Galois group of $x^p-a$ over $\mathbb{Q}$

Question

I have found that the Galois group $G$ of $f=x^p-a$ over $\mathbb{Q}$ is of order $p(p-1)$. I need to show that if $P$ is a subgroup of $G$ of order $p$, then $P$ is normal and $G/P$ is cyclic. Furthermore, I need to describe the fixed field of $P$ explicitly.

Ideas so far: Sylow's Theorem says that $P$ is the unique Sylow $p$-subgroup of $G$, and thus $P$ is normal. However, the quotient $G/P$ is of order $p-1$. I'm not sure how to show $G/P$ is cyclic.

Finally, to find the fixed field of $P$, I think I should use the fact that the splitting field of $f$ is $\mathbb{Q}(b,\zeta)$ where $b$ is some root of $f$ and $\zeta$ is a primitive $p$-th root of unity. I imagine the Fundamental Theorem of Galois Theory comes into play here, but I am not yet sure how. Any advice?

Answer 1

Using the fundamental theorem of Galois theory here sounds like a good idea to me. Let $K = \mathbb{Q}(b, \zeta)$ in your notation. Since $P$ is a normal subgroup of $G$, $E := K^{P}$ is a Galois extension of $\mathbb{Q}$ with $\mathrm{Gal}(E/\mathbb{Q}) \cong G/P$. Moreover, $E$ is the unique extension of $\mathbb{Q}$ contained in $K$ satisfying $[E:\mathbb{Q}] = p-1$. Indeed, for any subfield $E' \subset K$ with $[E':\mathbb{Q}] = p-1$, then writing $E' = K^{H}$ for some subgroup $H$ of $G$, we see that $[G:H] = [E':\mathbb{Q}] = p-1$, which implies that $|H| = p$, i.e. $H$ is another Sylow $p$-subgroup of $G$. This forces $P = H$, since $P$ is the unique Sylow $p$-subgroup of $G$, and so $E' = E$.

Thus, it suffices to exhibit a degree $p-1$ extension of $\mathbb{Q}$ contained in $K$ which has cyclic Galois group over $\mathbb{Q}$, since this must be the equal to $E$ by the reasoning above. I leave to you to check that $\mathbb{Q}(\zeta)/\mathbb{Q}$ is one such extension.

Answer 2

So $P$ is normal in $G$ and hence there exists a unique sub-field of index $K'|_\mathbb Q$ of $K$, such that $[K':\mathbb Q]=p-1$ by the Galois corrspondence. Since $Q(\zeta_p)$ is one such field, clearly we have $K'=\mathbb Q(\zeta_p)$. Again since $P$ is normal, we get $G/P\cong Gal(K|_\mathbb Q)/Gal (K|_{\mathbb Q(\zeta_p)})\cong Gal (\mathbb Q(\zeta_p)|_\mathbb Q)$ which is cyclic.

Answer 3

The most "economical" (not necessarily the shortest) solution seems to be the following. Fix a $p$-th root $\alpha$ of $a$ (you forgot to mention that $\alpha \notin \mathbf Q$) and a primitive $p$-th root of unity $\zeta$. Clearly the decomposition field of $X^p-a$ is $N=\mathbf Q(\zeta,\alpha)$, which is galois over $\mathbf Q$, with Galois group $G$ of order $p(p-1)$ ($p$ choices for the conjugates of $\alpha$ and $(p-1)$ for those of $\zeta$). By the description of the roots, it is straightforward that $K=\mathbf Q(\zeta)$ is galois over $\mathbf Q$, say with group $Q$ permuting the powers of $\zeta$ (hence cyclic of order$p-1$) and, by the fundamental theorem, $N$ is automatically normal over $K=\mathbf Q(\zeta)$, say with group $P$ permuting the conjugates of $\alpha$ (hence cyclic of order $p$). If $P$ were not unique, $G$ would contain a subgroup of order $p^2$, a contradiction.

We can do even better. The group $G$ is an extension of the quotient $Q$ by the subgroup $P$, and since the orders of $P, Q$ are coprime, a classical theorem (Frobenius ?) asserts that $G$ is a semi-direct product (here, $not$ a direct product). A complete multiplication table for $G$ could be written down by following the conjugates of $\alpha$ and $\zeta$, but it is simpler to describe $G$ by generators and relations.