Let \(p < q\) be two distinct odd primes such that \(p \nmid q - 1\). Let \(n = pq\) and \(\zeta_n\) be a primitive \(n\)-th root of unity in \(\mathbb{C}\).
Show that the splitting field \(E\) of \((x^p - 2)(x^q - 2)\) over \(\mathbb{Q}\) is \(\mathbb{Q}(\zeta_n, \sqrt[n]{2})\).
Determine \([E : \mathbb{Q}]\).
Find two subgroups of \(\mathrm{Gal}(E/\mathbb{Q})\) of order \((p-1)(q-1)\).
Show that the Sylow \(p\)-subgroup of \(\mathrm{Gal}(E/\mathbb{Q})\) is normal and determine its fixed field.
Hint
Since \(\gcd(p,q)=1\), there are integers \(a,b\) such that \(pa+qb=1\).
Solution
Put \(\alpha=\sqrt[n]{2}>0\), \(\zeta=\zeta_n\), and \(L=\mathbb Q(\zeta)\). Write \(m=(p-1)(q-1)\).
1. The splitting field is \(E=\mathbb Q(\zeta,\alpha)\).
The roots of \(x^p-2\) are \(\alpha^q(\zeta^q)^j\), \(0\leq j<p\); the roots of \(x^q-2\) are \(\alpha^p(\zeta^p)^k\), \(0\leq k<q\). All lie in \(\mathbb Q(\zeta,\alpha)\). Conversely, the splitting field contains \(\alpha^q\) and \(\alpha^p\), and their nonzero-root ratios give \(\zeta^q\) and \(\zeta^p\). Choose integers \(a,b\) with \(pa+qb=1\). Then \[ \alpha=(\alpha^p)^a(\alpha^q)^b, \qquad \zeta=(\zeta^p)^a(\zeta^q)^b \] also belong to the splitting field. Negative exponents are allowed because all four elements are nonzero. This proves equality of the two fields.
2. The degree is\[ \boxed{[E:\mathbb Q]=pq(p-1)(q-1)=nm}. \]
The polynomial \(x^n-2\) is Eisenstein at two, so \([\mathbb Q(\alpha):\mathbb Q]=n\) [@DF04]. The cyclotomic degree formula gives \([L:\mathbb Q]=\varphi(pq)=(p-1)(q-1)=m\) [@DF04]. Neither \(p\) nor \(q\) divides \(m\): for \(p\) this uses \(p\nmid q-1\), and for \(q\) it follows from \(p-1<q\) and \(q\nmid q-1\). Thus \(\gcd(n,m)=1\).
The degree \([E:\mathbb Q]\) is divisible by both \(n\) and \(m\), since \(E\) contains both subfields. It is consequently divisible by their product \(nm\). On the other hand, \(E=L(\alpha)\) and \(\alpha\) satisfies \(x^n-2\) over \(L\), so \([E:L]\leq n\) and \([E:\mathbb Q]\leq nm\). Both bounds force equality. In particular \(x^n-2\) remains irreducible over \(L\), and \([E:L]=n\).
3. The Galois group has the explicit form\[ G=\{\tau^a\sigma_b:a\in\mathbb Z/n\mathbb Z, \ b\in(\mathbb Z/n\mathbb Z)^\times\}, \]where\[ \tau(\alpha)=\zeta\alpha,\quad \tau(\zeta)=\zeta, \qquad \sigma_b(\alpha)=\alpha,\quad \sigma_b(\zeta)=\zeta^b. \]They satisfy \(\sigma_b\tau\sigma_b^{-1}=\tau^b\).
Since \(x^n-2\) is irreducible over \(L\), sending \(\alpha\) to its root \(\zeta\alpha\) defines an \(L\)-embedding of \(E\) into itself. Its image contains \(\alpha=\zeta^{-1}(\zeta\alpha)\), so it is the automorphism \(\tau\), of order exactly \(n\).
For each unit \(b\) modulo \(n\), the cyclotomic automorphism \(\gamma_b\) of \(L\) sends \(\zeta\) to \(\zeta^b\) [@DF04]. Every element of \(E\) has a unique expression \(\sum_{j=0}^{n-1}c_j\alpha^j\), \(c_j\in L\). Define \(\sigma_b\) by applying \(\gamma_b\) to each coefficient. This preserves addition. It preserves multiplication because products are reduced using \(\alpha^n=2\), whose right side is fixed by \(\gamma_b\). Its inverse is \(\sigma_{b^{-1}}\). Thus these are field automorphisms with the asserted values, and \(\sigma_b\sigma_c=\sigma_{bc}\).
The conjugation identity follows by evaluating both sides on \(\alpha\) and \(\zeta\), which generate \(E\). The maps \(\tau^a\sigma_b\) are all distinct: their values on \(\zeta\) determine \(b\), and their values on \(\alpha\) then determine \(a\). The splitting field is Galois in characteristic zero and has degree \(nm\), so these \(nm\) automorphisms exhaust its Galois group [@DF04].
4. Two distinct subgroups of order \(m\) are\[ H_0=\{\sigma_b:b\in(\mathbb Z/n\mathbb Z)^\times\}, \qquad H_1=\tau H_0\tau^{-1} =\{\tau^{1-b}\sigma_b:b\in(\mathbb Z/n\mathbb Z)^\times\}. \]
The composition rule in step 3 makes \(H_0\) a subgroup isomorphic to \((\mathbb Z/n\mathbb Z)^\times\), of order \(m\). Its conjugate \(H_1\) is a subgroup of the same order. The displayed description of \(H_1\) follows from \(\sigma_b\tau^{-1}=\tau^{-b}\sigma_b\). Every element of \(H_0\) fixes \(\alpha\), while every element of \(H_1\) fixes \(\tau(\alpha)=\zeta\alpha\). But \(\sigma_{-1}\in H_0\) sends \(\zeta\alpha\) to \(\zeta^{-1}\alpha\ne\zeta\alpha\), since \(\zeta\) has odd order \(n>2\). Hence \(\sigma_{-1}\notin H_1\), proving that these two subgroups are distinct.
5. The unique Sylow \(p\)-subgroup and its fixed field are\[ P=\langle\tau^q\rangle, \qquad \boxed{E^P=\mathbb Q(\zeta_n,\sqrt[q]{2})}. \]
Since \(|G|=pqm\) and \(p\nmid qm\), a Sylow \(p\)-subgroup has order \(p\). The element \(\tau^q\) has order \(p\), so its subgroup \(P\) is Sylow. Conjugation by \(\tau\) fixes \(\tau^q\), and conjugation by \(\sigma_b\) sends it to \(\tau^{bq}\), another generator of \(P\) because \(p\nmid b\). The elements \(\tau\) and all \(\sigma_b\) generate \(G\), so \(P\) is normal in \(G\). Sylow conjugacy then makes it the unique Sylow \(p\)-subgroup [@DF04].
Put \(M=\mathbb Q(\zeta,\alpha^p)\), noting that \(\alpha^p=\sqrt[q]{2}\). The subgroup \(P\) fixes both generators, because \((\tau^q)(\alpha^p)=\zeta^{pq}\alpha^p=\alpha^p\). Thus \(M\subseteq E^P\). Also \(E=M(\alpha)\) and \(\alpha\) satisfies \(T^p-\alpha^p\in M[T]\), so \([E:M]\leq p\). The \(p\) elements \(\tau^{qj}(\alpha)=\zeta^{qj}\alpha\), \(0\leq j<p\), are distinct, and each is a root of the minimal polynomial of \(\alpha\) over \(E^P\), because these automorphisms fix its coefficients. Hence \([E:E^P]\geq p\). The tower law yields \[ p\leq[E:E^P]\leq[E:M]\leq p, \] so equality holds throughout and \([E^P:M]=1\). Therefore \(E^P=M\), as claimed.