Mixing
Division algebra over rationals of dimension 9
$begingroup$
I want to understand about existence of some non-commutative division algebras over $mathbb{Q}$ of dimension $9$.
Q. Does there exist a division algebra $D$ such that
$D$ is non-commutative;
$D$ is of dimension $9$ over $mathbb{Q}$ and $Z(D)=mathbb{Q}$;
$K:=mathbb{Q}(2^{1/3})$ is a maximal sub-field of $D$?
My way towards solution: if $D$ is such algebra, then consider $xin D$ outside $K:=mathbb{Q}(2^{1/3})$. If conjugation by $x$ leaves $K$ invariant then it induces an automorphism of $K$ which fixed $mathbb{Q}$; the only possibility of this is trivial automorphism, which means $x$ centralizes $K$, contradiction.
In fact, we can show that $D=Koplus xK oplus x^2K$ as a vector space.
Further $x^3$ centralize all generators of $D$, so $x^3inmathbb{Q}setminus {0}$.
Next, how should I proceed to determine structure of $D$?
I never studied division algebras other than quaternions and fields. I do not know how this question will be, but I was trying to see whether after $2^2$, can we get non-commutative division algebra whose dimension over its center is $3^2$? So a simple case I thought is through above questions, I was unable to complete the solution of existence.
division-algebras
asked Jan 19 at 6:25
BeginnerBeginner
3,94111225
$endgroup$
add a comment |
$begingroup$
I want to understand about existence of some non-commutative division algebras over $mathbb{Q}$ of dimension $9$.
Q. Does there exist a division algebra $D$ such that
$D$ is non-commutative;
$D$ is of dimension $9$ over $mathbb{Q}$ and $Z(D)=mathbb{Q}$;
$K:=mathbb{Q}(2^{1/3})$ is a maximal sub-field of $D$?
My way towards solution: if $D$ is such algebra, then consider $xin D$ outside $K:=mathbb{Q}(2^{1/3})$. If conjugation by $x$ leaves $K$ invariant then it induces an automorphism of $K$ which fixed $mathbb{Q}$; the only possibility of this is trivial automorphism, which means $x$ centralizes $K$, contradiction.
In fact, we can show that $D=Koplus xK oplus x^2K$ as a vector space.
Further $x^3$ centralize all generators of $D$, so $x^3inmathbb{Q}setminus {0}$.
Next, how should I proceed to determine structure of $D$?
I never studied division algebras other than quaternions and fields. I do not know how this question will be, but I was trying to see whether after $2^2$, can we get non-commutative division algebra whose dimension over its center is $3^2$? So a simple case I thought is through above questions, I was unable to complete the solution of existence.
division-algebras
asked Jan 19 at 6:25
BeginnerBeginner
3,94111225
$endgroup$
1
$begingroup$
You can (all) construct dim $n^2$ central division algebras over the rationals as cyclic algebras (see math.stackexchange.com/q/133790/11323 for $n=3$). Suggestion: construct an appropriate cyclic algebra which contains a cube root of 2.
$endgroup$
– Kimball
Jan 19 at 17:47
1
$begingroup$
Matt Emerton describes such a cyclic algebra in this post. Take note of the relation $x^3=a=2$. $Ksimeq Bbb{Q}(x)$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:55
1
$begingroup$
In this post I describe the same algebra prof. Emerton gave, but using matrices with entries in the field $Bbb{Q}(cos(2pi/7))$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:01
add a comment |
$begingroup$
I want to understand about existence of some non-commutative division algebras over $mathbb{Q}$ of dimension $9$.
Q. Does there exist a division algebra $D$ such that
$D$ is non-commutative;
$D$ is of dimension $9$ over $mathbb{Q}$ and $Z(D)=mathbb{Q}$;
$K:=mathbb{Q}(2^{1/3})$ is a maximal sub-field of $D$?
My way towards solution: if $D$ is such algebra, then consider $xin D$ outside $K:=mathbb{Q}(2^{1/3})$. If conjugation by $x$ leaves $K$ invariant then it induces an automorphism of $K$ which fixed $mathbb{Q}$; the only possibility of this is trivial automorphism, which means $x$ centralizes $K$, contradiction.
In fact, we can show that $D=Koplus xK oplus x^2K$ as a vector space.
Further $x^3$ centralize all generators of $D$, so $x^3inmathbb{Q}setminus {0}$.
Next, how should I proceed to determine structure of $D$?
I never studied division algebras other than quaternions and fields. I do not know how this question will be, but I was trying to see whether after $2^2$, can we get non-commutative division algebra whose dimension over its center is $3^2$? So a simple case I thought is through above questions, I was unable to complete the solution of existence.
division-algebras
asked Jan 19 at 6:25
BeginnerBeginner
3,94111225
$endgroup$
I want to understand about existence of some non-commutative division algebras over $mathbb{Q}$ of dimension $9$.
Q. Does there exist a division algebra $D$ such that
$D$ is non-commutative;
$D$ is of dimension $9$ over $mathbb{Q}$ and $Z(D)=mathbb{Q}$;
$K:=mathbb{Q}(2^{1/3})$ is a maximal sub-field of $D$?
My way towards solution: if $D$ is such algebra, then consider $xin D$ outside $K:=mathbb{Q}(2^{1/3})$. If conjugation by $x$ leaves $K$ invariant then it induces an automorphism of $K$ which fixed $mathbb{Q}$; the only possibility of this is trivial automorphism, which means $x$ centralizes $K$, contradiction.
In fact, we can show that $D=Koplus xK oplus x^2K$ as a vector space.
Further $x^3$ centralize all generators of $D$, so $x^3inmathbb{Q}setminus {0}$.
Next, how should I proceed to determine structure of $D$?
I never studied division algebras other than quaternions and fields. I do not know how this question will be, but I was trying to see whether after $2^2$, can we get non-commutative division algebra whose dimension over its center is $3^2$? So a simple case I thought is through above questions, I was unable to complete the solution of existence.
division-algebras
division-algebras
asked Jan 19 at 6:25
BeginnerBeginner
3,94111225
asked Jan 19 at 6:25
BeginnerBeginner
3,94111225
asked Jan 19 at 6:25
BeginnerBeginner
3,94111225
asked Jan 19 at 6:25
BeginnerBeginner
3,94111225
asked Jan 19 at 6:25
BeginnerBeginner
3,94111225
3,94111225
1
$begingroup$
You can (all) construct dim $n^2$ central division algebras over the rationals as cyclic algebras (see math.stackexchange.com/q/133790/11323 for $n=3$). Suggestion: construct an appropriate cyclic algebra which contains a cube root of 2.
$endgroup$
– Kimball
Jan 19 at 17:47
1
$begingroup$
Matt Emerton describes such a cyclic algebra in this post. Take note of the relation $x^3=a=2$. $Ksimeq Bbb{Q}(x)$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:55
1
$begingroup$
In this post I describe the same algebra prof. Emerton gave, but using matrices with entries in the field $Bbb{Q}(cos(2pi/7))$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:01
add a comment |
1
$begingroup$
You can (all) construct dim $n^2$ central division algebras over the rationals as cyclic algebras (see math.stackexchange.com/q/133790/11323 for $n=3$). Suggestion: construct an appropriate cyclic algebra which contains a cube root of 2.
$endgroup$
– Kimball
Jan 19 at 17:47
1
$begingroup$
Matt Emerton describes such a cyclic algebra in this post. Take note of the relation $x^3=a=2$. $Ksimeq Bbb{Q}(x)$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:55
1
$begingroup$
In this post I describe the same algebra prof. Emerton gave, but using matrices with entries in the field $Bbb{Q}(cos(2pi/7))$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:01
1
1
$begingroup$
You can (all) construct dim $n^2$ central division algebras over the rationals as cyclic algebras (see math.stackexchange.com/q/133790/11323 for $n=3$). Suggestion: construct an appropriate cyclic algebra which contains a cube root of 2.
$endgroup$
– Kimball
Jan 19 at 17:47
$begingroup$
You can (all) construct dim $n^2$ central division algebras over the rationals as cyclic algebras (see math.stackexchange.com/q/133790/11323 for $n=3$). Suggestion: construct an appropriate cyclic algebra which contains a cube root of 2.
$endgroup$
– Kimball
Jan 19 at 17:47
1
1
$begingroup$
Matt Emerton describes such a cyclic algebra in this post. Take note of the relation $x^3=a=2$. $Ksimeq Bbb{Q}(x)$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:55
$begingroup$
Matt Emerton describes such a cyclic algebra in this post. Take note of the relation $x^3=a=2$. $Ksimeq Bbb{Q}(x)$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:55
1
1
$begingroup$
In this post I describe the same algebra prof. Emerton gave, but using matrices with entries in the field $Bbb{Q}(cos(2pi/7))$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:01
$begingroup$
In this post I describe the same algebra prof. Emerton gave, but using matrices with entries in the field $Bbb{Q}(cos(2pi/7))$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:01
add a comment |
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$begingroup$
You can (all) construct dim $n^2$ central division algebras over the rationals as cyclic algebras (see math.stackexchange.com/q/133790/11323 for $n=3$). Suggestion: construct an appropriate cyclic algebra which contains a cube root of 2.
$endgroup$
– Kimball
Jan 19 at 17:47
1
$begingroup$
Matt Emerton describes such a cyclic algebra in this post. Take note of the relation $x^3=a=2$. $Ksimeq Bbb{Q}(x)$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:55
1
$begingroup$
In this post I describe the same algebra prof. Emerton gave, but using matrices with entries in the field $Bbb{Q}(cos(2pi/7))$.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 20:01