I'm self-studying field extensions. I ran over an exercise which I can't completely solve. (I haven't yet started studying Galois theory, and I think this exercise isn't meant to be solved using it, just in case):
The problem is:
a) Prove $\sqrt{2}+\sqrt[3]{5}$ is algebraic over $\mathbb{Q}$ of degree 6.
Done: I know it has degree $\leq 6$ because $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ which has degree 6; then I explicitly found the polynomial by solving a 6-equation linear system, and Wolfram Alpha proved it irreducible (btw: how can I prove it by hand?). The polynomial is $t^6-6t^4-10t^3+12t^2-60t+17$.
b) What's its degree over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$?
It is this part b) which I can't solve. Of course its degree is $\leq 6$ in both cases, but I don't know what else to do.
Best Answer
Bruno: The degree over ${\mathbb Q}(\sqrt2)$ is 3: Consider $(x-\sqrt 2)^3-5$. Since $\root 3\of 5$ is in ${\mathbb Q}(\sqrt 2)(\sqrt 2+\root 3\of 5)$, the only other option is that it already belongs to ${\mathbb Q}(\sqrt 2)$. But this is impossible, since its minimal polynomial is $x^3-5$ over ${\mathbb Q}$ and ${\mathbb Q}(\sqrt 2)$ is an extension of degree 2.
[ By the way, one can use this and the tower law, to show that ${\mathbb Q}(\sqrt 2+\root 3\of 5)$ has degree 6 over ${\mathbb Q}$. The irreducibility of your polynomial then follows for free. ]
Similarly, the degree over ${\mathbb Q}(\root 3\of 5)$ is 2.