Minimum polynomial of $\sqrt{2} + \sqrt[3]{5}$ above $\mathbb{Q}$ (and a generalization)

Solution 1:

I kind of wonder how you found this polynomial, that must have been quite some work. Let me show you an alternative way. Let $\zeta$ be a $3$rd unit root. Then we examine the Galois extension $$ \mathbb{Q}(\sqrt{2},\sqrt[3]{5},\zeta) $$ of degree $2\cdot3\cdot 2 = 12$. This is Galois, as any automorphism must send $\sqrt{2}$ to either itself or $-\sqrt{2}$ and it must permutate the roots of $X^3-\sqrt[3]{5}$, giving us $2\cdot 3! = 12$ automorphisms. Now the minimal polynomial equals the polynomial whose roots are those elements that $\sqrt{2} + \sqrt[3]{5}$ can be mapped to by elements of the Galois group. In this case we get that the minimal polynomial equals $$ (X-(\sqrt{2} + \sqrt[3]{5}))(X-(\sqrt{2} + \zeta\sqrt[3]{5}))(X-(\sqrt{2} + \zeta^2\sqrt[3]{5}))(X-(-\sqrt{2} + \sqrt[3]{5}))(X-(-\sqrt{2} + \zeta\sqrt[3]{5}))(X-(-\sqrt{2} + \zeta^2\sqrt[3]{5})). $$