Determine the Galois groups of the following polynomials over \(\mathbb Q\).
\(f(x)=x^4+4x^2+1\)
\(f(x)=x^4+4x^2-5\).
Solution
1. For\[ f_1(x)=x^4+4x^2+1, \]let \(\alpha\) satisfy\[ \alpha^2=-2+\sqrt3. \]Then the four roots of \(f_1\) are\[ \pm\alpha,\qquad \pm\alpha^{-1}. \]
Solving the quadratic equation in \(y=x^2\) gives \[ y^2+4y+1=0, \qquad y=-2\pm\sqrt3. \] Since \[ (-2+\sqrt3)(-2-\sqrt3)=1, \] we have \[ \alpha^{-2}=-2-\sqrt3. \] Thus the square roots of the two values of \(y\) are exactly \(\pm\alpha\) and \(\pm\alpha^{-1}\).
2. The splitting field of \(f_1\) is\[ K_1=\mathbb Q(\alpha), \qquad [K_1:\mathbb Q]=4. \]
By 1, \(\mathbb Q(\alpha)\) already contains every root, so it is the splitting field. Moreover \[ \sqrt3=\alpha^2+2\in\mathbb Q(\alpha), \] so \[ \mathbb Q\subset \mathbb Q(\sqrt3)\subset \mathbb Q(\alpha). \] The first extension has degree \(2\). The element \(-2+\sqrt3\) is not a square in the real quadratic field \(\mathbb Q(\sqrt3)\): under the identity real embedding it is negative, whereas a square is nonnegative. Hence \[ [\mathbb Q(\alpha):\mathbb Q(\sqrt3)]=2. \] Therefore \([K_1:\mathbb Q]=4\).
3. We have\[ \operatorname{Gal}(K_1/\mathbb Q)\cong C_2\times C_2. \]
Since \(K_1\) is the splitting field of a separable polynomial over \(\mathbb Q\), it is Galois of degree \(4\). Two nontrivial automorphisms are determined by \[ \tau(\alpha)=-\alpha, \qquad \sigma(\alpha)=\alpha^{-1}. \] Both have order \(2\), and they commute because \[ \sigma\tau(\alpha)=(-\alpha)^{-1}=-\alpha^{-1} =\tau\sigma(\alpha). \] Thus the Galois group contains three nontrivial involutions and is therefore the Klein four group, not the cyclic group of order \(4\).
4. For\[ f_2(x)=x^4+4x^2-5, \]we have\[ f_2(x)=(x^2-1)(x^2+5). \]Hence its splitting field is\[ K_2=\mathbb Q(\sqrt{-5}). \]
The factorization is immediate: \[ (x^2-1)(x^2+5)=x^4+4x^2-5. \] The roots are \(\pm1\) and \(\pm\sqrt{-5}\), so adjoining \(\sqrt{-5}\) gives all roots.
5. Therefore\[ \operatorname{Gal}(K_2/\mathbb Q)\cong C_2. \]
Since \(-5\) is not a square in \(\mathbb Q\), \([K_2:\mathbb Q]=2\). Every quadratic extension in characteristic \(0\) is Galois, with nontrivial automorphism \[ \sqrt{-5}\longmapsto-\sqrt{-5}. \]