Fields for $\mathbb Q^3$?

Question

I am just a happy beginner in anything algebraic. It is discussed in this question why no fields exist for $\mathbb R^3$, but what about $\mathbb Q^3$? Can we build a division algebra by excluding the irrationals?

Answer 1

Suppose $A$ is a division $\mathbb{Q}$-algebra of dimension $3$, where we can think $\mathbb{Q}$ embedded in $A$.

Let $a\in A\setminus\mathbb{Q}$. I contend that $\{1,a,a^2\}$ is a basis.

Suppose $q_0+q_1a+q_2a^2=0$, with $q_0$, $q_1$ and $q_2$ not all zero. Then $q_2=0$ would imply $a\in\mathbb{Q}$.

In particular, the $\mathbb{Q}$-vector space endomorphism $l_a\colon x\mapsto ax$ satisfies a degree two polynomial, so its characteristic polynomial is reducible over $\mathbb{Q}$ and hence has a root $q$. If $b\ne0$ is an eigenvector, we conclude that $(a-q)b=0$, so $a=q$: a contradiction.

Since $A$ has a basis consisting of commuting elements, it is a field and $a$ satisfies a degree $3$ polynomial $f(X)$. On the other hand, the degree of $\mathbb{Q}(a)$ over $\mathbb{Q}$ has to be a divisor of $3$. Therefore $A=\mathbb{Q}(a)\cong \mathbb{Q}[X]/(f(X))$.

Conversely, if $f(X)\in\mathbb{Q}[X]$ is a degree $3$ irreducible polynomial, then $\mathbb{Q}[X]/(f(X))$ is a $3$-dimensional division algebra over $\mathbb{Q}$.