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Trivial intersection of all kernel implies the conductor is the least common multiple
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Let $ L/K$ be a finite abelian extension with Galois group $ G $, and let $ mathfrakf(L/K) $ denote the conductor. For a Dirichlet character $ chi: I^S to mathbbC^times $, there is a unique $ chi_1 $ defined on the smallest $ S $ equivalent to $ chi $, called a primitive character. The smallest modulus $ mathfrakm $ such that $ chi_1 $ is zero on $ K_m,1 $ is called the conductor $ mathfrakf(chi) $ of $ chi $. We have $$ mathfrakf(L/K) = textlcm_chi (mathfrakf(chi circ Psi_L/K)), $$
where $ Psi_L/K $ is the Artin map and $ chi $ ranges over all characters $ chi: G to mathbbC^times $. The proof of this fact is one line: clearly $ bigcap textKer (chi: G to mathbbC^times) = 0 $, from which the result follows. I'm not sure how this implies that the modulus of $ L/K$ is the least common multiple of this expression over all $ chi $. It would be great if someone can elaborate this, thank you.
class-field-theory
asked Jul 18 at 17:21
kgs
1707
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Let $ L/K$ be a finite abelian extension with Galois group $ G $, and let $ mathfrakf(L/K) $ denote the conductor. For a Dirichlet character $ chi: I^S to mathbbC^times $, there is a unique $ chi_1 $ defined on the smallest $ S $ equivalent to $ chi $, called a primitive character. The smallest modulus $ mathfrakm $ such that $ chi_1 $ is zero on $ K_m,1 $ is called the conductor $ mathfrakf(chi) $ of $ chi $. We have $$ mathfrakf(L/K) = textlcm_chi (mathfrakf(chi circ Psi_L/K)), $$
where $ Psi_L/K $ is the Artin map and $ chi $ ranges over all characters $ chi: G to mathbbC^times $. The proof of this fact is one line: clearly $ bigcap textKer (chi: G to mathbbC^times) = 0 $, from which the result follows. I'm not sure how this implies that the modulus of $ L/K$ is the least common multiple of this expression over all $ chi $. It would be great if someone can elaborate this, thank you.
class-field-theory
asked Jul 18 at 17:21
kgs
1707
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $ L/K$ be a finite abelian extension with Galois group $ G $, and let $ mathfrakf(L/K) $ denote the conductor. For a Dirichlet character $ chi: I^S to mathbbC^times $, there is a unique $ chi_1 $ defined on the smallest $ S $ equivalent to $ chi $, called a primitive character. The smallest modulus $ mathfrakm $ such that $ chi_1 $ is zero on $ K_m,1 $ is called the conductor $ mathfrakf(chi) $ of $ chi $. We have $$ mathfrakf(L/K) = textlcm_chi (mathfrakf(chi circ Psi_L/K)), $$
where $ Psi_L/K $ is the Artin map and $ chi $ ranges over all characters $ chi: G to mathbbC^times $. The proof of this fact is one line: clearly $ bigcap textKer (chi: G to mathbbC^times) = 0 $, from which the result follows. I'm not sure how this implies that the modulus of $ L/K$ is the least common multiple of this expression over all $ chi $. It would be great if someone can elaborate this, thank you.
class-field-theory
asked Jul 18 at 17:21
kgs
1707
Let $ L/K$ be a finite abelian extension with Galois group $ G $, and let $ mathfrakf(L/K) $ denote the conductor. For a Dirichlet character $ chi: I^S to mathbbC^times $, there is a unique $ chi_1 $ defined on the smallest $ S $ equivalent to $ chi $, called a primitive character. The smallest modulus $ mathfrakm $ such that $ chi_1 $ is zero on $ K_m,1 $ is called the conductor $ mathfrakf(chi) $ of $ chi $. We have $$ mathfrakf(L/K) = textlcm_chi (mathfrakf(chi circ Psi_L/K)), $$
where $ Psi_L/K $ is the Artin map and $ chi $ ranges over all characters $ chi: G to mathbbC^times $. The proof of this fact is one line: clearly $ bigcap textKer (chi: G to mathbbC^times) = 0 $, from which the result follows. I'm not sure how this implies that the modulus of $ L/K$ is the least common multiple of this expression over all $ chi $. It would be great if someone can elaborate this, thank you.
class-field-theory
asked Jul 18 at 17:21
kgs
1707
asked Jul 18 at 17:21
kgs
1707
asked Jul 18 at 17:21
kgs
1707
asked Jul 18 at 17:21
kgs
1707
1707
add a comment |Â
add a comment |Â
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