The Stacks project

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 $k''/k$ there exists a common field extension $k'''/k'$ and $k'''/k''$. 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 $k$. $\square$

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. $\square$

Proof. Immediate from the definitions. $\square$

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. $\square$

Proof. Assume $X_{\overline{k}}$ is connected. Let $k'/k$ be a field extension. There exists a field extension $\overline{k}'/\overline{k}$ 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. $\square$

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. $\square$

Proof. By Lemma 33.7.8 there exists a finite subextension $k'/k \subset \overline{k}$ 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. $\square$

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 $k'/k$ contained in $\overline{k}$ such that $x \in A \otimes _ k k'$. Set $G = \text{Gal}(k'/k)$. Set

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 $P(T) = T^ d + a_1T^{d - 1} + \ldots + a_ d$ the we see that $a_ i \in I := A \cap \overline{I}$. Combining $P(x) = 0$ and $a_ i \in I$ we find $x^ d = - a_1 x^{d - 1} - \ldots - a_ d \in I(A \otimes _ k \overline{k})$. Thus $x$ 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. $\square$

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 $\overline{k}/k'/k$ 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

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. $\square$

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 $\overline{k}/k'/k$ 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

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. $\square$

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

Proof. Set $T = \mathop{\mathrm{Spec}}(\kappa (x))$. Let $\overline{k}$ be a separable algebraic closure of $k$. The assumption on $\kappa (x)/k$ 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. $\square$

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 $C \subset T_ K$. $\square$

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. $\square$

Proof. When $K/k$ 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 15.105.2. 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 15.105.1. 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 15.104.9). 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

The maximal weakly étale $L'$-subalgebra of $A \otimes _ L L'$ is $L \otimes _ L L' = L'$ by More on Algebra, Lemma 15.105.4. 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

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

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. $\square$

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). $\square$

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 $\overline{k}/k$ 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). $\square$