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} \}.

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