t s (p). An Introduction

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1 Notes 6. Quadratic Gauss Sums Definition. Let a, b Z. Then we denote a b if a divides b. Definition. Let a and b be elements of Z. Then c Z s.t. a, b c, where c gcda, b max{x Z x a and x b }. 5, Chater1 Definition. We call integers t and s are congruent modulo, if t s, and we write; t s. Definition. Let R an integral domain. An element r R is irreducible if r ab imlies a or b is a unit in R. 9 Chater 1 Definition. An integral domain R is called euclidian domain, if there exists a function λ from the nonzero elements of R to the set {0, 1,..., } such that if a, b R, b 0, there exists c, d R with the roerty a cb + d and either d 0 or λd < λb. Definition. An integral domain R is called a uniue factorization domain if every nonzero element in R can be written as a uniue roduct of irreducible elements u to units and their orders. We use the fact that every euclidean domain is a uniue factorization domain: so C[x] and Z/Z[x] for a rime are uniue factorization domains. An Introduction For each rime and integer a, we consider the congruence x a. In the last chater, our goal was to find a way to determine the solvability of this congruence. We call a uadratic residue or nonresidue in the following manner. Definition. If a, m 1, a is called a uadratic residue nonresidue mod m if x a m has a solution no solutions. 51 Chater5 The answer to the uestion above was given by the value of the Legendre Symbol. Definition. The Legendre symbol is a uadratic nonresidue mod, and zero if a. 5 Chater 5 a will have the value 1 if a is a uadratic residue mod, -1 if a 1

2 And the value of Legendre Symbols can be comuted with law of uadratic recirocity. Law of uadratic recirocity Proosition 5,1, and Theorem 1 in Chater 5 a 1 a ab a a, a, b Z a b a b , where, and are odd rimes. Now our further interest is to generalize this law to that of higher degree. To achieve this, we need to study another method to rove the recirocity, which can be generalized in higher degree. In this chater we go back to the uadratic congruence and define the Quadratic Gauss sum to see how the Quadratic Gauss sum works to rove law of uadratic recirocity, together with how to comute them Algebraic Numbers and Algebraic Integers.. In this section, we define algebraic numbers/integers, whose set forms resectively a field and a ring. In this field and ring hold some imortant roositions we need in the later chaters. Definition. An algebraic number is a comlex number α that is a root of a olynomial a 0 x n + a 1 x n a n 0, where a 0, a 1, a,... a n Q, and a 0 0 An algebraic integer ω is a comlex number that is a root of a olynomial x n + b 1 x n b n 0, where b 1, b,..., b n Z. Proosition A rational number r Q is an algebraic integer iff r Z. Proof. If r Z, then r is a root of x r 0. Thus r is an algebraic integer. Suose that r Q and that r is an algebraic integer. Then r satisfies an euation x n + b 1 x n b n 0 with b 1,..., b n Z. Let r c d, where c, d Z and we may assume that c and d are relatively rime. Substituting c d is into the euation and multilying both sides by dn yields c n + b 1 c n 1 d + + b n d n 0. Since d divides the right side of the euation and other terms in the left side, it follows that d divides c n. Moreover, c d, actually, d c n imlies d y c n. Since c divides the both sides and d, c 1, c y. So we may rewrite the euation as d y 1 c n 1. By induction we have d y n 1 c. which follows that d c. This imlies also d d, c 1 so d ±1, and r c d is in Z. Definition. A subset V C of the comlex numbers is called a Q-module if a γ 1, γ V imlies that γ 1 + γ V.

3 3 b γ V and r Q imlies that rγ V. c There exist elements γ 1, γ,..., γ l V such that every γ V has the form l i1 r iγ i with r i Q More briefly, V C is a Q-module if it is a finite dimensional Q-subvector sace of C. If γ 1, γ,..., γ l C, the set of all exressions l r iγ i, r 1, r,..., r l Q is easily seen to be a Q-module. We denote this Q-module by [γ 1, γ,..., γ l ]. Proosition Let V [γ 1, γ,..., γ l ] 0, and suose that α C has the roerty that αγ V for all γ V. Then α is an algebraic number. Proof. αγ i V for i 1,,..., l. Thus αγ i l a ijγ j, where a ij Q. It follows that 0 l a ij δ ij αγ j, where δ ij 0 if i j and δ ij 1 if i j. This imlies that deta ij δ ij α 0. The euation above imlies that ranka ij δ ij α l 1, and hence deta ij δ ij α 0. By writing out the determinant we see that α satisfies a olynomial of degree l with rational coefficients. Thus α is an algebraic number. Proosition The set of algebraic numbers forms a field. Proof. Suose that α 1 and α are algebraic numbers. We shall show that α 1 α and α 1 +α are algebraic numbers. Suose that α1 n +r 1 α1 n 1 +r α1 n + +r n 0 and that α m +s 1 α m 1 +s α m + +s m 0, where r i, s j Q. Let V be the Q-module of all Q linear combinations of elements α1α i j, where 0 i < n and 0 j < m. Let γ a ij α1α i j, a ij Q. Then we have α 1 γ a ij α1 i+1 α j + a n 1 j α j r 1α1 n 1 + r α1 n + + r n b ij α1α i j V j α γ 0 i<n 1, 0 j<m 0 i<n, 0 j<m 1 a ij α i 1α j+1 + i a i m 1 α i 1 s 1 α m 1 + s α m + + s m c ij α i 1α j V, where b ij and c ij are some elements in Q. So for γ V we have α 1 γ V and α γ V. Thus we also have α 1 + α γ V and α 1 α γ V. By Proositon 6.1. it follows that both α 1 + α and α 1 α are algebraic numbers. Finally, if α is an algebraic number, not zero, we must show that α 1 is an algebraic number. Suose that a 0 α n + a 1 α n a n 0, where a i s are in Q. Then by multilying by α n, we have a n α n + a n 1 α n a 0 0, The result follows. Definition. A subset W C is called a Z-module if aγ 1, γ W imlies that γ 1 + γ W. b There exist elements γ 1, γ,..., γ l W such that every γ W has the form l i1 r iγ i with r i Z Proosition Let W [γ 1, γ,..., γ l ] 0 be a Z-module and suose that ω C has the roerty that ωγ W for all γ W. Then ω is an algebraic integer. Proof. ωγ i W for i 1,,..., l. Thus ωγ i l a ijγ j, where a ij Z. It follows that 0 l a ij δ ij ωγ j, where δ ij 0 if i j and δ ij 1 if i j. deta ij δ ij ω 0. The euation above imlies that ranka ij δ ij ω l 1, and deta ij δ ij ω 0. By writing out the determinant we see that ω satisfies a olynomial of degree l with the leading coefficient 1 and integer coefficients. Thus ω is an algebraic integer. Proosition The set of algebraic integers forms a ring.

4 4 Proof. Suose that ω 1 and ω are algebraic integers. We shall show that ω 1 ω and ω 1 +ω are algebraic integers. Suose that ω1 n +r 1 ω1 n 1 +r ω1 n + +r n 0 and that ω m +s 1 ω m 1 +s ω m + +s m 0, where r i, s j Z. Let W be the Z-module of all Z linear combinations of the elements ω1ω i j, where 0 i < n and 0 j < m. Let γ a ij ω1ω i j. Then we have ω 1 γ ω γ 0 i<n 1, 0 j<m 0 i<n, 0 j<m 1 a ij ω i+1 1 ω j + j a ij ω i 1ω j+1 + i a n 1 j ω j r 1ω n r ω n r n b ij ω i 1ω j W a i m 1 ω i 1 s 1 ω m 1 + s ω m + + s m c ij ω i 1ω j W. where b ij and c ij are some elements in Z. So for γ W we have ω 1 γ W and ω γ W. Thus we also have ω 1 + ω γ W and ω 1 ω γ W. By Proositon it follows that both ω 1 + ω and ω 1 ω are algebraic integers. Roots of unity are imortant examles of algebraic integers. Proosition If ω 1, ω Ω and Z is a rime, then Proof. ω 1 + ω k0 k ω k 1 ω k ω 1 + ω ω 1 + ω. ω 1 + 1! ω 1 1 ω + + k! k! ω k 1 ω k + + ω 1 ω 1 + ω For 1 k 1 no integers neither in the exression k! nor k! divide, since each of them is less than and relatively rime to. So divides k. Thus, it follows that; ω 1 + ω ω 1 + ω Proosition If α is an algebraic number, then α is the root of a uniue monic irreducible fx in Q. Furthermore if gx Q[x] and gα 0, then fx gx. Proof. Let fx be any monic irreducible with fα 0. We rove the second assertion first. If fx does not divide gx, then fx, gx 1, since fx is irrecucible and monic. So we may write fxhx + gxtx 1 by Lemma 4, Section, Chater 1., for olynomials hx, tx Q[x]. Putting x α gives a contradiction. If there exists another olynomial f x with the roerty, then by the conclution above fx f x and f x fx. Since fx is monic we conclude that fx f x. So the uniueness is roved. This uniuely determined olynomial fx of α is called the minimal olynomial of α. And if the degree of the minimal olynomial is n, then α is called an algebraic number of degree n. If α and γ are roots of fx then α und γ are said to be conjugate. Qα { gα hα gx, hx Q[x], hα 0} is a field. For such a field Qα holds an imortant roosition; Proosition If α Ω then Qα Q[α], the minimam ring containing α and Q.

5 5 Proof. Clearly Q[α] {gα gx Q[x]} Qα. Let an element γ gαhα 1 Qα. By the definition, hα Q[α], hα 0. Then by Proosition 6.1.7, fx does not divide hx, where fx is the minimal olynomial of α. Thus fx, hx 1 and fxsx + hxtx 1, for some olynomials sx, tx Q[x]. Put x α so that tαhα 1. Thus, tα hα 1 Q[α], and it follows immediately that hα 1 Q[α] and γ gαhα 1 Q[α]. Corollary. If α is an algebraic number of degree n then [Qα : Q] n. Proof. By Proositon it is enough to show that [Qα : Q] n. Since fα 0 it is easily seen that 1,..., α n 1 san Q[α]. On the other hand, if we have a 0 + a 1 α + + a n 1 α n 1 0, where a i Q, then gα 0 for gx a 0 + a 1 x + + a n 1 x n 1. Then, by Proosition fx gx. But deggx < degfx, which imlies that a 0 a 1 a n 1 0. Therefore 1,..., α n 1 are linealy indeendent over Q. 6.. The Quadratic Character of.. See also Proosition 5.1.3, 53. We want to show that Let ζ e πi 8, a rimitive eighth root of unity. 0 ζ 8 1 so ζ is an algebraic integer. Let τ ζ +ζ 1. Then we have τ ζ + ζ i i 0, thus τ is also an algebraic integer. So we may work with congruences in the ring of algebraic integers. Let be an odd rime in Z and by Proosition it follows that τ 1 τ 1 1 where is the Legendere Symbol. Hence τ τ On the other hand by Proosition 1.6, τ ζ + ζ 1 ζ + ζ. Remembering that ζ 8 1 we have ζ + ζ ζ 8 + ζ 8. Then ζ + ζ ζ + ζ 1, if ±1 8. and ζ + ζ ζ 3 + ζ 3, if ±3 8. The result in the latter case may be simlified by observing that ζ 4 1 imlies that ζ 3 ζ 1. Thus ζ + ζ ζ 1 + ζ 1 if ±3 8. Summarizing, { τ ζ + ζ τ, if ±1 8 Substituting this result into the relation τ 1 ϵ τ Multily both sides of the congruence by τ. Then imlying that 1 ϵ τ, if ±3 8 τ yields τ, where ϵ 1. 8, 1 ϵ. This last congruence imlies that 1 ϵ, which is the desired result.

6 6 Another examle of law of uadratic recirocity obtained by considering the sum of rimitive nth roots of unity Exercise 8 Let ω e πi 3, and notice that ω Then let σ ω + 1 and it follows by Proositon 5.1. that σ 1 1 σ 3 1 3, from which follows that 3 σ σ. On the other hand by Proositon 6.1.6, we have σ ω + 1 and notice that ω + 1 σ and ω ω ω ω + 1 ω + 1 ω + 1, which follows immediately that σ σ if 1 3, and σ σ if 3. By combining these conseuences we conclude that 1 if 1 3, and 1 if Quadratic Gauss Sums.. In this section, we define the Quadratic Gauss sum for an integer a. We see that by Proosition we only need to know the value of the Legendre symbol of a and that of the Quadratic Gauss sum of 1 to comute the Quadratic Gauss sum for some a. In the last of this section we rove another uadratic recirocity by using the Quadratic Gauss sum. Throughout the later sections we denote by ζ a rimitive th root of unity, where is an odd rime. Lemma 1. 1 t0 ζat is eual to if a 0. Otherwise it is zero. Proof. If a 0, then ζ at 1, t Z, and so 1 t0 ζat. If a 0, then ζ a 1 and 1 t0 ζat 1 + ζ a + + ζ 1a ζa 1 ζ a 1 0. *This exression is allowed because of the Proosition Corollary. 1 1 t0 ζtx y δx, y, where δx, y 1 if x y and δx, y 0 if x y Proof. By substituting a by x y and multilying the both sides of euation by 1 of Lemma 1, the result follows immediately. Lemma. 1 t0 t 0, where t is the Legendre symbol. Proof. By definition 0 0. Of the remaining 1 terms in the summation, half are +1 and half are -1, since by Corollary 1 to Proosition 5.1., there are as many uadratic residues as uadratic nonresidues mod. Definition. g a 1 t0 t ζat, a Z is called the Quadratic Gauss sum. 3

7 7 Proosition g a a g 1 Proof. If a 0, then ζ at 1 for all t, and g a 1 t0 the result in the case that a 0. Now suose that a 0. Then, from Proosition 5.1., a 1 at 1 g a ζ at t0 x0 t g a by Lemma. This gives x ζ x g 1. x We have used the fact that at runs over a comlete residue system mod when t does and that and ζ x deend only on the residue class of x modulo. a Since 1 if a 0, our result follows by multilying the both sides of the euation a g a g 1 by. a We denote the Quadratic Gauss sum of 1 by g instead of g 1. Proosition g 1 1 Proof. The idea of the roof is to evaluate the sum 1 a0 g ag a in two ways. If a 0, then a a g a g a g 1 a g 1 g. It follows that 1 1 g a g a 1g. Now, notice that a0 1 x g a g a x0 a0 a0 x0 y0 1 ζ ax y0 y ζ ay 1 1 x0 y0 x Summing both sides over a and using the corollary to Lemma 1 yields x y 1 1 g a g a ζ ax y x 1 1 y x0 y0 Putting these results together we obtain 1 δx, y 1 x0 x0 y0 x y ζ ax y. x y 1 1 a0 1g 1. Therefore, g See also Theorem1 in Chater5. We rove that for odd rimes, Let * 1 1. Then it holds Thus g 1 g 1 1 g g. ζ ax y holds.

8 8 Using Proosition and the fact is odd and the suare of the Legendre symbol take 0 or 1, we see 1 t 1 t 1 t g ζ t ζ t ζ t g. It follows that g g t0 Multily both sides by g, and use g : which imlies that t0 t0 g by Proosition 6.3.1, and so g g., and finally From this we may conclude immediately that 1 1 Multilying both sides by yields the euation above In each of the case in which we deduced the law of uadratic recirocity in the section and 3, we have seen the common methodology: we comute the th ower of the sum of rimitive th roots of unity in two ways with its uadratic roerty and Proosition and euate them. The author imlied this method can be generalized in higher degree The Sign of the uadratic gauss sum.. By the last roosition on the last section we see that g 1 1 holds, from which follows immediately the absolute value of the Quadratic Gauss sum is. The goal of this section is to determine its sign. From this section we denote the Legendre Symbol j by χj and the Quadratic Gauss sum g 1 j0 χjζj by gχ. Proosition The olynomial 1 + x + x 1 is irreducible in Q. Proof. By Exercise 4 and 5 at the end of this chate,r it is enough to show that 1 + x + + x 1 has no nontrivial factorization in Z[x]. Suose, on the contrary, that 1 + x + + x 1 fxgx where fx, gx Z[x] and each has degree greater than one. Putting x 1 gives f1g1. Therefore we may assume g1 1. Using a bar to denote reduction modulo we conclude that ḡ 1 0. On the other hand we have x 1 x 1, since it holds that k for 1 k 1 as in the roof of Proosition Using the fact that Z/Z[x] is a uniue factorlization domain we may divide the both sides of the euation by x 1, showing that 1 + x + + x 1 x 1 1 and hence gx x 1 s for some integer s. However, this contradicts the fact that ḡ 1 0.

9 9 Remark. Combining the above roosition with Proosition 6.1.7, we see that if gζ 0 for gx Q[x], then 1 + x + + x 1 gx: x 1 is the minimal olynomial of ζ. Proosition ζ k 1 ζ k Proof. One has x 1 x 1 1 x ζj. Divide the both sides by x 1 and ut x 1 to obtain 1 1 ζj. By taking another reresentatives of owers of a rimitive th root of 1, the index j can be rewritten by ±4k, k 1,..., 1. Thus 1 1 ζ j 1 ζ 4k 1 ζ 4k 1 1 ζ 4k ζ k 1 ζ k 1 1 ζ 4k 1 ζ k 1 ζ k 1 ζ k 1 ζ k ζ k 1 ζ k 1. 1 By multilying both sides by 1 1 we obtain the desired euation. * ζ ±4k ζ 0 iff ±4k 0 k 1, but it is imossible since is a rime and k 1 so k 1 <. Moreover, k s do not coincide mod. Suose 4k 1 ±4k, then it follows that k 1 k or k 1 + k 1, which is ossible only if k 1 k since k 1 k and 1 k 1 + k 1 by 1 k 1. Therefore k ±4k, k 1,,... 1 {ζk } has distinct 1 elements other than ζ 0, which imlies that {ζ k k ±4k, k 1,,... 1 } {ζs s 1,,..., 1} Proosition ζ k 1 ζ k 1 1 { if 1 4 i if 3 4.

10 10 Proof. By Proosition 6.4. we have to comute only the sign of the roduct. The roduct is ζ k 1 ζ k i 1 k 1π k 1π k 1π cos + i sin cos i sin i sin 1 sin 4k π 4k π. k 1π We consider how many terms in 1 4k π sin are negative over integers k. Since 1 k 1, sin 4k π < 0 iff π < 4k π < π + 4 < k < But since k, we count the number of integers in the interval + 4 < k 1 1. It follows that the roduct has + 4 negative terms.* *the number of integers x, which satisfy y < x, is y. In the case 1 4, i 1 i 4l l and l+1 1 4l+1+ 4 l l l, so 1 l 1 l 1. In the case 3 4, i 1 i 4l+3 1 i 1 l and l+3 1 4l+3+ 4 l +1 l +1 l, so i 1 l 1 l i. By Proosition 6.3. and Proosition 6.4. we know that 1 gχ ϵ where ϵ ±1. Proosition ϵ +1. Proof. Consider the olynomial ζ k 1 ζ k 1, 1 1 fx χjx j ϵ x k 1 x k 1. Then fζ 0 by 1 and f1 0 by Lemma. Actually, we have fζ 1 1 χjζ j ϵ gχ ϵ gχ ϵ 0 1 ζ k 1 ζ k 1 1 ζ k 1 ζ ζ k 1 ζ k 1 1 ζ k 1 1

11 11 and f1 1 χj1 j ϵ k 1 1 k 1 By Remark 1 it follows that 1 + x + + x 1 fx, so we may write fx h x1 + x + + x 1. From the fact that 1 + x + + x 1 and x 1 are relatively rime, we conclude that x 1 divides h x above and and we may rewrite this as fx x 1hx. Relace x by e z to obtain 1 χje jz ϵ 1 1 e k 1z e k 1z e z 1he z. We consider the coefficient of z 1 on the left-hand side of. Since 1 1 χje jz χj 1 + jz + 1! jz !jz 1 + the coefficiet of z 1 is 1 1 j χj. 1! On the other hand, notice that for each k, 1 k 1 e k 1z e k 1z has no constant terms because than 1 e k 1z e k 1z 1 + k 1z k 1z + 4k z +. And since the roduct is a roduct of 1 terms of degree at least 1, to comute the coefficient of z 1 we only need to focus on the terms of degree 1 the roduct of other terms are exclusively of degree more 1 1. Hence we have ϵ 4k as the coefficint of z. By Exercise 1 the coefficient of z 1 on the right-hand side of 3 is A B integers. To sum u, the euation of coefficients of z 1 is χjj! ϵ 4k A B. 1 Multilying by B 1! and reducing modulo shows that 1 1 χjj 1 ϵ 1! ϵ k k 1 1 ϵ ϵ 1! ϵ, where B, A and B being

12 1 *using Wilson s theorem Corollary to Proosition By Proosition 5.1. j 1 χj so one has 1 1 χj ±1 1 1 ϵ and therefore ϵ 1. Since ϵ ±1 we conclude finally that ϵ +1. This concludes the roof. Theorem 1. The value of the Quadratic Gauss sum gχ is given by { if 1 4 gχ i if 3 4. Solutions to imortant Exercises Excercise 4 Let fx a 0 +a 1 x+ +a n x n and gx b 0 +b 1 x+ +b m x m be rimitive olynomials in Z[x] and let c k, 0 k m + n denote the coefficients of fxgx : fxgx n+m k0 s+tk a sb t x k n+m k0 c kx k. Suose c 0, c 1,..., c m+n l, where l 1. Then there exist a rime which divides l, and suose that i and j are the smallest numbers such that a i and b j. Then it follows that c i+j a s b t a 0 b i+j + a 1 b i+j a i 1 b j+1 + a i b j + a i+1 b j a i+j b 0 s+ti+j a 0 b i+j + a 1 b i+j a i 1 b j+1 + a i+1 b j a i+j b 0 + a i b j a i b j 0 since divides a 0, a 1,..., a i 1 and b 0, b 1,..., b j 1 therefore a 0 b i+j +a 1 b i+j 1 + +a i 1 b j+1 +a i+1 b j a i+j b 0, and divides a i b j iff a i or b j. But it contradicts to the fact that divides c i+j and it follows that a i and b j for all i, j, which cannot be the case since fx and gx are rimitive. Hence c 0, c 1,..., c m+n 1 and fxgx is rimitive. Exercise 5 Let gx x n + a n 1 x n a 0 be a olynomial in Z[x] with gα 0. Then by Proosition fx gx. So for some hx Q[x], gx fxhx. Let c be the roduct of all denomitor of the coefficients of fx and hx. It follows that c gx f xh x, where f x and h x are in Z[x]. We may assume that f x is rimitive, by dividing f x by the greatest common diviser of its coefficients d and redefining h x as the roduct of h x and d. Let the greatest common divisor of coefficients of h x be n. Since the leading coefficient of gx is 1, no rime can devide gx, therefore n c and c n gx f x h x n. By Exercise 4, c n gx is rimitive, and since gx Z[x], c n must be ±1. Hence it follows that gx f x± h x n of gx is 1, those of f x and ± h x, where both f x and ± h x n are in Z[x]. Since the leading coefficient n are ±1. We may assume that of f x is 1. Again by Proosition fx f x with degfx degf x, which imlies fx f x Z[x].

13 13 Exercise 9 By Lemma 1, On the other hand, t 1 + ζ t ζ t + t 0 + g g. t 1 + ζ t ζ t ζ t t QR t QNR 1 + t QR 1 ζ t 1 + ζ t ζ t, where QR and QN R are resectively the set of the uadratic residues and that of uadratic nonresidues. The euation on the second line follows from the fact that a a and there exist 1 uadratic residues and uadratic nonresidues i.e. for each uadratic residue t there are exactly elements a, 1 a 1 such that t a. Exercise 16 We consider f x, the formal derivative of fx. Definition. In a olynomial ring with a commutative ring R, the formal derivative f x of fx a 0 + a 1 x + + a n x n R[x] is defined as t1 f x a 1 + a x i + 1a i+1 x i + + na n x n 1. Suose fx has β C as a reeated solution. Then, fx can be written as fx x β s x C[x], where s N. It follows that f x x β s 1 x + x β s x so f x has β as a root. On the other hand, f x itself is by the definition in Q[x]. fx is the minimal olynomial of β since if there exists other minimal olynomial other than f, it divides f and it contradicts to the fact that f is irreducible. Therefore by Proosition fx divide f x while degf > degf. A contradiction. Exercise 1 Let fx a n n0 n! x n and gx b n n0 n! x n, where a i, i 0, 1,,..., 1. Then it follows that 1 fxgx a 0 b 0 + a 0 b 1 + a 1 b 0 x +! a 0b + a 1 b 1 + 1! a b 0 x j! a 1 0b j + + i!j i! a ib j i j! a jb 0 x j ! a 1 0b i! 1 i! a 1 ib 1 i + + 1! a 1b 0 x 1 + B B a 0b 0 + B B a 0b 1 + a 1 b 0 x + 1 B B! a 0b + Ba 1 b 1 + B! a b 0 x B B j! a B 0b j + + i!j i! a ib j i + + B j! a jb 0 x j B! a B 0b i! 1 i! a ib 1 i + + 1!a 1b 0 x 1 +,

14 14 where B 1! 1 1!. does not divide 1!! since each of the integers in these exressions is B less than and relatively rime to. For every i j 1, i!j i! is an integer since eigher i or j i is eual or less than 1 and the another is also eual or less than 1, which follows that i!j i! B. And since divides a 0, a 1,..., a 1, divides 1 B B j! a 0b j + + B i!j i! a ib j i + + B j! a jb 0, 0 j 1. So the coefficients of x j, j 0, 1,..., 1 can be written as A B, where does not divide B.

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