## The bases of QQ(∛2)

Find a basis for $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$. Describe all of the bases for $\mathbb{Q}(\sqrt[3]{2})$ in terms of this basis.

Note that $\sqrt[3]{2}$ is a root of $p(x) = x^3 - 2$, which is irreducible over $\mathbb{Q}$ by Eisenstein’s criterion. In particular, $p(x)$ is the minimal polynomial for $\sqrt[3]{2}$ over $\mathbb{Q}$. As we saw in Theorem 4.6 in TAN, every element of $\mathbb{Q}(\sqrt[3]{2})$ is uniquely of the form $a_0 + a_1\theta + a_2 \theta^2$, where $\theta = \sqrt[3]{2}$. In particular, $\{1,\theta,\theta^2\}$ is a basis for $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$.

Suppose $\beta_1 = b_{1,1} + b_{2,1} \theta + b_{3,1} \theta^2$, $\beta_2 = b_{1,2} + b_{2,2} \theta + b_{3,2} \theta^2$, and $\beta_3 = b_{1,3} + b_{2,3} \theta + b_{3,3} \theta^2$ is a subset of $\mathbb{Q}(\sqrt[3]{2})$, and define $\varphi$ on $\mathbb{Q}(\sqrt[3]{2})$ by $1 \mapsto \beta_1$, $\theta \mapsto \beta_2$, and $\theta^2 \mapsto \beta_3$ and extending linearly. Now $B = \{\beta_1,\beta_2,\beta_3\}$ is a basis of $\mathbb{Q}(\sqrt[3]{2})$ if and only if $\theta$ is a $\mathbb{Q}$-isomorphism. Evidently, the matrix of $\varphi$ with respect to the basis $\{1,\theta,\theta^2\}$ (in both domain and codomain) is $A = [b_{i,j}]$. In turn, $\varphi$ is an isomorphism if and only if $A$ is invertible, which holds if and only if $\mathsf{det}(A)$ is nonzero.