By Fabrizio Catanese, Hélène Esnault, Alan Huckleberry, Klaus Hulek, Thomas Peternell

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**Example text**

Since ϕ normalizes the subgroup Gd,e = g we have ϕ−1 ◦g◦ϕ = g k for some k ∈ N and so g ◦ ϕ = ϕ ◦ g k . Hence g(ϕ(Cx )) = ϕ(g k (Cx )) = ϕ(Cx ) and, similarly, g(ϕ(Cy )) = ϕ(Cy ). This yields (25). e. ϕ(¯0) = ¯0. Now the last assertion follows easily. 10. If e2 ≡ 1 mod d then + − Nd,e = Nd,e , Nd,e , (27) while for e2 ≡ 1 mod d, (28) + + Nd,e = Nd,e , Nd,e = Nd,e ,τ . Proof. For ϕ ∈ Nd,e we let C = ϕ−1 (Cy ). 9 (see (25)) the cyclic group Gd,e stabilizes C and ¯0 ∈ C. 8 there is + − , Nd,e which sends C to one of the coordinate an automorphism ψ ∈ Nd,e −1 axes Cx , Cy .

10. (a) Let C = {f (y) = 0}, where f ∈ C[y] is a polynomial of degree ≥ 2 with simple roots. If K ⊆ C denotes the set of these roots, then Stab(C) = T1,0 · U + · Stab(K) , where U + ⊆ Aut(A2 ) is as in (6), T1,0 = {λ ∈ T | λ : (x, y) → (αx, y), α ∈ C× } , and the stabilizer Stab(K) ⊆ Aut(A1 ) → Aut(A2 ) acts naturally on the symbol y. (b) If C of type (II) is the coordinate cross {xy = 0} in A2 then Stab(C) = N (T) is the normalizer of the maximal torus T in the group GL(2, C). (c) If C of type (IV) is a union of r aﬃne lines through the origin, where r ≥ 3, then Stab(C) ⊆ GL(2, C) is a ﬁnite extension of the group T1,1 = C× · id of scalar matrices.

Hence this ﬁber cannot carry a curve with Euler characteristic 1. This leads to a contradiction, because d > 1 by our assumption. Thus the curve C 1 is smooth. It follows that every ﬁber of p is isomorphic to A1 . Hence there is an automorphism δ ∈ Aut(A2 ) sending the curves C i to the lines y = κi with distinct ki , where κ1 = 1. Moreover, we may suppose that δ(¯0) = ¯0. 8. 5). Hence the elements g = ρ(g) ∈ T and ρ(g ) ∈ T are conjugated in GL(2, C) and so either ρ(g ) = ρ(g) = g or ρ(g ) = τ gτ .