Let \(p\) be an odd prime. Show that \(\sin\!\left(\frac{2\pi}{p}\right)\) is algebraic over \(\mathbb{Q}\) and determine its degree over \(\mathbb{Q}\).
Show that \(\mathbb{Q}\!\left(\sin\!\left(\frac{2\pi}{p}\right)\right)\) is a Galois extension of \(\mathbb{Q}\) and determine its Galois group over \(\mathbb{Q}\).
Find all \(p\)’s such that \(p^{1/3} \in \mathbb{R}\) is contained in \(\mathbb{Q}\!\left(\sin\!\left(\frac{2\pi}{p}\right)\right)\).
Solution
Let \[ \xi=e^{2\pi i/(4p)}. \] Then \(\xi\) is a primitive \(4p\)-th root of unity.
1. The element \(\sin(2\pi/p)\) generates the maximal real subfield of \(\mathbb Q(\xi)\).
Using \(\sin\theta=\cos(\pi/2-\theta)\), \[ \sin\!\left(\frac{2\pi}{p}\right) = \cos\!\left(\frac{(p-4)\pi}{2p}\right) = \frac{\xi^{p-4}+\xi^{-(p-4)}}2. \] Because \(p\) is odd, \[ \gcd(p-4,4p)=1, \] so \(\eta:=\xi^{p-4}\) is again a primitive \(4p\)-th root of unity. Hence \[ \mathbb Q\!\left(\sin\frac{2\pi}{p}\right) = \mathbb Q(\eta+\eta^{-1}). \] Now \(\eta\) satisfies \[ T^2-(\eta+\eta^{-1})T+1=0, \] so \([\mathbb Q(\eta):\mathbb Q(\eta+\eta^{-1})]\le2\). Complex conjugation is a nontrivial automorphism of \(\mathbb Q(\eta)\) fixing \(\eta+\eta^{-1}\), so this degree is exactly \(2\). Thus \(\mathbb Q(\eta+\eta^{-1})\) is the maximal real subfield of \(\mathbb Q(\eta)=\mathbb Q(\xi)\).
2. Therefore\[ \left[\mathbb Q\!\left(\sin\frac{2\pi}{p}\right):\mathbb Q\right]=p-1. \]
Since \(p\) is odd, \[ [\mathbb Q(\xi):\mathbb Q]=\varphi(4p)=\varphi(4)\varphi(p)=2(p-1). \] By 1 the desired field has index \(2\) in \(\mathbb Q(\xi)\), so its degree over \(\mathbb Q\) is \(p-1\). This also proves algebraicity.
3. The extension \(\mathbb Q(\sin(2\pi/p))/\mathbb Q\) is Galois, with\[ \operatorname{Gal}\!\left(\mathbb Q\!\left(\sin\frac{2\pi}{p}\right)/\mathbb Q\right) \cong (\mathbb Z/4p\mathbb Z)^\times/\{\pm1\} \cong (\mathbb Z/p\mathbb Z)^\times \cong C_{p-1}. \]
The cyclotomic extension \(\mathbb Q(\xi)/\mathbb Q\) is Galois with abelian group \[ (\mathbb Z/4p\mathbb Z)^\times. \] Its maximal real subfield is the fixed field of complex conjugation, corresponding to the subgroup \(\{\pm1\}\). Since the ambient Galois group is abelian, this subgroup is normal, so the fixed field is Galois over \(\mathbb Q\), with quotient Galois group \[ (\mathbb Z/4p\mathbb Z)^\times/\{\pm1\}. \] By the Chinese remainder theorem, \[ (\mathbb Z/4p\mathbb Z)^\times \cong (\mathbb Z/4\mathbb Z)^\times\times(\mathbb Z/p\mathbb Z)^\times. \] Each coset modulo \(\{\pm1\}\) has a unique representative congruent to \(1\pmod4\), so restriction to the second factor identifies the quotient with \((\mathbb Z/p\mathbb Z)^\times\). The latter is cyclic of order \(p-1\).
4. There is no odd prime \(p\) such that \(p^{1/3}\in\mathbb Q(\sin(2\pi/p))\).
Set \[ K=\mathbb Q\!\left(\sin\frac{2\pi}{p}\right). \] By 3, \(K/\mathbb Q\) is cyclic Galois. Hence every intermediate field of \(K/\mathbb Q\) is Galois over \(\mathbb Q\), because every subgroup of a cyclic group is normal.
Suppose \(p^{1/3}\in K\). Then \(\mathbb Q(p^{1/3})\) is an intermediate field. The polynomial \[ T^3-p \] is irreducible over \(\mathbb Q\) by Eisenstein’s criterion at \(p\), so \([\mathbb Q(p^{1/3}):\mathbb Q]=3\). But \(\mathbb Q(p^{1/3})\subset\mathbb R\), while the other two roots \(\omega p^{1/3}\) and \(\omega^2p^{1/3}\) are nonreal. Therefore \(\mathbb Q(p^{1/3})/\mathbb Q\) is not normal, hence not Galois, a contradiction.
Thus no such prime exists.