The Stacks project

Proof. If $X$ is geometrically irreducible over $k$, then it is clear that $X_{k'}$ is geometrically irreducible 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 irreducibility of $X_{k'''}$ implies the irreducibility of $X_{k''}$. Thus if $X_{k'}$ is geometrically irreducible over $k'$ then $X$ is geometrically irreducible over $k$. $\square$

Proof. Use Properties, Lemma 28.3.3 and Algebra, Lemma 10.47.2. $\square$

Proof. First, note that the scheme theoretic fibres of $p$ are irreducible, since they are base changes of the geometrically irreducible 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 5.8.15 to conclude. $\square$

Proof. An affine scheme $\mathop{\mathrm{Spec}}(A)$ over $k$ is geometrically irreducible if and only if $A$ is geometrically irreducible over $k$; this is immediate from the definitions. Recall that if a scheme is irreducible so is every nonempty open subscheme of $X$, any two nonempty open subsets have a nonempty intersection. Also, if every affine open is irreducible then the scheme is irreducible, see Properties, Lemma 28.3.3. Hence the final statement of the lemma is clear, as well as the implications (1) $\Rightarrow $ (2), (2) $\Rightarrow $ (3), and (3) $\Rightarrow $ (4). If (4) holds, then for any field extension $k'/k$ the scheme $X_{k'}$ has a covering by irreducible opens which pairwise intersect. Hence $X_{k'}$ is irreducible. Hence (4) implies (1). $\square$

Proof. Assume (1). Recall that $\mathcal{O}_{X, \xi }$ is the filtered colimit of $\mathcal{O}_ X(U)$ where $U$ runs over the nonempty open affine subschemes of $X$. Combining Lemma 33.8.5 and Algebra, Lemma 10.47.6 we see that $\mathcal{O}_{X, \xi }$ is geometrically irreducible over $k$. Since $\mathcal{O}_{X, \xi } \to \kappa (\xi )$ is a surjection with locally nilpotent kernel (see Algebra, Lemma 10.25.1) it follows that $\kappa (\xi )$ is geometrically irreducible, see Algebra, Lemma 10.46.7.

Assume (2). We may assume that $X$ is reduced. Let $U \subset X$ be a nonempty affine open. Then $U = \mathop{\mathrm{Spec}}(A)$ where $A$ is a domain with fraction field $\kappa (\xi )$. Thus $A$ is a $k$-subalgebra of a geometrically irreducible $k$-algebra. Hence by Algebra, Lemma 10.47.6 we see that $A$ is geometrically irreducible over $k$. By Lemma 33.8.5 we conclude that $X$ is geometrically irreducible over $k$. $\square$

Proof. Since $k$ is separably algebraically closed we see that $k'$ is geometrically irreducible over $k$, see Algebra, Lemma 10.47.5. Hence $Z = \mathop{\mathrm{Spec}}(k')$ is geometrically irreducible over $k$. by Lemma 33.8.5 above. Since $X' = Z \times _ k X$ the result is a special case of Lemma 33.8.4. $\square$

Proof. Assume $X_{\overline{k}}$ is irreducible, i.e., assume (3). 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 33.8.7 we see that $X_{\overline{k}'}$ is irreducible. Since $X_{\overline{k}'} \to X_{k'}$ is surjective we conclude that $X_{k'}$ is irreducible. Hence (1) holds.

Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. Assume not (3), i.e., assume $X_{\overline{k}}$ is reducible. Our goal is to show that also $X_{k'}$ is reducible for some finite subextension $\overline{k}/k'/k$. Let $X = \bigcup _{i \in I} U_ i$ be an affine open covering with $U_ i$ not empty. If for some $i$ the scheme $U_ i$ is reducible, or if for some pair $i \not= j$ the intersection $U_ i \cap U_ j$ is empty, then $X$ is reducible (Properties, Lemma 28.3.3) and we are done. In particular we may assume that $U_{i, \overline{k}} \cap U_{j, \overline{k}}$ for all $i, j \in I$ is nonempty and we conclude that $U_{i, \overline{k}}$ has to be reducible for some $i$. According to Algebra, Lemma 10.47.3 this means that $U_{i, k'}$ is reducible for some finite separable field extension $k'/k$. Hence also $X_{k'}$ is reducible. Thus we see that (2) implies (3).

The implication (1) $\Rightarrow $ (2) is immediate. This proves the lemma. $\square$

Proof. Let $T \subset X$ be an irreducible component of $X$. The morphism $T_ K \to T$ is flat, so generalizations lift along $T_ K \to T$. Hence every $\xi \in T_ K$ which is a generic point of an irreducible component of $T_ K$ maps to the generic point $\eta $ of $T$. If $\xi ' \leadsto \xi $ is a specialization in $X_ K$ then $\xi '$ maps to $\eta $ since there are no points specializing to $\eta $ in $X$. Hence $\xi ' \in T_ K$ and we conclude that $\xi = \xi '$. In other words $\xi $ is the generic point of an irreducible component of $X_ K$. This means that the irreducible components of $T_ K$ are all irreducible components of $X_ K$. $\square$

Proof. Consider the diagram

where $\overline{K}$ is the separable algebraic closure of $K$, and where $\overline{k}$ is the separable algebraic closure of $k$. By Lemma 33.8.7 the morphism $X_{\overline{K}} \to X_{\overline{k}}$ induces a bijection between irreducible components. Hence it suffices to show the lemma for the morphisms $X_{\overline{k}} \to X$ and $X_{\overline{K}} \to X_ K$. In other words we may assume that $K = \overline{k}$.

The morphism $p : X_{\overline{k}} \to X$ is integral, flat and surjective. Flatness implies that generalizations lift along $p$, see Morphisms, Lemma 29.25.9. Hence generic points of irreducible components of $X_{\overline{k}}$ map to generic points of irreducible components of $X$. Integrality implies that $p$ is universally closed, see Morphisms, Lemma 29.44.7. Hence we conclude that the image $p(\overline{T})$ of an irreducible component is a closed irreducible subset which contains a generic point of an irreducible component of $X$, hence $p(\overline{T})$ is an irreducible component of $X$. This proves the first assertion. If $T \subset X$ is an irreducible component, then $p^{-1}(T) =T_ K$ is a nonempty union of irreducible components, see Lemma 33.8.9. Each of these necessarily maps onto $T$ by the first part. Hence the map is surjective. $\square$

Proof. Let $k'/k$ be a finite extension of fields and let $Z, Z' \subset X_{k'}$ be irreducible components. It suffices to show $Z = Z'$, see Lemma 33.8.8. By Lemma 33.8.10 we have $p(Z) = p(Z') = X$ where $p : X_{k'} \to X$ is the projection. If $Z \not= Z'$ then $Z \cap Z'$ is nowhere dense in $X_{k'}$ and hence $p(Z \cap Z')$ is not dense by Morphisms, Lemma 29.48.7; here we also use that $p$ is a finite morphism as the base change of the finite morphism $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$, see Morphisms, Lemma 29.44.6. Thus we can pick a $k$-rational point $x \in X$ with $x \not\in p(Z \cap Z')$. Since the residue field of $x$ is $k$ we see that $p^{-1}(\{ x\} ) = \{ x'\} $ where $x' \in X_{k'}$ is a point whose residue field is $k'$. Since $x \in p(Z) = p(Z')$ we conclude that $x' \in Z \cap Z'$ which is the contradiction we were looking for. $\square$

Proof. The action (33.7.8.1) of $\text{Gal}(\overline{k}/k)$ on $X_{\overline{k}}$ induces an action on its irreducible components. Irreducible 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 irreducible over $k$, see Lemma 33.8.8. To see that $T$ is an irreducible component of $X$, suppose that $T \subset T'$, $T \not= T'$ where $T'$ is an irreducible component of $X$. Let $\overline{\eta }$ be the generic point of $\overline{T}$. It maps to the generic point $\eta $ of $T$. Then the generic point $\xi \in T'$ specializes to $\eta $. As $X_{\overline{k}} \to X$ is flat there exists a point $\overline{\xi } \in X_{\overline{k}}$ which maps to $\xi $ and specializes to $\overline{\eta }$. It follows that the closure of the singleton $\{ \overline{\xi }\} $ is an irreducible closed subset of $X_{\overline{\xi }}$ which strictly contains $\overline{T}$. This is the desired contradiction.

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

Proof. Let $T \subset X$ be an irreducible component of $X$. Let $\eta \in T$ be its generic point. By Lemmas 33.8.9 and 33.8.10 the generic points of irreducible components of $\overline{T}$ which map into $T$ map to $\eta $. By Algebra, Lemma 10.47.14 the Galois group acts transitively on all of the points of $X_{\overline{k}}$ mapping to $\eta $. Hence the lemma follows. $\square$

Proof. Let $\overline{T}$ be an irreducible component of $X_{\overline{k}}$. We may choose an affine open $U \subset X$ such that $\overline{T} \cap U_{\overline{k}}$ is not empty. Write $U = \mathop{\mathrm{Spec}}(A)$, so $A$ is a finite type $k$-algebra, see Morphisms, Lemma 29.15.2. Hence $A_{\overline{k}}$ is a finite type $\overline{k}$-algebra, and in particular Noetherian. Let $\mathfrak p = (f_1, \ldots , f_ n)$ be the prime ideal corresponding to $\overline{T} \cap U_{\overline{k}}$. Since $A_{\overline{k}} = A \otimes _ k \overline{k}$ we see that there exists a finite subextension $\overline{k}/k'/k$ such that each $f_ i \in A_{k'}$. It is clear that $\text{Gal}(\overline{k}/k')$ fixes $\overline{T}$, which proves (1).

Part (2) follows by applying Lemma 33.8.12 (1) to the situation over $k'$ which implies the irreducible component $\overline{T}$ is of the form $T'_{\overline{k}}$ for some irreducible $T' \subset X_{k'}$.

To prove (3), let $T \subset X$ be an irreducible component. Choose an irreducible component $\overline{T} \subset X_{\overline{k}}$ which maps to $T$, see Lemma 33.8.10. By the above the orbit of $\overline{T}$ is finite, say it is $\overline{T}_1, \ldots , \overline{T}_ n$. Then $\overline{T}_1 \cup \ldots \cup \overline{T}_ n$ is a $\text{Gal}(\overline{k}/k)$-invariant closed subset of $X_{\overline{k}}$ hence of the form $W_{\overline{k}}$ for some $W \subset X$ closed by Lemma 33.7.10. Clearly $W = T$ and we win. $\square$

Proof. Let $\overline{k}$ be a separable algebraic closure of $k$. The assumption that $X$ has finitely many irreducible components combined with Lemma 33.8.14 (3) shows that $X_{\overline{k}}$ has finitely many irreducible components $\overline{T}_1, \ldots , \overline{T}_ n$. By Lemma 33.8.14 (2) there exists a finite extension $\overline{k}/k'/k$ and irreducible components $T_ i \subset X_{k'}$ such that $\overline{T}_ i = T_{i, \overline{k}}$ and we win. $\square$

Proof. This is clear because a connected component is a union of irreducible components. Details omitted. $\square$