By Mura R.B., Rhemtulla A.

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Symmetric groups 41 Case 2: {x, y, z} ∩ Ω has 1 element, say x. Choose two elements y , z ∈ Ω distinct from x; it is easy to see that (xyz) and (xy z ) generate the alternating group A5 on {x, y, z, y , z }. In particular, the cycle (xy z) is in G; since this 3-cycle meets Ω in two elements, we are reduced to case 1, QED. 39). Th. 5 If K is any ﬁeld of characteristic 0, or of characteristic p not dividing n, and f (X) ∈ K[X] is Morse, then Gal(f (X) − T ) = Sn over K(T ). Proof: We may assume K to be algebraically closed.

By replacing V by a dense open subset, we may assume that the ai have no poles, that V is smooth and that the discriminant ∆ of f is invertible. 4. Hilbert’s irreducibility theorem 25 is an ´etale covering of degree n. Its Galois closure W = Vfgal has Galois group G. The proposition follows by applying prop. 1 to W . Examples: • G = S3 . Let f (X) = X 3 + a1 X 2 + a2 X + a3 be irreducible, with Galois group S3 over K(V ). The specialization at t has Galois group S3 if the following properties are satisﬁed: 1.

3 can be extended to aﬃne nspace An . More precisely, let A be a thin set in An (Q), and let IntA (N ) be the number of integral points (x1 , . . , xn ) ∈ A with |xi | ≤ N . D. Cohen) IntA (N ) = O(N n− 2 log N ). One can replace the log N term in this inequality by (log N )γ , where γ < 1 is a constant depending on A. The proof is based on the large sieve inequality: one combines th. 2 with cor. 2 of the appendix, cf. [Coh] and [Se9]. Problem: Let X ⊂ Pn be an absolutely irreducible variety of dimension r.