What is the Galois group of \(x^4 - 2x^2 + 9\)?
Solution
Let \[ f(x)=x^4-2x^2+9. \]
1. The polynomial \(f\) is irreducible over \(\QQ\).
By Gauss’s lemma, if \(f\) were reducible over \(\QQ\), then since it is monic it would factor into monic polynomials in \(\ZZ[x]\). It has no rational root, so any factorization would be into quadratics: \[ f=(x^2+ax+b)(x^2+cx+d), \qquad a,b,c,d\in\ZZ. \] Comparing the \(x^3\)-coefficient gives \(c=-a\), and comparing the \(x\)-coefficient gives \[ a(d-b)=0. \]
If \(a=0\), then \(b+d=-2\) and \(bd=9\), which has no integer solution. If \(d=b\), then \(b^2=9\). For \(b=3\), the \(x^2\)-coefficient gives \(6-a^2=-2\), so \(a^2=8\), impossible in \(\ZZ\); for \(b=-3\), it gives \(-6-a^2=-2\), so \(a^2=-4\), also impossible. Hence \(f\) is irreducible.
2. If \(\alpha\) is any root of \(f\), then \(3/\alpha\) is also a root.
Since \(\alpha\ne0\) and \[ \alpha^4-2\alpha^2+9=0, \] we have \[ \left(\frac3\alpha\right)^4 -2\left(\frac3\alpha\right)^2+9 = \frac{9(\alpha^4-2\alpha^2+9)}{\alpha^4}=0. \]
3. The field \(\QQ(\alpha)\) already contains all four roots of \(f\).
Because \(f\) is even, \(-\alpha\) is a root. By 2, \(3/\alpha\) is a root, and then so is \(-3/\alpha\). These four roots are distinct: equality \(\alpha=\pm3/\alpha\) would imply \(\alpha^2=\pm3\), neither of which satisfies \(f\).
Thus the four roots are \[ \pm\alpha, \qquad \pm\frac3\alpha, \] all of which lie in \(\QQ(\alpha)\).
4. Therefore the splitting field is \(L=\QQ(\alpha)\) and\[ [L:\QQ]=4. \]
By 1, the minimal polynomial of \(\alpha\) has degree \(4\), so \([\QQ(\alpha):\QQ]=4\). By 3 this field contains every root, hence is the splitting field.
5. The Galois group has at least two distinct nontrivial involutions.
Since \(L\) is the splitting field of a separable polynomial over \(\QQ\), \(L/\QQ\) is Galois of degree \(4\).
The embedding determined by \[ \alpha\mapsto-\alpha \] is an automorphism \(\tau\) of order \(2\). Likewise, \[ \alpha\mapsto\frac3\alpha \] defines an automorphism \(\sigma\) because \(3/\alpha\) is another root and generates the same field. Moreover, \[ \sigma^2(\alpha) = \frac{3}{3/\alpha} = \alpha, \] so \(\sigma\) also has order \(2\). These automorphisms are distinct.
6. Hence\[ \operatorname{Gal}(f/\QQ)\cong C_2\times C_2. \]
A group of order \(4\) is either \(C_4\) or \(C_2\times C_2\). The cyclic group \(C_4\) has exactly one element of order \(2\), while 5 exhibits two distinct nontrivial involutions. Therefore the Galois group is the Klein four group.