0

Consider the extension, $L=K(\sqrt{-1})(a^{1/2})$, for $a\in K(\sqrt{-1}),\; \sqrt{-1}\notin K$. I am trying to find an $a$ for which the extension $L:K$ becomes cyclic with cyclic group isomorphic to $Z/4Z$. What could be such an $a$?

If it is cyclic then what would be the generator of the cyclic Galois group?

roydiptajit
  • 738
  • 1
    Depends on $K$, surely. The first example that occurred to me was $K=\Bbb{Q}(\sqrt{-2})$, $a=e^{2\pi i/8}$ when $L=\Bbb{Q}(\zeta), \zeta=e^{2\pi i/16}$ is the sixteenth cyclotomic field. The Galois group $Gal(L/K)$ is then cyclic, generated by $\sigma:\zeta\mapsto\zeta^3$. Observe that $\sqrt{-2}=\zeta^2+\zeta^6$ is a fixed point of $\sigma$. – Jyrki Lahtonen Jan 12 '21 at 11:49
  • $K$ cannot be real because then complex conjugation would be an order two automorphism of $L$, but $\sqrt{-1}$ must be a fixed point of the unique automorphism of order two in the Galois group. – Jyrki Lahtonen Jan 12 '21 at 11:52
  • I've given an answer in https://math.stackexchange.com/questions/3980776/show-there-are-elements-x-y-in-k-s-t-x2y2-1 –  Jan 12 '21 at 14:03
  • Thank you sir @franzlemmermeyer – roydiptajit Jan 12 '21 at 14:05
  • Thanks @JyrkiLahtonen – roydiptajit Jan 12 '21 at 14:06
  • @JyrkiLahtonen In my example can you give a $Z/4Z$ cyclotomic extension over it? I need to prove that this type of extensions are indeed Galois.. – roydiptajit Jan 14 '21 at 17:06
  • Sorry I need to prove it cyclic – roydiptajit Jan 14 '21 at 17:13
  • For an arbitrary choice of $a$ the extension will fail to be Galois. We want $\sigma(a)/a$ to be a square in $K(\sqrt{-1})$ without $a$ being one (reusing the idea from the other question). I think the minimal polynomial of $\sqrt{a}$ over $K$ is biquadratic when this old thread may prove useful to you. – Jyrki Lahtonen Jan 14 '21 at 17:14
  • This may also help with the construction. With obvious corrections. – Jyrki Lahtonen Jan 14 '21 at 17:18

0 Answers0