Consider the splitting field \(E\) of the polynomial \(f(x) = x^4 - 5\) over \(\mathbb{Q}\).
Find the degree \([E : \mathbb{Q}]\).
Determine the Galois group of \(E\) over \(\mathbb{Q}\) as a subgroup of the symmetric group \(S_4\).
Solution
Let \[ \alpha:=\sqrt[4]{5}>0. \] The roots of \(x^4-5\) are \[ \alpha, \qquad i\alpha, \qquad -\alpha, \qquad -i\alpha. \]
1. The splitting field is\[ E=\mathbb Q(\alpha,i). \]
The field \(\mathbb Q(\alpha,i)\) contains all four roots displayed above, so it contains the splitting field. Conversely, any splitting field contains both \(\alpha\) and \(i\alpha\), hence also their quotient \[ \frac{i\alpha}{\alpha}=i. \] Therefore it contains \(\mathbb Q(\alpha,i)\). Thus the two fields are equal.
2. One has\[ [E:\mathbb Q]=8. \]
The polynomial \[ x^4-5 \] is Eisenstein at \(5\), so it is irreducible over \(\mathbb Q\). Hence \[ [\mathbb Q(\alpha):\mathbb Q]=4. \] Since \(\alpha\) is real, \[ \mathbb Q(\alpha)\subseteq\mathbb R. \] Therefore \[ i\notin\mathbb Q(\alpha). \] The element \(i\) has degree \(2\) over \(\mathbb Q\), and its polynomial \(x^2+1\) remains irreducible over the real field \(\mathbb Q(\alpha)\). Thus \[ [E:\mathbb Q(\alpha)]=2. \] By the tower law, \[ [E:\mathbb Q] =[E:\mathbb Q(\alpha)]\,[\mathbb Q(\alpha):\mathbb Q] =2\cdot4=8. \]
3. There is an automorphism \(r\in\operatorname{Gal}(E/\mathbb Q)\) satisfying\[ r(\alpha)=i\alpha, \qquad r(i)=i, \]and it has order \(4\).
The fields \(\mathbb Q(\alpha)\) and \(\mathbb Q(i)\) have intersection \(\mathbb Q\): their intersection has degree dividing \(2\), and a nontrivial intersection would equal the imaginary quadratic field \(\mathbb Q(i)\), impossible because \(\mathbb Q(\alpha)\subseteq\mathbb R\). Therefore \[ [E:\mathbb Q(i)]=[\mathbb Q(\alpha):\mathbb Q]=4. \] Hence the minimal polynomial of \(\alpha\) over \(\mathbb Q(i)\) still has degree \(4\) and is \(x^4-5\). Sending the root \(\alpha\) to the root \(i\alpha\) while fixing \(\mathbb Q(i)\) therefore defines a \(\mathbb Q(i)\)-automorphism \(r\) of \(E\). Iterating gives \[ \alpha\mapsto i\alpha\mapsto-\alpha\mapsto-i\alpha\mapsto\alpha, \] so \(r\) has order \(4\).
4. Complex conjugation gives an automorphism \(s\) satisfying\[ s(\alpha)=\alpha, \qquad s(i)=-i, \]and\[ srs=r^{-1}. \]
Because \(E=\mathbb Q(\alpha,i)\) is stable under complex conjugation, conjugation restricts to a \(\mathbb Q\)-automorphism \(s\) of \(E\). Clearly \(s^2=1\). On the generators, \[ srs(\alpha) =sr(\alpha) =s(i\alpha) =-i\alpha =r^{-1}(\alpha), \] and \[ srs(i)=i=r^{-1}(i). \] Thus \[ srs=r^{-1}. \]
5. With the roots labeled\[ 1:\alpha, \qquad 2:i\alpha, \qquad 3:-\alpha, \qquad 4:-i\alpha, \]one has\[ \operatorname{Gal}(E/\mathbb Q) =\langle(1234),(24)\rangle \cong D_4. \]
Under the chosen labeling, \(r\) acts by \[ (1234), \] while complex conjugation \(s\) fixes \(\alpha\) and \(-\alpha\) and exchanges \(i\alpha\) with \(-i\alpha\), so it acts by \[ (24). \] The subgroup they generate is the dihedral group \[ \langle r,s\mid r^4=s^2=1,\ srs=r^{-1}\rangle \] of order \(8\). By 2, the Galois group of the splitting field has order \[ [E:\mathbb Q]=8. \] Hence the subgroup generated by \(r\) and \(s\) is the entire Galois group. Therefore, as a subgroup of \(S_4\), \[ \operatorname{Gal}(E/\mathbb Q) =\langle(1234),(24)\rangle. \]