Definition 33.8.1. Let $X$ be a scheme over the field $k$. We say $X$ is *geometrically irreducible* over $k$ if the scheme $X_{k'}$ is irreducible^{1} for any field extension $k'$ of $k$.

## 33.8 Geometrically irreducible schemes

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

Lemma 33.8.2. Let $X$ be a scheme over the field $k$. Let $k'/k$ be a field extension. Then $X$ is geometrically irreducible over $k$ if and only if $X_{k'}$ is geometrically irreducible over $k'$.

**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 $k$.
$\square$

Lemma 33.8.3. Let $X$ be a scheme over a separably closed field $k$. If $X$ is irreducible, then $X_ K$ is irreducible for any field extension $K/k$. I.e., $X$ is geometrically irreducible over $k$.

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

Lemma 33.8.4. Let $k$ be a field. Let $X$, $Y$ be schemes over $k$. Assume $X$ is geometrically irreducible over $k$. Then the projection morphism

induces a bijection between irreducible components.

**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.
$\square$

Lemma 33.8.5. Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent

$X$ is geometrically irreducible over $k$,

for every nonempty affine open $U$ the $k$-algebra $\mathcal{O}_ X(U)$ is geometrically irreducible over $k$ (see Algebra, Definition 10.47.4),

$X$ is irreducible and there exists an affine open covering $X = \bigcup U_ i$ such that each $k$-algebra $\mathcal{O}_ X(U_ i)$ is geometrically irreducible, and

there exists an open covering $X = \bigcup _{i \in I} X_ i$ with $I \not= \emptyset $ such that $X_ i$ is geometrically irreducible for each $i$ and such that $X_ i \cap X_ j \not= \emptyset $ for all $i, j \in I$.

Moreover, if $X$ is geometrically irreducible so is every nonempty open subscheme of $X$.

**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).
$\square$

Lemma 33.8.6. Let $X$ be an irreducible scheme over the field $k$. Let $\xi \in X$ be its generic point. The following are equivalent

$X$ is geometrically irreducible over $k$, and

$\kappa (\xi )$ is geometrically irreducible over $k$.

**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 $k$. $\square$

Lemma 33.8.7. Let $k'/k$ be an extension of fields. Let $X$ be a scheme over $k$. Set $X' = X_{k'}$. Assume $k$ separably algebraically closed. Then the morphism $X' \to X$ induces a bijection of irreducible components.

**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.
$\square$

Lemma 33.8.8. Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent:

$X$ is geometrically irreducible over $k$,

for every finite separable field extension $k'/k$ the scheme $X_{k'}$ is irreducible, and

$X_{\overline{k}}$ is irreducible, where $k \subset \overline{k}$ is a separable algebraic closure of $k$.

**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. $\square$

Lemma 33.8.9. Let $K/k$ be an extension of fields. Let $X$ be a scheme over $k$. For every irreducible component $T$ of $X$ the inverse image $T_ K \subset X_ K$ is a union of irreducible components of $X_ K$.

**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 $X_ K$.
$\square$

For a scheme $X$ we denote $\text{IrredComp}(X)$ the set of irreducible components of $X$.

Lemma 33.8.10. Let $K/k$ be an extension of fields. Let $X$ be a scheme over $k$. For every irreducible component $\overline{T} \subset X_ K$ the image of $\overline{T}$ in $X$ is an irreducible component in $X$. This defines a canonical map

which is surjective.

**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. $\square$

Lemma 33.8.11. Let $k$ be a field. Let $X$ be a scheme over $k$. If $X$ is irreducible and has a dense set of $k$-rational points, then $X$ is geometrically irreducible.

**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.
$\square$

Lemma 33.8.12. Let $k$ be a field, with separable algebraic closure $\overline{k}$. Let $X$ be a scheme over $k$. There is an action

with the following properties:

An element $\overline{T} \in \text{IrredComp}(X_{\overline{k}})$ is fixed by the action if and only if there exists an irreducible component $T \subset X$, which is geometrically irreducible over $k$, such that $T_{\overline{k}} = \overline{T}$.

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

\[ \xymatrix{ \text{Gal}(\overline{k}'/k') \times \text{IrredComp}(X_{\overline{k}'}) \ar[r] \ar[d] & \text{IrredComp}(X_{\overline{k}'}) \ar[d] \\ \text{Gal}(\overline{k}/k) \times \text{IrredComp}(X_{\overline{k}}) \ar[r] & \text{IrredComp}(X_{\overline{k}}) } \]is commutative (where the right vertical arrow is a bijection according to Lemma 33.8.7).

**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). $\square$

Lemma 33.8.13. Let $k$ be a field, with separable algebraic closure $\overline{k}$. Let $X$ be a scheme over $k$. The fibres of the map

of Lemma 33.8.10 are exactly the orbits of $\text{Gal}(\overline{k}/k)$ under the action of Lemma 33.8.12.

**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.
$\square$

Lemma 33.8.14. Let $k$ be a field. Assume $X \to \mathop{\mathrm{Spec}}(k)$ locally of finite type. In this case

the action

\[ \text{Gal}(\overline{k}/k)^{opp} \times \text{IrredComp}(X_{\overline{k}}) \longrightarrow \text{IrredComp}(X_{\overline{k}}) \]is continuous if we give $\text{IrredComp}(X_{\overline{k}})$ the discrete topology,

every irreducible component of $X_{\overline{k}}$ can be defined over a finite extension of $k$, and

given any irreducible component $T \subset X$ the scheme $T_{\overline{k}}$ is a finite union of irreducible components of $X_{\overline{k}}$ which are all in the same $\text{Gal}(\overline{k}/k)$-orbit.

**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. $\square$

Lemma 33.8.15. Let $k$ be a field. Let $X \to \mathop{\mathrm{Spec}}(k)$ be locally of finite type. Assume $X$ has finitely many irreducible components. Then there exists a finite separable extension $k'/k$ such that every irreducible component of $X_{k'}$ is geometrically irreducible over $k'$.

**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.
$\square$

Lemma 33.8.16. Let $X$ be a scheme over the field $k$. Assume $X$ has finitely many irreducible components which are all geometrically irreducible. Then $X$ has finitely many connected components each of which is geometrically connected.

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

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