Why is, for a group scheme of finite type, “smooth” (resp. irreducible) equivalent to “geometrically...
$begingroup$
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 $mathcal{O}_{A,x}$ 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 $A×_K bar{K}$ 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 $bar{K}/K$ 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 8 hours ago
Qfwfq
10.2k1083170
asked 9 hours ago
Karl_PeterKarl_Peter
463
New contributor
Karl_Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 $mathcal{O}_{A,x}$ 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 $A×_K bar{K}$ 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 $bar{K}/K$ 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 8 hours ago
Qfwfq
10.2k1083170
asked 9 hours ago
Karl_PeterKarl_Peter
463
New contributor
Karl_Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
8 hours ago
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 $mathcal{O}_{A,x}$ 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 $A×_K bar{K}$ 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 $bar{K}/K$ 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 8 hours ago
Qfwfq
10.2k1083170
asked 9 hours ago
Karl_PeterKarl_Peter
463
New contributor
Karl_Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 $mathcal{O}_{A,x}$ 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 $A×_K bar{K}$ 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 $bar{K}/K$ 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 8 hours ago
Qfwfq
10.2k1083170
asked 9 hours ago
Karl_PeterKarl_Peter
463
New contributor
Karl_Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
Qfwfq
10.2k1083170
asked 9 hours ago
Karl_PeterKarl_Peter
463
New contributor
Karl_Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
Qfwfq
10.2k1083170
edited 8 hours ago
Qfwfq
10.2k1083170
edited 8 hours ago
Qfwfq
10.2k1083170
10.2k1083170
asked 9 hours ago
Karl_PeterKarl_Peter
463
New contributor
Karl_Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Karl_PeterKarl_Peter
463
asked 9 hours ago
Karl_PeterKarl_Peter
463
463
New contributor
Karl_Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Karl_Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Karl_Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
8 hours ago
add a comment |
$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
8 hours ago
$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
8 hours ago
$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
8 hours ago
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 8 hours ago
Piotr AchingerPiotr Achinger
8,37712853
$endgroup$
$begingroup$
Thank you for your answer. One penible question: When you talk about the ${rm Gal}(bar K/K)$-action on $bar G = G otimes bar{K}$ do you implicitely mean the action on "points" $bar{G}(bar{K})= Hom(bar{K}, bar{G})$ 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 $bar{G}$ via $id otimes g$?
$endgroup$
– Karl_Peter
7 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
But what is concretely the "canonical" action of ${rm Gal}(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
4 hours ago
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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 8 hours ago
Piotr AchingerPiotr Achinger
8,37712853
$endgroup$
$begingroup$
Thank you for your answer. One penible question: When you talk about the ${rm Gal}(bar K/K)$-action on $bar G = G otimes bar{K}$ do you implicitely mean the action on "points" $bar{G}(bar{K})= Hom(bar{K}, bar{G})$ 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 $bar{G}$ via $id otimes g$?
$endgroup$
– Karl_Peter
7 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
But what is concretely the "canonical" action of ${rm Gal}(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
4 hours ago
add a comment |
$begingroup$
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 8 hours ago
Piotr AchingerPiotr Achinger
8,37712853
$endgroup$
$begingroup$
Thank you for your answer. One penible question: When you talk about the ${rm Gal}(bar K/K)$-action on $bar G = G otimes bar{K}$ do you implicitely mean the action on "points" $bar{G}(bar{K})= Hom(bar{K}, bar{G})$ 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 $bar{G}$ via $id otimes g$?
$endgroup$
– Karl_Peter
7 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
But what is concretely the "canonical" action of ${rm Gal}(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
4 hours ago
add a comment |
$begingroup$
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 8 hours ago
Piotr AchingerPiotr Achinger
8,37712853
$endgroup$
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 8 hours ago
Piotr AchingerPiotr Achinger
8,37712853
answered 8 hours ago
Piotr AchingerPiotr Achinger
8,37712853
answered 8 hours ago
Piotr AchingerPiotr Achinger
8,37712853
answered 8 hours ago
Piotr AchingerPiotr Achinger
8,37712853
8,37712853
$begingroup$
Thank you for your answer. One penible question: When you talk about the ${rm Gal}(bar K/K)$-action on $bar G = G otimes bar{K}$ do you implicitely mean the action on "points" $bar{G}(bar{K})= Hom(bar{K}, bar{G})$ 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 $bar{G}$ via $id otimes g$?
$endgroup$
– Karl_Peter
7 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
But what is concretely the "canonical" action of ${rm Gal}(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
4 hours ago
add a comment |
$begingroup$
Thank you for your answer. One penible question: When you talk about the ${rm Gal}(bar K/K)$-action on $bar G = G otimes bar{K}$ do you implicitely mean the action on "points" $bar{G}(bar{K})= Hom(bar{K}, bar{G})$ 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 $bar{G}$ via $id otimes g$?
$endgroup$
– Karl_Peter
7 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
But what is concretely the "canonical" action of ${rm Gal}(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
4 hours ago
$begingroup$
Thank you for your answer. One penible question: When you talk about the ${rm Gal}(bar K/K)$-action on $bar G = G otimes bar{K}$ do you implicitely mean the action on "points" $bar{G}(bar{K})= Hom(bar{K}, bar{G})$ 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 $bar{G}$ via $id otimes g$?
$endgroup$
– Karl_Peter
7 hours ago
$begingroup$
Thank you for your answer. One penible question: When you talk about the ${rm Gal}(bar K/K)$-action on $bar G = G otimes bar{K}$ do you implicitely mean the action on "points" $bar{G}(bar{K})= Hom(bar{K}, bar{G})$ 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 $bar{G}$ via $id otimes g$?
$endgroup$
– Karl_Peter
7 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
But what is concretely the "canonical" action of ${rm Gal}(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
4 hours ago
$begingroup$
But what is concretely the "canonical" action of ${rm Gal}(bar K/K)$ on the underlying topological space $|bar G|$?
$endgroup$
– Karl_Peter
4 hours ago
add a comment |
Karl_Peter is a new contributor. Be nice, and check out our Code of Conduct.
Karl_Peter is a new contributor. Be nice, and check out our Code of Conduct.
Karl_Peter is a new contributor. Be nice, and check out our Code of Conduct.
Karl_Peter is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I took the liberty of changing the title of your question.
$endgroup$
– Piotr Achinger
8 hours ago