0

I was reading these lecture notes of Miles Reid: https://homepages.warwick.ac.uk/~masda/MA3D5/Galois.pdf

on page 47, he writes example 3.21 of $\mathbb{Q}(\sqrt{a+\sqrt{b}},\sqrt{a-\sqrt{b}})$, but he doesn't write the Galois group of this field.

So it triggered me the question how would one find the Galois group of the splitting field: $\mathbb{Q}(\sqrt{a+d\sqrt{b}},\sqrt{a-d\sqrt{b}})$?

Thanks in advance!

Kevin.S
  • 3,706
MathematicalPhysicist
  • 4,191
  • 1
    It will depend on $a$ and $b$, but there might be an answer that holds for all appropriately generic $a$ and $b$, or something like that. – pancini May 07 '22 at 05:29
  • 1
    @pancini a lot of work I guess. – MathematicalPhysicist May 07 '22 at 05:33
  • You can obviously do this by at most three quadratic extensions - add successively $\sqrt b$, $\sqrt{a+\sqrt b}$ and $\sqrt{a-\sqrt b}$. The Galois group therefore won't have any elements of order 3 (as an example of a very easy deduction). – Mark Bennet May 07 '22 at 06:47
  • Deleted my comment as I misread the page I linked. Mark is right; the Galois group will be a $2$-group of order at most $8$. Note that it can still be nonabelian: https://math.stackexchange.com/questions/3653776/galois-group-of-splitting-field-of-x4-6x27-is-non-abelian – pancini May 07 '22 at 07:23
  • See https://math.stackexchange.com/questions/4435634/galois-group-of-certain-quartic-polynomial –  May 08 '22 at 05:11

0 Answers0