Given the twentieth root of unity a splitting for it's minimal polynomial is clearly $\Bbb Q(w_{20})$ where $w_{20}$ is the twentieth primitive root of unity. As the isomorphisms only map to roots of the minimal polynomial it only maps to primitive roots and so the Galois group is $G\cong \Bbb U_{20}$ I have a question about one of the Galois correspondences between the subfields of $\Bbb Q(w_{20})$ and the subgroups of $G$ in particular though . Consider the subgroup $\langle 9 \rangle$.

It seems to me that the element $w+w^9$ would be fixed by this group as (denoting $\sigma$ the automorphism which generates this group ) $\sigma(w+w^9)=w^9+w^{81}=w^9+w^1$. However in my course notes it says that the element fixed by this subgroup is $w^2+w^{18}$. I can see from applying the automorphism that this is indeed fixed but I don't understand why my choice of element wasn't valid .. Could anyone please explain this to me ?

  • $\begingroup$ There are infinitely many elements fixed by $\sigma$. Who said your choice is not correct? $\endgroup$ – Jyrki Lahtonen Aug 10 at 5:51
  • $\begingroup$ @JyrkiLahtonen ah okay I had it in my head that there could only be one valid choice . So basically there are infinitely many elements fixed by $\sigma$ but also infinitely many choices not fixed by alpha and we just have to be sure not to mistakenly choose the latter ? Is the fact that there are infinitely many elements fixed due to the fact that this is a cyclotomic extension ? $\endgroup$ – excalibirr Aug 10 at 5:56
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    $\begingroup$ It depends on what the notes want to do with this fixed value. If the notes just need $w^2+w^{18}$ to be fixed, that's fine. However, while many elements are fixed, there is only one maximal subfield that is fixed, and as far as I can tell, $w+w^9\notin \Bbb Q(w^2+w^{18})$. $\endgroup$ – Arthur Aug 10 at 5:58
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    $\begingroup$ $(\omega+\omega^9)^2=\omega^2+\omega^{18}-2$. $\endgroup$ – Gerry Myerson Aug 10 at 7:04
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    $\begingroup$ You are right that the fixed field of $\langle 9\rangle$ is generated by $\omega+\omega^9$. It contains but is not generated by $\omega^2 +\omega^{18}$ - this latter element is also fixed by $\langle -1\rangle$, and $\langle 9, -1\rangle\not=\langle 9\rangle$. $\endgroup$ – ancientmathematician Aug 10 at 7:15

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