NON-LINEAR ORDERS ON COMPLEX NUMBERS
TOMÁŠ KEPKA AND MIROSLAV KORBELÁŘ
Abstract. The ring (and field) C of complex numbers cannot be ordered in a linear way and this is because of the archetypical way that
i2 = −1 6= 1 = i4 . On the other hand, there are many non-linear (alias
partial) orders on C and the present note is a brief essay on the thème.
1. Introduction
Let be a relation of (partial) order defined on the field C of complex
numbers such that the pair (C, ) makes an ordered ring. That is, is a
reflexive, antisymmetric and transitive relation satisfying the following two
conditions for all α, β, γ ∈ C:
(1) α β ⇒ α + γ β + γ
(2) (α β & 0 γ) ⇒ αγ βγ.
Of course, α β if and only if 0 β − α and the order is uniquely
determined by the set (or cone) P() = {α ∈ C | 0 ≺ α} of positive
elements. This set is closed under addition and multiplication, it does not
contain zero and it is non-empty if and only if the order is non-identical
(so that P() is a subsemiring of C in that case). Consequently, we have a
biunique correspondence between the non-identical orders and subsemirings
of the field C that do not contain 0. Now, the following natural observation
comes up:
1.1. Proposition. The following conditions are equivalent for a complex
number α:
(i) There exists at least one order relation defined on C and satisfying
(1), (2) such that 0 ≺ α (or α ∈ P()).
(ii) n0 + P
n1 α + · · · + nk αk 6= 0 for all non-negative integers k and ni such
that
ni > 0.
(iii) q0 + q1 α + · · · + qk αk 6= 0 for everyPnon-negative integer k and all
non-negative rationals qi such that
qi > 0.
(iv) α 6= 0 and m0 + m1 α + · · · + ml αl 6= 0 for every non-negative integer
l and all positive integers mi .
1991 Mathematics Subject Classification. 16Y60.
Key words and phrases. complex number, algebraic number, order, parasemifield.
The work is a part of the research project MSM0021620839, financed by MŠMT.
The first author was supported by the Grant Agency of the Czech Republic, grant
#201/09/0296. The second author was supported by the project LC 505 of Eduard Čech’s
Center for Algebra and Geometry.
1
2
T. KEPKA AND M. KORBELÁŘ
(v) α 6= 0 and p0 + p1 α + · · · + pl αl 6= 0 for every non-negative integer l
and all positive rationals pi .
Locally to this paper, let us call a complex number α quasipositive if
α satisfies the equivalent conditions 1.1(i),. . . ,(v). It is immediately clear
that all transcendental complex numbers are quasipositive and all (in the
usual sense) positive real numbers are quasipositive. A rational number is
quasipositive if and only if it is positive. An algebraic complex number α is
quasipositive if and only if all the conjugates
of α are such. In particular,
√
the negative algebraic real number − 2 is quasipositive. If α is a complex
number that is algebraic of degree two, then α is quasipositive if and only
√
if (α is real and) α = p ± q, where p, q ∈ Q (the field of rationals), q > 0,
√
√ √
√
√
q ∈
/ Q and q > −p. The complex number − 3 2 + i 3 2 3 is algebraic
√
√ √
of degree three and quasipositive. The complex number 3 2 + i 3 2 3 is
algebraic of degree three, but it is not quasipositive. In [3], a characterization
is given for the quasipositive algebraic complex
√ numbers
√
√α of degree
√ four
such that |α| = 1. The conjugate numbers 4 2, − 4 2, i 4 2 and −i 4 2 are
algebraic of degree four and quasipositive.
Let 1 be a non-identical order satisfying (1) and (2) and let P = P(1 )
and R = P P −1 = {αβ −1 | α, β ∈ P }. Then R is a subsemiring of C,
0 ∈
/ R, P ⊆ R and the multiplicative semigroup of R is a subgroup of
the multiplicative group of non-zero complex numbers. In other words, R
is a subparasemifield of the field C. The corresponding order relation 2
contains 1 and satisfies the following condition:
(3) 0 ≺2 α ⇒ 0 ≺2 α−1 .
2. A characterization of quasipositive algebraic complex
numbers
It follows immediately from the definition that all transcendental complex
numbers and all positive real numbers are quasipositive. If α is an algebraic
complex number that is conjugate to a positive (algebraic) real number, then
α is quasipositive. The converse statement was conjectured and partially
solved in [3]. A full solution was given in [1]. Thus:
2.1. Theorem. ([1]) A complex number α is algebraic and quasipositive if
and only if α is conjugate to a positive algebraic real number.
The original proof of 2.1 is not completely constructive. A more straightforward proof can be found in [2] (see the next three sections for an outline).
3. Technical observations (a)
n
First, put sc,n = 2n
n c for all c ∈ R (the field of real numbers) and all
n ∈ N0 (the semiring of non-negative integers). It follows from Stirling’s
formula that limn→∞ sc,n = +∞ for c > 1/4, limn→∞ sc,n = 0 for |c| ≤ 1/4
and limn→∞ sc,n does not exist for c < −1/4. In particular, if c > 1/4,
NON-LINEAR ORDERS ON COMPLEX NUMBERS
3
then we are able to find m ∈ N (the semiring of positive integers) such that
sc,k > 1 for every k ≥ 2m .
Now, let a, b ∈ R. For all k, m ∈ N0 define real numbers (rk,m,a,b =)rk,m
by means of the following rules:
m+1 m+1
(4) rk,m = rk+1,m = 2k/2 b2 −k/2 if k is even, 0 ≤ k ≤ 2m+2 and
k 6= 2m+1 ;
m+1
(5) r2m+1 ,m = r2m+1 +1,m = sb,2m − a2
;
m+2
(6) rk,m = 0 for k ≥ 2
+ 2.
If b > 0 then, quite evidently, rk,m > 0 for every k such that k ≤ 2m+2 + 1,
k∈
/ {2m+1 , 2m+1 + 1}. The only possibly non-positive number among these
numbers is r2m+1 ,m = r2m+1 +1,m . However, if 4b > a2 > 0, then we can
always find m ∈ N with sb/a2 ,2m > 1. Then we get r2m+1 ,m > 0.
P
k
Finally, put (fm,a,b =)fa,b = ∞
k=0 rk,m x ∈ R[x]. Then fa,b is a monic
m+2
2
polynomial of degree 2
+ 1. If 4b > a > 0, then there exists at least
one m ∈ N such that all the coefficients (more precisely, the first 2m+2 + 2
coefficients) of fa,b are positive real numbers. If, moreover, a, b ∈ Q, then
the coefficients of fa,b are positive rationals.
4. Technical observations (b)
i
i
Put (hn (x, y, z) =)hn = Πni=0 ((xy)2 + (x2 + z)2 ) ∈ N0 [x, y, z] for every
n ∈ N0 . Clearly, degx hn = 2n+2 − 2, h0 = x2 + xy + z and h1 = x6 + x5 y +
2x4 z + x3 y 3 + 2x2 yz + x2 y 2 + x2 y 2 z + 3x2 z 2 + xyz 2 + z 3 . Furthermore, put
(gn,a,b (x, y, z) =)gn,a,b = (x3 + (1 − a)x2 + (b − a)x + b)hn ∈ R[x, y, z] for all
n ∈ N0 and a, b ∈ R. Now, degx gn,a,b = 2n+2 +1 and g0,a,b = x5 +(1−a)x4 +
x4 y+(b−a)x3 +(1−a)x3 y+bx2 +(b−a)x2 y+(2−a)x2 z +bxy+(b−a)xz +bz.
Of course, there are polynomials (sk,n,a,b (y, z) =)sk,n,a,b ∈ R[y, z] such that
P n+2
n+2
gn,a,b = 2k=0 +1 sk,n,a,b xk . We have s2n+2 +1,n,a,b = 1, s0,n,a,b = bz 2 −1 and
s5,0,a,b = 1, s4,0,a,b = y, s3,0,a,b = (1−a)y+b−a, s2,0,a,b = (b−a)y+(2−a)z+b,
s1,0,a,b = by + (b − a)z and s0,0,a,b = bz. Besides, one checks easily that:
(7) sk,n,a,b (a, b) = rk,n,a,b
for all n ∈ N and k ∈ N0 , 0 ≤ k ≤ 2n+2 + 1 (see (4) and (5)). (On
the other hand, s5,0,a,b (a, b) = 1 = r5,0,a,b , s4,0,a,b (a, b) = a, 1 = r4,0,a,b ,
s3,0,a,b (a, b) = b − a2 = r3,0,a,b , s2,0,a,b (a, b) = 3b − a2 , b − a2 = r2,0,a,b ,
s1,0,a,b (a, b) = b2 = r1,0,a,b , s0,0,a,b (a, b) = b2 = r0,0,a,b . Thus (7) is true for
n = 0 only if a = 1 and b = 0.)
Now, assume that 4b > a2 > 0. Then we find m ∈ N such that all the real
numbers r0,m,a,b , r1,m,a,b , . . . , r2m+2 ,m,a,b , r2m+2 +1,m,a,b are positive. By (7),
sk,m,a,b (a, b) > 0 for all k = 0, 1, . . . , 2m+2 + 1. Next, we find p, q ∈ Q such
that
(8) sk,m,a,b (p, q) > 0
4
T. KEPKA AND M. KORBELÁŘ
for all 0 ≤ k ≤ 2m+2 + 1. Now, the monic polynomial
(tm,a,b,p,q (x) =)tm,a,b,p,q = gm,a,b (x, p, q) =
2m+2
X+1
sk,m,a,b (p, q)rk
k=0
is of degree
2m+2
+ 1 (≥ 9) and has positive real coefficients.
5. Partial summary
If r ∈ R is not quasipositive, then r ≤ 0 and the monic polynomial
x − r ∈ R[x] has non-negative coefficients. The polynomial has positive
coefficients if and only if r < 0.
Now, consider a complex number α = r + si, r, s ∈ R, s 6= 0, and put
a = −2r, b = r2 + s2 and f = (x − α)(x − ᾱ) = x2 + ax + b ∈ R[x]. If
r < 0, then f has positive coefficients. If r = 0, then a = 0 and the monic
polynomial (x + 1)f = x3 + x2 + bx + b has positive coefficients. Finally, if
0 < r, then a < 0 and there exist m ∈ N and p, q ∈ Q such that the monic
polynomial tm,a,b,p,q = (x + 1)f hm (x, p, q) ∈ R[x] has positive coefficients
(see the preceding section). Notice that (x + 1)hm (x, p, q) ∈ Q[x] is a monic
polynomial of degree 2m+2 − 1 (≥ 7).
Let f ∈ R[x] be a polynomial such that f has no non-negative real root.
We have f = tf1 . . . fn where n ∈ N0 , t ∈ R, t 6= 0, and fi ∈ R[x] are
monic irreducible polynomials. Then, of course, degfi ∈ {1, 2} and taking
into account the preceding part of this section, we can find monic rational
polynomials hi ∈ Q[x] such that the product hi fi has only positive coefficients. Now, the product h = h1 . . . hn is a monic rational polynomial and
the polynomial hf for t > 0 (−hf for t < 0, resp.) has positive coefficients.
If f ∈ Q[x], then hf ∈ Q[x] as well.
6. Algorithmic questions
Let α ∈ C be not quasipositive. Then f (α) = 0 for a polynomial f ∈
Q+ [x] (Q+ being the parasemifield of positive rationals), i.e., a non-zero
polynomial with non-negative rational coefficients. (Of course, f can be
chosen to possess only positive integral coefficients.) The technical notes
collected in the preceding three sections suggest a more or less efficient
way how to find at least one sample of such a polynomial. On the other
hand, quite inefficient ways are immediately at hand. It is enough to find a
recursive bijection n 7→ fn of the set N onto the set N0 [x]. One familiar way
is the following one: let p0 < p1 < . . . be the sequence of all prime numbers
and put fn = e0 + e1 x + · · · + ek xk where ei are the non-negative integers
with n = pe00 . . . pekk .
7. Ultraquasipositive complex numbers
+
n
Let
P α ∈ C and P = Q [α] (= {q0 +q1 α+· · ·+qn α | n ∈ N0 , qi ∈ Q, qi ≥
0,
qi > 0}). Then P is a subsemiring of C and, if α is transcendental,
then 0 ∈
/ P and α−1 ∈
/ P . If α = 0, then P = Q+ is a subparasemifield of C.
NON-LINEAR ORDERS ON COMPLEX NUMBERS
5
Now, assume that α is algebraic, α 6= 0 and α is not quasipositive and put
R = P ∩Q. Then R is a subsemiring of Q, 0 ∈ R and Q+ ⊆ R. Furthermore,
we see easily that R contains at least one negative rational number and we
conclude that R = Q. Consequently, P = Q[α] is a subfield of C. Finally,
assume that α is algebraic and quasipositive. Then 0 ∈
/ P and it may (but
need not) happen that P is a subparasemifield of C (i.e., the multiplicative
semigroup of P is a group). In such a case, the number α will be called
ultraquasipositive. We do not know much about these numbers. Evidently,
all positive rationals are ultraquasipositive. According to Lemma 7.3 of [2],
if α is algebraic of degree two, then α is ultraquasipositive if and only if
√
√
√
α = p ± q for some p, q ∈ Q such that p < 0, 0 < q, q ∈
/ Q and q > −p.
√
√ +
√
√
Moreover, if α = p + q, then P = Q[ q] = {r + s q | r, s ∈ Q, r + s q >
√
√
√
0} and, if α = p − q, then P = {r − s q | r, s ∈ Q, r + s q > 0}. In
√
particular, α =√ 2 is quasipositive,
but not√ultraquasipositive (in spite of
√
the fact that √
( 2)−1 = 2/2 ∈ P = Q+ [ 2]). On the other hand, the
number −1 − 2 is ultraquasipositive.
8. Minimal and maximal orders
Let P be a subsemiring of C such that 0 ∈
/ P . Take α ∈ P
P and put
R = {k1 α2 + k2 α4 + k3 α6 + · · · + kn α2n | n ∈ N, ki ∈ N0 ,
ki > 0}.
Clearly, R ⊆ P and R is just the subsemiring of C generated by the square
α2 . If R = P , then 1 ∈ P , and therefore N ⊆ P . In view of this easy fact, it is
clear that there are no minimal non-identical orders on the ring C (i.e., those
satisfying (1) and (2)). On the other hand, if P were a subparasemifield,
then Q+ ⊆ P . It follows that the order corresponding to Q+ is the least
order of the field C (i.e., satisfying (1), (2) and (3)).
Every order relation satisfying (1) and (2) is contained in a maximal one
and the maximal orders satisfy (3). That is, the maximal orders correspond
to maximal subparasemifields of C. Unfortunately, the nature of such subparasemifields escapes better understanding so far. Easy observations are
available, however.
First, notice that R+ is a maximal subparasemifield of C. Indeed, proceeding by contradiction, assume that R+ ⊆ P where P is a parasemifield
containing a number α = a + bi, a, b ∈ R, b 6= 0. If a ≤ 0, then bi ∈ P ,
−b2 ∈ P and, finally, −1 ∈ P , a contradiction. If a 6= 0 and b2 ≥ a2 , then
2abi ∈ P , −4a2 b2 ∈ P and −1 ∈ P again. If a > 0, then a2 + 2abi ∈ P
implies a + 2bi ∈ P . Similarly, a + 4bi ∈ P , etc. But a2 ≤ (2k b)2 for some
k ∈ N and this is the final contradiction.
Let P be a maximal subparasemifield of C and let R = P − P (the
difference ring of P ). We claim that R is a subfield of C. Indeed, let
α, β ∈ P , α 6= β and γ = α − β. If γ ∈ P , then γ −1 ∈ P ⊆ R. If γ ∈
/ P , then
0 ∈ P [γ] and there exist n ∈ N and π0 , π1 , . . . , πn ∈ P such that π0 6= 0 =
π0 + π1 γ + · · · + πn γ n . Now, γ −1 = −π0−1 π1 − π0−1 π2 γ − · · · − π0−1 πn γ n−1 ∈ R.
6
T. KEPKA AND M. KORBELÁŘ
We have thus shown that R is a subfield of C. Evidently, C is algebraic
over R.
References
[1] A. Dubickas, On roots of polynomials with positive coefficients. Manuscripta Math.
123 (2007), 353–356.
[2] T. Kepka and M. Korbelář, Various examples of parasemifields. Acta Univ. Carol.
Math. Phys. 50/1 (2009), 61–72.
[3] G. Kuba, Several types of algebraic numbers on the unit circle. Arch. Math. 85 (2005),
70–78.
Charles University, Faculty of Mathematics and Physics, Department of
Algebra, Sokolovská 83, 186 75 Prague 8, Czech Republic
E-mail address: [email protected]
Department of Mathematics and Statistics, Faculty of Science, Masaryk
University, Kotlářská 2, 611 37 Brno, Czech Republic
E-mail address: [email protected]