## 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 $\mathbb{Q}$.