Artin reciprocity
$begingroup$
The Artin reciprocity says that if $L/mathbb{Q}$ is a finite abelian extension with defining modulus $m$, then the sequence of groups
$$
1to I_{L,m}to (mathbb{Z}/mmathbb{Z})^timesto Gal(L/mathbb{Q})to 1
$$
is exact.
I do not quite see why this is a substantial theorem, because it is simply saying that the Artin map (the third arrow) is surjective; $I_{L,m}$ is defined as the kernel of the Artin map, so the second arrow being injective is trivial. Also, why is it called "reciprocity"? Does this theorem have any interesting consequences?
class-field-theory
asked Dec 31 '18 at 7:47
saisai
1376
$endgroup$
add a comment |
$begingroup$
The Artin reciprocity says that if $L/mathbb{Q}$ is a finite abelian extension with defining modulus $m$, then the sequence of groups
$$
1to I_{L,m}to (mathbb{Z}/mmathbb{Z})^timesto Gal(L/mathbb{Q})to 1
$$
is exact.
I do not quite see why this is a substantial theorem, because it is simply saying that the Artin map (the third arrow) is surjective; $I_{L,m}$ is defined as the kernel of the Artin map, so the second arrow being injective is trivial. Also, why is it called "reciprocity"? Does this theorem have any interesting consequences?
class-field-theory
asked Dec 31 '18 at 7:47
saisai
1376
$endgroup$
$begingroup$
Well it's supposed to generalize quadratic reciprocity and cubic reciprocity...
$endgroup$
– Kenny Lau
Dec 31 '18 at 8:02
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@Kenny Lau: How does it generalize them?
$endgroup$
– sai
Dec 31 '18 at 8:13
2
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The most substantive part is that there is a defining modulus.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:22
$begingroup$
@Lord Shark the Unknown: Isn't that the Kronecker-Weber theorem, rather than Artin reciprocity?
$endgroup$
– sai
Dec 31 '18 at 8:34
add a comment |
$begingroup$
The Artin reciprocity says that if $L/mathbb{Q}$ is a finite abelian extension with defining modulus $m$, then the sequence of groups
$$
1to I_{L,m}to (mathbb{Z}/mmathbb{Z})^timesto Gal(L/mathbb{Q})to 1
$$
is exact.
I do not quite see why this is a substantial theorem, because it is simply saying that the Artin map (the third arrow) is surjective; $I_{L,m}$ is defined as the kernel of the Artin map, so the second arrow being injective is trivial. Also, why is it called "reciprocity"? Does this theorem have any interesting consequences?
class-field-theory
asked Dec 31 '18 at 7:47
saisai
1376
$endgroup$
The Artin reciprocity says that if $L/mathbb{Q}$ is a finite abelian extension with defining modulus $m$, then the sequence of groups
$$
1to I_{L,m}to (mathbb{Z}/mmathbb{Z})^timesto Gal(L/mathbb{Q})to 1
$$
is exact.
I do not quite see why this is a substantial theorem, because it is simply saying that the Artin map (the third arrow) is surjective; $I_{L,m}$ is defined as the kernel of the Artin map, so the second arrow being injective is trivial. Also, why is it called "reciprocity"? Does this theorem have any interesting consequences?
class-field-theory
class-field-theory
asked Dec 31 '18 at 7:47
saisai
1376
asked Dec 31 '18 at 7:47
saisai
1376
asked Dec 31 '18 at 7:47
saisai
1376
asked Dec 31 '18 at 7:47
saisai
1376
asked Dec 31 '18 at 7:47
saisai
1376
1376
$begingroup$
Well it's supposed to generalize quadratic reciprocity and cubic reciprocity...
$endgroup$
– Kenny Lau
Dec 31 '18 at 8:02
$begingroup$
@Kenny Lau: How does it generalize them?
$endgroup$
– sai
Dec 31 '18 at 8:13
2
$begingroup$
The most substantive part is that there is a defining modulus.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:22
$begingroup$
@Lord Shark the Unknown: Isn't that the Kronecker-Weber theorem, rather than Artin reciprocity?
$endgroup$
– sai
Dec 31 '18 at 8:34
add a comment |
$begingroup$
Well it's supposed to generalize quadratic reciprocity and cubic reciprocity...
$endgroup$
– Kenny Lau
Dec 31 '18 at 8:02
$begingroup$
@Kenny Lau: How does it generalize them?
$endgroup$
– sai
Dec 31 '18 at 8:13
2
$begingroup$
The most substantive part is that there is a defining modulus.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:22
$begingroup$
@Lord Shark the Unknown: Isn't that the Kronecker-Weber theorem, rather than Artin reciprocity?
$endgroup$
– sai
Dec 31 '18 at 8:34
$begingroup$
Well it's supposed to generalize quadratic reciprocity and cubic reciprocity...
$endgroup$
– Kenny Lau
Dec 31 '18 at 8:02
$begingroup$
Well it's supposed to generalize quadratic reciprocity and cubic reciprocity...
$endgroup$
– Kenny Lau
Dec 31 '18 at 8:02
$begingroup$
@Kenny Lau: How does it generalize them?
$endgroup$
– sai
Dec 31 '18 at 8:13
$begingroup$
@Kenny Lau: How does it generalize them?
$endgroup$
– sai
Dec 31 '18 at 8:13
2
2
$begingroup$
The most substantive part is that there is a defining modulus.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:22
$begingroup$
The most substantive part is that there is a defining modulus.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:22
$begingroup$
@Lord Shark the Unknown: Isn't that the Kronecker-Weber theorem, rather than Artin reciprocity?
$endgroup$
– sai
Dec 31 '18 at 8:34
$begingroup$
@Lord Shark the Unknown: Isn't that the Kronecker-Weber theorem, rather than Artin reciprocity?
$endgroup$
– sai
Dec 31 '18 at 8:34
add a comment |
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$begingroup$
Well it's supposed to generalize quadratic reciprocity and cubic reciprocity...
$endgroup$
– Kenny Lau
Dec 31 '18 at 8:02
$begingroup$
@Kenny Lau: How does it generalize them?
$endgroup$
– sai
Dec 31 '18 at 8:13
2
$begingroup$
The most substantive part is that there is a defining modulus.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 8:22
$begingroup$
@Lord Shark the Unknown: Isn't that the Kronecker-Weber theorem, rather than Artin reciprocity?
$endgroup$
– sai
Dec 31 '18 at 8:34