# Division algebra over rationals of dimension 9

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.

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