Bourbaki

Exercise 18, § 10

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 (aT+bcT+d)\left( \frac{aT + b}{cT + d} \right), for y = (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} ∈ 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=0h(−1)iSi(T)Xh−i H(X) = \prod_{s \in H} (X - s(T)) = \sum_{i=0}^{h} (-1)^i S_i(T) X^{h-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 (asbscsds)\begin{pmatrix} a_s & b_s \\ c_s & d_s \end{pmatrix} 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)\prod_s (c_s T + d_s) 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)\begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}, 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)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. 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−(1T)n)\left( X^n - \left( \frac{1}{T} \right)^n \right) = 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)\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}. We have S(X)=(X−T)(X−1T−1)(X−T+1T−1)S(X) = (X - T) \left( X - \frac{1}{T-1} \right) \left( X - \frac{T+1}{T-1} \right) and Rs(T)=T3−3T+1T2−1R^s(T) = \frac{T^3 - 3T + 1}{T^2 - 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 (10b1)\begin{pmatrix} 1 & 0 \\ b & 1 \end{pmatrix}, b∈Kb \in K. Then

U(X)=(X−T)q−(X−T)=Xq−X−(Tq−T)andRU(T)=Tq−T. U(X) = (X - T)^q - (X - T) = X^q - X - (T^q - T) \quad \text{and} \quad R^U(T) = T^q - T .

Let B be the group generated by Cq−1C_{q-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 B(X) = (X^q - X)^{q-1} - (T^q - T)^{q-1} \quad \text{and} \quad R^H(T) = (T^q - T)^{q-1}
Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.