How many subfields does the splitting field of \(x^5-2\) over \(\mathbb Q\) have?
Solution
Let \[ \alpha\coloneqq2^{1/5}, \qquad \zeta\coloneqq e^{2\pi i/5}, \] and let \(L\) be the splitting field of \(x^5-2\) over \(\QQ\).
1. One has\[ L=\QQ(\alpha,\zeta) \]and\[ [L:\QQ]=20. \]
The five roots of \(x^5-2\) are \[ \alpha,\zeta\alpha,\zeta^2\alpha,\zeta^3\alpha,\zeta^4\alpha, \] so the splitting field is \(\QQ(\alpha,\zeta)\).
The polynomial \(x^5-2\) is Eisenstein at \(2\), hence \[ [\QQ(\alpha):\QQ]=5. \] Also \[ [\QQ(\zeta):\QQ]=\varphi(5)=4. \] The degree of \[ \QQ(\alpha)\cap\QQ(\zeta) \] over \(\QQ\) divides both \(5\) and \(4\), hence is \(1\). Therefore the two extensions are linearly disjoint over \(\QQ\), and \[ [L:\QQ] = 5\cdot4 = 20. \]
2. The Galois group \(G\coloneqq\Gal(L/\QQ)\) is generated byautomorphisms \(\sigma\) and \(\tau\) satisfying\[ \sigma^5=\tau^4=1, \qquad \tau\sigma\tau^{-1}=\sigma^2. \]In particular,\[ G\cong C_5\rtimes C_4. \]
Define \[ \sigma(\alpha)=\zeta\alpha, \qquad \sigma(\zeta)=\zeta, \] and \[ \tau(\alpha)=\alpha, \qquad \tau(\zeta)=\zeta^2. \] By step 1, \[ [L:\QQ(\zeta)]=5 \qquad\text{and}\qquad [L:\QQ(\alpha)]=4. \] Thus \(x^5-2\) remains the minimal polynomial of \(\alpha\) over \(\QQ(\zeta)\), so the assignment \(\alpha\mapsto\zeta\alpha\) defines a \(\QQ(\zeta)\)-automorphism \(\sigma\) of \(L\). Likewise, the linear disjointness in step 1 lets the cyclotomic automorphism \[ \zeta\longmapsto\zeta^2 \] extend to an automorphism \(\tau\) of \(L\) that fixes \(\QQ(\alpha)\). Their orders are respectively \(5\) and \(4\). Moreover, \[ \tau\sigma\tau^{-1}(\alpha) = \zeta^2\alpha = \sigma^2(\alpha), \] and both sides fix \(\zeta\), so \[ \tau\sigma\tau^{-1}=\sigma^2. \]
The subgroup generated by \(\sigma\) and \(\tau\) contains the \(20\) distinct elements \[ \sigma^i\tau^j, \qquad 0\leq i<5, \quad 0\leq j<4. \] Indeed, equality of two such elements, evaluated first on \(\zeta\), forces the two exponents \(j\) to agree modulo \(4\); evaluation on \(\alpha\) then forces the exponents \(i\) to agree modulo \(5\). By step 1, the whole Galois group has order \(20\), so these automorphisms generate all of \(G\).
3. The group\[ N\coloneqq\langle\sigma\rangle \]is the unique subgroup of \(G\) of order \(5\).
The number \(n_5\) of Sylow \(5\)-subgroups satisfies \[ n_5\equiv1\pmod5 \] and \[ n_5\mid4. \] Hence \(n_5=1\). Since \(N\) has order \(5\), it is this unique subgroup.
4. There is exactly one subgroup of each of the orders\[ 5,\ 10,\ 20. \]
3. By the correspondencetheorem, subgroups of \(G\) containing \(N\) correspond bijectively tosubgroups of\[ G/N\cong C_4. \]The cyclic group \(C_4\) has exactly one subgroup of each order \(1,2,4\).Their inverse images have orders \(5,10,20\), respectively, so there isexactly one subgroup of each of those orders.
5. The group \(G\) has exactly five subgroups of order \(4\).
These are the Sylow \(2\)-subgroups. Their number \(n_2\) satisfies \[ n_2\equiv1\pmod2 \] and \[ n_2\mid5, \] so \(n_2\) is either \(1\) or \(5\).
3. Since their orders are coprime,\[ N\cap P=\{1\}, \]and normality would give\[ [N,P]\subseteq N\cap P=\{1\}. \]Thus \(N\) and \(P\) would commute. Since \(\langle\tau\rangle\) has order\(4\), uniqueness would give \(P=\langle\tau\rangle\). But the relation instep 2 gives\[ \tau\sigma\tau^{-1}=\sigma^2\neq\sigma, \]so the action is nontrivial. Therefore \(n_2\neq1\), and hence\[ n_2=5. \]
6. The group \(G\) has exactly five subgroups of order \(2\).
Every element of \(G\) has a unique form \[ \sigma^i\tau^j \] with \(0\leq i<5\) and \(0\leq j<4\). An element of order \(2\) must map to an element of order dividing \(2\) in \[ G/N\cong C_4. \] It cannot lie in \(N\), whose nonidentity elements have order \(5\), so it must have the form \[ \sigma^i\tau^2. \]
Since conjugation by \(\tau^2\) sends \(\sigma\) to \(\sigma^4=\sigma^{-1}\), \[ (\sigma^i\tau^2)^2 = \sigma^i \bigl(\tau^2\sigma^i\tau^{-2}\bigr) \tau^4 = \sigma^i\sigma^{-i} = 1. \] Thus the five elements \[ \sigma^i\tau^2, \qquad 0\leq i<5, \] are precisely the five involutions of \(G\). Each generates a distinct subgroup of order \(2\).
7. The total number of subgroups of \(G\) is\[ \boxed{14}. \]
4. Therefore\[ 1+5+5+1+1+1=14. \]
8. The splitting field \(L\) has exactly\[ \boxed{14} \]subfields containing \(\QQ\).
The extension \(L/\QQ\) is finite Galois. By the fundamental theorem of Galois theory, its intermediate fields are in bijection with the subgroups of \(G=\Gal(L/\QQ)\). Step 7 counts those subgroups.
9. Q.E.D.
Step 8 gives the requested number of subfields.