Could you help me showing that for a field $F$ the degree of the extension $F(x^{2}+\frac{1}{x^{2}})\subset F(x)$ is $4$?

I have found the polynomial $y^{4}-(x^{2}+\frac{1}{x^{2}})y^{2}+1$ such that $x$ is its root, but how can I prove that it is the minimal one?

Thanks in advance.


Observe that $F(x^2+\frac{1}{x^2})\subset F(x^2)\subset F(x)$. Now prove that each extension is of degree two.

  • $\begingroup$ I have tried that way either but I had a problem with the formal proof that $x^{2}\notin F(x^{2}+\frac{1}{x^{2}})$. $\endgroup$ – DeltaDelta Sep 6 '14 at 16:54
  • $\begingroup$ For that proof $F(x^2+\frac{1}{x^2})$ is invariant under the last two automorphisms on my list but $x^2$ is not. $\endgroup$ – Rene Schipperus Sep 6 '14 at 16:58

The solution by Brett is great. I just what to add that you can explicitly write the four automorphisms:

$$x\mapsto x$$ $$x\mapsto -x$$ $$x\mapsto \frac{1}{x}$$ $$x\mapsto -\frac{1}{x}$$

  • 1
    $\begingroup$ This works well ... unless $F$ happens to have characteristic two. In that case the extension isn't Galois (it's no longer separable), so it is pointless to look for more than two automorphisms. $\endgroup$ – Jyrki Lahtonen Sep 6 '14 at 18:01
  • $\begingroup$ @JyrkiLahtonen maybe you would care to add an answer for the characteristic 2 case ? $\endgroup$ – Rene Schipperus Sep 6 '14 at 18:03
  • $\begingroup$ In characteristic two your third automorphism still tells that $F(x+1/x)$ is a proper subfield of $F(x)$. Then $$x^2+\frac1{x^2}=(x+\frac1x)^2,$$ so $F(x+1/x)$ is a purely inseparable extension of $F(x^2+1/x^2)$. $\endgroup$ – Jyrki Lahtonen Sep 6 '14 at 18:06

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