# Minimal polynomial using Galois theory

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:

$$(T-\zeta_{13})(T-\zeta_{13}^{-1})=T^2-\theta_4T+1$$

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 $\tau^6(\mu_{13}) = \mu_{13}^{-1}$.

• 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? Jun 1, 2016 at 14:21
• 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 $\frac{n}{m}$. Jun 1, 2016 at 14:34
• The above facts follow from: The Fundamental Theorem of Galois Theory Jun 1, 2016 at 14:35