Why is, for a group scheme of finite type, “smooth” (resp. irreducible) equivalent to “geometrically reduced” (resp. geometrically irreducible)?Definition of abelian varietyPro-affine varieties over a local fieldIs the pushout of smooth varieties along a smooth closed subvariety again a variety?If $X,Y$ are regular and of finite type over $S$, can $Xtimes _S Y$ be embedded into a regular $S$-scheme? Conjugate surfaces: informations about the orbitsExamples of étale covers of arithmetic surfaces$Bbb A^1$-Localisation of Schemes, and $Bbb A^1$-Rigid SchemesAlbanese variety over non-perfect fieldsGalois invariant line bundles on a product of varietiesschemes vs varieties in abelian varieties and maximal subscheme where line bundle is trivial
Why is, for a group scheme of finite type, “smooth” (resp. irreducible) equivalent to “geometrically reduced” (resp. geometrically irreducible)?
Definition of abelian varietyPro-affine varieties over a local fieldIs the pushout of smooth varieties along a smooth closed subvariety again a variety?If $X,Y$ are regular and of finite type over $S$, can $Xtimes _S Y$ be embedded into a regular $S$-scheme? Conjugate surfaces: informations about the orbitsExamples of étale covers of arithmetic surfaces$Bbb A^1$-Localisation of Schemes, and $Bbb A^1$-Rigid SchemesAlbanese variety over non-perfect fieldsGalois invariant line bundles on a product of varietiesschemes vs varieties in abelian varieties and maximal subscheme where line bundle is trivial
$begingroup$
I have some questions about two statements from Bosch's "Algebraic Geometry and Commutative Algebra" about algebraic varieties (page 479). Since I still don't have the permission to add images I quote the relevant excerpt:
...The notion of properness has been introduced in 9.5/4. It means that the
structural morphism $p: A to Spec(K)$ is of finite type, separated, and universally
closed. For the property of smoothness see 8.5/1. It follows from 8.5/15
in conjunction with 2.4/19 that all stalks $mathcalO_A,x$ of a smooth $K$-group scheme $A$
are integral domains. Since abelian varieties are required to be irreducible, they
give rise to integral schemes. Also let us mention that for $K$-group schemes of
finite type smooth is equivalent to geometrically reduced, which means that all
stalks of the structure sheaf of $A×_K barK$ are reduced. In addition, let us point out
that for $K$-group schemes of finite type the property irreducible can be checked
after base change with $barK/K$ so that we may replace irreducible by geometrically
irreducible...
We fix an abelian variety $A$ over field $K$. By definition an abelian variety over $K$ is a proper smooth $K$-group scheme that is irreducible.
Following two questions:
Why is for a $K$-group scheme of finite type smooth equivalent to geometrically reduced?
Why under same conditions as in 1. (so $K$-group scheme of finite type) the property irreducible is equivlaent to geometrically irreducible?
Remark: Here I previously asked this question in MSE: https://math.stackexchange.com/questions/3136827/abelian-varieties
ag.algebraic-geometry abelian-varieties schemes
edited Mar 7 at 19:51
Qfwfq
10.3k1085171
asked Mar 7 at 18:57
Karl_PeterKarl_Peter
926
$endgroup$
add a comment |
$begingroup$
I have some questions about two statements from Bosch's "Algebraic Geometry and Commutative Algebra" about algebraic varieties (page 479). Since I still don't have the permission to add images I quote the relevant excerpt:
...The notion of properness has been introduced in 9.5/4. It means that the
structural morphism $p: A to Spec(K)$ is of finite type, separated, and universally
closed. For the property of smoothness see 8.5/1. It follows from 8.5/15
in conjunction with 2.4/19 that all stalks $mathcalO_A,x$ of a smooth $K$-group scheme $A$
are integral domains. Since abelian varieties are required to be irreducible, they
give rise to integral schemes. Also let us mention that for $K$-group schemes of
finite type smooth is equivalent to geometrically reduced, which means that all
stalks of the structure sheaf of $A×_K barK$ are reduced. In addition, let us point out
that for $K$-group schemes of finite type the property irreducible can be checked
after base change with $barK/K$ so that we may replace irreducible by geometrically
irreducible...
We fix an abelian variety $A$ over field $K$. By definition an abelian variety over $K$ is a proper smooth $K$-group scheme that is irreducible.
Following two questions:
Why is for a $K$-group scheme of finite type smooth equivalent to geometrically reduced?
Why under same conditions as in 1. (so $K$-group scheme of finite type) the property irreducible is equivlaent to geometrically irreducible?
Remark: Here I previously asked this question in MSE: https://math.stackexchange.com/questions/3136827/abelian-varieties
ag.algebraic-geometry abelian-varieties schemes
edited Mar 7 at 19:51
Qfwfq
10.3k1085171
asked Mar 7 at 18:57
Karl_PeterKarl_Peter
926
$endgroup$
$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
Mar 7 at 19:43
add a comment |
$begingroup$
I have some questions about two statements from Bosch's "Algebraic Geometry and Commutative Algebra" about algebraic varieties (page 479). Since I still don't have the permission to add images I quote the relevant excerpt:
...The notion of properness has been introduced in 9.5/4. It means that the
structural morphism $p: A to Spec(K)$ is of finite type, separated, and universally
closed. For the property of smoothness see 8.5/1. It follows from 8.5/15
in conjunction with 2.4/19 that all stalks $mathcalO_A,x$ of a smooth $K$-group scheme $A$
are integral domains. Since abelian varieties are required to be irreducible, they
give rise to integral schemes. Also let us mention that for $K$-group schemes of
finite type smooth is equivalent to geometrically reduced, which means that all
stalks of the structure sheaf of $A×_K barK$ are reduced. In addition, let us point out
that for $K$-group schemes of finite type the property irreducible can be checked
after base change with $barK/K$ so that we may replace irreducible by geometrically
irreducible...
We fix an abelian variety $A$ over field $K$. By definition an abelian variety over $K$ is a proper smooth $K$-group scheme that is irreducible.
Following two questions:
Why is for a $K$-group scheme of finite type smooth equivalent to geometrically reduced?
Why under same conditions as in 1. (so $K$-group scheme of finite type) the property irreducible is equivlaent to geometrically irreducible?
Remark: Here I previously asked this question in MSE: https://math.stackexchange.com/questions/3136827/abelian-varieties
ag.algebraic-geometry abelian-varieties schemes
edited Mar 7 at 19:51
Qfwfq
10.3k1085171
asked Mar 7 at 18:57
Karl_PeterKarl_Peter
926
$endgroup$
I have some questions about two statements from Bosch's "Algebraic Geometry and Commutative Algebra" about algebraic varieties (page 479). Since I still don't have the permission to add images I quote the relevant excerpt:
...The notion of properness has been introduced in 9.5/4. It means that the
structural morphism $p: A to Spec(K)$ is of finite type, separated, and universally
closed. For the property of smoothness see 8.5/1. It follows from 8.5/15
in conjunction with 2.4/19 that all stalks $mathcalO_A,x$ of a smooth $K$-group scheme $A$
are integral domains. Since abelian varieties are required to be irreducible, they
give rise to integral schemes. Also let us mention that for $K$-group schemes of
finite type smooth is equivalent to geometrically reduced, which means that all
stalks of the structure sheaf of $A×_K barK$ are reduced. In addition, let us point out
that for $K$-group schemes of finite type the property irreducible can be checked
after base change with $barK/K$ so that we may replace irreducible by geometrically
irreducible...
We fix an abelian variety $A$ over field $K$. By definition an abelian variety over $K$ is a proper smooth $K$-group scheme that is irreducible.
Following two questions:
Why is for a $K$-group scheme of finite type smooth equivalent to geometrically reduced?
Why under same conditions as in 1. (so $K$-group scheme of finite type) the property irreducible is equivlaent to geometrically irreducible?
Remark: Here I previously asked this question in MSE: https://math.stackexchange.com/questions/3136827/abelian-varieties
ag.algebraic-geometry abelian-varieties schemes
ag.algebraic-geometry abelian-varieties schemes
edited Mar 7 at 19:51
Qfwfq
10.3k1085171
asked Mar 7 at 18:57
Karl_PeterKarl_Peter
926
edited Mar 7 at 19:51
Qfwfq
10.3k1085171
asked Mar 7 at 18:57
Karl_PeterKarl_Peter
926
edited Mar 7 at 19:51
Qfwfq
10.3k1085171
edited Mar 7 at 19:51
Qfwfq
10.3k1085171
edited Mar 7 at 19:51
Qfwfq
10.3k1085171
10.3k1085171
asked Mar 7 at 18:57
Karl_PeterKarl_Peter
926
asked Mar 7 at 18:57
Karl_PeterKarl_Peter
926
asked Mar 7 at 18:57
Karl_PeterKarl_Peter
926
926
$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
Mar 7 at 19:43
add a comment |
$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
Mar 7 at 19:43
$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
Mar 7 at 19:43
$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
Mar 7 at 19:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $G/K$ be a group scheme of finite type.
$G/K$ is smooth if and only if $bar G / bar K$ is smooth. Suppose $bar G$ is reduced, then it has a smooth $bar K$-point $x$ (because we are over an algebraically closed field). But $bar G(bar K)$ acts transitively on itself, so now every closed point of $bar G$ is smooth, so $bar G$ is smooth. (And of course if $bar G$ is smooth then it is reduced.)
The point is that $G$ comes with a section, the neutral element $ein G(K)$. Suppose that $bar G$ is reducible, then since $bar G^rm red$ is reduced and hence smooth, we see that $bar G$ is disconnected. If $bar G^circ$ is the connected component of the neutral element $e$, then since the Galois group $rm Gal(bar K/K)$ acts on $bar G$ preserving $e$, it has to preserve $bar G^circ$, and so $bar G^circ$ descends to give a component of $G$, so $G$ is disconnected.
answered Mar 7 at 19:18
Piotr AchingerPiotr Achinger
8,40212854
$endgroup$
$begingroup$
Thank you for your answer. One penible question: When you talk about the $rm Gal(bar K/K)$-action on $bar G = G otimes barK$ do you implicitely mean the action on "points" $barG(barK)= Hom(barK, barG)$ via composing $phi mapsto phi circ g$ for a $g in rm Gal(bar K/K)$ or the action via "base change" namely that $g$ induces automorphism on $barG$ via $id otimes g$?
$endgroup$
– Karl_Peter
Mar 7 at 21:06
$begingroup$
@Karl_Peter I think the latter action induces the former on $bar K$-points. Here it is enough to say that the profinite group $rm Gal(bar K/K)$ acts continuously on the underlying topological space $|bar G|$ of $bar G$ (or just the subspace of closed points $bar G(bar K)$) and the quotient space is identified with $|G|$.
$endgroup$
– Piotr Achinger
Mar 7 at 22:56
$begingroup$
But what is concretely the "canonical" action of $rm Gal(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
Mar 7 at 23:36
$begingroup$
The argument in the answer that the identity component of $G$ over the algebraic closure of $K$ is defined over $K$ applies only in characteristic zero. Otherwise the proof is more difficult.
$endgroup$
– anon
Mar 9 at 7:01
1
$begingroup$
@Karl_Peter See Section b of Chapter 1 of Milne: Algebraic Groups, The theory of group schemes of finite type over a field, Cambridge UP, 2017.
$endgroup$
– anon
Mar 9 at 19:20
|
show 3 more comments
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $G/K$ be a group scheme of finite type.
$G/K$ is smooth if and only if $bar G / bar K$ is smooth. Suppose $bar G$ is reduced, then it has a smooth $bar K$-point $x$ (because we are over an algebraically closed field). But $bar G(bar K)$ acts transitively on itself, so now every closed point of $bar G$ is smooth, so $bar G$ is smooth. (And of course if $bar G$ is smooth then it is reduced.)
The point is that $G$ comes with a section, the neutral element $ein G(K)$. Suppose that $bar G$ is reducible, then since $bar G^rm red$ is reduced and hence smooth, we see that $bar G$ is disconnected. If $bar G^circ$ is the connected component of the neutral element $e$, then since the Galois group $rm Gal(bar K/K)$ acts on $bar G$ preserving $e$, it has to preserve $bar G^circ$, and so $bar G^circ$ descends to give a component of $G$, so $G$ is disconnected.
answered Mar 7 at 19:18
Piotr AchingerPiotr Achinger
8,40212854
$endgroup$
$begingroup$
Thank you for your answer. One penible question: When you talk about the $rm Gal(bar K/K)$-action on $bar G = G otimes barK$ do you implicitely mean the action on "points" $barG(barK)= Hom(barK, barG)$ via composing $phi mapsto phi circ g$ for a $g in rm Gal(bar K/K)$ or the action via "base change" namely that $g$ induces automorphism on $barG$ via $id otimes g$?
$endgroup$
– Karl_Peter
Mar 7 at 21:06
$begingroup$
@Karl_Peter I think the latter action induces the former on $bar K$-points. Here it is enough to say that the profinite group $rm Gal(bar K/K)$ acts continuously on the underlying topological space $|bar G|$ of $bar G$ (or just the subspace of closed points $bar G(bar K)$) and the quotient space is identified with $|G|$.
$endgroup$
– Piotr Achinger
Mar 7 at 22:56
$begingroup$
But what is concretely the "canonical" action of $rm Gal(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
Mar 7 at 23:36
$begingroup$
The argument in the answer that the identity component of $G$ over the algebraic closure of $K$ is defined over $K$ applies only in characteristic zero. Otherwise the proof is more difficult.
$endgroup$
– anon
Mar 9 at 7:01
1
$begingroup$
@Karl_Peter See Section b of Chapter 1 of Milne: Algebraic Groups, The theory of group schemes of finite type over a field, Cambridge UP, 2017.
$endgroup$
– anon
Mar 9 at 19:20
|
show 3 more comments
$begingroup$
Let $G/K$ be a group scheme of finite type.
$G/K$ is smooth if and only if $bar G / bar K$ is smooth. Suppose $bar G$ is reduced, then it has a smooth $bar K$-point $x$ (because we are over an algebraically closed field). But $bar G(bar K)$ acts transitively on itself, so now every closed point of $bar G$ is smooth, so $bar G$ is smooth. (And of course if $bar G$ is smooth then it is reduced.)
The point is that $G$ comes with a section, the neutral element $ein G(K)$. Suppose that $bar G$ is reducible, then since $bar G^rm red$ is reduced and hence smooth, we see that $bar G$ is disconnected. If $bar G^circ$ is the connected component of the neutral element $e$, then since the Galois group $rm Gal(bar K/K)$ acts on $bar G$ preserving $e$, it has to preserve $bar G^circ$, and so $bar G^circ$ descends to give a component of $G$, so $G$ is disconnected.
answered Mar 7 at 19:18
Piotr AchingerPiotr Achinger
8,40212854
$endgroup$
$begingroup$
Thank you for your answer. One penible question: When you talk about the $rm Gal(bar K/K)$-action on $bar G = G otimes barK$ do you implicitely mean the action on "points" $barG(barK)= Hom(barK, barG)$ via composing $phi mapsto phi circ g$ for a $g in rm Gal(bar K/K)$ or the action via "base change" namely that $g$ induces automorphism on $barG$ via $id otimes g$?
$endgroup$
– Karl_Peter
Mar 7 at 21:06
$begingroup$
@Karl_Peter I think the latter action induces the former on $bar K$-points. Here it is enough to say that the profinite group $rm Gal(bar K/K)$ acts continuously on the underlying topological space $|bar G|$ of $bar G$ (or just the subspace of closed points $bar G(bar K)$) and the quotient space is identified with $|G|$.
$endgroup$
– Piotr Achinger
Mar 7 at 22:56
$begingroup$
But what is concretely the "canonical" action of $rm Gal(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
Mar 7 at 23:36
$begingroup$
The argument in the answer that the identity component of $G$ over the algebraic closure of $K$ is defined over $K$ applies only in characteristic zero. Otherwise the proof is more difficult.
$endgroup$
– anon
Mar 9 at 7:01
1
$begingroup$
@Karl_Peter See Section b of Chapter 1 of Milne: Algebraic Groups, The theory of group schemes of finite type over a field, Cambridge UP, 2017.
$endgroup$
– anon
Mar 9 at 19:20
|
show 3 more comments
$begingroup$
Let $G/K$ be a group scheme of finite type.
$G/K$ is smooth if and only if $bar G / bar K$ is smooth. Suppose $bar G$ is reduced, then it has a smooth $bar K$-point $x$ (because we are over an algebraically closed field). But $bar G(bar K)$ acts transitively on itself, so now every closed point of $bar G$ is smooth, so $bar G$ is smooth. (And of course if $bar G$ is smooth then it is reduced.)
The point is that $G$ comes with a section, the neutral element $ein G(K)$. Suppose that $bar G$ is reducible, then since $bar G^rm red$ is reduced and hence smooth, we see that $bar G$ is disconnected. If $bar G^circ$ is the connected component of the neutral element $e$, then since the Galois group $rm Gal(bar K/K)$ acts on $bar G$ preserving $e$, it has to preserve $bar G^circ$, and so $bar G^circ$ descends to give a component of $G$, so $G$ is disconnected.
answered Mar 7 at 19:18
Piotr AchingerPiotr Achinger
8,40212854
$endgroup$
Let $G/K$ be a group scheme of finite type.
$G/K$ is smooth if and only if $bar G / bar K$ is smooth. Suppose $bar G$ is reduced, then it has a smooth $bar K$-point $x$ (because we are over an algebraically closed field). But $bar G(bar K)$ acts transitively on itself, so now every closed point of $bar G$ is smooth, so $bar G$ is smooth. (And of course if $bar G$ is smooth then it is reduced.)
The point is that $G$ comes with a section, the neutral element $ein G(K)$. Suppose that $bar G$ is reducible, then since $bar G^rm red$ is reduced and hence smooth, we see that $bar G$ is disconnected. If $bar G^circ$ is the connected component of the neutral element $e$, then since the Galois group $rm Gal(bar K/K)$ acts on $bar G$ preserving $e$, it has to preserve $bar G^circ$, and so $bar G^circ$ descends to give a component of $G$, so $G$ is disconnected.
answered Mar 7 at 19:18
Piotr AchingerPiotr Achinger
8,40212854
answered Mar 7 at 19:18
Piotr AchingerPiotr Achinger
8,40212854
answered Mar 7 at 19:18
Piotr AchingerPiotr Achinger
8,40212854
answered Mar 7 at 19:18
Piotr AchingerPiotr Achinger
8,40212854
8,40212854
$begingroup$
Thank you for your answer. One penible question: When you talk about the $rm Gal(bar K/K)$-action on $bar G = G otimes barK$ do you implicitely mean the action on "points" $barG(barK)= Hom(barK, barG)$ via composing $phi mapsto phi circ g$ for a $g in rm Gal(bar K/K)$ or the action via "base change" namely that $g$ induces automorphism on $barG$ via $id otimes g$?
$endgroup$
– Karl_Peter
Mar 7 at 21:06
$begingroup$
@Karl_Peter I think the latter action induces the former on $bar K$-points. Here it is enough to say that the profinite group $rm Gal(bar K/K)$ acts continuously on the underlying topological space $|bar G|$ of $bar G$ (or just the subspace of closed points $bar G(bar K)$) and the quotient space is identified with $|G|$.
$endgroup$
– Piotr Achinger
Mar 7 at 22:56
$begingroup$
But what is concretely the "canonical" action of $rm Gal(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
Mar 7 at 23:36
$begingroup$
The argument in the answer that the identity component of $G$ over the algebraic closure of $K$ is defined over $K$ applies only in characteristic zero. Otherwise the proof is more difficult.
$endgroup$
– anon
Mar 9 at 7:01
1
$begingroup$
@Karl_Peter See Section b of Chapter 1 of Milne: Algebraic Groups, The theory of group schemes of finite type over a field, Cambridge UP, 2017.
$endgroup$
– anon
Mar 9 at 19:20
|
show 3 more comments
$begingroup$
Thank you for your answer. One penible question: When you talk about the $rm Gal(bar K/K)$-action on $bar G = G otimes barK$ do you implicitely mean the action on "points" $barG(barK)= Hom(barK, barG)$ via composing $phi mapsto phi circ g$ for a $g in rm Gal(bar K/K)$ or the action via "base change" namely that $g$ induces automorphism on $barG$ via $id otimes g$?
$endgroup$
– Karl_Peter
Mar 7 at 21:06
$begingroup$
@Karl_Peter I think the latter action induces the former on $bar K$-points. Here it is enough to say that the profinite group $rm Gal(bar K/K)$ acts continuously on the underlying topological space $|bar G|$ of $bar G$ (or just the subspace of closed points $bar G(bar K)$) and the quotient space is identified with $|G|$.
$endgroup$
– Piotr Achinger
Mar 7 at 22:56
$begingroup$
But what is concretely the "canonical" action of $rm Gal(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
Mar 7 at 23:36
$begingroup$
The argument in the answer that the identity component of $G$ over the algebraic closure of $K$ is defined over $K$ applies only in characteristic zero. Otherwise the proof is more difficult.
$endgroup$
– anon
Mar 9 at 7:01
1
$begingroup$
@Karl_Peter See Section b of Chapter 1 of Milne: Algebraic Groups, The theory of group schemes of finite type over a field, Cambridge UP, 2017.
$endgroup$
– anon
Mar 9 at 19:20
$begingroup$
Thank you for your answer. One penible question: When you talk about the $rm Gal(bar K/K)$-action on $bar G = G otimes barK$ do you implicitely mean the action on "points" $barG(barK)= Hom(barK, barG)$ via composing $phi mapsto phi circ g$ for a $g in rm Gal(bar K/K)$ or the action via "base change" namely that $g$ induces automorphism on $barG$ via $id otimes g$?
$endgroup$
– Karl_Peter
Mar 7 at 21:06
$begingroup$
Thank you for your answer. One penible question: When you talk about the $rm Gal(bar K/K)$-action on $bar G = G otimes barK$ do you implicitely mean the action on "points" $barG(barK)= Hom(barK, barG)$ via composing $phi mapsto phi circ g$ for a $g in rm Gal(bar K/K)$ or the action via "base change" namely that $g$ induces automorphism on $barG$ via $id otimes g$?
$endgroup$
– Karl_Peter
Mar 7 at 21:06
$begingroup$
@Karl_Peter I think the latter action induces the former on $bar K$-points. Here it is enough to say that the profinite group $rm Gal(bar K/K)$ acts continuously on the underlying topological space $|bar G|$ of $bar G$ (or just the subspace of closed points $bar G(bar K)$) and the quotient space is identified with $|G|$.
$endgroup$
– Piotr Achinger
Mar 7 at 22:56
$begingroup$
@Karl_Peter I think the latter action induces the former on $bar K$-points. Here it is enough to say that the profinite group $rm Gal(bar K/K)$ acts continuously on the underlying topological space $|bar G|$ of $bar G$ (or just the subspace of closed points $bar G(bar K)$) and the quotient space is identified with $|G|$.
$endgroup$
– Piotr Achinger
Mar 7 at 22:56
$begingroup$
But what is concretely the "canonical" action of $rm Gal(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
Mar 7 at 23:36
$begingroup$
But what is concretely the "canonical" action of $rm Gal(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
Mar 7 at 23:36
$begingroup$
The argument in the answer that the identity component of $G$ over the algebraic closure of $K$ is defined over $K$ applies only in characteristic zero. Otherwise the proof is more difficult.
$endgroup$
– anon
Mar 9 at 7:01
$begingroup$
The argument in the answer that the identity component of $G$ over the algebraic closure of $K$ is defined over $K$ applies only in characteristic zero. Otherwise the proof is more difficult.
$endgroup$
– anon
Mar 9 at 7:01
1
1
$begingroup$
@Karl_Peter See Section b of Chapter 1 of Milne: Algebraic Groups, The theory of group schemes of finite type over a field, Cambridge UP, 2017.
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– anon
Mar 9 at 19:20
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@Karl_Peter See Section b of Chapter 1 of Milne: Algebraic Groups, The theory of group schemes of finite type over a field, Cambridge UP, 2017.
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– anon
Mar 9 at 19:20
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I took the liberty of changing the title of your question.
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– Piotr Achinger
Mar 7 at 19:43