0

I'm asked to decide whether the extension $K = \mathbb Q(\gamma) / \mathbb Q$ is Galois where $\gamma = \sqrt{5 + \sqrt{15}}$. I know that it's not the case by general results about biquadratic quartic.

I proved that $\gamma$ is a root of the irreducible polynomial $p(x) = x^4 - 10 x^2 + 10$. And that the other roots are $\pm \gamma$ and $\pm \delta$ where $\delta = \sqrt{5 - \sqrt{15}}$. Therefore $K$ is Galois over $\mathbb Q$ if and only if $\delta \in \mathbb Q(\gamma)$.

Is there a straight argument to prove that $\delta \notin \mathbb Q(\gamma)$? Otherwise, is there a straight argument to prove that if $\delta \in \mathbb Q(\gamma)$, there is a non-rational number fixed by all the elements of the Galois group of $K$? I suspect that $\sqrt{10}$ would be a good candidate as $\gamma \delta = \sqrt{10}$!

More generally, what would you do to prove that $K$ is not Galois?

mathcounterexamples.net
  • 71,758
  • Compare with this post, for $b=1$, $a=5$ and $d=21$. Since $a^2-db^2$ is not a square, the extension is Galois. – Dietrich Burde Apr 13 '24 at 15:33
  • If true, this is bad news as it means that I was not applying properly the result for biquadratic extensions to this specific case! – mathcounterexamples.net Apr 13 '24 at 15:37
  • @DietrichBurde Sorry, there was a typo in my initial question: replacing 21 by 15. I edited the question. – mathcounterexamples.net Apr 13 '24 at 15:44
  • 1
    Sorry, in the first comment we have $a^2-db^2=4=2^2$ for $d=21$. In general, the duplicate says that $\mathbb Q(\sqrt{a+b\sqrt{d}} / \mathbb Q$ is Galois if and only if $a^2-db^2$ is a square. For $b=1$, $a=5$, $d=15$ this is not a square, so the extension is not Galois. – Dietrich Burde Apr 13 '24 at 16:27
  • @DietrichBurde This answers the question. However, I was expecting, maybe wrongly, something more straightforward. I'll let the question open for a while and close it if no specific answer appears. – mathcounterexamples.net Apr 13 '24 at 16:35

0 Answers0