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

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?

• I’m convinced by what you did! May 8, 2020 at 20:26

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 $$\zeta_{3}$$.