Show $\sqrt[3]{5}$ is not contained in any cyclotomic extension of $\mathbb{Q}$.

Question

Find the Galois group of $x^3-5$ over $\mathbb{Q}$, then show $\sqrt[3]{5}$ is not contained in any cyclotomic extension of $\mathbb{Q}$.

My attempt:

The roots of $x^3-5$ are $\sqrt[3]{5},\zeta_3\sqrt[3]{5},\zeta_3^2\sqrt[3]{5}$. So the splitting field for $x^3-5$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[3]{5},\zeta_3)$, where $\zeta_3$ is a primitive $3^\text{rd}$ root of unity. By the Degree Formula for field extensions, we have $$ [\mathbb{Q}(\sqrt[3]{5},\zeta_3):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{5},\zeta_3):\mathbb{Q}(\zeta_3)][\mathbb{Q}(\zeta_3):\mathbb{Q}]=3\varphi(3)=3\cdot2=6, $$ where $\varphi$ is Euler's totient function. Define the automorphisms $$ \sigma_{ij}:=\begin{cases} \sqrt[3]{5}&\longmapsto\quad\zeta_3^i\sqrt[3]{5}\\ \zeta_3&\longmapsto\quad\zeta_3^j \end{cases} $$ where $0\leq i\leq 2$ and $1\leq j\leq 2$. Counting the $\sigma_{ij}$, we see we have found $6$ automorphisms, so we have found all elements of the Galois group. Since there is no element of order $6$, we know the Galois group is $S_3$. Finally, suppose $\sqrt[3]{5}$ is contained in a cyclotomic extension of $\mathbb{Q}$, call it $\mathbb{Q}(\zeta_n)$. By the Fundamental Theorem of Galois Theory, $S_3$ is a subgroup of $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$. Since $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is abelian, this implies $S_3$ is abelian, a contradiction. Hence $\sqrt[3]{5}$ is not contained in any cyclotomic extension of $\mathbb{Q}$. Is this correct?

Answer 1

The splitting field is correct and the reasoning seems the right path to take, but notice that in order to affirm that $\mathbb{Q}(\sqrt[3]5) \not\subset \mathbb{Q}(\zeta_{n})$ you don't have to put in play $S_{3}$ : If $\mathbb{Q}(\sqrt[3]5) \subset \mathbb{Q}(\zeta_{n})$ since Gal($\mathbb{Q}(\zeta_{n})/\mathbb{Q}$) is in particolar abelian, every subgroup is normal. Thanks to the fundamental theorem of Galois this condition translates into : every subextension is normal over $\mathbb{Q}$. But $\mathbb{Q}(\sqrt[3]5)$ can't be since is not the splitting field of $x^{3}-5$, infact he's missing $\zeta_{3}$.