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.

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