Find the Galois group of \(x^4 + x^3 + x^2 + x + 1\)
over \(\mathbb{Q}\).
over \(\mathbb{F}_5\).
Solution
Put \(f(x)=x^4+x^3+x^2+x+1\).
1. Over \(\mathbb Q\), the Galois group is cyclic of order \(4\).
Let \(\zeta=e^{2\pi i/5}\). The identity \((x-1)f(x)=x^5-1\) shows that the roots of \(f\) are \(\zeta,\zeta^2,\zeta^3,\zeta^4\). They are distinct and all belong to \(L=\mathbb Q(\zeta)\), which they generate. Thus \(L\) is the splitting field.
The translated polynomial is \[ f(y+1)=y^4+5y^3+10y^2+10y+5. \] It is Eisenstein at \(5\), so it is irreducible over \(\mathbb Q\) [@DF04]. Translation is an automorphism of the polynomial ring, so \(f\) is irreducible as well and \([L:\mathbb Q]=4\).
For each \(a\in\{1,2,3,4\}\), sending \(\zeta\) to \(\zeta^a\) defines an embedding of \(\mathbb Q(\zeta)\) into \(L\), because \(\zeta^a\) has the same minimal polynomial \(f\). It is onto: choose \(b\) with \(ab\equiv1\pmod5\), and then \((\zeta^a)^b=\zeta\). Denote this automorphism by \(\sigma_a\). Conversely, every automorphism must send \(\zeta\) to a root of \(f\), so these are all the automorphisms. Their composition is \[ \sigma_a\sigma_b(\zeta)=\zeta^{ab}. \] Therefore \(\operatorname{Gal}(L/\mathbb Q)\) is isomorphic to \((\mathbb Z/5\mathbb Z)^\times\). The residues \(1,2,2^2=4,2^3=3\) are all its elements and \(2^4=1\) modulo \(5\). Hence the group is \(C_4\), generated by \(\zeta\mapsto\zeta^2\).
2. Over \(\mathbb F_5\), the Galois group is trivial.
In characteristic \(5\), \[ (x-1)^4=x^4-4x^3+6x^2-4x+1 =x^4+x^3+x^2+x+1=f(x). \] The only root is \(1\), already in \(\mathbb F_5\). The splitting field is therefore \(\mathbb F_5\) itself, and its only automorphism fixing the base field pointwise is the identity. The trivial extension \(\mathbb F_5/\mathbb F_5\) is Galois, with Galois group \(\{1\}\).