$mathbb Q$-simple algebraic groups and restriction of scalars.
$begingroup$
I'm trying to understand the following statement:
Any simply connected $mathbb{Q}$-simple (algebraic) group has the form $mathbf{R}_{F/mathbb{Q}}(G)$ where $F$ is some field containing $mathbb{Q}$, $G$ is an almost simple $F$-group and $mathbf{R}_{F/mathbb{Q}}$ is the restriction of scalars functor.
This statement appears in the book 'Algebraic groups and number theory' by Platunov and Rapinchuk on page 77. It seems to be stated there as a kind of trivial observation. To me it seems like a non-trivial fact for which I'd like justification for. I can imagine trying to justify it by considering the action of the Galois group of the field extension (assuming it is Galois) and showing that the action on the factors of a $mathbb{Q}$-simple group must be transitive. Is this the right idea? Or is there something simpler?
group-theory number-theory algebraic-groups
edited Dec 5 '18 at 23:19
YCor
7,353829
asked Dec 5 '18 at 13:16
OrbitOrbit
62
$endgroup$
add a comment |
$begingroup$
I'm trying to understand the following statement:
Any simply connected $mathbb{Q}$-simple (algebraic) group has the form $mathbf{R}_{F/mathbb{Q}}(G)$ where $F$ is some field containing $mathbb{Q}$, $G$ is an almost simple $F$-group and $mathbf{R}_{F/mathbb{Q}}$ is the restriction of scalars functor.
This statement appears in the book 'Algebraic groups and number theory' by Platunov and Rapinchuk on page 77. It seems to be stated there as a kind of trivial observation. To me it seems like a non-trivial fact for which I'd like justification for. I can imagine trying to justify it by considering the action of the Galois group of the field extension (assuming it is Galois) and showing that the action on the factors of a $mathbb{Q}$-simple group must be transitive. Is this the right idea? Or is there something simpler?
group-theory number-theory algebraic-groups
edited Dec 5 '18 at 23:19
YCor
7,353829
asked Dec 5 '18 at 13:16
OrbitOrbit
62
$endgroup$
$begingroup$
As stated, it's a tautology (take $F=Q$). I think the correct statement includes "... $G$ is an almost absolutely simple $F$-group...". Also $F$ has to be a finite extension.
$endgroup$
– YCor
Dec 5 '18 at 23:21
$begingroup$
Well you are right that in Platanov and Rapinchuk they say absolutely simple. But they say that absolutely simple means 'no nontrvial normal connected subgroups', but I thought I'd also heard this property called almost simple so I thought they meant the same thing... The fact that F is a finite extension should follow from the construction.
$endgroup$
– Orbit
Dec 6 '18 at 11:42
$begingroup$
This is because there are two ways of putting the algebraic geometry language. In most modern algebraic geometry treaties, including algebraic group treaties, one considers object primarily over an algebraic closed field, and then considers things (subgroups, etc) defined over some given subfield $F$. So it can happen that there are some nontrivial normal subgroup (which is the failure of being "absolutely simple" or even "simple"), but there are no nontrivial $F$-defined normal subgroup (which is called being "$F$-simple", and also "simple"). Whence the confusion.
$endgroup$
– YCor
Dec 6 '18 at 12:33
$begingroup$
Yes I understand that this is what can happen. But I'm still not sure why the original statement is a triviality. Another way to put it: why is any $mathbb{Q}$-simple group (no nontrivial normal $mathbb{Q}$-subgroups) is just a direct product of the Galois conjugates of an absolutely simple group defined over another field $F$ (which is a finite extension of $mathbb{Q}$)?
$endgroup$
– Orbit
Dec 6 '18 at 13:38
add a comment |
$begingroup$
I'm trying to understand the following statement:
Any simply connected $mathbb{Q}$-simple (algebraic) group has the form $mathbf{R}_{F/mathbb{Q}}(G)$ where $F$ is some field containing $mathbb{Q}$, $G$ is an almost simple $F$-group and $mathbf{R}_{F/mathbb{Q}}$ is the restriction of scalars functor.
This statement appears in the book 'Algebraic groups and number theory' by Platunov and Rapinchuk on page 77. It seems to be stated there as a kind of trivial observation. To me it seems like a non-trivial fact for which I'd like justification for. I can imagine trying to justify it by considering the action of the Galois group of the field extension (assuming it is Galois) and showing that the action on the factors of a $mathbb{Q}$-simple group must be transitive. Is this the right idea? Or is there something simpler?
group-theory number-theory algebraic-groups
edited Dec 5 '18 at 23:19
YCor
7,353829
asked Dec 5 '18 at 13:16
OrbitOrbit
62
$endgroup$
I'm trying to understand the following statement:
Any simply connected $mathbb{Q}$-simple (algebraic) group has the form $mathbf{R}_{F/mathbb{Q}}(G)$ where $F$ is some field containing $mathbb{Q}$, $G$ is an almost simple $F$-group and $mathbf{R}_{F/mathbb{Q}}$ is the restriction of scalars functor.
This statement appears in the book 'Algebraic groups and number theory' by Platunov and Rapinchuk on page 77. It seems to be stated there as a kind of trivial observation. To me it seems like a non-trivial fact for which I'd like justification for. I can imagine trying to justify it by considering the action of the Galois group of the field extension (assuming it is Galois) and showing that the action on the factors of a $mathbb{Q}$-simple group must be transitive. Is this the right idea? Or is there something simpler?
group-theory number-theory algebraic-groups
group-theory number-theory algebraic-groups
edited Dec 5 '18 at 23:19
YCor
7,353829
asked Dec 5 '18 at 13:16
OrbitOrbit
62
edited Dec 5 '18 at 23:19
YCor
7,353829
asked Dec 5 '18 at 13:16
OrbitOrbit
62
edited Dec 5 '18 at 23:19
YCor
7,353829
edited Dec 5 '18 at 23:19
YCor
7,353829
edited Dec 5 '18 at 23:19
YCor
7,353829
7,353829
asked Dec 5 '18 at 13:16
OrbitOrbit
62
asked Dec 5 '18 at 13:16
OrbitOrbit
62
asked Dec 5 '18 at 13:16
OrbitOrbit
62
62
$begingroup$
As stated, it's a tautology (take $F=Q$). I think the correct statement includes "... $G$ is an almost absolutely simple $F$-group...". Also $F$ has to be a finite extension.
$endgroup$
– YCor
Dec 5 '18 at 23:21
$begingroup$
Well you are right that in Platanov and Rapinchuk they say absolutely simple. But they say that absolutely simple means 'no nontrvial normal connected subgroups', but I thought I'd also heard this property called almost simple so I thought they meant the same thing... The fact that F is a finite extension should follow from the construction.
$endgroup$
– Orbit
Dec 6 '18 at 11:42
$begingroup$
This is because there are two ways of putting the algebraic geometry language. In most modern algebraic geometry treaties, including algebraic group treaties, one considers object primarily over an algebraic closed field, and then considers things (subgroups, etc) defined over some given subfield $F$. So it can happen that there are some nontrivial normal subgroup (which is the failure of being "absolutely simple" or even "simple"), but there are no nontrivial $F$-defined normal subgroup (which is called being "$F$-simple", and also "simple"). Whence the confusion.
$endgroup$
– YCor
Dec 6 '18 at 12:33
$begingroup$
Yes I understand that this is what can happen. But I'm still not sure why the original statement is a triviality. Another way to put it: why is any $mathbb{Q}$-simple group (no nontrivial normal $mathbb{Q}$-subgroups) is just a direct product of the Galois conjugates of an absolutely simple group defined over another field $F$ (which is a finite extension of $mathbb{Q}$)?
$endgroup$
– Orbit
Dec 6 '18 at 13:38
add a comment |
$begingroup$
As stated, it's a tautology (take $F=Q$). I think the correct statement includes "... $G$ is an almost absolutely simple $F$-group...". Also $F$ has to be a finite extension.
$endgroup$
– YCor
Dec 5 '18 at 23:21
$begingroup$
Well you are right that in Platanov and Rapinchuk they say absolutely simple. But they say that absolutely simple means 'no nontrvial normal connected subgroups', but I thought I'd also heard this property called almost simple so I thought they meant the same thing... The fact that F is a finite extension should follow from the construction.
$endgroup$
– Orbit
Dec 6 '18 at 11:42
$begingroup$
This is because there are two ways of putting the algebraic geometry language. In most modern algebraic geometry treaties, including algebraic group treaties, one considers object primarily over an algebraic closed field, and then considers things (subgroups, etc) defined over some given subfield $F$. So it can happen that there are some nontrivial normal subgroup (which is the failure of being "absolutely simple" or even "simple"), but there are no nontrivial $F$-defined normal subgroup (which is called being "$F$-simple", and also "simple"). Whence the confusion.
$endgroup$
– YCor
Dec 6 '18 at 12:33
$begingroup$
Yes I understand that this is what can happen. But I'm still not sure why the original statement is a triviality. Another way to put it: why is any $mathbb{Q}$-simple group (no nontrivial normal $mathbb{Q}$-subgroups) is just a direct product of the Galois conjugates of an absolutely simple group defined over another field $F$ (which is a finite extension of $mathbb{Q}$)?
$endgroup$
– Orbit
Dec 6 '18 at 13:38
$begingroup$
As stated, it's a tautology (take $F=Q$). I think the correct statement includes "... $G$ is an almost absolutely simple $F$-group...". Also $F$ has to be a finite extension.
$endgroup$
– YCor
Dec 5 '18 at 23:21
$begingroup$
As stated, it's a tautology (take $F=Q$). I think the correct statement includes "... $G$ is an almost absolutely simple $F$-group...". Also $F$ has to be a finite extension.
$endgroup$
– YCor
Dec 5 '18 at 23:21
$begingroup$
Well you are right that in Platanov and Rapinchuk they say absolutely simple. But they say that absolutely simple means 'no nontrvial normal connected subgroups', but I thought I'd also heard this property called almost simple so I thought they meant the same thing... The fact that F is a finite extension should follow from the construction.
$endgroup$
– Orbit
Dec 6 '18 at 11:42
$begingroup$
Well you are right that in Platanov and Rapinchuk they say absolutely simple. But they say that absolutely simple means 'no nontrvial normal connected subgroups', but I thought I'd also heard this property called almost simple so I thought they meant the same thing... The fact that F is a finite extension should follow from the construction.
$endgroup$
– Orbit
Dec 6 '18 at 11:42
$begingroup$
This is because there are two ways of putting the algebraic geometry language. In most modern algebraic geometry treaties, including algebraic group treaties, one considers object primarily over an algebraic closed field, and then considers things (subgroups, etc) defined over some given subfield $F$. So it can happen that there are some nontrivial normal subgroup (which is the failure of being "absolutely simple" or even "simple"), but there are no nontrivial $F$-defined normal subgroup (which is called being "$F$-simple", and also "simple"). Whence the confusion.
$endgroup$
– YCor
Dec 6 '18 at 12:33
$begingroup$
This is because there are two ways of putting the algebraic geometry language. In most modern algebraic geometry treaties, including algebraic group treaties, one considers object primarily over an algebraic closed field, and then considers things (subgroups, etc) defined over some given subfield $F$. So it can happen that there are some nontrivial normal subgroup (which is the failure of being "absolutely simple" or even "simple"), but there are no nontrivial $F$-defined normal subgroup (which is called being "$F$-simple", and also "simple"). Whence the confusion.
$endgroup$
– YCor
Dec 6 '18 at 12:33
$begingroup$
Yes I understand that this is what can happen. But I'm still not sure why the original statement is a triviality. Another way to put it: why is any $mathbb{Q}$-simple group (no nontrivial normal $mathbb{Q}$-subgroups) is just a direct product of the Galois conjugates of an absolutely simple group defined over another field $F$ (which is a finite extension of $mathbb{Q}$)?
$endgroup$
– Orbit
Dec 6 '18 at 13:38
$begingroup$
Yes I understand that this is what can happen. But I'm still not sure why the original statement is a triviality. Another way to put it: why is any $mathbb{Q}$-simple group (no nontrivial normal $mathbb{Q}$-subgroups) is just a direct product of the Galois conjugates of an absolutely simple group defined over another field $F$ (which is a finite extension of $mathbb{Q}$)?
$endgroup$
– Orbit
Dec 6 '18 at 13:38
add a comment |
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$begingroup$
As stated, it's a tautology (take $F=Q$). I think the correct statement includes "... $G$ is an almost absolutely simple $F$-group...". Also $F$ has to be a finite extension.
$endgroup$
– YCor
Dec 5 '18 at 23:21
$begingroup$
Well you are right that in Platanov and Rapinchuk they say absolutely simple. But they say that absolutely simple means 'no nontrvial normal connected subgroups', but I thought I'd also heard this property called almost simple so I thought they meant the same thing... The fact that F is a finite extension should follow from the construction.
$endgroup$
– Orbit
Dec 6 '18 at 11:42
$begingroup$
This is because there are two ways of putting the algebraic geometry language. In most modern algebraic geometry treaties, including algebraic group treaties, one considers object primarily over an algebraic closed field, and then considers things (subgroups, etc) defined over some given subfield $F$. So it can happen that there are some nontrivial normal subgroup (which is the failure of being "absolutely simple" or even "simple"), but there are no nontrivial $F$-defined normal subgroup (which is called being "$F$-simple", and also "simple"). Whence the confusion.
$endgroup$
– YCor
Dec 6 '18 at 12:33
$begingroup$
Yes I understand that this is what can happen. But I'm still not sure why the original statement is a triviality. Another way to put it: why is any $mathbb{Q}$-simple group (no nontrivial normal $mathbb{Q}$-subgroups) is just a direct product of the Galois conjugates of an absolutely simple group defined over another field $F$ (which is a finite extension of $mathbb{Q}$)?
$endgroup$
– Orbit
Dec 6 '18 at 13:38