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On the Discriminant of a Certain Quartinomial and Its Totally Complexness

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On the Discriminant of a Certain Quartinomial and Its Totally Complexness ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS SHUICHI OTAKE AND TONY SHASKA Abstract. In this paper, we compute the discriminant of a quartinomial n 2 of the form f(b;a;1)(t; x) = x + t(x + ax + b) by using the Bezoutian. Then, by using this result and another theorem of our previous paper, we construct a family of totally complex polynomials of the form f(b;a;1)(ξ; x) (ξ 2 R; (a; b) 6= (0; 0)). 1. Introduction The discriminant of a polynomial has been a major objective of research in algebra because of its importance. For example, the discriminant of a polynomial allows us to know whether the polynomial has multiple roots or not and it also plays an important role when we compute the discriminant of a number field. Moreover the discriminant of a polynomial also tells us whether the Galois group of the polynomial is contained in the alternating group or not. This is why there are so many papers focusing on studying the discriminant of a polynomial ([B-B-G], [D-S], [Ked], [G-D], [Swa]). In the last two papers, the authors concern the computation of the discriminant of a trinomial and it has been carried out in different ways. n In this paper, we compute the discriminant of a quartinomial f(b;a;1)(t; x) = x + t(x2 + ax + b) by using the Bezoutian (Theorem2). Let F be a field of characteristic zero and f1(x), f2(x) be polynomials over F . Then, for any integer n such that n ≥ maxfdegf1; degf2g, we put n f1(x)f2(y) − f1(y)f2(x) X B (f ; f ) : = = α xn−iyn−j 2 F [x; y]; n 1 2 x − y ij i;j=1 Mn(f1; f2) : = (αij)1≤i;j≤n: 0 The n × n matrix Mn(f1; f2) is called the Bezoutian of f1 and f2. When f2 = f1, the formal derivative of f1 with respect to the indeterminate x, we often write 0 0 Bn(f1) := Bn(f1; f1), Mn(f1) := Mn(f1; f1) and the matrix Mn(f1) is called the Bezoutian of f1. We denote by Nf1 the number of distinct real roots of f1(x) and ∆(f1) the discriminant of f1(x). Moreover, for any real symmetric matrix M, we denote by σ(M) the index of inertia of M. The following are the list of important properties of Bezoutians. Proposition 1. Notations as above, we have (i) Mn(f1; f2) is an n × n symmetric matrix over F . (Mn(f1; f2) 2 Symn(F )). (ii) Bn(f1; f2)(Mn(f1; f2)) is linear in f1 and f2, separately. (iii) Bn(f1; f2) = −Bn(f2; f1)(Mn(f1; f2) = −Mn(f2; f1)). (iv) Nf1 = σ(Mn(f1)). 1 2 SHUICHI OTAKE AND TONY SHASKA 1 (v) ∆(f1) = 2 det Mn(f1), where led(f1) is the leading coefficient of f1. led(f1) In particular, if f1 is monic, we have ∆(f1) = det Mn(f1). µ ν (vi) Let λ, µ, ν be integers such that λ ≥ µ > ν ≥ 0. Then Mλ(x ; x ) = (mij)1≤i;j≤λ, where ( 1 i + j = 2λ − (µ + ν) + 1 (λ − µ + 1 ≤ i; j ≤ λ − ν); mij = 0 otherwise: Proof. For (i){(iv), see [Fuh, Theorem 8.25, 9.2]. For (v), see equation (4) of [Fro, p.217]. For (vi), see [O-S, Lemma 2] s+1 For any vector r = (r0; ··· ; rs) 2 R , let us put s X s−k gr(x) = g(r0; ··· ; rs; x) = rs−kx 2 R[x]; k=0 (n) n fr(t; x) = f (r0; ··· ; rs; t; x) = x + t · gr(x) 2 R(t)[x]: In the previous paper [O-S], by using the above properties of Bezoutians, we ob- tained the next theorem s+1 Theorem 1. [O-S, Theorem 2] Let r = (r0; ··· ; rs) 2 R be a vector such that gr(x) is a degree s separable polynomial satisfying Ngr (x) = γ (0 ≤ γ ≤ s). Let us (n) consider fr(t; x) = f (r0; ··· ; rs; t; x) as a polynomial over R(t) in x and put 0 Pr(t) = det Mn(fr(t; x)) = det Mn(fr(t; x); fr(t; x)); 0 where fr(t; x) is a derivative of fr(t; x) with respect to x. Then, for any real number ξ > αr = maxfα 2 R j Pr(α) = 0g, we have 8 γ + 1 n − s : odd <> Nfr (ξ;x) = γ n − s : even, rs > 0 > :γ + 2 n − s : even, rs < 0: Then, by using Theorem1 and Theorem2, we construct a certain family of totally complex polynomials of the form f(b;a;1)(ξ; x) (Theorem3). Here, note that a real polynomial f(x) to be called totally complex if it has no real roots, that is, Nf = 0. 2. Main results Let n ≥ 3 be an integer and a, b be real numbers. Let us put c = (b; a; 1) 2 R3 and 2 n gc(x) = x + ax + b 2 R[x]; fc(t; x) = x + tgc(x) 2 R(t)[x]: n n We denote by m (n; m 2 Z≥0) the binomial coefficient. Note that 0 = 1 for any n n integer n ≥ 0 and m = 0 if n < m. Moreover, we define m = 0 for any n 2 Z≥0 and m 2 Z<0. In the following, we prove the next two theorems. ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS3 Theorem 2. Put m0 = b(n − 3)=2c, m1 = d(n − 3)=2e. Moreover, for any integers n ≥ 4 and k (0 ≤ k ≤ m0), put ! 2 ! n − k − 3 n(n − 1) 5n − (6k + 23)n + 10k + 24 n − k − 3 S =(n − 1)3 a4 − a2b k k n − k − 3 k ! n − k − 4 + 4n2(n − 2) b2: k Here, b·c is the floor function and d·e is the ceiling function. (1) Suppose n = 3. Then, we have ( ) 2 2 2 3 2 ∆ (fc(t; x)) = t (a − 4b)t − (4a − 18ab)t − 27b : (2) Suppose n ≥ 4. Then, we have ( ) m1 n−1 n−2 2 2 n n−1 ∆ (fc(t; x)) = (−1) t (n − 2) (a − 4b)t + γct − n b ; where 8 m0 >X n+k k n−2k−4 k n−2k−4 k <> (−1) n (n − 1) (n − 2) a b Sk (a 6= 0 or a = 0, n : even); γc = k=0 > :>0 (a = 0, n : odd): Theorem 3. Suppose n ≥ 4 is an even integer. Moreover, let us suppose b 6= 0 and 2 2 (n − 1) a =4n(n − 2) ≤ b. Then we have Nfc(ξ;x) = 0 for any positive real number ξ. By a direct computation, we have ∆ x3 + t(x2 + ax + b) = t2 (a2 − 4b)t2 − (4a3 − 18ab)t − 27b2 ; ∆ x4 + t(x2 + ax + b) = −t3 (4a2 − 16b)t2 + (27a4 − 144a2b + 128b2)t − 256b3 and we get Theorem2 for n = 3, 4. Moreover, if n = 4 and (n−1)2a2=4n(n−2) = b, we have 1 27 2 P (t) = ∆ x4 + t(x2 + ax + b) = a2 t + a2 t3 c 2 8 and hence αc = 0, which implies Nfc(ξ;x) = 0 for any ξ > 0 by Theorem1 since x2 + ax + b has no real root in this case. Then, by considering the graph of the 4 2 2 2 function y = x +ξ(x +ax+b), we have Nfc(ξ;x) = 0 whenever (n−1) a =4n(n−2) ≤ b and ξ > 0, which is the claim of Theorem3 for n = 4. Therefore, let us assume n ≥ 5 hereafter. In the following, we put (c) Ac(t) = (aij (t))1≤i;j≤n = Mn(fc(t; x)) 2 Symn(R(t)); (c) Bc = (bij )1≤i;j≤2 = M2(gc(x)) 2 Sym2(R); Pc(t) = det Ac(t) = ∆(fc(t; x)): Then, by Theorem1, we have the next lemma. 2 Lemma 1. Suppose n is even and a =4 < b. Put αc = maxfα 2 R j Pc(α) = 0g. Then, for any real number ξ > αc, we have Nfc(ξ;x) = 0. 4 SHUICHI OTAKE AND TONY SHASKA To prove Theorem2, we will use the equality ∆( fc(t; x)) = det Ac(t). Also, to 2 2 prove Theorem3, it is enough to prove αc = 0 when (n − 1) a =4n(n − 2) ≤ b and (a; b) 6= (0; 0), which implies we need to compute the polynomial Pc(t) = det Ac(t) precisely. Here, let Qm(k; c) = (qij)1≤i;j≤m, Rm(k; l; c) = (rij)1≤i;j≤m be m × m elementary matrices such that 2 3 2 1 3 1 . 6 .. 7 6 .. 7 6 . 7 6 7 6 7 6 7 6 1 c 7 6 1 7 6 7 6 7 6 .. 7 Qm(k; c)= 6 c 7, Rm(k; l; c)= 6 . 7 ; 6 7 6 7 6 1 7 6 1 7 6 . 7 6 7 6 .. 7 6 .. 7 4 5 4 . 5 1 1 where qkk = c, rkl = c, respectively and, for any m × m matrices M1, M2, ··· , Ml, Ql put k=1 Mk = M1M2 ··· Ml. Moreover, put lk = n + k. Then, as is the case in [O-S], we inductively define the matrix Ac(t)k (1 ≤ k ≤ n − 2) as follows; (c) t (1) Ac(t)1 = (aij (t)1)1≤i;j≤n = Sc(t)1Ac(t)Sc(t)1, where 1 p Y (c) p S (t) = Q (1; 1= n) R (1; l − 1; −a (t)= n): c 1 n n k 1;lk−1 k=0 (2) Put ( (n − 1)=2; n : odd; n0 = n=2; n : even: (c) t Then, Ac(t)k = (aij (t)k)1≤i;j≤n = Sc(t)kAc(t)k−1Sc(t)k, where 8 n (c) ! > Y akm(t)k−1 > Rn l0 − k; m; − (2 ≤ k ≤ n0) > −(n − 2)t >m=l0−k+1 < (c) ! n (c) ! Sc(t)k = akk (t)k−1 Y akm(t)k−1 >Rn l0 − k; k; − Rn l0 − k; m; − > −2(n − 2)t −(n − 2)t > m=k+1 > : (n0 < k ≤ n − 2): By the definition of Ac(t)k (1 ≤ k ≤ n − 2), we have det Ac(t) = n det Ac(t)n−2.
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