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I was reading this question earlier and I was quite confused by Ege's answer when he said 'Moreover, they have zero trace.' I was wondering how could we tell that? So my question is how did we tell that $ \sqrt{a + b \sqrt{c}} + \sqrt{a - b \sqrt{c}} $ has trace $0$?

I am assuming by the word 'trace', we are talking about the sum of all the roots of the minimal polynomial of $ \sqrt{a + b \sqrt{c}} + \sqrt{a - b \sqrt{c}} $ but I was really unsure on how to find its minimal polynomial. So I was wondering is there a slick way of seeing this comment by Ege?

Many thanks in advance!!

UnsinkableSam
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  • The explanation of the zero trace is there in the answer, considering those elements of the form $ x + y \sqrt{d} $ for $ x, y, d \in \mathbf Q $ . What beyond this are you not following? –  May 26 '21 at 09:23
  • @fundamentalform Thank you for your comment. Ege said that $ \sqrt{a + b \sqrt{c}} + \sqrt{a - b \sqrt{c}} $ has trace $0$ and so $x=0$. In particular why does $ \sqrt{a + b \sqrt{c}} + \sqrt{a - b \sqrt{c}} $ have trace $0$ is my question. I think you are understanding this the wrong way round? – UnsinkableSam May 26 '21 at 09:43

1 Answers1

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You don't need to calculate the minpoly. The conjugates are

  • $\sqrt{a + b \sqrt{c}} + \sqrt{a - b \sqrt{c}}$
  • $\sqrt{a - b \sqrt{c}} + \sqrt{a + b \sqrt{c}}$
  • $- \sqrt{a + b \sqrt{c}} - \sqrt{a - b \sqrt{c}}$
  • $- \sqrt{a - b \sqrt{c}} - \sqrt{a + b \sqrt{c}}$

these all cancel so you get 0.

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    Thank you for your answer! how did we determine that these four (well two actually) are the conjugates? Do we just apply Galois action on $ \sqrt{a + b \sqrt{c}} + \sqrt{a - b \sqrt{c}} $ individually? And we are allowed to do this because the Galois group also permutes the roots of the min poly of $ \sqrt{a + b \sqrt{c}} + \sqrt{a - b \sqrt{c}} $? – UnsinkableSam May 26 '21 at 10:50