By L. R. A. Casse, W. D. Wallis
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Extra info for Combinatorial Mathematics IV
Hence t R 2 (t) = λD2n+1 (t/γ ). Lemma 19. If r = 2 the second term of the alternative in Lemma 11 gives (21). Proof. The second term of the alternative in Lemma 11 gives for r = 2 G(y) = (y − η)Q 2 (y) + λ∗ , H (x) = (x − ξ )P 2 (x) + λ∗ , where Q(η), P(ξ ) = 0, p, Q = 2. By Lemma 12 we have y03 Q (y0 + η) = x03 P 2 (x0 + ξ ) = λ for all x0 , y0 satisfying Q(y0 + η) = P(x0 + ξ ) = 0. Hence polynomials Q(t + η), P(t + ξ ) satisfy the assumptions of Lemma 18 and by that lemma t Q(t + η)2 = λDm (t/γ ), t Pn (t + ξ )2 = λDn (t/β), where β, γ ∈ k∗ .
However a0 F˜ r2 − C r = a0 r −1/r ( F˜2 − ζrν a0 C) ν=1 and at most one factor has degree < n. −1/r It follows that F˜2 = ζrν a0 C for some ν ≤ r . Setting F˜1 (x) = a0 x r + r i=1 i −1 a˜ i x r −i we infer from F = F˜1 ◦ F˜2 by induction on i that a˜ i ζr−νi a0r ∈ k, −1/r whence F˜1 (ζrν a0 x) ∈ k[x]. But then F is decomposable over k. 3 Ritt’s first theorem Example 3. Let k = F2 . Then F(x) = x 4 + x 2 + x = (x 2 + αx)2 + α −1 (x 2 + αx) where α 2 − α + 1 = 0, α ∈ F4 shows that the assumption ∂ F ≡ 0(mod char k) cannot be omitted.
Also ∂ A = r µ, ∂ B = r ν. By Theorem 4, there exists a polynomial generating k(F) ∩ k(G); by Lemma 1 we may assume it without loss of generality to be H . We shall prove r = 1. By Lemma 3 there exist monic polynomials C, D ∈ k[x] such that ∂C = µ, ∂ D = ν, ∂(A − C r ) < µ(r − 1), ∂(B − Dr ) < ν(r − 1). Hence ∂ A(F) − C r (F) < dµν(r − 1), ∂ B(G) − Dr (G) < dµν(r − 1), ∂ C r (F) − Dr (G) < dµν(r − 1). But C r (F) − Dr (G) = C(F) − D(G) C r −1 (F) + · · · + Dr −1 (G) . Since C, D, F, G are monic and r ≡ 0(mod char k), the second factor has degree (r − 1) dµν and therefore C(F) = D(G) ∈ k(F) ∩ k(G) = k(H ).