if there exists a connected component $T \subset X$, which is geometrically connected over $k$, such that $T_{\overline{k}} = \overline{T}$.

2. For any field extension $k \subset k'$ with separable algebraic closure $\overline{k}'$ the diagram

$\xymatrix{ \text{Gal}(\overline{k}'/k') \times \pi _0(X_{\overline{k}'}) \ar[r] \ar[d] & \pi _0(X_{\overline{k}'}) \ar[d] \\ \text{Gal}(\overline{k}/k) \times \pi _0(X_{\overline{k}}) \ar[r] & \pi _0(X_{\overline{k}}) }$

is commutative (where the right vertical arrow is a bijection according to Lemma 33.7.6).

Proof. The action (33.7.8.1) of $\text{Gal}(\overline{k}/k)$ on $X_{\overline{k}}$ induces an action on its connected components. Connected components are always closed (Topology, Lemma 5.7.3). Hence if $\overline{T}$ is as in (1), then by Lemma 33.7.10 there exists a closed subset $T \subset X$ such that $\overline{T} = T_{\overline{k}}$. Note that $T$ is geometrically connected over $k$, see Lemma 33.7.7. To see that $T$ is a connected component of $X$, suppose that $T \subset T'$, $T \not= T'$ where $T'$ is a connected component of $X$. In this case $T'_{k'}$ strictly contains $\overline{T}$ and hence is disconnected. By Lemma 33.7.12 this means that $T'$ is disconnected! Contradiction.

We omit the proof of the functoriality in (2). $\square$

Lemma 33.7.19. Let $k$ be a field, with separable algebraic closure $\overline{k}$. Let $X$ be a scheme over $k$. Assume

1. $X$ is quasi-compact, and

2. the connected components of $X_{\overline{k}}$ are open.

Then

1. $\pi _0(X_{\overline{k}})$ is finite, and

2. the action of $\text{Gal}(\overline{k}/k)$ on $\pi _0(X_{\overline{k}})$ is continuous.

Moreover, assumptions (1) and (2) are satisfied when $X$ is of finite type over $k$.

Proof. Since the connected components are open, cover $X_{\overline{k}}$ (Topology, Lemma 5.7.3) and $X_{\overline{k}}$ is quasi-compact, we conclude that there are only finitely many of them. Thus (a) holds. By Lemma 33.7.8 these connected components are each defined over a finite subextension of $k \subset \overline{k}$ and we get (b). If $X$ is of finite type over $k$, then $X_{\overline{k}}$ is of finite type over $\overline{k}$ (Morphisms, Lemma 29.15.4). Hence $X_{\overline{k}}$ is a Noetherian scheme (Morphisms, Lemma 29.15.6). Thus $X_{\overline{k}}$ has finitely many irreducible components (Properties, Lemma 28.5.7) and a fortiori finitely many connected components (which are therefore open). $\square$

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