## Compute the degree of a field extension and find a basis

Let $K = \mathbb{Q}(\sqrt[3]{2}, \sqrt{7}, i\sqrt{5})$. Compute the degree of $K$ over $\mathbb{Q}$ and find a basis.

Note that $\sqrt{7}$ is a root of $a(x) = x^2 - 7$, which is irreducible over $\mathbb{Q}$ by Eisenstein and thus is the minimal polynomial of $\sqrt{7}$ over $\mathbb{Q}$. So $\mathbb{Q}(\sqrt{7})$ has degree 2 over $\mathbb{Q}$, with $\{1, \sqrt{7}\}$ a basis. Suppose there exist $\alpha, \beta \in \mathbb{Q}$ such that $(\alpha + \beta\sqrt{7})^2 = -5$; comparing coefficients, we see that $\alpha^2 + 7\beta^2 = -5$. However, the left hand side of this equation is positive, while the right hand side is negative- a contradiction. In particular, $b(x) = x^2 + 5$ is irreducible over $\mathbb{Q}(\sqrt{7})$, and hence is the minimal polynomial of $i\sqrt{5}$ over $\mathbb{Q}(\sqrt{7})$. So $\mathbb{Q}(\sqrt{7}, i\sqrt{5})$ has degree 4. We may take as a basis the set $\{1, \sqrt{7}, i\sqrt{5}, i\sqrt{35} \}$.

Note that the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$ is $c(x) = x^3 - 2$. (Use Eisenstein to show irreducibility.) By this previous exercise, $c(x)$ is also irreducible over $\mathbb{Q}(\sqrt{7}, i\sqrt{5})$. Thus $\mathbb{Q}(\sqrt[3]{2}, \sqrt{7}, i\sqrt{5})$ has degree 12 over $\mathbb{Q}$. We may take as a basis the set $\{ 1, \sqrt[3]{2}, \sqrt[3]{4}, \sqrt{7},$ $\sqrt{7}\sqrt[3]{2}, \sqrt{7}\sqrt[3]{4},$ $i\sqrt{5}, i\sqrt{5}\sqrt[3]{2},$ $i\sqrt{5}\sqrt[3]{4}, i\sqrt{35},$ $i\sqrt{35}\sqrt[3]{2}, i\sqrt{35}\sqrt[3]{4} \}$.