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Let $K=Q(\zeta_{13})$ a Number field of degree $12$ over $Q$ which is cyclic and has unique sub-fields $K_2,K_3,K_4$ and $K_6$ of degree $2,3,4$ and $6$ respectively. I need a single generator $\beta\in K$ such that whose powers should generate these sub-fields.

I was able to find an element $\alpha=(\zeta+\zeta^{5}+\zeta^{8}+\zeta^{12})^{1/2}$ which generates the degree $6$ field $K_6$ and $\alpha^2$ generates the degree $3$ field $K_3$. I don't know whether such a $\beta$ exists so that I can generate all of the $4$ sub-fields by raising $\beta$ to some power $k\in Z$.

Krithika M
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  • See this post, which also answers it for $\zeta_{13}$ in the text. – Dietrich Burde Mar 02 '22 at 08:59
  • Anyway, this idea it is not possible. So the question comes back to finding generators simply. – Dietrich Burde Mar 02 '22 at 11:07
  • I'm curious to know if there is a more elementary way – reuns Mar 02 '22 at 11:08
  • @reuns You mean to show that this is not possible? Your answer looks quite convincing and is elementary, not? – Dietrich Burde Mar 02 '22 at 11:13
  • @DietrichBurde Yes, but still using a few results including that subfields of an abelian extension are Galois. I wondered about finding $e_3,e_3,v_p(\beta^{e_2e_3})$ directly through elementary considerations. $\overline{\beta}$ is in $\Bbb{Q}(\zeta_{13})$ so we know that $b^2 \in \Bbb{Q}(\zeta_{13})$ and $b^{2e_2e_3}\in \Bbb{Q}_{>0}$. – reuns Mar 02 '22 at 11:19

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It is not possible to find $\beta\in \Bbb{Q}(\zeta_{13})$ and $e_j\in \Bbb{Z}$ such that $K_j=\Bbb{Q}(\beta^{e_j})$ for both $j=2,3$.

This is because $[\Bbb{Q}(\beta^{e_2e_3}):\Bbb{Q}]$ divides $2$ and $3$ so $\beta^{e_2e_3}\in \Bbb{Q}$.

Let $b=|\beta|$ and $n$ the least integer such that $b^n\in \Bbb{Q}$, so that (Eisenstein criterion) $x^n-b$ is the minimal polynomial of $b$ and $\beta=b\xi$ with $\xi^{e_2e_3}=\pm 1$.

$b\in \Bbb{Q}(\zeta_{13},\xi)$ so $\Bbb{Q}(b)/\Bbb{Q}$ must Galois which implies that $n\in 1,2$.

$[\Bbb{Q}(\beta^{e_3}):\Bbb{Q}]=3$ gives that $\Bbb{Q}(\beta^{e_3})=\Bbb{Q}(\beta^{2e_3})=\Bbb{Q}(\xi^{2e_3})$.

This is impossible, the only cyclotomic fields contained in $\Bbb{Q}(\zeta_{13})$ are $\Bbb{Q}(\zeta_{13})$ and $\Bbb{Q}$.

reuns
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