Polynomials with Special Regard to Reducibility by A. Schinzel

By A. Schinzel

This treatise covers lots of the identified effects on reducibility of polynomials over arbitrary fields, algebraically closed fields, and finitely generated fields. the writer contains a number of theorems on reducibility of polynomials over quantity fields which are both completely genuine or complicated multiplication fields. a few of these effects are in keeping with the hot paintings of E. Bombieri and U. Zannier, offered the following through Zannier in an appendix. The ebook additionally treats different matters corresponding to Ritt's conception of composition of polynomials, and homes of the Mahler degree and concludes with a bibliography of over three hundred goods.

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E. dµν ≥ r dµν, where r = 1. Hence [k(F): k(F) ∩ k(G)] = = [k(F): k(H )] = [k(F): k(A(F))] [∂ F, ∂G] . ∂A = µ = ∂F Similarly [∂ F, ∂G] ∂G and since the right hand sides of the above equalities are coprime [k(G): k(F) ∩ k(G)] = [k(F, G): k(F) ∩ k(G)] = [∂ F, ∂G] . (∂ F, ∂G) The theorem follows. The following examples show that the assumption ∂ H ≡ 0(mod char k) cannot be omitted. Example 1. k = F2 , F = x 2 , G = x 2 + x, H = x 4 + x 2 = F 2 + F = G 2 ; k(F) ∩ k(G) = k(H ), k(F, G) = k(x). Example 2.

Let k be a field and f ∈ k[x] be a tame polynomial of degree n > 1. Then the following assertions are equivalent ¯ (i) ( f (x) − f (y))/(x − y) is irreducible over k, (ii) ( f (x) − f (y))/(x − y) is irreducible over k(ζn ), (iii) f (x) is indecomposable and if n is an odd prime then we do not have f (x) = α Dn (x + b, a) + c for α, a, b, c ∈ k with a = 0 if n = 3. Lemma 1. Let n be a positive integer not divisible by char k. Put αk = ζnk + ζn−k , βk = ζnk − ζn−k . Then for every a ∈ k we have (n−1)/2 (x 2 − αk x y + y 2 + βk2 a) Dn (x, a) − Dn (y, a) = (x − y) k=1 if n is odd and otherwise (n−2)/2 (x 2 − αk x y + y 2 + βk2 a).

If F1 = a0 x r +a1 x r −1 + · · · + ar we can write F = F˜1 ◦ F˜2 , where F˜1 (x) = F1 x − aa01r , F˜2 (x) = F2 (x) + a1 and the coefficient of x r −1 in F˜1 (x) is 0. a0 r By Lemma 3 there exists C ∈ k[x] such that ∂C = n and ∂(F − C r ) < n(r −1), so ∂( F˜1 ◦ F˜2 −C r ) < n(r −1). It follows that ∂(a0 F˜ r2 −C r ) < n(r −1). 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].

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