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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?

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    $\begingroup$ A cubic extension of $\mathbb{Q}$ (for instance $\mathbb{Q}(\sqrt[3]{2})$) is isomorphic to $\mathbb{Q}^3$ as a $\mathbb{Q}$-vector space. $\endgroup$ – carmichael561 Sep 24 '17 at 21:37
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    $\begingroup$ Easily. My favorite as of late is $$\Bbb{Q}\oplus \Bbb{Q}\cos(2\pi/7)\oplus\Bbb{Q}\cos(4\pi/7)\subset\Bbb{R}.$$ But you can construct one such field starting with any cubic polynomial with rational coefficients and no rational zeros. $\endgroup$ – Jyrki Lahtonen Sep 24 '17 at 21:40
  • $\begingroup$ en.wikipedia.org/wiki/Cubic_field $\endgroup$ – Qiaochu Yuan Sep 24 '17 at 22:12
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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 $\mathbb{Q}$.

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