How can I compute the discriminant of $\mathbb{Q}(\sqrt{2}+\sqrt{5})$?

Question

How can I compute the discriminant of $\mathbb{Q}(\sqrt{2}+\sqrt{5})$?

I get stuck in this exercise of chapter 12 of textbook "A classical introduction to modern number theory" very long time...

How can I determine an integral basis in this situation?

I guess that basis $\{1, \sqrt{2}, \frac{-1+\sqrt{5}}{2}, \frac{-\sqrt{2}+\sqrt{10}}{2}\}$ may answer this question with discriminant 1600. But I can not prove it.

Answer 1

I have found a solution to my question. But it requires some advanced tools: If we use relative discriminant formula for K/L/F, $K=\mathbb{Q}(\sqrt{2}+\sqrt{5})=L(\sqrt{2}), L=\mathbb{Q}(\sqrt{5}), F=\mathbb{Q}$

and $D_{K/F}=N_{L/F}(D_{K/L})\cdot D_{L/F}^{[K:L]}$ which is $8^2\cdot 5^2=1600$. Then finally we reach the result mentioned above.

Any one can give some much more elementary proof here? Appreciate your help.