# Why isn't my choice of element fixed by the automorphism of this subgroup of $Gal(\Bbb Q(w_{20})/Q)$?

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 ?

• There are infinitely many elements fixed by $\sigma$. Who said your choice is not correct? – Jyrki Lahtonen Aug 10 at 5:51
• @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 ? – excalibirr Aug 10 at 5:56
• 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, $w+w^9\notin \Bbb Q(w^2+w^{18})$. – Arthur Aug 10 at 5:58
• $(\omega+\omega^9)^2=\omega^2+\omega^{18}-2$. – Gerry Myerson Aug 10 at 7:04
• 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 $\langle 9, -1\rangle\not=\langle 9\rangle$. – ancientmathematician Aug 10 at 7:15