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The question I have on my homework is to prove the following theorem.

If $n \in \{3, 4, 6\}$ and $p \in \mathbb{Z}_{>0}$ is prime, then the Galois group of $x^n − p$ is the dihedral group $D_n$.

So far I have the following:

Let us define the following fields $F = \mathbb{Q}$, $L = \mathbb{Q}(\zeta_n)$, and $K= L(\sqrt[n]{p}) = \mathbb{Q}(\zeta_n,\sqrt[n]{p})$. As we have shown in part (a). $|\text{Gal}(L/F)| = 2$. The nontrivial element of the Galois group sends $\zeta_n \mapsto -\zeta_n$. When we draw this in the complex plane this is a reflection about the real axis. Now, looking at $K/L$, we can see that $|\textrm{Gal}(K/L)| = [K:L] = n$ because $x^n-p$ has degree $n$ and is irreducible in $L$. This tells us that $|\text{Gal}(K/F)| = [K:L][L:F] = 2n = |D_n|$. It now remains to show that $\text{Gal}(K/L)$ corresponds to the rotations of $D_n$.

I'm not entirely sure where to go from here. Intuitively, it makes sense because $\alpha_1 = \sqrt[n]{p}$ can go to any root of $x^n-p$ and then the rest of the roots are fixed because they are just multiples of $\alpha_1$ in $L$ but I don't know how to show this rigorously.

I have seen two similar questions but the answers don't seem to lead me any closer to the answer of my question. Here is one, and here is the other. Thank you in advance.

suckling pig
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Nic
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  • In part (a) I showed that $x^2+x+1$, $x^2+1$, and $x^2-x+1$ have splitting fields $\mathbb{Q}(\zeta_n)$ for $n=3,4,6$, respectively. And that the only nontrivial element of the Galois group is given by complex conjugation – Nic May 02 '24 at 21:42

2 Answers2

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Your second link leads to a rigorous answer of your question. Indeed, in the first response of this link, the following proposition is stated:

Proposition [Jacobson, Velez]: One has $≅ℤ/⋊(ℤ/)^{\times}$ if and only if $$ is odd, or $$ is even and $\sqrt a∉ℚ(_)$.

You need now to prove, that if $n \in \{4, 6\}$ and $p$ is a prime number, then $\sqrt p∉ℚ(_)$. Indeed this follows from the statement in this link.

Since $n \in \{3, 4, 6\}$, $(ℤ/n)^{\times}=ℤ/2$, and we obtain $$≅ℤ/⋊(ℤ/)^{\times}≅ℤ/⋊ℤ/2=D_n.$$

nor
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I asked this a few weeks ago and didn't get an answer that made sense to me so I'll leave my solution here:

We know that roots of $x^n-p$ in $K$ are $\alpha_j = \zeta_n^j\sqrt[n]{p}$ for $j \in \{1,...,n\}$. For $\phi\in \textrm{Aut}(K/L)$, we know that $\phi(\alpha_1)$ can equal any $\alpha_j$ let us call $\phi(\alpha_1) = \zeta_n^k\alpha_1$. However, after that, the rest of the $\alpha_j$'s are fixed because $\alpha_j = \zeta_n^{j-1}\alpha_1$. Then $\phi(\alpha_j) = \phi(\zeta_n^{j-1})\phi(\alpha_1)$ and because the $\zeta_n$'s are fixed we have that $\phi(\alpha_j) = \zeta_n^{j+k-1}\alpha_1 = \zeta_n^k \alpha_j$. This means that every $\alpha$ is rotated in the complex plane by a phase $\zeta_n^k$. We have now shown that every element of $\text{Gal}(K/F)$ can be described by a unique element of $D_n$ acting on the complex plane with the roots of the $x^n-p$ as the corners of a polygon.

Nic
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