Polynomial imaginary decompositions for finite extensions of fields of characteristic zero

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Polynomial imaginary decompositions for finite extensions of fields of characteristic zero Article in Bulletin of the Polish Academy of Sciences Mathematics · January 2003

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Polynomial imaginary decompositions for finite extensions of fields of characteristic zero Andrzej Nowicki and Stanislaw Spodzieja∗ December 27, 2001

Abstract Let k be a field of characteristic zero, L = k[ξ] a finite field extension of degree m > 1, and f (z) a polynomial in one variable z over L. Then there exist unique polynomials u0 , . . . , um−1 belonging to k[x0 , . . . , xm−1 ] such that f (x0 + ξx1 + · · · + ξ m−1 xm−1 ) = u0 + ξu1 + · · · + ξ m−1 um−1 . We prove that if f (z) 6∈ L, then the polynomials u0 , . . . , um1 are algebraically independent over k and they have no common divisors in k[x0 , . . . , xm−1 ] of positive degrees. Some other properties of polynomials u0 , . . . , um−1 are also given.

1

Introduction

If z, x, y are variables and f (z) is a polynomial from C[z], then there exist unique polynomials u(x, y), v(x, y) belonging to R[x, y] such that f (x + iy) = u(x, y) + iv(x, y). We will show that if f is nonzero, then the polynomials u(x, y) and v(x, y) are coprime. We will also show that the same is true if instead of the extension R ⊂ C we consider a finite field extension of characteristic zero. More exactly, assume that k is a field and L = k[ξ] is a finite field extension of degree m > 1. Let z and x0 , . . . , xn−1 be variables and let f (z) be a polynomial from L[z]. Then there exist unique polynomials u0 , . . . , um−1 , belonging to k[x] := k[x0 , . . . , xm−1 ], such that f (x0 + ξx1 + · · · + ξ m−1 xm−1 ) = u0 + ξu1 + · · · + ξ m−1 um−1 . ∗

Supported by KBN Grant 2 PO3A 007 18

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This representation we call the imaginary decomposition of f , and the polynomials u0 , . . . , um−1 we call the imaginary parts of f . We will show that if f is nonzero, then the imaginary parts of f have no common divisors in k[x] of positive degrees. Moreover, we prove that a sequence (u0 , . . . , um−1 ) of polynomials from k[x] forms imaginary parts of a polynomial f (z) ∈ L[z] if and only if the polynomials u0 , . . . , um−1 satisfy some generalizations of Cauchy - Riemann equations. Some other properties concerning the divisibility of imaginary parts are also given.

2

Notations and preliminaries

Throughout the paper k is a field of characteristic zero and L = k[ξ] is a finite field extension of degree m > 1. Assume that ϕ(t) = tm − am−1 tm−1 − · · · − a1 t − a0 (with a0 , . . . , am−1 ∈ k) is the minimal polynomial of ξ over k. Let x = (x0 , x1 , . . . , xm−1 ), where x0 , . . . , xm−1 are variables, and let k[x] := k[x0 , . . . , xm−1 ], L[x] := L[x0 , . . . , xm−1 ] be the polynomial rings. We denote by M the set k[X]m , that is, M := {(u0 , u1 , . . . , um−1 ); u0 , . . . , um−1 ∈ k[x]} . Let u = (u0 , . . . , um−1 ) ∈ M . We use the following notations. We denote by u = (u0 , . . . , um−1 ) the element from M defined by:   u0 = a0 um−1 ,     = u0 + a1 um−1 ,  u1 u2 = u1 + a2 um−1 ,  ..   .    u m−1 = um−2 + am−1 um−1 . Moreover, we denote by [u] the polynomial from L[x] defined as [u] := u0 + ξu1 + · · · + ξ m−1 um−1 . In particular, [x] := x0 +ξx1 +· · ·+ξ m−1 xm−1 . If the polynomials u0 , . . . , um−1 are imaginary parts of a polynomial f (z) ∈ L[z], then f ([x]) = [u]. Observe that the equality ξ m = a0 + a1 ξ + · · · + am−1 ξ m−1 implies that [u]ξ = [u] . ∂u If i ∈ {0, . . . , m − 1}, then we denote by ∂x the element from M defined i as   ∂um−1 ∂u0 ∂u := , . . . , . ∂xi ∂xi ∂xi 2

Lemma 2.1. If u0 , . . . , um−1 are imaginary parts of a polynomial f (z) ∈ L[z], then h i ∂u = f 0 ([x])ξ i ∂xi for i ∈ {0, . . . , m − 1}, where f 0 (z) means the ordinary derivative of f with respect to z. h i ∂u = ∂x∂ i [u] = ∂x∂ i f ([x]) = f 0 ([x]) ∂x∂ i [x] = f 0 ([x])ξ i .  Proof. ∂xi As a consequence of this lemma we obtain the following proposition. Proposition 2.2. If u0 , . . . , um−1 ∈ k[x] are imaginary parts of a polynomial f (z) ∈ L[z] r L, then u0 , . . . , um−1 are algebraically independent over k. Proof. Assume that f ([x]) = [u], where u := (u0 , . . . , um−1 ) and f (z) ∈ L[z] r L. Suppose that u0 , . . . , um−1 are algebraically dependent over k. ∂u , . . . , ∂x∂u are linearly dependent over the field k(x) := Then the vectors ∂x 0 m−1 k(x0 , . . . , xm−1 ). Hence, there exist a nonzero sequence α = (α0 , . . .h, αm−1 i ) of Pm−1 Pm−1 ∂u ∂u polynomials from k[x] such that i=0 αi ∂xi = 0, that is, i=0 αi ∂xi = 0. Now, by Lemma 2.1 we get: m−1  m−1 P h ∂u i m−1 P P 0 i 0 i 0= αi ∂xi = αi f ([x])ξ = f ([x]) αi ξ = f 0 ([x])[α]. i=0

i=0

i=0

Hence, in the polynomial ring L[x] we have the equality f 0 ([x])[α] = 0. But f 0 ([x]) 6= 0 and [α] 6= 0. So, we have a contradiction. 

3

Generalization of the Cauchy-Riemann equation

We introduce the following generalization of the Cauchy - Riemann equation. Definition 3.1. Let u ∈ M . We say that u is a ξ-sequence if ∂u ∂u = , ∂xi ∂xi−1 for all i = 1, 2, . . . , m − 1. The following proposition show that the imaginary parts of a polynomial from L[z] form a ξ-sequence. 3

Proposition 3.2. Let f (z) ∈ L[z] and let u be the element from M such that f ([x]) = [u]. Then u is a ξ-sequence. Proof. If i ∈ {1, . . . , m − 1} then, by Lemma 2.1, we have: h i h i h i h i ∂u ∂u ∂u ∂u 0 i 0 i−1 = f ([x])ξ = f ([x])ξ · ξ = · ξ = = , ∂xi ∂xi−1 ∂xi−1 ∂xi−1 and hence

∂u ∂xi

=

∂u . ∂xi−1



We will show that the converse of the above proposition is also true. For a proof of this fact we need several lemmas. Lemma 3.3. If u ∈ M is a ξ-sequence, then each partial derivative j = 0, . . . , m − 1, is also a ξ-sequence.  ∂u 0 Proof. Let j ∈ {0, . . . , m − 1} and put w := ∂xj = ∂u , ..., ∂xj Then for every i ∈ {1, . . . , m − 1} we have: ∂w ∂xi

=

∂2u ∂xi ∂xj

=

∂ ∂u ∂xj ∂xi

=

∂ ∂u ∂xj ∂xi−1

=

∂ ∂u ∂xi−1 ∂xj

=

∂u , ∂xj

for

∂um−1 ∂xj

 .

∂w , ∂xi−1

and hence, w is a ξ-sequence.  Lemma 3.4. If u ∈ M is a ξ-sequence, then h i h i ∂u ∂u ξ = ∂xi+1 , ∂xi for i = 0, 1, . . . , m − 2. In particular, h i h i h i h i ∂u ∂u ∂u ∂u 2 ξ = , ξ = , ∂x0 ∂x1 ∂x0 ∂x2 Proof.

h

∂u ∂xi

i

ξ=

h

∂u ∂xi

i

=

h

∂u ∂xi

i

=

h

...,

∂u ∂xi+1

h

∂u ∂x0

i

ξ m−1 =

h

∂u ∂xm−1

i

.

i . 

L Consider the usual gradation k[x] = s>0 k[x]s , where each k[x]s is the subgroup of k[x] containing zero and all homogeneous polynomials from k[x] of degree s. We say that an element u = (u0 , . . . , um−1 ) from M is homogeneous of degree s, if all the polynomials u0 , . . . , um−1 belong to k[x]s . Every polynomial h ∈ k[x] has a presentation of the form h = h(0) + (1) h + · · · + h(r) , where each h(j) , for j = 0, . . . , r, is a unique homogeneous polynomial belonging to k[x]j . 4

Let u = (u0 , . . . , um−1 ) ∈ M . Then there exists a common integer r > 0 such that (0) (1) (r) ui = ui + ui + · · · + ui , for all i = 0, . . . , m − 1, and we have u = u(0) + u(1) + · · · + u(r) ,   (j) (j) (j) (j) where u = u0 , u1 , . . . , um−1 for j = 0, 1, . . . , r. Then each u(j) , for j = 0, . . . , r, is a homogeneous element of M of degree j. We call it the homogeneous component of u of degree j. Since ∂ (k[x]s ) ⊆ k[x]s−1 , ∂xi for every s > 0 and i = 0, 1, . . . , m − 1 (where k[x]−1 = 0), we obtain the following lemma. Lemma 3.5. Let u ∈ M . If u is a ξ-sequence, then each homogeneous component of u is also a ξ-sequence.  Note also: Lemma 3.6. Let u be a homogeneous element of M of degree s > 0. If u is a ξ-sequence, then h i ∂u [x] ∂x = s[u]. 0 Proof. As a consequence of Lemma 3.4 we have: h i h i ∂u ∂u 2 m−1 [x] ∂x = (x + ξx + ξ x + · · · + ξ x ) 0 1 2 m−1 ∂x0 0 h i h i h i h i ∂u ∂u ∂u ∂u 2 m−1 = x0 ∂x + x ξ + x ξ + · · · + x 1 ∂x0 2 ∂x0 m−1 ∂x0 ξ 0 h i h i h i h i ∂u ∂u ∂u ∂u = x0 ∂x + x + x + · · · + x 1 ∂x1 2 ∂x2 m−1 ∂xm−1 0 h i ∂u ∂u ∂u ∂u = x0 ∂x + x + x + · · · + x 1 2 m−1 ∂x1 ∂x2 ∂xm−1 0 = [su] = s[u]. We used also the well-known Euler equality.  Lemma 3.7. Let u be a homogeneous element of M of degree s > 0. If u is a ξ-sequence, then there exists a unique b ∈ L such that [u] = b[x]s . 5

Proof. We use an induction with respect to s. It is obvious for s = 0. Let s > 0 and assume that it is true for all homogeneous ξ-sequences of degree s − 1. Let u be a homogeneous ξ-sequence of degree s. Then, by Lemma 3.3, ∂u is a homogeneous ξ-sequence of degree s − 1 and the partial derivative ∂x 0 hence, by induction, there exists an element c ∈ L such that h i ∂u = c[x]s−1 . ∂x0 Put b := 1s c. Then, by Lemma 3.6, we have: h i ∂u [x] = 1s s[u] = [u]. b[x]s = 1s c[x]s−1 [x] = 1s ∂x 0 The uniqueness is obvious.  Now we are ready to prove the following theorem. Theorem 3.8. Let k be a field of characteristic zero and let L = k[ξ] be a finite field extension of degree m > 1. Let z, x0 , x1 , . . . , xm−1 be variables and let u = (u0 , . . . , um− ) be a sequence of polynomials belonging to k[x] := k[x0 , . . . , xm−1 ]. Then the following two conditions are equivalent. (1) u is a ξ-sequence. (2) There exists a polynomial f (z) ∈ L[z] such that the polynomials u0 , . . . , um−1 are the imaginary parts of f (z), that is, f (x0 + ξx1 + · · · + ξ m−1 xm−1 ) = u0 + ξu1 + · · · + ξ m−1 um−1 . Proof. The implication (2) ⇒ (1) we already proved (see Proposition 3.2). Assume now that u is a ξ-sequence. Let u = u(0) + u(1) + · · · + u(r) be the homogeneous decomposition of u. Then each u(j) , for j = 0, . . . , r, is (by Lemma 3.5) a homogeneous ξ-sequence of degree j and so, by Lemma 3.7,  (j) there exists bj ∈ L such that u = bj [x]j . Put f (z) := b0 + b1 z + · · · + br z r . Then f ([x]) = b0 [x]0 + b1 [x]1 + · · · + br [x]r       = u(0) + u(1) + · · · + u(r)   = u(0) + u(1) + · · · + u(r) = [u]. This completes the proof.  Corollary 3.9. If u ∈ M is a ξ-sequence, then u is also a ξ-sequence. Proof. By Theorem 3.8 there exists a polynomial f (z) ∈ L[z] such that f ([x]) = [u]. Consider the polynomial g(z) := ξf (z) ∈ L[z]. Since g([x]) = ξf (z) = ξ[u] = [u], the sequence u is, again by Theorem 3.8, a ξ-sequence.  6

4

Divisibility In this section we use the same notations as in the previous sections.

Proposition 4.1. Let u = (u0 , . . . , um−1 ) be a ξ-sequence. Assume that the polynomials u1 , u2 , . . . , um−1 belong to k. Then u0 ∈ k. Proof. we have

Since

∂u ∂xi

=

∂u ∂xi−1

∂u0 ∂xi

=

for all i = 1, . . . , m − 1 (see Definition 3.1), ∂u0 ∂xi−1

=

∂a0 um−1 ∂xi−1

= 0,

(for i = 1, . . . , m − 1) and moreover, 0=

∂u1 ∂x1

=

∂u1 ∂x0

=

∂(u0 +a1 um−1 ) ∂x0

=

∂u0 ∂x0

m−1 + a1 ∂u∂x = 0

∂u0 . ∂x0

0 0 0 Therefore, ∂u = ∂u = · · · = ∂x∂u = 0 and hence, since char(k) = 0, the ∂x0 ∂x1 m−1 polynomial u0 belongs to k. 

Proposition 4.2. Let u = (u0 , . . . , um−1 ) be a nonzero homogeneous ξsequence. Then gcd(u0 , . . . , um−1 ) = 1. Proof. We know, by Lemma 3.7, that [u] = b[x]s for some b ∈ L and s > 0. Let 0 6= h ∈ k[x] := k[x0 , . . . , xm−1 ] be a common divisor of all the polynomials u0 , . . . , um−1 . We will show that h ∈ k. Put u0 = v0 h, . . . , um−1 = vm−1 h with v0 , . . . , vm−1 ∈ k[x]. Then in the polynomial ring L[x] := L[x0 , . . . , xm−1 ] we have the equality b[x]s = [v]h. But L[x] is a UFD and [x] = x0 + ξx1 + · · · + ξ m−1 xm−1 is an irreducible polynomial in L[x], so h = c[x]r for some nonzero c ∈ L and 0 6 r 6 s. This means (by Theorem 3.8) that (h, 0, 0, . . . , 0) is a ξ-sequence. Now Proposition 4.1 implies that h ∈ k.  Theorem 4.3. Let k be a field of characteristic zero and let L = k[ξ] be a finite field extension of degree m > 1. Let z, x0 , x1 , . . . , xm−1 be variables and let u = (u0 , . . . , um−1 ) be a sequence of polynomials belonging to k[x] := k[x0 , . . . , xm−1 ]. If the polynomials u0 , . . . , um−1 are the imaginary parts of a nonzero polynomial f (z) ∈ L[z], then gcd(u0 , . . . , um−1 ) = 1.

7

Proof. Let 0 6= h ∈ k[x] := k[x0 , . . . , xm−1 ] be a common divisor of all the polynomials u0 , . . . , um−1 . Denote by h∗ the homogeneous component of highest degree of h, and let u∗0 , . . . , u∗m−1 be the homogeneous components of highest degree of u0 , . . . , um−1 , respectively. Then 0 6= h∗ is a common divisor of all the polynomials u∗0 , . . . , u∗m−1 and moreover, by Lemma 3.5, (u∗0 , . . . , u∗m−1 ) is a homogeneous ξ-sequence. This implies, by Proposition 4.2, that h∗ ∈ k. Therefore, h ∈ k and so, gcd(u0 , . . . , um−1 ) = 1.  As a consequence of theorems 3.8 and 4.3 we have: Theorem 4.4. Let k be a field of characteristic zero and let k ( L be a finite field extension. If (u0 , . . . , um−1 ) is a nonzero ξ-sequence, then the polynomials u0 , . . . , um−1 have no common divisors of positive degrees. 

5

Quadratic extensions

Throughout this section L = k[ξ] is a quadratic field extension of k. We assume that ξ 2 = r, where r is a nonzero element from k such that the polynomial t2 − r is irreducible in k[t]. Every element of L has a unique presentation of the form a + bξ with a, b ∈ k. Let x, y, z be variables. If w ∈ k[x, y], then we denote by wx and wy the partial derivatives ∂w and ∂w , respectively. In this case a pair (u, v) of ∂x ∂y polynomials from k[x, y] is a ξ-sequence iff uy = rvx and vy = ux . Such a pair we will call a ξ-pair. By Theorem 3.8 we have: Proposition 5.1. Let (u, v) be a pair of polynomials from k[x, y]. The following two conditions are equivalent. (1) There exists a polynomial f (z) ∈ L[z] such that f (x + ξy) = u + ξv. (2) uy = rvx and vy = ux .  Let ∆ : k[x, y] → k[x, y] be a generalization of the Laplace operator defined by ∂2 ∂2 ∆ := r ∂x 2 − ∂y 2 , that is, ∆(w) = rwxx − wyy for w ∈ k[x, y]. It is easy to check that if (u, v) is a ξ-pair, then ∆(u) = ∆(v) = 0. Note the following observation. Proposition 5.2. Let u ∈ k[x, y]. The following two conditions are equivalent. (1) There exists a polynomial v ∈ k[x, y] such that (u, v) is a ξ-pair. (2) ∆(u) = 0. 8

Proof. The implication (1) ⇒ (2) is obvious. We prove the implication (2) ⇒ (1). Let f := 1r uy and g := ux . Since fy = gx and char(k) = 0, there exists a polynomial v ∈ k[x, y] such that vx = f and vy = g. Then uy = rvx and vy = ux and hence, by Proposition 5.1, (u, v) is a ξ-pair.  Consider two sequences (pn ) and (qn ) of polynomials from k[x, y] defined as follows:   p0 = 1, q0 = 0, pn+1 = xpn + ryqn ,  qn+1 = ypn + xqn . In particular, p1 = x, p2 = x2 +ry 2 , p3 = x(x2 +3ry 2 ), p4 = x4 +6rx2 y 2 +r2 y 4 , q1 = y, q2 = 2xy, q3 = y(3x2 + ry 2 ), q4 = 4xy(x2 + ry 2 ). Note the following matrix presentation of these sequences.     x ry pn rqn n Proposition 5.3. If A = , then A = , for all n > 0. qn pn y x  It is easy to check that, for any nonnegative n, the pair (pn , qn ) is a ξ-pair such that (∗)

pn + ξqn = (x + ξy)n .

We present some observations concerning the sequence (pn ) and (qn ). As a consequence of (∗) and Proposition 2.2 we obtain Proposition 5.4. If n > 1, then the polynomials pn and qn are algebraically independent over k.  As a consequence of (∗) and Theorem 4.3 we obtain Proposition 5.5. gcd(pn , qn ) = 1.  The equality (∗) implies the following proposition. Proposition 5.6. If n and m are nonnegative integers, then pn+m = pn pm + rqn qm and qn+m = pn qm + pm qn .  In particular, for n = m, we get Proposition 5.7. p2n = p2n + rqn2 and q2n = 2pn qn .  Using a simple induction and the above propositions it is easy to prove the next proposition. 9

Proposition 5.8. (1) y | qn , for n ∈ N. (2) If n | m, then qn | qm . (3) pn | q2kn , for n, k ∈ N. (4) pn | p(2k+1)n , for n, k ∈ N. (5) gcd(pkn , qn ) = 1, for n, k ∈ N. (6) gcd(qkn+r , qn ) = gcd(qr , qn ), for n, k, r ∈ N.  Now, by Proposition 5.8 and the Euclid algorithm, we have: Proposition 5.9. If n, m ∈ N, then gcd (qn , qm ) = qgcd(n,m) .  Note that a similar proposition is well known for some classical sequences of integers. Such a property have the sequences of Fibonacci numbers, Mersenne numbers and others (see for example, [1], [2]).

References [1] W. Sierpi´ nski, Elementary Theory of Numbers, Hafner Publishing Company, New York, 1964. [2] N. N. Vorobiev, The Fibonacci numbers, (Russian), Nauka, Moskow, 1978.

Faculty of Mathematics and Computer Science, Nicolaus Copernicus University, 87-100 Toru´ n, Poland e-mail: [email protected] Faculty of Mathematics University of L´od´z S. Banacha 22 90-238 L´od´z, Poland e-mail: [email protected]

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