What is the condoctor of $ mathbb Q(sqrt 2 )/mathbb Q?$ $4mathbb Z or 8 mathbb Z?$
0
By execise 6.8 of Childress's Algebraic Number Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considerring the norm of the local field about the place corespondent with 2, it should be $4mathbb Z$ and there is contraction to the ording theorem.
algebraic-number-theory
asked 12 hours ago
lorry
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By execise 6.8 of Childress's Algebraic Number Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considerring the norm of the local field about the place corespondent with 2, it should be $4mathbb Z$ and there is contraction to the ording theorem.
algebraic-number-theory
asked 12 hours ago
lorry
1
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lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
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.
– Jyrki Lahtonen
11 hours ago
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?
– Jyrki Lahtonen
11 hours ago
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.
– lorry
10 hours ago
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.
– Jyrki Lahtonen
8 hours ago
add a comment |
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By execise 6.8 of Childress's Algebraic Number Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considerring the norm of the local field about the place corespondent with 2, it should be $4mathbb Z$ and there is contraction to the ording theorem.
algebraic-number-theory
asked 12 hours ago
lorry
1
New contributor
lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
By execise 6.8 of Childress's Algebraic Number Theory, $mathbb f(mathbb Q(sqrt 2)/ mathbb Q)=8mathbb Z$. But considerring the norm of the local field about the place corespondent with 2, it should be $4mathbb Z$ and there is contraction to the ording theorem.
algebraic-number-theory
algebraic-number-theory
asked 12 hours ago
lorry
1
New contributor
lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 12 hours ago
lorry
1
New contributor
lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 12 hours ago
lorry
1
New contributor
lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 12 hours ago
lorry
1
asked 12 hours ago
lorry
1
1
New contributor
lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
lorry is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
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.
– Jyrki Lahtonen
11 hours ago
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?
– Jyrki Lahtonen
11 hours ago
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.
– lorry
10 hours ago
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.
– Jyrki Lahtonen
8 hours ago
add a comment |
1
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.
– Jyrki Lahtonen
11 hours ago
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?
– Jyrki Lahtonen
11 hours ago
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.
– lorry
10 hours ago
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.
– Jyrki Lahtonen
8 hours ago
1
1
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.
– Jyrki Lahtonen
11 hours ago
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.
– Jyrki Lahtonen
11 hours ago
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?
– Jyrki Lahtonen
11 hours ago
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?
– Jyrki Lahtonen
11 hours ago
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.
– lorry
10 hours ago
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.
– lorry
10 hours ago
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.
– Jyrki Lahtonen
8 hours ago
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.
– Jyrki Lahtonen
8 hours ago
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1
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.
– Jyrki Lahtonen
11 hours ago
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?
– Jyrki Lahtonen
11 hours ago
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.
– lorry
10 hours ago
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.
– Jyrki Lahtonen
8 hours ago