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 $x\in D$ 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 $D=K\oplus xK \oplus x^2K$ as a vector space.

Further $x^3$ centralize all generators of $D$, so $x^3\in\mathbb{Q}\setminus \{0\}$. 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.

  • 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, 2019 at 17:47
  • 1
    $\begingroup$ Matt Emerton describes such a cyclic algebra in this post. Take note of the relation $x^3=a=2$. $K\simeq \Bbb{Q}(x)$. $\endgroup$ Jan 22, 2019 at 19:55
  • 1
    $\begingroup$ In this post I describe the same algebra prof. Emerton gave, but using matrices with entries in the field $\Bbb{Q}(\cos(2\pi/7))$. $\endgroup$ Jan 22, 2019 at 20:01


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