Polynomial imaginary decompositions for finite extensions of fields
of characteristic zero
Andrzej Nowicki and Stanis law Spodzieja
∗December 27, 2001
Abstract
Let k be a field of characteristic zero, L = k[ξ] a finite field ex- tension of degree m > 1, and f (z) a polynomial in one variable z over L. Then there exist unique polynomials u0, . . . , um−1 belong- ing to k[x0, . . . , xm−1] such that f (x0 + ξx1 + · · · + ξm−1xm−1) = u0+ ξu1+ · · · + ξm−1um−1. We prove that if f (z) 6∈ L, then the poly- nomials 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 exten- sion 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+ · · · + ξm−1xm−1) = u0+ ξu1 + · · · + ξm−1um−1.
∗Supported by KBN Grant 2 PO3A 007 18
This representation we call the imaginary decomposition of f , and the poly- nomials 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 poly- nomials 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−1tm−1 − · · · − a1t − a0
(with a0, . . . , am−1 ∈ k) is the minimal polynomial of ξ over k. Let x = (x0, x1, . . . , xm−1), where x0, . . . , xm−1are 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 = a0um−1, u1 = u0+ a1um−1, u2 = u1+ a2um−1,
...
um−1 = um−2+ am−1um−1.
Moreover, we denote by [u] the polynomial from L[x] defined as [u] := u0+ ξu1+ · · · + ξm−1um−1.
In particular, [x] := x0+ξx1+· · ·+ξm−1xm−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 ξm = a0+ a1ξ + · · · + am−1ξm−1 implies that [u]ξ = [u] .
If i ∈ {0, . . . , m − 1}, then we denote by ∂x∂u
i the element from M defined as
∂u
∂xi :=
∂u0
∂xi, . . . , ∂u∂xm−1
i
.
Lemma 2.1. If u0, . . . , um−1 are imaginary parts of a polynomial f (z) ∈ L[z], then
h∂u
∂xi
i
= f0([x])ξi
for i ∈ {0, . . . , m − 1}, where f0(z) means the ordinary derivative of f with respect to z.
Proof. h
∂u
∂xi
i
= ∂x∂
i[u] = ∂x∂
if ([x]) = f0([x])∂x∂
i[x] = f0([x])ξi. 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.
Then the vectors ∂x∂u
0, . . . ,∂x∂u
m−1 are linearly dependent over the field k(x) :=
k(x0, . . . , xm−1). Hence, there exist a nonzero sequence α = (α0, . . . , αm−1) of polynomials from k[x] such thatPm−1
i=0 αi∂x∂u
i = 0, that is,Pm−1 i=0 αih
∂u
∂xi
i
= 0.
Now, by Lemma 2.1 we get:
0 =
m−1
P
i=0
αih
∂u
∂xi
i
=
m−1
P
i=0
αif0([x])ξi = f0([x])
m−1 P
i=0
αiξi
= f0([x])[α].
Hence, in the polynomial ring L[x] we have the equality f0([x])[α] = 0. But f0([x]) 6= 0 and [α] 6= 0. So, we have a contradiction.
3 Generalization of the Cauchy-Riemann equa- tion
We introduce the following generalization of the Cauchy - Riemann equa- tion.
Definition 3.1. Let u ∈ M . We say that u is a ξ-sequence if
∂u
∂xi = ∂u
∂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.
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∂u
∂xi
i
= f0([x])ξi = f0([x])ξi−1· ξ = h
∂u
∂xi−1
i· ξ = h
∂u
∂xi−1
i
=h
∂u
∂xi−1
i , and hence ∂x∂u
i = ∂x∂u
i−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 ∂x∂u
j, for j = 0, . . . , m − 1, is also a ξ-sequence.
Proof. Let j ∈ {0, . . . , m − 1} and put w := ∂x∂u
j =
∂u0
∂xj, . . . , ∂u∂xm−1
j
. Then for every i ∈ {1, . . . , m − 1} we have:
∂w
∂xi = ∂x∂2u
i∂xj = ∂x∂
j
∂u
∂xi = ∂x∂
j
∂u
∂xi−1 = ∂x∂
i−1
∂u
∂xj = ∂x∂w
i−1, and hence, w is a ξ-sequence.
Lemma 3.4. If u ∈ M is a ξ-sequence, then h∂u
∂xi
i ξ =h
∂u
∂xi+1
i , for i = 0, 1, . . . , m − 2. In particular,
h∂u
∂x0
i ξ =h
∂u
∂x1
i , h
∂u
∂x0
i
ξ2 =h
∂u
∂x2
i
, . . . , h
∂u
∂x0
i
ξm−1 =h
∂u
∂xm−1
i . Proof. h
∂u
∂xi
i ξ =h
∂u
∂xi
i
=h
∂u
∂xi
i
=h
∂u
∂xi+1
i . Consider the usual gradation k[x] = L
s>0k[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 homoge- neous 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) + h(1) + · · · + h(r), where each h(j), for j = 0, . . . , r, is a unique homogeneous polynomial belonging to k[x]j.
Let u = (u0, . . . , um−1) ∈ M . Then there exists a common integer r > 0 such that
ui = u(0)i + u(1)i + · · · + u(r)i , for all i = 0, . . . , m − 1, and we have
u = u(0)+ u(1)+ · · · + u(r), where u(j) =
u(j)0 , u(j)1 , . . . , u(j)m−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
∂
∂xi (k[x]s) ⊆ k[x]s−1,
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
[x]h
∂u
∂x0
i
= s[u].
Proof. As a consequence of Lemma 3.4 we have:
[x]h
∂u
∂x0
i
= (x0+ ξx1+ ξ2x2+ · · · + ξm−1xm−1)h
∂u
∂x0
i
= x0h
∂u
∂x0
i
+ x1h
∂u
∂x0
i
ξ + x2h
∂u
∂x0
i
ξ2+ · · · + xm−1h
∂u
∂x0
i ξm−1
= x0h
∂u
∂x0
i
+ x1h
∂u
∂x1
i
+ x2h
∂u
∂x2
i
+ · · · + xm−1h
∂u
∂xm−1
i
= h
x0 ∂u
∂x0 + x1 ∂u
∂x1 + x2 ∂u
∂x2 + · · · + xm−1 ∂u
∂xm−1
i
= [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.
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, the partial derivative ∂x∂u
0 is a homogeneous ξ-sequence of degree s − 1 and hence, by induction, there exists an element c ∈ L such that
h∂u
∂x0
i
= c[x]s−1. Put b := 1sc. Then, by Lemma 3.6, we have:
b[x]s = 1sc[x]s−1[x] = 1sh
∂u
∂x0
i
[x] = 1ss[u] = [u].
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 vari- ables 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+ · · · + ξm−1xm−1) = u0+ ξu1 + · · · + ξm−1um−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, there exists bj ∈ L such thatu(j) = bj[x]j. Put f (z) := b0+ b1z + · · · + brzr. 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.
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. Since ∂x∂u
i = ∂x∂u
i−1 for all i = 1, . . . , m − 1 (see Definition 3.1), we have
∂u0
∂xi = ∂x∂u0
i−1 = ∂a∂x0um−1
i−1 = 0, (for i = 1, . . . , m − 1) and moreover,
0 = ∂u∂x1
1 = ∂u∂x1
0 = ∂(u0+a∂x1um−1)
0 = ∂u∂x0
0 + a1∂u∂xm−1
0 = ∂u∂x0
0. Therefore, ∂u∂x0
0 = ∂u∂x0
1 = · · · = ∂x∂u0
m−1 = 0 and hence, since char(k) = 0, the polynomial u0 belongs to k.
Proposition 4.2. Let u = (u0, . . . , um−1) be a nonzero homogeneous ξ- sequence. 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 = v0h, . . . , um−1 = vm−1h 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 + · · · + ξm−1xm−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.
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 ξ2 = 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∂x and ∂w∂y, respectively. In this case a pair (u, v) of 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 fol- lowing 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
∆ := r∂x∂22 − ∂y∂22,
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 equiva- lent.
(1) There exists a polynomial v ∈ k[x, y] such that (u, v) is a ξ-pair.
(2) ∆(u) = 0.
Proof. The implication (1) ⇒ (2) is obvious. We prove the implication (2) ⇒ (1). Let f := 1ruy 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+ry2, p3 = x(x2+3ry2), p4 = x4+6rx2y2+r2y4, q1 = y, q2 = 2xy, q3 = y(3x2+ ry2), q4 = 4xy(x2+ ry2). Note the following matrix presentation of these sequences.
Proposition 5.3. If A = x ry y x
, then An= pn rqn qn pn
, for all n > 0.
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 = pnpm+ rqnqm and qn+m= pnqm+ pmqn.
In particular, for n = m, we get
Proposition 5.7. p2n = p2n+ rq2n and q2n = 2pnqn.
Using a simple induction and the above propositions it is easy to prove the next proposition.
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 Com- pany, 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: anow@mat.uni.torun.pl Faculty of Mathematics
University of L´od´z S. Banacha 22 90-238 L´od´z, Poland
e-mail: spodziej@imul.uni.lodz.pl
