Find the Galois groups, over \(\mathbb{Q}\), of the following polynomials.
\(x^4 + 4\).
\(x^3 + x^2 - 2x - 1\). Hint: show that if \(r\) is a root, then \(r^2 - 2\) also is a root.
Solution
1. The Galois group of \(x^4+4\) over \(\mathbf Q\) is \(C_2\).
The factorization \[ x^4+4=(x^2-2x+2)(x^2+2x+2) \] gives roots \(1+i\), \(1-i\), \(-1+i\), and \(-1-i\). They all lie in \(\mathbf Q(i)\), and a field containing the roots contains \(i=(1+i)-1\). Hence the splitting field is exactly \(\mathbf Q(i)\). The polynomial \(x^2+1\) has no rational root, so this field has degree \(2\) over \(\mathbf Q\). Its automorphisms are the identity and complex conjugation, giving \(C_2\).
2. The Galois group of \(h(x)=x^3+x^2-2x-1\) over \(\mathbf Q\) is\(C_3\), acting cyclically on its roots.
2.1. The only possible rational roots are \(\pm1\), but\(h(1)=-1\) and \(h(-1)=1\). A reducible cubic has a linear factor, so\(h\) is irreducible. If \(\alpha\) is any root, then\([\mathbf Q(\alpha):\mathbf Q]=3\).
2.2. Direct multiplication gives\[ \begin{aligned} h(x^2-2)&=x^6-5x^4+6x^2-1\\ &=h(x)(x^3-x^2-2x+1). \end{aligned} \]Thus \(\beta=\alpha^2-2\) is another root in \(\mathbf Q(\alpha)\).It is distinct from \(\alpha\): otherwise \(\alpha\) would satisfy\(x^2-x-2=0\), contrary to its degree \(3\). Dividing \(h\) by\((x-\alpha)(x-\beta)\) over \(\mathbf Q(\alpha)\) gives a linear factor.Its root is\[ \gamma=-1-\alpha-\beta=1-\alpha-\alpha^2, \]by the coefficient of \(x^2\) in \(h\). All three roots are distinct,since an irreducible polynomial over a characteristic-zero field isseparable. Therefore the splitting field is exactly\(L=\mathbf Q(\alpha)\), of degree \(3\) over \(\mathbf Q\).
2.3. A splitting field in characteristic zero is Galois and itsautomorphism group has order equal to its degree [@DF04]. Hence\(\operatorname{Gal}(L/\mathbf Q)\) has order \(3\) and is cyclic.More explicitly, the isomorphism\(\mathbf Q(\alpha)\to\mathbf Q(\beta)\) determined by\(\alpha\mapsto\beta\) is an automorphism of \(L\), since\(\mathbf Q(\beta)\subseteq L\) also has degree \(3\) over \(\mathbf Q\).The relation \(h(\alpha)=0\) gives\(\alpha^4=3\alpha^2-\alpha-1\), and consequently\[ \beta^2-2=\alpha^4-4\alpha^2+2 =1-\alpha-\alpha^2=\gamma. \]Thus this automorphism sends \(\alpha\) to \(\beta\) and \(\beta\) to\(\gamma\); being a permutation of the three roots, it sends \(\gamma\)back to \(\alpha\). It is the required generator of \(C_3\).