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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.

戴星宇
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  • You mean the discriminant of the minimal polynomial I take it? – Zelos Malum Jun 02 '15 at 04:23
  • I mean the discriminant of algebraic number fields. Most of the time I think it does not equal the discriminant of the minimal polynomial which is obvious. – 戴星宇 Jun 02 '15 at 04:47
  • For an elementary proof that your proposed basis is indeed a basis, see here. The basis in this question can be easily transformed to your basis. – Fnark Man Feb 10 '25 at 20:05

1 Answers1

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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.

戴星宇
  • 171