Algebra, V, p. 158
Let K be a field and F the field of rational fractions K(T). The group GL₂(K) operates on F by the law (yf)(T) = f (cT+daT+b), for y = (acbd) ∈ GL₂(K) and f ∈ F. The centre K* . I of GL₂(K) operates trivially, whence we obtain, by passage to quotients, a homomorphism of PGL₂(K) = GL₂(K)/K* . I into the automorphism group of the extension F of K. This homomorphism is bijective (V, p. 148, Ex. 11). a) Let H be a subgroup of PGL₂(K) of finite order h, and let E be the field of invariants of H, so that F = E(T) is a Galois extension of E with Galois group H (V, p. 66, Th. 3). Let
H(X)=s∈H∏(X−s(T))=i=0∑h(−1)iSi(T)Xh−i be the minimal polynomial of T over E. For each i such that 0 ≤ i ≤ h, either S_i(T) ∈ K or E = K(S_i(T)). (Use Ex. 11, a) of V, p. 157.) b) For each s ∈ H let
(ascsbsds) be a representative in GL₂(K) and let H₀ be the subgroup of H consisting of those s for which cₛ = 0; let h₀ = Card H₀ and put R^H(T) = S_{h₀}(T). Show that E = K(R^H(T)) and that R^H(T) = P(T)/Q(T)^{h₀} where P is a polynomial of degree h and Q(T) =
∏s(csT+ds) where s runs over a system of representatives of the cosets H₀s different from H₀; the degree of Q is thus (h/h₀) − 1. c) Suppose that K* contains a subgroup μₙ of finite order n. Let C, be the image in PGL₂(K) of the subgroup of GL₂(K) consisting of the matrices
(a001), a ∈ μₙ. Then C, (X) = Xⁿ − Tⁿ and R^{Cₙ}(T) = (−1)^{n+1} Tⁿ. d) Let w be the element of PGL₂(K) represented by
(0110). Then w² = 1 and wsw⁻¹ = s⁻¹ if s is represented by a diagonal matrix. The subgroup Dₙ of PGL₂(K) generated by Cₙ and w is therefore of order 2n. We have Dₙ(X) = (Xⁿ − Tⁿ)
(Xn−(T1)n) = X^{2n} + (−1) R^{Dₙ}(T) + 1 where R^{Dₙ}(T) = − ((−T)ⁿ + (−T)⁻ⁿ).
e) Let S be the group of order 3 in PGL_2(K) generated by the image of the matrix (0111). We have S(X)=(X−T)(X−T−11)(X−T−1T+1) and Rs(T)=T2−1T3−3T+1
f) Suppose that K is a finite field with q elements. Let U be the image in PGL_2(K) of the group of all matrices (1b01), b∈K. Then
U(X)=(X−T)q−(X−T)=Xq−X−(Tq−T)andRU(T)=Tq−T.
Let B be the group generated by Cq−1 and U; it is of order q(q - 1). We have
B(X)=(Xq−X)q−1−(Tq−T)q−1andRH(T)=(Tq−T)q−1