Is there any conditions on a finite abelian group so that it cannot be class group of any number field?Is there a “purely algebraic” proof of the finiteness of the class number?Cohen-Lenstra Heuristics referenceIs there an excplicit number field of definition for an Abelian Variety $A/mathbbC$ with CM?Freeness of the group of principal ideals of a number fieldclass number of prime degree field with prime conductorrelation between class number of an algebraic number field and its Galois closureIs every group an ideal class group of a number field?class field theory for K-groups of number fieldsClass field theory and the class groupIs there an elementary proof that there are infinitely many primes that are *not* completely split in an abelian extension?
Is there any conditions on a finite abelian group so that it cannot be class group of any number field?
Is there a “purely algebraic” proof of the finiteness of the class number?Cohen-Lenstra Heuristics referenceIs there an excplicit number field of definition for an Abelian Variety $A/mathbbC$ with CM?Freeness of the group of principal ideals of a number fieldclass number of prime degree field with prime conductorrelation between class number of an algebraic number field and its Galois closureIs every group an ideal class group of a number field?class field theory for K-groups of number fieldsClass field theory and the class groupIs there an elementary proof that there are infinitely many primes that are *not* completely split in an abelian extension?
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The Cohen-Lenstra paper says the probability that the odd part of a class group being cyclic is close to 0.98. So I was thinking: can we find any conditions on a finite abelian group so that it cannot be a class group of any number field?
algebraic-number-theory
edited 6 hours ago
YCor
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asked 8 hours ago
user594846user594846
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The Cohen-Lenstra paper says the probability that the odd part of a class group being cyclic is close to 0.98. So I was thinking: can we find any conditions on a finite abelian group so that it cannot be a class group of any number field?
algebraic-number-theory
edited 6 hours ago
YCor
30.2k4 gold badges91 silver badges146 bronze badges
asked 8 hours ago
user594846user594846
362 bronze badges
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1
$begingroup$
related discussion math.stackexchange.com/questions/10949/…
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– Matthias Wendt
8 hours ago
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@YCor I'd think "old part" probably should be "odd part"...
$endgroup$
– paul garrett
6 hours ago
1
$begingroup$
Actually, the passage from Cohen--Lenstra that the question refers to concerns class groups of imaginary quadratic fields. The paper makes no statements about the statistics of class groups of all number fields.
$endgroup$
– Alex B.
5 hours ago
add a comment |
$begingroup$
The Cohen-Lenstra paper says the probability that the odd part of a class group being cyclic is close to 0.98. So I was thinking: can we find any conditions on a finite abelian group so that it cannot be a class group of any number field?
algebraic-number-theory
edited 6 hours ago
YCor
30.2k4 gold badges91 silver badges146 bronze badges
asked 8 hours ago
user594846user594846
362 bronze badges
$endgroup$
The Cohen-Lenstra paper says the probability that the odd part of a class group being cyclic is close to 0.98. So I was thinking: can we find any conditions on a finite abelian group so that it cannot be a class group of any number field?
algebraic-number-theory
algebraic-number-theory
edited 6 hours ago
YCor
30.2k4 gold badges91 silver badges146 bronze badges
asked 8 hours ago
user594846user594846
362 bronze badges
edited 6 hours ago
YCor
30.2k4 gold badges91 silver badges146 bronze badges
asked 8 hours ago
user594846user594846
362 bronze badges
edited 6 hours ago
YCor
30.2k4 gold badges91 silver badges146 bronze badges
edited 6 hours ago
YCor
30.2k4 gold badges91 silver badges146 bronze badges
edited 6 hours ago
YCor
30.2k4 gold badges91 silver badges146 bronze badges
30.2k4 gold badges91 silver badges146 bronze badges
asked 8 hours ago
user594846user594846
362 bronze badges
asked 8 hours ago
user594846user594846
362 bronze badges
asked 8 hours ago
user594846user594846
362 bronze badges
362 bronze badges
1
$begingroup$
related discussion math.stackexchange.com/questions/10949/…
$endgroup$
– Matthias Wendt
8 hours ago
$begingroup$
@YCor I'd think "old part" probably should be "odd part"...
$endgroup$
– paul garrett
6 hours ago
1
$begingroup$
Actually, the passage from Cohen--Lenstra that the question refers to concerns class groups of imaginary quadratic fields. The paper makes no statements about the statistics of class groups of all number fields.
$endgroup$
– Alex B.
5 hours ago
add a comment |
1
$begingroup$
related discussion math.stackexchange.com/questions/10949/…
$endgroup$
– Matthias Wendt
8 hours ago
$begingroup$
@YCor I'd think "old part" probably should be "odd part"...
$endgroup$
– paul garrett
6 hours ago
1
$begingroup$
Actually, the passage from Cohen--Lenstra that the question refers to concerns class groups of imaginary quadratic fields. The paper makes no statements about the statistics of class groups of all number fields.
$endgroup$
– Alex B.
5 hours ago
1
1
$begingroup$
related discussion math.stackexchange.com/questions/10949/…
$endgroup$
– Matthias Wendt
8 hours ago
$begingroup$
related discussion math.stackexchange.com/questions/10949/…
$endgroup$
– Matthias Wendt
8 hours ago
$begingroup$
@YCor I'd think "old part" probably should be "odd part"...
$endgroup$
– paul garrett
6 hours ago
$begingroup$
@YCor I'd think "old part" probably should be "odd part"...
$endgroup$
– paul garrett
6 hours ago
1
1
$begingroup$
Actually, the passage from Cohen--Lenstra that the question refers to concerns class groups of imaginary quadratic fields. The paper makes no statements about the statistics of class groups of all number fields.
$endgroup$
– Alex B.
5 hours ago
$begingroup$
Actually, the passage from Cohen--Lenstra that the question refers to concerns class groups of imaginary quadratic fields. The paper makes no statements about the statistics of class groups of all number fields.
$endgroup$
– Alex B.
5 hours ago
add a comment |
1 Answer
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It follows from the Cohen-Lenstra heuristic that every finite abelian group is isomorphic to infinitely many class groups of real quadratic fields (even to a positive proportion of real quadratics, which is stronger), but nothing like this is known.
However, if you take the Galois action into account, then things get interesting: there are Galois modules that are not isomorphic to any class group of a Galois number field with the respective Galois group. See Corollary 4.8 and the bottom of page 17 in this paper of mine with Lenstra: https://arxiv.org/abs/1803.06903v2. You have to read a bit between the lines, but it is shown there that if $G$ is a cyclic group of order degree $58$, then there are finite $mathbbQ[G]$-modules that cannot be realised as the class group of a $G$-extension (the "almost all" in the last line of page 17 can be strengthened to "all but one, namely the one for the field $mathbbQ_zeta_59$").
Of course, there are cheap ways of doing that, by demanding that the fixed submodule is something silly, contradicting the fact that the class group of $mathbbQ$ is trivial, but that is not what is happening in our paper. For example our obstruction cannot be seen by looking at any particular $p$-Sylow of the class group.
answered 7 hours ago
Alex B.Alex B.
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$begingroup$
It follows from the Cohen-Lenstra heuristic that every finite abelian group is isomorphic to infinitely many class groups of real quadratic fields (even to a positive proportion of real quadratics, which is stronger), but nothing like this is known.
However, if you take the Galois action into account, then things get interesting: there are Galois modules that are not isomorphic to any class group of a Galois number field with the respective Galois group. See Corollary 4.8 and the bottom of page 17 in this paper of mine with Lenstra: https://arxiv.org/abs/1803.06903v2. You have to read a bit between the lines, but it is shown there that if $G$ is a cyclic group of order degree $58$, then there are finite $mathbbQ[G]$-modules that cannot be realised as the class group of a $G$-extension (the "almost all" in the last line of page 17 can be strengthened to "all but one, namely the one for the field $mathbbQ_zeta_59$").
Of course, there are cheap ways of doing that, by demanding that the fixed submodule is something silly, contradicting the fact that the class group of $mathbbQ$ is trivial, but that is not what is happening in our paper. For example our obstruction cannot be seen by looking at any particular $p$-Sylow of the class group.
answered 7 hours ago
Alex B.Alex B.
9,5101 gold badge42 silver badges74 bronze badges
$endgroup$
add a comment |
$begingroup$
It follows from the Cohen-Lenstra heuristic that every finite abelian group is isomorphic to infinitely many class groups of real quadratic fields (even to a positive proportion of real quadratics, which is stronger), but nothing like this is known.
However, if you take the Galois action into account, then things get interesting: there are Galois modules that are not isomorphic to any class group of a Galois number field with the respective Galois group. See Corollary 4.8 and the bottom of page 17 in this paper of mine with Lenstra: https://arxiv.org/abs/1803.06903v2. You have to read a bit between the lines, but it is shown there that if $G$ is a cyclic group of order degree $58$, then there are finite $mathbbQ[G]$-modules that cannot be realised as the class group of a $G$-extension (the "almost all" in the last line of page 17 can be strengthened to "all but one, namely the one for the field $mathbbQ_zeta_59$").
Of course, there are cheap ways of doing that, by demanding that the fixed submodule is something silly, contradicting the fact that the class group of $mathbbQ$ is trivial, but that is not what is happening in our paper. For example our obstruction cannot be seen by looking at any particular $p$-Sylow of the class group.
answered 7 hours ago
Alex B.Alex B.
9,5101 gold badge42 silver badges74 bronze badges
$endgroup$
add a comment |
$begingroup$
It follows from the Cohen-Lenstra heuristic that every finite abelian group is isomorphic to infinitely many class groups of real quadratic fields (even to a positive proportion of real quadratics, which is stronger), but nothing like this is known.
However, if you take the Galois action into account, then things get interesting: there are Galois modules that are not isomorphic to any class group of a Galois number field with the respective Galois group. See Corollary 4.8 and the bottom of page 17 in this paper of mine with Lenstra: https://arxiv.org/abs/1803.06903v2. You have to read a bit between the lines, but it is shown there that if $G$ is a cyclic group of order degree $58$, then there are finite $mathbbQ[G]$-modules that cannot be realised as the class group of a $G$-extension (the "almost all" in the last line of page 17 can be strengthened to "all but one, namely the one for the field $mathbbQ_zeta_59$").
Of course, there are cheap ways of doing that, by demanding that the fixed submodule is something silly, contradicting the fact that the class group of $mathbbQ$ is trivial, but that is not what is happening in our paper. For example our obstruction cannot be seen by looking at any particular $p$-Sylow of the class group.
answered 7 hours ago
Alex B.Alex B.
9,5101 gold badge42 silver badges74 bronze badges
$endgroup$
It follows from the Cohen-Lenstra heuristic that every finite abelian group is isomorphic to infinitely many class groups of real quadratic fields (even to a positive proportion of real quadratics, which is stronger), but nothing like this is known.
However, if you take the Galois action into account, then things get interesting: there are Galois modules that are not isomorphic to any class group of a Galois number field with the respective Galois group. See Corollary 4.8 and the bottom of page 17 in this paper of mine with Lenstra: https://arxiv.org/abs/1803.06903v2. You have to read a bit between the lines, but it is shown there that if $G$ is a cyclic group of order degree $58$, then there are finite $mathbbQ[G]$-modules that cannot be realised as the class group of a $G$-extension (the "almost all" in the last line of page 17 can be strengthened to "all but one, namely the one for the field $mathbbQ_zeta_59$").
Of course, there are cheap ways of doing that, by demanding that the fixed submodule is something silly, contradicting the fact that the class group of $mathbbQ$ is trivial, but that is not what is happening in our paper. For example our obstruction cannot be seen by looking at any particular $p$-Sylow of the class group.
answered 7 hours ago
Alex B.Alex B.
9,5101 gold badge42 silver badges74 bronze badges
answered 7 hours ago
Alex B.Alex B.
9,5101 gold badge42 silver badges74 bronze badges
answered 7 hours ago
Alex B.Alex B.
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answered 7 hours ago
Alex B.Alex B.
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$begingroup$
related discussion math.stackexchange.com/questions/10949/…
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– Matthias Wendt
8 hours ago
$begingroup$
@YCor I'd think "old part" probably should be "odd part"...
$endgroup$
– paul garrett
6 hours ago
1
$begingroup$
Actually, the passage from Cohen--Lenstra that the question refers to concerns class groups of imaginary quadratic fields. The paper makes no statements about the statistics of class groups of all number fields.
$endgroup$
– Alex B.
5 hours ago