What is the conductor of $ mathbb Q(sqrt 2 )/mathbb Q?$ $4mathbb Z$ or $8 mathbb Z?$
2
$begingroup$
By execise 6.8 of Childress's Class Field Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considering the norm of the local field about the place correspondent with 2, it should be $4mathbb Z$ and there is contraction to the ordering theorem.
algebraic-number-theory
asked Jan 3 at 6:36
lorry lorry
112
$endgroup$
|
show 2 more comments
2
$begingroup$
By execise 6.8 of Childress's Class Field Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considering the norm of the local field about the place correspondent with 2, it should be $4mathbb Z$ and there is contraction to the ordering theorem.
algebraic-number-theory
asked Jan 3 at 6:36
lorry lorry
112
$endgroup$
1
$begingroup$
Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 6:50
$begingroup$
Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:18
$begingroup$
Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
$endgroup$
– lorry
Jan 3 at 8:09
$begingroup$
I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 10:29
$begingroup$
Let $k_1/ mathbb Q_p$ and $k_2/k_1$ be finite extensions and $Gal(k_2/k_1)$ is cyclic. Let $mathbb U_j$ denote the units of the valuation ring of $k_j$.Then $[mathbb U_1 : mathbb N_{k_2/k_1}mathbb U_2]=e(k_2/k_1)$. In this case, $[1+2mathbb Z_2:mathbb N_{k_2/mathbb Q_2}mathbb U_2]=2$, then the image of the norm map is $1+2^2mathbb Z_2$
$endgroup$
– lorry
Jan 4 at 3:34
|
show 2 more comments
2
2
2
$begingroup$
By execise 6.8 of Childress's Class Field Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considering the norm of the local field about the place correspondent with 2, it should be $4mathbb Z$ and there is contraction to the ordering theorem.
algebraic-number-theory
asked Jan 3 at 6:36
lorry lorry
112
$endgroup$
By execise 6.8 of Childress's Class Field Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considering the norm of the local field about the place correspondent with 2, it should be $4mathbb Z$ and there is contraction to the ordering theorem.
algebraic-number-theory
algebraic-number-theory
asked Jan 3 at 6:36
lorry lorry
112
asked Jan 3 at 6:36
lorry lorry
112
edited Jan 7 at 0:32
lorry
asked Jan 3 at 6:36
lorry lorry
112
asked Jan 3 at 6:36
lorry lorry
112
asked Jan 3 at 6:36
lorry lorry
112
112
1
$begingroup$
Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 6:50
$begingroup$
Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:18
$begingroup$
Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
$endgroup$
– lorry
Jan 3 at 8:09
$begingroup$
I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 10:29
$begingroup$
Let $k_1/ mathbb Q_p$ and $k_2/k_1$ be finite extensions and $Gal(k_2/k_1)$ is cyclic. Let $mathbb U_j$ denote the units of the valuation ring of $k_j$.Then $[mathbb U_1 : mathbb N_{k_2/k_1}mathbb U_2]=e(k_2/k_1)$. In this case, $[1+2mathbb Z_2:mathbb N_{k_2/mathbb Q_2}mathbb U_2]=2$, then the image of the norm map is $1+2^2mathbb Z_2$
$endgroup$
– lorry
Jan 4 at 3:34
|
show 2 more comments
1
$begingroup$
Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 6:50
$begingroup$
Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:18
$begingroup$
Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
$endgroup$
– lorry
Jan 3 at 8:09
$begingroup$
I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 10:29
$begingroup$
Let $k_1/ mathbb Q_p$ and $k_2/k_1$ be finite extensions and $Gal(k_2/k_1)$ is cyclic. Let $mathbb U_j$ denote the units of the valuation ring of $k_j$.Then $[mathbb U_1 : mathbb N_{k_2/k_1}mathbb U_2]=e(k_2/k_1)$. In this case, $[1+2mathbb Z_2:mathbb N_{k_2/mathbb Q_2}mathbb U_2]=2$, then the image of the norm map is $1+2^2mathbb Z_2$
$endgroup$
– lorry
Jan 4 at 3:34
1
1
$begingroup$
Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 6:50
$begingroup$
Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 6:50
$begingroup$
Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:18
$begingroup$
Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:18
$begingroup$
Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
$endgroup$
– lorry
Jan 3 at 8:09
$begingroup$
Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
$endgroup$
– lorry
Jan 3 at 8:09
$begingroup$
I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 10:29
$begingroup$
I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 10:29
$begingroup$
Let $k_1/ mathbb Q_p$ and $k_2/k_1$ be finite extensions and $Gal(k_2/k_1)$ is cyclic. Let $mathbb U_j$ denote the units of the valuation ring of $k_j$.Then $[mathbb U_1 : mathbb N_{k_2/k_1}mathbb U_2]=e(k_2/k_1)$. In this case, $[1+2mathbb Z_2:mathbb N_{k_2/mathbb Q_2}mathbb U_2]=2$, then the image of the norm map is $1+2^2mathbb Z_2$
$endgroup$
– lorry
Jan 4 at 3:34
$begingroup$
Let $k_1/ mathbb Q_p$ and $k_2/k_1$ be finite extensions and $Gal(k_2/k_1)$ is cyclic. Let $mathbb U_j$ denote the units of the valuation ring of $k_j$.Then $[mathbb U_1 : mathbb N_{k_2/k_1}mathbb U_2]=e(k_2/k_1)$. In this case, $[1+2mathbb Z_2:mathbb N_{k_2/mathbb Q_2}mathbb U_2]=2$, then the image of the norm map is $1+2^2mathbb Z_2$
$endgroup$
– lorry
Jan 4 at 3:34
|
show 2 more comments
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1
$begingroup$
Certainly $m=8$ is the smallest integer such that $Bbb{Q}(sqrt2)subseteqBbb{Q}(zeta_m)$. What does the ording theorem say. I couldn't find it.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 6:50
$begingroup$
Also, it seems to me that $1+8Bbb{Z}$ is contained in the image of the norm map, but $1+4Bbb{Z}$ certainly is not: $x^2-2y^2notequiv5pmod8$. So the local conductor is $3$ resulting in a factor $2^3$. What result suggests that it should $4$ only?
$endgroup$
– Jyrki Lahtonen
Jan 3 at 7:18
$begingroup$
Yes, you are right. By the lemme 5.3 of Childress's Class Field Theory, I get that. And the ordering theorem is on P136. It seems that I am wrong about the image of the norm map.
$endgroup$
– lorry
Jan 3 at 8:09
$begingroup$
I don't have Childress, that's why I asked you to quote the theorem. Anyway, integers congruent to $1$ modulo $8$ have square roots in $Bbb{Q}_2$, so they are trivially included in the image of the norm.
$endgroup$
– Jyrki Lahtonen
Jan 3 at 10:29
$begingroup$
Let $k_1/ mathbb Q_p$ and $k_2/k_1$ be finite extensions and $Gal(k_2/k_1)$ is cyclic. Let $mathbb U_j$ denote the units of the valuation ring of $k_j$.Then $[mathbb U_1 : mathbb N_{k_2/k_1}mathbb U_2]=e(k_2/k_1)$. In this case, $[1+2mathbb Z_2:mathbb N_{k_2/mathbb Q_2}mathbb U_2]=2$, then the image of the norm map is $1+2^2mathbb Z_2$
$endgroup$
– lorry
Jan 4 at 3:34