I have a couple of questions, given below, about the following problem from a course in Galois Theory.

Let $K=\mathbb{Q}(\zeta_{13})$. $K$ contains a unique subfield $L_4$ such that $[L_4 : \mathbb{Q}]=6$

Find the minimal polynomial for $\zeta_{13}$ over $L_4$

My lecturer provided a solution to this problem as follows:

$Gal(\mathbb{Q}(\zeta_{13})/\mathbb{Q})=\{e, \tau^6\}$. Therefore, $\zeta_{13}$ has two conjugates over $L_4$. Therefore, the appropriate minimal polynomial equals:


I have a few questions:

  1. Why does: $[L: \mathbb{Q}]=p$ where $L$ is a subfield of $\mathbb{Q}(\zeta_{q}) \implies Gal(\mathbb{Q}(\zeta_{q})/\mathbb{Q})=\{e, \tau^p\}$?
  2. Why does $Gal(\mathbb{Q}(\zeta_{q})/\mathbb{Q})=\{e, \tau^p\}$ imply the existence of two conjugates? Is it because it is a cylic group?

Thank you


You have some typos:

$Gal(\mathbb{Q}(\mu_{13})/\mathbb{Q})\cong \left(\mathbb{Z}_{13}\right)^* \cong \mathbb{Z}_{12}$.

By the basics of the Galois correspondence, $[L_4:\mathbb{Q}]=6$ tells you that the subgroup which has $L_4$ as its fixed field would be index $6$. And so the subgroup we are looking for would be order 2. If $\tau$ represents the generator of $Gal(\mathbb{Q}(\mu_{13})/\mathbb{Q})$, then we see that the subgroup of order 2 is exactly $\{e, \tau^6\}$. So $Gal(\mathbb{Q}(\mu_{13})/L_4) = \{e,\tau^6\}$.

The reason that $\mu_{13}$ has two conjugates over $L_4$ is that the conjugates are precisely the image of $\mu_{13}$ under the elements of $Gal(\mathbb{Q}(\mu_{13})/L_4)$. That is the conjugates of $\mu_{13}$ are $e(\mu_{13})=\mu_{13}$ and $\tau^6(\mu_{13}) = \mu_{13}^{-1}$.

Another way to think about it is that $[\mathbb{Q}(\mu_{13}):L_4]=2$. So the minimal polynomial of $\mu_{13}$ over $L_4$ has degree dividing 2. However the degree cannot be $1$ since $\mu_{13} \notin L_4$. So we have a quadratic minimal polynomial and we already know one root. The other root is found as above to be $\tau^6(\mu_{13}) = \mu_{13}^{-1}$.

  • $\begingroup$ thank you this clarified things a lot for me. Just one thing - If in general the Galois group is $\mathbb{Z_n}$ and $[L, \mathbb{Q}]=m$ then the order of the subgroup with $L$ as its fixed field is $\frac{n}{m}$? Is because of Lagrange's theorem? $\endgroup$ – thinker Jun 1 '16 at 14:21
  • $\begingroup$ For each $m$ that divides $n$, there is exactly one subgroup of $\mathbb{Z}_n$ of order $\frac{n}{m}$. This is a straigtforward group theory exercise. (And Lagrange's theorem tells you that $m$ must divide $n$.) Let $\mathbb{Q} \subset K$ be a Galois extension with Galois group $\mathbb{Z}_n$, $m$ divide $n$, and $L$ be the fixed field corresponding to the subgroup of order $\frac{n}{m}$. Then $[L:\mathbb{Q}]=m$. Conversely for each $m$ dividing $n$ there is a unique subfield with $[L:\mathbb{Q}]=m$. It's corresponding subgroup has order $\frac{n}{m}$. $\endgroup$ – Ken Duna Jun 1 '16 at 14:34
  • $\begingroup$ The above facts follow from: The Fundamental Theorem of Galois Theory $\endgroup$ – Ken Duna Jun 1 '16 at 14:35

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