Compute the degree of a given extension of QQ

Compute the degree of \mathbb{Q}(\alpha) over \mathbb{Q}, where \alpha is 2+\sqrt{3} or 1 + \sqrt[3]{2} + \sqrt[3]{4}.


Since 2+\sqrt{3} \in \mathbb{Q}(\sqrt{3}), the degree of \mathbb{Q}(2+\sqrt{3}) over \mathbb{Q} is at most 2. We can solve the linear system \alpha^2 + a\alpha + b = 0 in \mathbb{Q}(\sqrt{3}) (as a vector space over \mathbb{Q}) to find a polynomial satisfied by \alpha; evidently 2+\sqrt{3} is a root of p(x) = x^2 - 4x + 1. (WolframAlpha agrees.) Evidently, p(x+1) = x^2-2x-2 (WolframAlpha agrees), which is irreducible by Eisenstein’s criterion; so p(x) is irreducible. Thus p(x) is the minimal polynomial of 2+\sqrt{3} over \mathbb{Q}, and so the degree of \mathbb{Q}(2+\sqrt{3}) over \mathbb{Q} is 2.

In a similar fashion, \beta = 1+\sqrt[3]{2} + \sqrt[3]{4}, as an element of \mathbb{Q}(\sqrt[3]{2}), has degree at most 3 over \mathbb{Q}. Evidently, \beta is a root of q(x) = x^3 - 3x^2 - 3x - 1 (WolframAlpha agrees). Evidently, q(x+1) = x^3-6x-6, which is irreducible by Eisenstein. So q(x) is the minimal polynomial of \beta, and thus \mathbb{Q}(\beta) has degree 3 over \mathbb{Q}.

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