1

Find primitive element of $Q(\xi, \sqrt[6]{5})$, where $\xi^3 = -1 \neq - \xi$.
Find Galois group of this extension and describe it`s subfields.

First of all, finding primitive element. The only idea I have is take $\gamma, \gamma^2, \gamma^3$, where $\gamma \in Q(\xi, \sqrt[6]{5})$ and try to derive $\xi$ and $\sqrt[6]{5}$ from this linear system.

I know what is Galois group, but not understand how I can build it in this case.

wuzzapcom
  • 469
  • The primitive element theorem is a constructive proof, so you can apply it to find a primitive element in this case. – David Reed Nov 19 '17 at 20:44
  • It is the Galois group of $x^6-5$, which has been computed above. The subgroups are easy to see then. – Dietrich Burde Nov 19 '17 at 20:44
  • @DietrichBurde Could you explain, why my question is duplicate? I checked those proof and didn`t understood. Thank you. – wuzzapcom Nov 19 '17 at 20:51
  • The main part of your question is the computation of the Galois group and its subgroups. This is proved in all detail in Jacobson and Velez in "The Galois group of a radical extension of the rationals", in 1990; see the duplicate. Have a look at this beautiful paper. – Dietrich Burde Nov 19 '17 at 21:05

0 Answers0