Mixing
Hasse's classification of Quaternion algebras (over number fields)
2
$begingroup$
I am looking for the reference to Hasse's original publication classifying the Quaternion algebras over a number field $K$. The statement in the paper or book should read something like:
``Two Quaternion algebras over a number field $K$ are isomorphic iff they are ramified at the same number of places"
reference-request ring-theory algebraic-number-theory quaternions
asked Feb 1 at 14:47
Sam HughesSam Hughes
741114
$endgroup$
add a comment |
2
$begingroup$
I am looking for the reference to Hasse's original publication classifying the Quaternion algebras over a number field $K$. The statement in the paper or book should read something like:
``Two Quaternion algebras over a number field $K$ are isomorphic iff they are ramified at the same number of places"
reference-request ring-theory algebraic-number-theory quaternions
asked Feb 1 at 14:47
Sam HughesSam Hughes
741114
$endgroup$
2
$begingroup$
Ramification is for quaternion algebras over number fields, not over arbitrary fields?
$endgroup$
– Dietrich Burde
Feb 1 at 15:11
$begingroup$
oops corrected the question
$endgroup$
– Sam Hughes
Feb 1 at 15:12
3
$begingroup$
Theorem 4.8. here says:"Let $H$ and $H'$ be quaternion algebras over a number field $F$. Then $Hcong H'$ if and only if $Ram(H) = Ram(H')$.
$endgroup$
– Dietrich Burde
Feb 1 at 15:15
1
$begingroup$
It says $H_v = H otimes_K K_v$, or $H_v$ is a division algebra (and $v$ is said ramified) or $H_v cong M_2(K_v)$. Theorem 1.13 says it is about the equation $ax^2+by^2=1, (x,y) in K_v$ where $H=K[i,j], i^2=a,j^2=b,ij = -ji$
$endgroup$
– reuns
Feb 1 at 22:17
add a comment |
2
2
2
$begingroup$
I am looking for the reference to Hasse's original publication classifying the Quaternion algebras over a number field $K$. The statement in the paper or book should read something like:
``Two Quaternion algebras over a number field $K$ are isomorphic iff they are ramified at the same number of places"
reference-request ring-theory algebraic-number-theory quaternions
asked Feb 1 at 14:47
Sam HughesSam Hughes
741114
$endgroup$
I am looking for the reference to Hasse's original publication classifying the Quaternion algebras over a number field $K$. The statement in the paper or book should read something like:
``Two Quaternion algebras over a number field $K$ are isomorphic iff they are ramified at the same number of places"
reference-request ring-theory algebraic-number-theory quaternions
reference-request ring-theory algebraic-number-theory quaternions
asked Feb 1 at 14:47
Sam HughesSam Hughes
741114
asked Feb 1 at 14:47
Sam HughesSam Hughes
741114
edited Feb 3 at 20:03
Sam Hughes
asked Feb 1 at 14:47
Sam HughesSam Hughes
741114
asked Feb 1 at 14:47
Sam HughesSam Hughes
741114
asked Feb 1 at 14:47
Sam HughesSam Hughes
741114
741114
2
$begingroup$
Ramification is for quaternion algebras over number fields, not over arbitrary fields?
$endgroup$
– Dietrich Burde
Feb 1 at 15:11
$begingroup$
oops corrected the question
$endgroup$
– Sam Hughes
Feb 1 at 15:12
3
$begingroup$
Theorem 4.8. here says:"Let $H$ and $H'$ be quaternion algebras over a number field $F$. Then $Hcong H'$ if and only if $Ram(H) = Ram(H')$.
$endgroup$
– Dietrich Burde
Feb 1 at 15:15
1
$begingroup$
It says $H_v = H otimes_K K_v$, or $H_v$ is a division algebra (and $v$ is said ramified) or $H_v cong M_2(K_v)$. Theorem 1.13 says it is about the equation $ax^2+by^2=1, (x,y) in K_v$ where $H=K[i,j], i^2=a,j^2=b,ij = -ji$
$endgroup$
– reuns
Feb 1 at 22:17
add a comment |
2
$begingroup$
Ramification is for quaternion algebras over number fields, not over arbitrary fields?
$endgroup$
– Dietrich Burde
Feb 1 at 15:11
$begingroup$
oops corrected the question
$endgroup$
– Sam Hughes
Feb 1 at 15:12
3
$begingroup$
Theorem 4.8. here says:"Let $H$ and $H'$ be quaternion algebras over a number field $F$. Then $Hcong H'$ if and only if $Ram(H) = Ram(H')$.
$endgroup$
– Dietrich Burde
Feb 1 at 15:15
1
$begingroup$
It says $H_v = H otimes_K K_v$, or $H_v$ is a division algebra (and $v$ is said ramified) or $H_v cong M_2(K_v)$. Theorem 1.13 says it is about the equation $ax^2+by^2=1, (x,y) in K_v$ where $H=K[i,j], i^2=a,j^2=b,ij = -ji$
$endgroup$
– reuns
Feb 1 at 22:17
2
2
$begingroup$
Ramification is for quaternion algebras over number fields, not over arbitrary fields?
$endgroup$
– Dietrich Burde
Feb 1 at 15:11
$begingroup$
Ramification is for quaternion algebras over number fields, not over arbitrary fields?
$endgroup$
– Dietrich Burde
Feb 1 at 15:11
$begingroup$
oops corrected the question
$endgroup$
– Sam Hughes
Feb 1 at 15:12
$begingroup$
oops corrected the question
$endgroup$
– Sam Hughes
Feb 1 at 15:12
3
3
$begingroup$
Theorem 4.8. here says:"Let $H$ and $H'$ be quaternion algebras over a number field $F$. Then $Hcong H'$ if and only if $Ram(H) = Ram(H')$.
$endgroup$
– Dietrich Burde
Feb 1 at 15:15
$begingroup$
Theorem 4.8. here says:"Let $H$ and $H'$ be quaternion algebras over a number field $F$. Then $Hcong H'$ if and only if $Ram(H) = Ram(H')$.
$endgroup$
– Dietrich Burde
Feb 1 at 15:15
1
1
$begingroup$
It says $H_v = H otimes_K K_v$, or $H_v$ is a division algebra (and $v$ is said ramified) or $H_v cong M_2(K_v)$. Theorem 1.13 says it is about the equation $ax^2+by^2=1, (x,y) in K_v$ where $H=K[i,j], i^2=a,j^2=b,ij = -ji$
$endgroup$
– reuns
Feb 1 at 22:17
$begingroup$
It says $H_v = H otimes_K K_v$, or $H_v$ is a division algebra (and $v$ is said ramified) or $H_v cong M_2(K_v)$. Theorem 1.13 says it is about the equation $ax^2+by^2=1, (x,y) in K_v$ where $H=K[i,j], i^2=a,j^2=b,ij = -ji$
$endgroup$
– reuns
Feb 1 at 22:17
add a comment |
1 Answer
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As detailed here, Hasse introduced an invariant, now known as the Hasse invariant in two papers (here and here) to classify central simple algebras.
answered Feb 2 at 15:02
Sam HughesSam Hughes
741114
$endgroup$
1
$begingroup$
In particular, for rank $4$ (quaternion algebras), the local invariant is either $0$ or $1/2$, the latter being the nonsplit (ramified) case. These local invariants must add up to $0pmod{Bbb Z}$. So to specify the isomorphism class it’s enough to specify which local invariants are $1/2$.
$endgroup$
– Lubin
Feb 2 at 17:58
add a comment |
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1 Answer
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1 Answer
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1
$begingroup$
As detailed here, Hasse introduced an invariant, now known as the Hasse invariant in two papers (here and here) to classify central simple algebras.
answered Feb 2 at 15:02
Sam HughesSam Hughes
741114
$endgroup$
1
$begingroup$
In particular, for rank $4$ (quaternion algebras), the local invariant is either $0$ or $1/2$, the latter being the nonsplit (ramified) case. These local invariants must add up to $0pmod{Bbb Z}$. So to specify the isomorphism class it’s enough to specify which local invariants are $1/2$.
$endgroup$
– Lubin
Feb 2 at 17:58
add a comment |
1
$begingroup$
As detailed here, Hasse introduced an invariant, now known as the Hasse invariant in two papers (here and here) to classify central simple algebras.
answered Feb 2 at 15:02
Sam HughesSam Hughes
741114
$endgroup$
1
$begingroup$
In particular, for rank $4$ (quaternion algebras), the local invariant is either $0$ or $1/2$, the latter being the nonsplit (ramified) case. These local invariants must add up to $0pmod{Bbb Z}$. So to specify the isomorphism class it’s enough to specify which local invariants are $1/2$.
$endgroup$
– Lubin
Feb 2 at 17:58
add a comment |
1
1
1
$begingroup$
As detailed here, Hasse introduced an invariant, now known as the Hasse invariant in two papers (here and here) to classify central simple algebras.
answered Feb 2 at 15:02
Sam HughesSam Hughes
741114
$endgroup$
As detailed here, Hasse introduced an invariant, now known as the Hasse invariant in two papers (here and here) to classify central simple algebras.
answered Feb 2 at 15:02
Sam HughesSam Hughes
741114
answered Feb 2 at 15:02
Sam HughesSam Hughes
741114
answered Feb 2 at 15:02
Sam HughesSam Hughes
741114
answered Feb 2 at 15:02
Sam HughesSam Hughes
741114
741114
1
$begingroup$
In particular, for rank $4$ (quaternion algebras), the local invariant is either $0$ or $1/2$, the latter being the nonsplit (ramified) case. These local invariants must add up to $0pmod{Bbb Z}$. So to specify the isomorphism class it’s enough to specify which local invariants are $1/2$.
$endgroup$
– Lubin
Feb 2 at 17:58
add a comment |
1
$begingroup$
In particular, for rank $4$ (quaternion algebras), the local invariant is either $0$ or $1/2$, the latter being the nonsplit (ramified) case. These local invariants must add up to $0pmod{Bbb Z}$. So to specify the isomorphism class it’s enough to specify which local invariants are $1/2$.
$endgroup$
– Lubin
Feb 2 at 17:58
1
1
$begingroup$
In particular, for rank $4$ (quaternion algebras), the local invariant is either $0$ or $1/2$, the latter being the nonsplit (ramified) case. These local invariants must add up to $0pmod{Bbb Z}$. So to specify the isomorphism class it’s enough to specify which local invariants are $1/2$.
$endgroup$
– Lubin
Feb 2 at 17:58
$begingroup$
In particular, for rank $4$ (quaternion algebras), the local invariant is either $0$ or $1/2$, the latter being the nonsplit (ramified) case. These local invariants must add up to $0pmod{Bbb Z}$. So to specify the isomorphism class it’s enough to specify which local invariants are $1/2$.
$endgroup$
– Lubin
Feb 2 at 17:58
add a comment |
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2
$begingroup$
Ramification is for quaternion algebras over number fields, not over arbitrary fields?
$endgroup$
– Dietrich Burde
Feb 1 at 15:11
$begingroup$
oops corrected the question
$endgroup$
– Sam Hughes
Feb 1 at 15:12
3
$begingroup$
Theorem 4.8. here says:"Let $H$ and $H'$ be quaternion algebras over a number field $F$. Then $Hcong H'$ if and only if $Ram(H) = Ram(H')$.
$endgroup$
– Dietrich Burde
Feb 1 at 15:15
1
$begingroup$
It says $H_v = H otimes_K K_v$, or $H_v$ is a division algebra (and $v$ is said ramified) or $H_v cong M_2(K_v)$. Theorem 1.13 says it is about the equation $ax^2+by^2=1, (x,y) in K_v$ where $H=K[i,j], i^2=a,j^2=b,ij = -ji$
$endgroup$
– reuns
Feb 1 at 22:17