Generalizing circles over algebraic extensions
7 Generalizing circles over algebraic extensions
T. Recio∗ J.R. Sendra∗† L.F. Tabera∗‡ C. Villarino∗†
August 29, 2021
Abstract
This paper deals with a family of spatial rational curves that were
introduced in [4], under the name of hypercircles, as an algorithmic
cornerstone tool in the context of improving the rational parametriza-
tion (simplifying the coefficients of the rational functions, when pos-
sible) of algebraic varieties. A real circle can be defined as the image
of the real axis under a Moebius transformation in the complex field.
Likewise, and roughly speaking, a hypercircle can be defined as the
image of a line (“the K-axis”) in a n-degree finite algebraic extension
K(α) ≈ Kn under the transformation at+b
: K(α) → K(α).
The aim of this article is to extend, to the case of hypercircles,
some of the specific properties of circles. We show that hypercircles
are precisely, via K-projective transformations, the rational normal
curve of a suitable degree. We also obtain a complete description of
the points at infinity of these curves (generalizing the cyclic structure
at infinity of circles). We characterize hypercircles as those curves of
degree equal to the dimension of the ambient affine space and with
infinitely many K-rational points, passing through these points at in-
finity. Moreover, we give explicit formulae for the parametrization
and implicitation of hypercircles. Besides the intrinsic interest of this
very special family of curves, the understanding of its properties has
a direct application to the simplification of parametrizations problem,
as shown in the last section.
∗The authors are partially supported by the project MTM2005-08690-CO2-01/02 “Min-
isterio de Educación y Ciencia”.
†Partially supported by CAM-UAH2005/053 “Dirección General de Universidades de
la Consejeŕıa de Educación de la CAM y la Universidad de Alcalá”.
‡L.F.Tabera also supported by a FPU research grant.
http://arxiv.org/abs/0704.1384v1
1 Introduction
The problem of obtaining a real parametrization of a rational planar curve
given by a complex parametrization has been studied –from an algorithmic
point of view– in [9]. There, the problem is reduced to determining that a
certain curve obtained after manipulating the given parametrization is a real
line or a real circle. From a real parametrization of this circle (or line), a real
parametrization of the original curve is then achieved. This auxiliary circle is
found by an analogous to Weil descente’s method [16] applied to the complex
parametrization of the originally given curve. In [4], the same approach has
been extended to the general case of planar or spatial rational curves C given
by a parametrization over K(α), where α is an algebraic element over K. In
order to obtain, whenever possible, a parametrization over K of C, another
rational curve, with remarkable properties, is associated to C. In [4] it is
shown that this associated curve is, in the relevant cases, a generalization of
a circle, in the sense we will discuss below, deserving to be named hypercircle.
The simplest hypercircles should be the circles themselves. We can think
of the real plane as the field of complex numbers C, an algebraic extension
of the reals R of degree 2. Analogously, we can consider a characteristic zero
base field K and an algebraic extension of degree n, K(α). Let us identify
K(α) as the vector space Kn, via the choice of a suitable base, such as the
one given by the powers of α. This is the framework in which hypercircles
are defined.
Now let us look to the different, equivalent, ways of defining a common
circle on the real plane, with the purpose of taking the most convenient one
for generalization. The first definition of a circle is the set of points in the
real plane that are equidistant from a fixed point. This approach does not
extend well to more general algebraic extensions, because we do not have an
immediate notion of metric over Kn. On the other hand, algebraically, a real
planar circle is a conic such that its homogeneous degree two form is x2 + y2
and such that it contains an infinite number of real points. Even if we will
prove in Section 6 that we can show an analogous definition for a hypercircle,
this is not an operative way to start defining them.
Finally, from another point of view, we see that circles are real rational
curves. This means that there are two real rational functions (φ1(t), φ2(t))
whose image cover almost all the points of the circle. For instance, the circle
x2+y2 = 1 is parametrized by φ(t) = ( t
). Every proper (almost one-
to-one [14]) rational parametrization of a circle verifies that φ1(t) + iφ2(t) =
∈ C(t) \ C, which defines a conformal mapping u : C → C. Moreover,
if we identify C with R2, the image of the real axis (t, 0) under u is exactly
the circle parametrized by φ(t). Conversely, let u(t) = at+b
∈ C(t) be a unit
of the near-ring C(t) under the composition operator (see [17]). If c 6= 0
and d/c /∈ R then, the closure of the image by u of the real axis is a circle.
Otherwise, it is a line. This method to construct circles generalizes easily
to algebraic extensions. Namely, let u(t) = at+b
be a unit of K(α)(t) (i.e.
verifying that ad − bc 6= 0). Let us identify K(α) with Kn and let u be the
u : K(α) ≈ Kn → K(α) ≈ Kn
t 7→ u(t)
Then, the Zariski-closure of the image of the axis (t, 0, . . . , 0) under the map
u is a rational curve in Kn. These curves are, by definition, our hypercircles.
Roughly speaking, it happens (see [4]) that a parametrization over K of
the hypercircle associated to a given rational curve C (whose parametrization
we want to simplify) can be used to get –in a straightforward manner– a
parametrization of C over K. As pointed in [4], it seems that, due to the
geometric properties of hypercircles, it is algorithmically simpler to obtain
such parametrization for this type of curves than it is for C. In fact, it is
shown in [10] how to get this in some cases. Therefore, the reparametrization
problem is behind our increasing interest in the study of hypercircles on its
The structure of this paper is as follows. In Section 2 we formally intro-
duce the notion of hypercircle. We study the influence on a hypercircle when
adding and multiplying the defining unit u(t) by elements of K(α), reduc-
ing the affine classification of hypercircles to those defined by some simpler
units. Next we characterize the units associated to lines. In Section 3 we
show how to transform, projectively, a hypercircle into the rational normal
curve (see [6]). From this, we derive the main geometric properties of hy-
percircles (smoothness, degree, affine equivalence, etc.) and we reduce the
study of hypercircles to the subclass of primitive hypercircles (See Definition
3.5). In Section 4 the behavior of hypercircles at infinity is analyzed, showing
its precise and rich structure. In Section 5, exploiting the stated geometric
features, we present ad hoc parametrization and implicitization methods for
hypercircles. In Section 6 we characterize hypercircles among curves of de-
gree equal to the dimension of the ambient affine space, passing through
the prescribed points of infinity described in Section 4 and having infinitely
many rational points. Finally, Section 7 is devoted to show how the insight
gained throughout this paper can be applied to derive heuristics for solving
the problem of simplifying the parametrization of curves with coefficients
involving algebraic elements.
Throughout this paper the following notation and terminology will be
used.
• K will be a field of characteristic zero, K ⊆ L a finite algebraic extension
of degree n and F the algebraic closure of K.
• α will be a primitive element of L over K.
• u(t) will be a unit under composition of L(t). That is, u(t) = at+b
ad− bc 6= 0. Its inverse −dt+b
is denoted by v(t).
• For u(t) = at+b
and c 6= 0, M(t) = tr + kr−1t
r−1 + · · · + k0 ∈ K[t]
denotes the minimal polynomial of −d/c over K.
• We will denote as m(t) the polynomial obtained by dividing M(t) by
ct+ d. That is, m(t) =
ct+ d
= lr−1t
r−1 + lr−2t
r−2 + · · ·+ l0 ∈ L[t].
• Sometimes we will represent u(t) as
u(t) =
(at+ b)m(t)
p0(t) + p1(t)α + · · ·+ pn−1(t)α
where pi(t) ∈ K[t].
• By {σ1 = Id, σ2, . . . , σs}, s ≥ n we will denote the group of K-
automorphisms of the normal closure of K ⊆ L.
• We will represent by {α1 = α, . . . , αn} the conjugates of α. We assume
without loss of generality that σi(α) = αi for i = 1, . . . , n.
2 Definition and First Properties
In this section we begin with the formal definition of a hypercircle.
Definition 2.1. Let u(t) be a unit in L(t), where L = K(α). Let
u(t) =
φi(t)α
where φi(t) ∈ K(t), for i = 0, . . . , n − 1. The α-hypercircle U generated by
u(t) is the rational curve in Fn parametrized by φ(t) = (φ0(t), . . . , φn−1(t)).
Observe that the expansion of u(t) in powers of α is unique, because
{1, α, . . . , αn−1} is a basis ofK(α)(t) as aK(t)−vector space. The parametriza-
tion can be obtained by rationalizing the denominator as follows: suppose
given the unit u(t) = at+b
, c 6= 0 (remark that, if c = 0, it is straightforward
to obtain φ(t)), and the extension K ⊆ K(α). Let M(t) be the minimal
polynomial of −d/c over K. Compute the quotient m(t) =
∈ K(α)[t]
and develop the unit as
at+ b
ct+ d
(at+ b)m(t)
p0(t) + p1(t)α + · · ·+ pn−1(t)α
where pi(t) ∈ K[t]. From this, φ(t) =
p0(t)
, . . . ,
pn−1(t)
is the parametriza-
tion associated to u(t). Remark that gcd(p0(t), . . . , pn−1(t),M(t)) = 1. More-
over, it is clear that F(φ0(t), . . . , φn−1(t)) = F(t). So this parametrization is
proper in F, and it follows from the results in [1] that alsoK(φ0(t), . . . , φn−1(t)) =
K(t).
Example 2.2. Let us consider the algebraic extension Q ⊆ Q(α), where
α3 + 2α+ 2 = 0. The unit t−α
has an associated hypercircle parametriced by
φ(t) =
t3 + 2t+ 2
t3 + 2t− 2
t3 + 2t− 2
t3 + 2t− 2
A picture of the spatial real curve is shown in Figure 1
As it stands, the definition of a hypercircle U depends on a given unit
u(t) ∈ L(t) and on a primitive generator α of an algebraic extension L. In
what follows we will analyze the effect on U when varying some of these
items, searching for a simple representation of a hypercircle to ease studying
its geometry.
First notice that, given a unit u(t) ∈ L(t) and two different primitive
elements α and β of the extension K ⊆ L, we can expand the unit in
Figure 1: A hypercircle in R3
two different ways u(t) =
i=0 α
iφi(t) =
i=0 β
iψi(t). The hypercircles
Uα ≃ (φ0(t), . . . , φn−1(t)) and Uβ ≃ (ψ0(t), . . . , ψn−1(t)) generated by u(t)
are different curves in Fn, see Example 2.3. Nevertheless, let A ∈ Mn×n(K)
be the matrix of change of basis from {1, α, . . . , αn−1} to {1, β, . . . , βn−1}.
Then, A(φ0(t), . . . , φn−1(t))
t = (ψ0(t), . . . , ψn−1(t))
t. That is, it carries one
of the curve onto the other. Thus, Uα and Uβ are related by the affine trans-
formation induced by the change of basis and, so, they share many important
geometric properties.
In the sequel, if there is no confusion about the algebraic extension and
the primitive element, we will simply call U a hypercircle.
Example 2.3. Let us consider the algebraic extension Q ⊆ Q(α), where
α4 + 1 = 0. Let us take the unit u(t) = t−α
. By normalizing u(t), we obtain
the parametrization φ(t) associated to u(t):
φ(t) =
t4 − 1
t4 + 1
t4 + 1
t4 + 1
t4 + 1
This hypercircle Uα is the zero set of {X1X2 − X3X0 − X3, X
1 + X
2X2, X1X0+X2X3−X1, X
0 +X3X1− 1}. Now, we take β = α
3+1, instead
of α, as the primitive element of Q(α) = Q(β). The same unit u(t) generates
the β-hypercircle Uβ parametrized by
ψ(t) =
t4 + 2t3 − 2t2 + 2t− 1
t4 + 1
−6t3 + 4t2 − 2t
t4 + 1
6t3 − 2t2
t4 + 1
t4 + 1
which is different to Uα; note that ψ(1) = (1,−2, 2,−1) that does not satisfy
the equation X20 +X3X1 − 1 = 0 of Uα.
On the other hand it is well known that a given parametric curve can be
parametrized over a given field S by different proper parametrizations, pre-
cisely, those obtained by composing to the right a given proper parametriza-
tion by a unit in S(t). In this way, we have a bijection between α-hypercircles
and the equivalence classes of units of K(α)(t) under the equivalence relation
“u ∼ v iff u(t) = v(τ(t)) for a unit τ(t) ∈ K(t)” (fixing the correspondence,
between a unit in K(α)(t) and a hypercircle, by means of the expansion of
the unit in terms of powers of α).
More interesting is to analyze, on a hypercircle defined by a unit u(t),
the effect of composing it to the left with another unit τ(t) ∈ K(α)(t), that
is, of getting τ(u(t)). For instance, τ(t) could be τ(t) = t + λ or τ(t) = λt,
or τ(t) = 1/t, with λ ∈ K(α)∗. Every unit is a sequence of compositions of
these three simpler cases, for instance, when c 6= 0, we have
t 7−→ ct 7−→ ct + d 7−→
ct + d
bc− ad
ct+ d
bc− ad
ct+ d
at + b
ct+ d
= u(t).
Therefore, studying their independent effect is all we need to understand
completely the behavior of a hypercircle under left composition by units.
For circles, adding a complex number to the unit that defines the circle
correspond to a translation of the circle. Multiplying it by a complex number
acts as the composition of a rotation and a dilation. And the application
τ(t) = 1/t gives an inversion. The following lemma analyzes what happens
in the general case.
Lemma 2.4. Let U be the α-hypercircle generated by u(t), and λ =
i ∈ K(α)∗,
where λi ∈ K. Then,
1. λ + u(t) is a unit generating the hypercircle obtained from U by the
translation of vector (λ0, . . . , λn−1).
2. λu(t) is a unit generating the hypercircle obtained from U by the affine
transformation over K given by the matrix of change of basis from
B⋆ = {λ, λα, . . . , λαn−1} to B = {1, α, . . . , αn−1}.
Proof. To prove (1), let φ(t) = (φ0(t), . . . , φn−1(t)) ∈ K(t)
n be the parametriza-
tion of U obtained from u(t). Then, λ+u(t) =
i=0 (λi +φi(t))α
i generates
the hypercircle parametrized by (λ0 + φ0(t), . . . , λn−1 + φn−1(t)) ∈ K(t)
which is the translation of U of vector (λ0, . . . , λn−1). For the second asser-
tion, let φ⋆(t) ∈ K(t)n be the parametrization of the hypercircle associated
to the unit λu(t). The rational coordinates φ⋆i (t) of φ
⋆(t) are obtained from
the matrix A = (ai,j) ∈ Mn×n(K) of change of basis from B
⋆ to B, for
i, j = 0, . . . , n− 1. Indeed,
λu(t) =
φi(t)λα
φi(t)
ajiφi(t)
Then φ⋆(t)t = A φ(t)t.
Finally, the following lemma uses the previous results to transform affinely
one hypercircle into another one whose unit is simpler.
Lemma 2.5. Let u(t) = at+b
be a unit and U its associated hypercircle.
1. If c = 0 then U is affinely equivalent over K to the line generated by
u⋆(t) = t.
2. If c 6= 0 then U is affinely equivalent over K to the hypercircle U⋆
generated by u⋆(t) = 1
t+d/c
Proof. This lemma follows from Lemma 2.4, taking into account that u(t) is
obtained from u⋆(t) by the following composition:
u⋆(t) 7→ λ1u
⋆(t) 7→ λ1u
⋆(t) + λ2 = u(t)
with suitable λ1, λ2, u
⋆. If c = 0, then λ1 =
6= 0 and λ2 =
for u⋆(t) = t.
Analogously, if c 6= 0, then u(t) is obtained from u⋆(t) = 1
t+d/c
taking λ1 =
bc−ad
6= 0 and λ2 =
Therefore the (affine) geometry of hypercircles can be reduced to those
generated by a unit of type 1
(then we say the unit is in reduced form).
The simplest hypercircle of this kind is given by 1
, when d ∈ K. It is the
line parametrized by ( 1
, 0, . . . , 0). In the complex case, Moebius transfor-
mations defining lines are precisely those given either by a polynomial unit
in t (i.e. a unit without t at the denominator) or by a unit such that the root
of the denominator is in R. The same property holds for hypercircles.
Theorem 2.6. Let U be the α-hypercircle associated to u(t). Then, the
following statements are equivalent:
1. U is a line.
2. U is associated to a polynomial unit.
3. The root of the denominator of every non polynomial unit generating
U belongs to K.
4. U is polynomially parametrizable (over F).
5. U has one and only one branch (over F ) at infinity.
6. U is polynomially parametrizable over K.
7. U has one and only one branch (over K ) at infinity.
Proof. (1)⇔ (2). By definition, we know that hypercircles have a parametriza-
tion overK. Thus, if U is a line, it can be parametrized as (a0t+b0, . . . , an−1t+
bn−1), where ai , bi ∈ K. Therefore, u(t) =
(∑n−1
i=0 aiα
i=0 biα
i is a
polynomial unit associated to U . Conversely, let u(t) = at+ b ∈ L(t), a 6= 0,
be a polynomial unit associated to U . Then U is the line parametrized
by P(t) = (a0t + b0, . . . , an−1t + bn−1) ∈ K[t]
n, where a =
i=0 ai α
i and
i=0 bi α
(2) ⇔ (3). Let u(t) = at+b be a polynomial unit associated to U , and let
u⋆(t) be another non polynomial unit associated to U . Then, u⋆(t) = u(τ(t)),
where τ(t) is a unit of K (t). Therefore, the root of u⋆(t) belongs to K.
Conversely, by Lemma 2.5, (3) implies (1), and we know that (1) implies (2).
(3) ⇔ (4). Indeed, (3) implies (2) and therefore (4). Conversely, let u(t)
be a non-polynomial unit generating U , and let φ(t) = (φi)i=1,...,n ∈ K(t)
the associated parametrization of U . Then, φ(t) is proper, φi(t) =
pi(t)
deg(pi) ≤ deg(M) and gcd(p0(t) . . . pn−1(t),M(t)) = 1. Thus, the fact that
U admits a polynomial parametrization, implies, by Abhyankar-Manocha-
Canny’s criterion of polynomiality (see [8]), that the denominator M(t) is
either constant or has only one root. Now, M(t) can not be constant, since
it is a minimal polynomial. Thus, M has only one root, and since it is
irreducible, it must be linear. Moreover, since M ∈ K[t], its root is an
element in K.
(4) ⇔ (5) This is, again, the geometric version of Abhyankar-Manocha-
Canny’s criterion. Same for (6) ⇔ (7).
(4) ⇔ (6) Obviously (6) implies (4). Conversely, if we have a polynomial
parametrization over F, it happens [2] that any proper parametrization must
be either polynomial or in all its components the degree of the numerator
must be smaller or equal than the degree of the denominator and, then, this
denominator has only one single root over F. So, since the parametrization
φ(t) induced by the unit is proper, and by hypothesis U is polynomial, then
φ(t) must be either polynomial (in which case we are done because φ(t) is
over K) or its denominator M(t) has a single root a ∈ F. Now, reasoning
as above one gets that a ∈ K. So, a change of parameter, such as t 7→ 1+as
turns φ(t) into a K-polynomial parametrization.
As a corollary of this theorem, we observe that a parabola can never be
a hypercircle, since it is polynomially parametrizable, but it is not a line.
Nevertheless, it is easy to check that the other irreducible conics are indeed
hypercircles for certain algebraic extensions of degree 2.
3 Main Geometric Properties.
This section is devoted to the analysis on the main geometric properties of
hypercircles. The key idea, when not dealing with lines, will be to use the
reduction to units of the form u(t) = 1
, where d /∈ K (see Lemma 2.5).
Theorem 3.1. Let U be the α-hypercircle associated to the unit u(t) = at+b
K(α)(t) and let r = [K(−d) : K]. Then,
1. there exists an affine transformation χ : Fn −→ Fn defined over K such
that the curve χ(U) is parametrized by
χ̃(t) =
, . . . ,
, 0, . . . , 0
2. there exists a projective transformation ρ : P(F)
−→ P(F)
, defined
over K, such that the curve ρ(U) is the rational normal curve of degree
r in P(F)
, parametrized by
ρ̃(t : s) = [sr : sr−1t : · · · : str−1 : tr : 0 : · · · : 0].
Proof. For the case of lines the result is trivial. By Lemma 2.5, we can
consider that U is the hypercircle associated to u(t) = 1
and r ≥ 2. Let
M(t) = tr + kr−1t
k−1 + · · ·+ k0 ∈ K[t], m(t) =
i=0 lit
i ∈ L[t], as indicated
in Section 1 and, since the numerator of u(t) is 1, it holds that m(t) =∑n−1
i=0 pi(t)α
i, pi(t) ∈ K[t]. Also, note that both M(t) and the denominator
of u(t) are monic, and hence lr−1 = 1. First of all, we prove that there
are exactly r polynomials in {pi(t), i = 0, . . . , n − 1} ⊂ K[t] being linearly
independent. For this purpose, we observe that the coefficients of m(t),
{1, lr−2, . . . , l0} ⊂ L, are linearly independent over K. Indeed, from the
equalityM(t) = (t+d)m(t), one has that lr−i = (−d)
i−1+(−d)i−2kr−1+· · ·+
kr−i+1, for i = 2, . . . , r. So, {1, lr−2, . . . , l0} ⊂ L are K–linearly independent,
since otherwise one would find a non-zero polynomial of degree smaller than
r vanishing at −d. Now, let ~li = (li,0, . . . , li,n−1)
t be the vector of coordinates
of li in the base {1, α, . . . , α
n−1}. Then, {~1,~lr−2, . . . ,~l0} ⊂ K
n are K–linearly
independent. Moreover, since (p0(t), . . . , pn−1(t))
t = ~1tr−1+~lr−2t
r−2+· · ·+~l0,
there are r polynomials pij , 0 ≤ i1 < · · · < ir ≤ n− 1, linearly independent.
By simplicity, we assume w.l.o.g. that the first r polynomials are linearly
independent. Observe that this is always possible through a permutation
matrix. The new curve, that we will continue denoting by U , is not, in
general, a hypercircle. In this situation, we proceed to prove (1) and (2).
In order to prove (1), let A ∈ Mn−r×r(K) be the matrix providing the
linear combinations of the n − r last polynomials in terms of the first r
polynomials; i.e. (pr(t), . . . , pn−1(t))
t = A(p0(t), . . . , pr−1(t))
t. Now, given
the bases B = {1, . . . , tr−1} and B⋆ = {p0(t), . . . , pr−1(t)}, let M ∈ Mr×r(K)
be the transpose matrix of change of bases from B to B⋆. Finally, the n× n
matrix
M Or,n−r
−A In−r
defines, under the previous assumptions, the affine transformation χ. Note
that if r = n then Q = M.
The proof of (2) is analogous to (1). Now, let consider the basis B =
{1, . . . , tr−1, tr} and B⋆ = {p0(t), . . . , pr−1(t),M(t)}. Let A ∈ Mn−r×r+1(K)
be the matrix providing the linear combinations of the n−r last polynomials
in terms of basis B⋆; i.e. (pr(t), . . . , pn−1(t))
t = A(p0(t), . . . , pr−1(t),M(t))
Let M ∈ Mr+1×r+1(K) be the transpose matrix of change of bases from B
to B⋆. Finally, the n + 1× n + 1 matrix
M Or+1,n−r
−A In−r
defines, under the previous assumptions, the projective transformation ρ.
Note that if r = n then Q = M.
As a direct consequence, we derive the following geometric properties of
hypercircles.
Corollary 3.2. In the hypothesis of Theorem 3.1
1. U defines a curve of degree r.
2. U is contained in a linear variety of dimension r and it is not contained
in a variety of dimension r − 1.
3. U is a regular curve in P(F)
4. The Hilbert function of U is equal to its Hilbert polynomial and hU(m) =
mn+ 1.
Proof. All these properties are well known to hold for the rational normal
curve of degree r e.g. [6], [7], [15]).
In the following theorem, we classify the hypercircles that are affinely
equivalent over K. We will assume that the denominator of the generat-
ing units are not constant. The case where the units are polynomials are
described in Theorem 2.6.
Theorem 3.3. Let Ui, i = 1, 2, be α-hypercircles associated to ui(t) =
ait+bi
and let Mi(t) be the minimal polynomial of −di over K. Then, the following
statements are equivalent:
1. U1 and U2 are affinely equivalent over K.
2. There exists a unit τ(t) ∈ K(t) such that it maps a root (and hence all
roots) of M1(t) onto a root (resp. all roots) of M2(t).
Proof. First of all note that, because of Theorem 2.6, the result for lines
is trivial. For dealing with the general case, we observe that, by Lemma
2.5, we can assume that ui(t) = 1/(t + di). Next, suposse that U1 and U2
are affinely equivalent over K. By Theorem 3.1, statement (1), [K(d1) :
K] = [K(d2) : K] = r and the curves U
1 := χ(U1) and U
2 := χ(U2)
parametrized by χ̃1(t) = (
M1(t)
, . . . , t
M1(t)
) and χ̃2(t) = (
M2(t)
, . . . , t
M2(t)
respectively, are affinely equivalent over K; note that, for simplicity we
have omitted the last zero components in these parametrizations. There-
fore, there exists A = (ai,j) ∈ GL(r,K) and ~v ∈ Mr×1(K), such that
ϕ(t) := A χ̃1(t)
t+~v parametrizes U⋆2 . In consequence, since ϕ(t) and χ̃2(t) are
proper parametrizations of the same curve, there exists a unit τ(t) ∈ K(t)
such that ϕ(t) = χ̃2(τ(t)). Then, considering the first component in the
above equality, one gets that
(a1,1 + · · ·+ a1,rt
r−1 + v1M1(t))M2(τ(t)) =M1(t).
Now, substituting t by −d1, we obtain
(a1,1 + · · ·+ a1,r(−d1)
r−1 + v1M1(−d1))M2(τ(−d1)) =M1(−d1) = 0.
Note that a1,1 + · · ·+ a1,r(−d1)
r−1 6= 0, because [K(d1) : K] = r. Also, note
that τ(−d1) is well defined, because −d1 does not belong to K. This implies
that M2(τ(−d1)) = 0. So, τ(−d1) is a root of M2(t).
Conversely, let τ(t) = k1t+k2
k3t+k4
∈ K(t) be a unit that maps the root γ of
M1(t) onto the root β of M2(t), i.e. τ(γ) = β. This relation implies that
K(γ) = K(β) and that deg (M1(t)) = deg (M2(t)) = r. Therefore, because
of Theorem 3.1, it is enough to prove that the curves U⋆1 := χ(U1) and
U⋆2 := χ(U2) are affinely equivalent over K. Recall that U
i is parametrized
by ϕi(t) := χ̃(t) =
Mi(t)
, . . . , t
Mi(t)
; here again, we omit the last zero
components of the parametrization. In order to prove the result, we find
an invertible matrix A ∈ GL(r,K) and a vector ~v ∈ Mr×1(K), such that
Aϕt1(t)+~v = ϕ
2(τ(t)). For this purpose, we consider the polynomialM(t) =
M2(τ(t))(k3t + k4)
r ∈ K[t]. Now, since τ(t) is a unit of K(t), and the roots
of M2(t) are not in K, one gets that deg(M) = deg(M2) = r. Moreover,
since γ is a root of M(t), and taking into account that M1(t) is the minimal
polynomial of γ over K and that deg(M) = r = deg(M1), one has that there
exists c ∈ K∗ such that M(t) = cM1(t). Now, in order to determine A and
~v, let us substitute τ(t) in the i-th component of ϕ2(t):
τ(t)i
M2(τ(t))
τ(t)i(k3t+ k4)
M2(τ(t))(k3t+ k4)r
(k1t + k2)
i(k3t+ k4)
cM1(t)
Since numerator and denominator in the above rational function have the
same degree, taking quotients and remainders, ϕ2(t) can be expressed as
(ϕ2(τ(t)))i=1,...,r = (vi +
ai,1 + · · ·+ ai,rt
M1(t)
)i=1,...,r,
for some vi, ai,j ∈ K. Take A = (ai,j) and ~v = (vi). Then, A(ϕ1(t))
t + ~v =
(ϕ2(τ(t))
t. Finally, let us see that A is regular. Indeed, suppose that
A is singular and that there exists a non trivial linear relation λ1F1 +
· · · + λrFr = ~0, where Fi denotes the i-th row of A. This implies that(
M2(t)
+ · · ·+ λr
M2(t)
◦ τ(t) = λ1v1 + · · · + λrvr is constant, which is
impossible because λ1+···+λrt
M2(t)
is not constant and τ(t) is a unit of K(t).
For two true circles, there is always a real affine transformation relating
them. We have seen that this is not the case of hypercircles. However, for
algebraic extensions of degree 2 (where the circle case fits), we recover this
property for hypercircles that are not lines.
Corollary 3.4. Let K(α) be an extension of degree 2. Then all α-hypercircles,
that are not lines, are affinely equivalent over K.
Proof. By Lemma 2.5, we may assume that the hypercircles are associated
to units of the form 1
. Now, we consider two α-hypercircles not being
lines, namely, let Ui be the α-hypercircle associated to
for i = 1, 2,
and di 6∈ K. Let di = λi + µiα, with λi, µi ∈ K and µi 6= 0. Then, the
unit τ(t) = τ0 + τ1t ∈ K[t] where τ0 =
µ2λ1−µ1λ2
and τ1 =
, verifies that
τ(−d1) = −d2. By Theorem 3.3, U1 and U2 are affinely equivalent over K.
In Corollary 3.2 we have seen that the degree of a hypercircle is given by
the degree of the field extension provided by the pole of any non polynomial
generating unit. Lines are curves of degree one, a particular case of this
phenomenon. Now, we consider other kind of hypercircles of degree smaller
than n. This motivates the following concept.
Definition 3.5. Let U be an α-hypercircle. If the degree of U is [K(α) : K],
we say that it is a primitive hypercircle. Otherwise, we say that U is a non-
primitive hypercircle.
Regarding the complex numbers as an extension of the reals, lines may
be considered as circles when we define them through a Moebius transforma-
tion. Lines are the only one curves among these such that its degree is not
[C : R]. The situation is more complicated in the general case. Apart from
lines, which have been thoroughly studied in Theorem 2.6, there are other
non-primitive hypercircles. This is not a big challenge because, as we will
see, non-primitive hypercircles are primitive on another extension. Moreover,
these cases reflect some algebraic aspects of the extension K ⊆ K(α) = L in
the geometry of the hypercircles. Actually, we will see that there is a cor-
respondence between non-primitive hypercircles and the intermediate fields
of K ⊆ L. More precisely, let U be a non-primitive hypercircle associated to
u(t) = 1
, where r = [K(d) : K] < [L : K] = n. In this case, we have the
algebraic extensions K ⊆ K(d) ( L. We may consider u(t) as a unit either
in the extension K ⊆ K(d) with primitive element d or in K(d) ( L with
primitive element α. In the first case, u(t) defines a primitive hypercircle in
Fr. In the second case, as u(t) is a K(d) unit, it defines a line. The analysis
of U can be reduced to the case of the primitive hypercircle associated to
u(t) in the extension K ⊆ K(d).
Theorem 3.6. Let U be the non-primitive hypercircle associated to u(t) =
∈ K(α)(t). Let V be the hypercircle generated by the unit 1
in the
extension K ⊆ K(d). Then, there is an affine inclusion from Fr to Fn,
defined over K, that maps the hypercircle V onto U .
Proof. Taking into account Lemma 2.5, we may assume that u(t) = 1
. Let
φ(t) = (φ0(t), . . . , φn−1(t)) ∈ K(t)
n be the parametrization of U , obtained
from u(t), with respect to the basis B = {1, α, . . . , αn−1}. Similarly, let
ψ(t) = (ψ0(t), . . . , ψr−1(t)) ∈ K
r(t) be the parametrization of the hypercircle
V, associated to u(t), with respect to the basis B⋆ = {1, d, . . . , dr−1}, where
r = [K(d) : K]. The matrix D = (dji) ∈ Mn×r(K) whose columns are the co-
ordinates of di with respect to B induces a K-linear transformation χ : Fr 7→
Fn that maps V onto U . Indeed, as u(t) =
i=0 ψi(t)d
j=0 φj(t)α
j, one
has that
ψi(t)d
ψi(t)
dj,iα
dj,iψi(t)
φj(t)α
Then φ(t)t = D ψ(t)t. Moreover, χ is one to one, because rank(D) = r.
As a consequence of this theorem, every hypercircle is affinely equivalent,
over K, to a primitive hypercircle. Therefore, the study of hypercircles can
be reduced to the study of primitives hypercircles.
4 Properties at Infinity of a Hypercircle
Circles have a very particular structure at infinity, namely, they pass through
the cyclic points, i.e. [±i : 1 : 0], which are related to the minimal polynomial
defining the circle as a hypercircle as remarked in the introduction. In this
section, we will see that a similar situation occurs for more general primitive
hypercircles. More precisely, let U be the primitive hypercircle defined by the
unit u(t) = at+b
. By Corollary 3.2, U is a parametric affine curve of degree
n. So, there are at most n different points in the hyperplane at infinity. Let
φ(t) = (φ0(t), . . . , φn−1(t)) be the parametrization of U generated by u(t);
recall that φi(t) =
pi(t)
. Thus, projective coordinates of the points attained
by φ(t) are given by [p0(t) : · · · : pn−1(t) : M(t)]. Now, substituting t by
every conjugate σ(−d) of −d, we obtain
[p0(σ(−d)) : · · · : pn−1(σ(−d)) : 0] = [σ(p0(−d)) : · · · : σ(pn−1(−d)) : 0]
We prove next that these points are the points of the hypercircle at infinity.
Lemma 4.1. Let U be a primitive hypercircle associated to the unit u(t) =
. The n points at infinity are
Pj = [σj(p0(−d)) : · · · : σj(pn−1(−d)) : 0], 1 ≤ j ≤ n
where σj are the K-automorphisms of the normal closure of L = K(α) over
Proof. First of all, observe that gcd(p0, . . . , pn−1,M) = 1, and hence Pj
are well defined. Moreover, pi(−d) 6= 0, for every i ∈ {0, . . . , n − 1},
since pi(t) ∈ K[t] is of degree at most n and, thus, if pi(−d) = 0, then
pi(t)
= c ∈ K and the hypercircle would be contained in a hyperplane.
But this is impossible since U is primitive (see Corollary 3.2). It remains
to prove that they are different points. Suppose that two different tuples
define the same projective point. We may suppose that P1 = Pj. P1
verifies that
i=0 pi(−d)α
i = (−ad + b)m(−d) 6= 0 and Pj verifies that∑n−1
i=0 pi(σj(−d))α
i = (aσj(−d) + b)m(σj(−d)) = 0. Thus, Pj is contained in
the projective hyperplane
i=0 α
iXi = 0, but not P1. Hence, P1 6= Pj.
Let us check that, as in the case of circles, the points at infinity of prim-
itive α-hypercircles do not depend on the particular hypercircle.
Theorem 4.2. For a fixed extension K ⊆ K(α) of degree n, the set of points
at the infinity P = {P1, . . . , Pn} of any primitive hypercircle does not depend
on the particular α-hypercircle U , but only on the algebraic extension and on
the primitive element α. Moreover, the set P is characterized by the following
property:
{X0 + αjX1 + · · ·+ α
j Xn−1 = 0} ∩ U = P \ {Pj},
where αj = σj(α) are the conjugates of α in F, 1 ≤ j ≤ n, and U is the
projective closure of U .
Proof. Let U be the primitive α-hypercircle generated be a unit u(t) = at+b
U has the projective parametrization [p0(t) : · · · : pn−1(t) : M(t)]. Let
Pj = [σj(p0(−d)) : · · · : σj(pn−1(−d)) : 0]. Its evaluation in the equation of
hyperplane X0 + αkX1 + . . .+ α
k Xn−1, yields:
σj(pi(−d))α
k = σk
σ−1k ◦ σj(pi(−d))α
(a(σ−1k ◦ σj(−d)) + b)m(σ
k ◦ σj(−d))
If j = k, the previous expression equals σk ((−ad + b)m(−d)) 6= 0. If j 6= k,
then σ−1k ◦ σj(−d) is a conjugate of −d, different from −d, because −d is a
primitive element. So m(σ−1k ◦ σj(−d)) = 0.
In order to show that this point does not depend on a particular hyper-
circle, take the n hyperplanes X0 + αkX1 + · · ·+ α
k Xn−1 = 0, k = 1 . . . n.
Every point at infinity of a hypercircle is contained in exactly n− 1 of those
hyperplanes. Also, any of these hyperplanes contains exactly n− 1 points at
infinity of the hypercircle. One point at infinity may be computed by solving
the linear system given by any combination of n−1 hyperplanes. The matrix
of the linear system is a Vandermonde matrix, each row depending on the
corresponding αk, so there is only one solution.
Remark 4.3. Notice that this theorem provides a n-simplex combinatorial
structure of the points at infinity of any primitive hypercircle.
The following result shows that the points at infinity can be read directly
from the minimal polynomial of α.
Proposition 4.4. Let Mα(t) be the minimal polynomial of α over K. Let
mα(t) =
Mα(t)
i ∈ K(α)[t], where ln−1 = 1. Then, the points at
infinity of every primitive α-hypercircle are [l0 : l1 : · · · : ln−2 : ln−1 : 0] and
its conjugates.
Proof. We consider the symmetric polynomial r(x, y) =
Mα(x)−Mα(y)
. Substi-
tuting (x, y) by (t, α) we obtain that
r(t, α) =
Mα(t)−Mα(α)
Mα(t)
= mα(t).
That is, mα(t) is symmetric in t and α. Take now the hypercircle induced
by the unit 1
mα(t)
Mα(t)
. By Lemma 4.1, we already know that one point
at infinity is [p0(α) : · · · : pn−1(α) : 0], where mα(t) =
pi(t)α
i. By
symmetry,
i=0 pi(t)α
i=0 pi(α)t
i. That is, pi(α) = li. Thus, the
points at infinity are [l0 : l1 : · · · : ln−2 : 1 : 0] and its conjugates.
Next result deals with the tangents of a hypercircle at infinity, and it
explains again why parabolas can not be hypercircles.
Proposition 4.5. The tangents to a primitive hypercircle at the points at
infinity are not contained in the hyperplane at infinity.
Proof. Let U be the primitive α-hypercircle generated by at+b
, and [p0(t) :
· · · : pn−1(t) :M(t)] the projective parametrization generated by the unit. In
the proof of Lemma 4.1, we have seen that pn−1(t) is not identically 0, because
pn−1(−d) 6= 0. So, we can dehomogenize w.r.t. the variable Xn−1, obtaining
the affine parametrization (
p0(t)
pn−1(t)
, . . . ,
pn−2(t)
pn−1(t)
pn−1(t)
) of U on another affine
chart. We have to check that the tangents to the curve at the intersection
points with the hyperplane Xn−1 = 0 are not contained in this hyperplane.
The points of C in the hyperplane Xn−1 = 0 are obtained by substituting t
by σ(−d). The last coordinate of the tangent vector is
M ′(t)pn−1(t)−M(t)p
n−1(t)
pn−1(t)2
We evaluate this expression at σ(−d). M(σ(−d)) = 0 and, as all its roots are
different in F, M ′(σ(−d)) 6= 0. We also know that σ(pn−1(−d)) 6= 0. Hence,
the last coordinate of the tangent vector is non-zero. Thus, the tangent line
is not contained in the hyperplane at infinity.
Finally, we present a property of hypercircles that can be derived from
the knowledge of its behavior at infinity. We remark a property of circles
stating that given three different points in the plane, there is exactly one
circle passing through them (which is a line if they are collinear). The result
is straightforward if we recall that there is only one conic passing throught
five points. In the case of circles, we have the two points at infinity already
fixed, so, given three points in the affine plane there will only be a conic
(indeed a circle if it passes through the cyclic points at infinity) through them.
Even if hypercircles are curves in n-space, surprisingly, the same occurs for
hypercircles.
We are going to prove that, given 3 different points in Kn, there is ex-
actly one hypercircle passing through them. If the points are not in general
position, the resulting hypercircle needs not to be a primitive one. First, we
need a lemma that states what are the points over K of the hypercircle that
are reachable by the parametrization.
Lemma 4.6. Let U be the α–hypercircle, non necessarily primitive, associ-
ated to u(t) = at+b
with induced parametrization Φ(t). Φ(K) = U ∩Kn \ {ā}
with a =
i=0 aiα
i, ā = (a0, . . . , an−1).
Proof. We already know that Φ(t) is proper and, obviously, Φ(K) ⊆ U ∩Kn,
also, ā is not reachable by Φ(t), since otherwise one would have that a = u(λ)
for some λ, and this implies that ad − b = 0, which is impossible since u(t)
is a unit. In order to prove the other inclusion, write as before φi(t) =
pi(t)
where M(t) is the minimal polynomial of −d over K. Then, we consider the
ideal I over F[t, X̄ ] generated by (p0(t)−X0M(t), . . . , pn−1(t)−Xn−1M(t)),
where X̄ = (X0, . . . , Xn−1), and the ideal J = I + (ZM(t)− 1) ⊆ F[Z, t, X̄].
Let I1 be the first elimination ideal of I; i.e. I1 = I ∩ F[X̄ ] and let J2 be
the second elimination ideal of J ; i.e. J2 = J ∩ F[X̄]. Observe that I ⊆ J
and therefore I1 ⊆ J2. Note that U = V (J2); i.e. U is the variety defined
by J2 over F. Thus U ⊆ V (I1). Now, let us take x̄ ∈ (U ∩ K
n) \ {ā}.
Then x̄ ∈ V (I1). Observe that, by construction, the leading coefficient of
pi(t) − XiM(t) w.r.t. t is ai − Xi. Therefore, since x̄ 6= ā one has that at
least one of the leading coefficients of the polynomials in I w.r.t. t does
not vanish at x̄. Thus, applying the Extension Theorem (see Theorem 3,
pp. 117 in [5]), there exists t0 ∈ F such that (t0, x̄) ∈ V (I). This implies
that pi(t0) − xiM(t0) = 0 for i = 1 . . . n − 1. Let us see that M(t0) 6= 0.
Indeed, if M(t0) = 0 then pi(t0) is also zero for every index and therefore
gcd(p0(t), . . . , pn−1(t),M(t)) 6= 1, which is impossible. Hence Φ is defined at
t0 and Φ(t0) = x̄. To end up, we only need to show that t0 ∈ K. For this
purpose, we note that the inverse of Φ(t) is given by
P (X̄) =
i + b∑
Xiαi − a
Now, since x̄ 6= ā one deduces that P (x̄) is well defined, and the only pa-
rameter value generating x̄ is t0 = P (x̄). Hence, the gcd of the polynomi-
als pi(t) − xiM(t) is a power of (t − t0). Thus, taking into account that
pi,M ∈ K[t], one deduces that t0 ∈ K. Finally, it only remains to state that
ā is generated when t takes the value of the infinity of K. But this follows
taking Φ(1/t) and substituting by t = 0.
Proposition 4.7. Let Xi = (Xi0, . . . , Xi,n−1) ∈ K
n ⊆ Fn, 1 ≤ i ≤ 3 be three
different points. Then, there exists only one α–hypercircle passing through
them.
Proof. Let Yi =
j=0 Xijα
j ∈ K(α), 1 ≤ i ≤ 3. Consider the following
linear homogeneous system in a, b, c, d:
b = Y1d, a+ b = Y2(c+ d), a = Y3c
Observe that, if the three points are different, there is only one projective
solution, namely [a : b : c : d] where a = Y1Y3 − Y3Y2, b = Y1Y2 − Y1Y3,
c = Y1 − Y2, d = Y2 − Y3.
Take the unit u(t) = at+b
. It verifies that u(0) = Y1, u(1) = Y2, u(∞) =
Y3. Then, the hypercircle associated to u passes through X1, X2, X3. In
order to prove that this hypercircle is unique, let v be the unit associated to
a hypercircle passing through the three points and ψ(t) the parametrization
induced by v(t). By Lemma 4.6, as Xi ∈ K
n, the point Xi is reached for a
parameter value ti in K∪{∞}. So, there are three values t1, t2, t3 ∈ K∪{∞}
such that v(ti) = Yi. Let τ(t) ∈ K(t) be the unique unit associated to the
transformation of the projective line P(F) into itself given by τ(0) = t1,
τ(1) = t2, τ(∞) = t3. Then v(τ(t)) = u(t) and both units represents the
same hypercircle.
5 Parametrization and Implicitation of a Hy-
percircle
In this section, we will provide specific methods to parametrize and implic-
itate hypercircles. These methods show the power of the rich structure of
hypercircles, simplifying problems that are usually much harder in general.
Given a unit u(t) defining U , it is immediate to obtain a parametrization
of U (see Section 2). If U is given by implicit equations (as it is usually
the case in Weil’s descente method), the next proposition shows how to
parametrize it.
Proposition 5.1. The pencil of hyperplanes X0+X1α+ · · ·+Xn−1α
n−1 = t
parametrizes the primitive α–hypercircle U .
Proof. Let I be the implicit ideal of U . Note that, since U is K−rational it is
K-definable, and hence a set of generators of I can be taken inK[X0, . . . , Xn−1].
Let u(t) be any unit associated with U and (φ0(t), . . . , φn−1(t)) the induced
parametrization. Let v(t) be the inverse unit of u(t), u(v(t)) = v(u(t)) = t.
Then (φ0(v(t)), . . . , φn−1(v(t))) = (ψ0(t), . . . , ψn−1(t)) = Ψ(t) is another
parametrization of U which is no more defined over K but over K(α). The
later parametrization is in standard form [10], that is
ψi(t)α
φi(t)α
◦ v(t) = u ◦ v(t) = t.
This implies that the pencil of hyperplanes Ht ≡ X0+X1α+· · ·+Xn−1α
n−1−t
parametrizes U . Indeed, if Ψ(t) is defined, Ht ∩U consists in n− 1 points at
infinity of U (Theorem 4.2) and Ψ(t) itself. We deduce that ψi(t)−Xi belongs
to the ideal I +Ht, which has a set of generators in K(α)(t)[X0, . . . , Xn−1].
So, the parametrization Ψ(t) can be computed from I.
Notice that the obtained parametrization Ψ(t) has coefficients over K(α).
Thus, it is not the parametrization induced by any associated unit u(t).
The interest of obtaining a unit associated to a hypercircle is that it helps
us to solve the problem of reparametrizing a curve over an optimal field
extension of K, see [4]. There, it is shown that given a parametrization
Ψ(t) ∈ K(α)r of a curve there is a hypercircle associated to it. Any unit
associated to the hypercircle reparametrizes the original curve over K. To
get a parametrization φ(t) over K or, equivalently, a unit u(t) associated to
U , we refer to [10]. In addition, note that the proof of Proposition 4.7 shows
how to construct a unit associated to a hypercircle, when points over K are
known, and therefore a parametrization of it.
The inverse problem, computing implicit equations of a hypercircle from
the parametrization induced by an associated unit, can be performed using
classic implicitation methods. However, the special structure of hypercircles
provides specific methods that might be more convenient.
Proposition 5.2. Let U be a hypercircle associated to the unit u(t), and let
v(t) be the inverse of u(t). Let
ri(X0, . . . , Xn−1)
s(X0, . . . , Xn−1)
where ri, s ∈ K[X0, . . . , Xn−1]. Then, the ideal of U is the elimination ideal
with respect to Z:
I(U) = (r1(X̄), . . . , rn(X̄), s(X̄)Z − 1) ∩ F[X0, . . . , Xn−1].
Proof. Let u(t) = at+b
, then v(t) = −dt+b
. Now, consider
ξi(X0, . . . , Xn−1)α
ηi(X0, . . . , Xn−1)α
where ξi, ηj ∈ K(X0, . . . , Xn−1) and ηi =
ri(X0,...,Xn−1)
s(X0,...,Xn−1)
. The map ξ : Fn −→
Fn, ξ = (ξ0, . . . , ξn−1) is birational and its inverse is η = (η0, . . . , ηn−1).
Indeed:
ηi(ξ0(X̄), . . . , ξn−1(X̄))α
i = v
αjξj(X̄)
is an equality in K(α)(X0, . . . , Xn−1). We deduce that
ξ0(X0, . . . , Xn−1), . . . , ξn−1(X0, . . . , Xn−1)
It is clear that U is the image of the line L ≡ {X1 = 0, . . . , Xn−1 = 0} under
the map ξ, U = ξ(L). The set of points where ξ is not defined is the union of
the hyperplanes
i=0 σj(α)
iXi + σj(d) = 0, 1 ≤ j ≤ n. The intersection of
these hyperplanes with L is the set of points (−σ(d)j, 0, . . . , 0), 1 ≤ j ≤ n.
Thus, for a generic p ∈ L, ξ(p) is defined and belongs to U . The result is
similar for the inverse map η. The set of points where η is not defined is the
union of the hyperplanes
i=0 σj(α)
iXi − σj(a) = 0, 1 ≤ j ≤ n. These n
hyperplanes intersect U in at most one affine point, see Proposition 5.1. So,
for a generic p ∈ U , η(p) is again defined and belongs to L. Let us compute
now the points X̄ such that η(X̄) is defined, but it does not belong to the
domain of ξ. If X̄ is such a point, then
σj(α)
iηi(X̄) + σj(d) = 0.
As ηi is defined over K, applying σj to the definition of η, we obtain that
σj(v)
σj(α)
= −σj(d)
But σj(v) =
−σj(d)t+σj (b)
t−σj(a)
. It follows from Lemma 4.6 that the value −σj(d)
cannot be reached, even in F. Thus, the image of η is contained in the domain
of ξ.
We are ready to prove the theorem, by verifying that the set U \ {s = 0},
which is just eliminating a finite number of points in U , is the set of points X̄
such that ri(X̄) = 0, i ≥ 1 and s(X̄) 6= 0. If X̄ ∈ U\{s = 0}, then η is defined
and η(X̄) = (η0(X̄), 0, . . . , 0). Hence ηi(X̄) = ri(X̄) = 0. Conversely, if X̄ is
a point such that ri(X̄) = 0 and s(X̄) 6= 0, then η(X̄) is defined and belongs
to L. It is proven that ξ is defined in η(X̄), so X̄ = ξ(η(X̄)) ∈ ξ(L) = U . The
thesis of the theorem follows taking the Zariski closure of U \ {s = 0}.
This method to compute the implicit equations of U is not free from
elimination techniques, as it has to eliminate the variable Z. However, it has
the advantage that it yields already an ideal in F[X0, . . . , Xn−1] defined over
K and such that it describes a non trivial variety containing the hypercircle.
Namely, (r1(X̄), . . . , rn−1(X̄)) are polynomials over K whose zero set contains
the hypercircle. The following example shows that the elimination step is
necessary in some cases.
Example 5.3. Let Q ⊆ Q(α) be the algebraic extension defined by α3+α2−
3 = 0. Let us consider the unit u(t) =
(2+α)t+α
t+1−α
. Its inverse is v(t) =
(α−1)t+α
t−2−α
A parametrization of U is
φ(t) =
2t3 + 6t2 + 7t+ 3
t3 + 4t2 + 5t− 1
t3 + 6t2 + 9t+ 2
t3 + 4t2 + 5t− 1
t2 + 4t+ 1
t3 + 4t2 + 5t− 1
A Gröbner basis of the ideal of the curve is
I := {x21 − x2x0 − x2x1 − x1 + x2, x0x1 − x2x0 − 3x
2 − 2x1 + 4x2,
x20 − 3x2x1 − 2x0 + 2x1 + 3x2 − 2}.
Then, proposition 5.2 states that this ideal is
I = (r1(x0, x1, x2), r2(x0, x1, x2), s(x0, x1, x2)Z − 1) ∩ F[x0, x1, x2]
where
r1 = 2 − 8x2 + 4x2x0 + 6x
2x0 + 17x2x1 + x2x
0 + 3x1 − 3x
1x2 + x
0 − x
0x1 +
4x0x1 − 12x
2 − 8x
1 + 9x
2 + 3x
1 − 3x
0 − 9x0x1x2,
r2 = −2 − 7x2 + 4x2x0 − x2x1 + 8x1 − 2x0 − 2x0x1 + 6x
2 − 2x
1 + x
s = 9x32 + 6x
2x0 − 12x
2 + 5x2x0 − 17x2 − 3x
1x2 − 9x0x1x2 + x2x
0 + 24x2x1 +
3x31 + 8x0 + 4x0x1 − 5x
0 − x
0x1 + 5x1 − 9x
1 − 7 + x
But, if we take J = (r1, r2), then J ( I. The saturation of J with respect to
I is J : I∞ = (x21 − x0x2 − x1x2 − 2x1 + 3x2 + 1, x0x1 − x0x2 − 3x
2 − x0 −
2x1 + 2x2 + 2, x
0 − 3x1x2 − 4x0 + 3x2 + 4)
This ideal corresponds to the union of the line
−αx0 +3x2 = −2α
(α + α2)x0 −3x1 = −3 + 2α + 2α
and its conjugates.
Next theorem shows an alternative method to implicitate a hypercircle
without using any elimination techniques. It is based on properties of the
normal rational curve of degree n.
Theorem 5.4. Let ϕ(t) = (
q0(t)
, . . . ,
qn−1(t)
) be a proper parametrization of
a primitive hypercircle U with coefficients in F. Let I be the homogeneous
ideal of the rational normal curve of degree n in P(F)
given by a set of
homogeneous generators h1(Ȳ ), . . . , hr(Ȳ ). Let Q ∈ Mn+1×n+1(F) be the
matrix that carries {q0(t), . . . , qn−1(t), N(t)} onto {1, t, . . . , t
n}. Let
fi(X̄) = hi
Q0jXj , . . . ,
QnjXj
, 1 ≤ i ≤ r.
Then {f1, . . . , fr} is a set of generators of the homogeneous ideal of U .
Proof. If the parametrization is proper, {q0(t), . . . , qn−1(t), N(t)} is a basis
of the polynomials of degree at most n. This follows from the fact shown in
Corollary 3.2 that a primitive hypercircle is not contained in any hyperplane.
Note that a projective point X̄ belongs to U if and only if Q(X̄) belongs to
the rational normal curve, if and only if hi(Q(X̄)) = 0, 1 ≤ i ≤ r.
Remark 5.5.
• It is well known that the set of polynomials {YiYj−1−Yi−1Yj | 1 ≤ i, j ≤
n} is a generator set of I (see [6]).
• Notice that it is straightforward to compute Q from the parametrization.
Therefore, we have an effective method to compute the implicit ideal of
the projective closure of U . The affine ideal of U can be obtained by
dehomogenization Xn = 1.
• If the parametrization is given by polynomials over an algebraic exten-
sion K(β) of K, then the coefficients of fi belongs to K(β). Moreover,
if we write fi(X̄) =
j=0 fij(X̄)β
j, with fij ∈ K[X̄ ], then, {fij} is a
set of generators over K of the hypercircle U .
• In practice, this method is much more suited to compute an implicita-
tion of a hypercircle than the method presented in Proposition 5.2.
Example 5.6. The implicit equations of a hypercircle can be computed by
classical implizitation methods, for example Gröbner basis or with the two
methods presented in Proposition 5.2 and Theorem 5.4. Here, we present two
cases that show the practical behavior of these methods. The first example
considers the algebraic extension Q ⊆ Q(α), where α4 + α2 − 3 and the unit
(1−α3)t+α2
t+1+2α−3α2
. The parametrization of the hypercircle is given by
t4 + 15t3 + 22t2 + 101t− 195
t4 + 10t3 − 17t2 − 366t+ 233
, φ1 =
−11t3 − 73t2 + 65t− 114
t4 + 10t3 − 17t2 − 366t+ 233
2t3 + 57t2 − 25t− 59
t4 + 10t3 − 17t2 − 366t+ 233
, φ3 =
−t4 − 6t3 + 4t2 + 17t− 56
t4 + 10t3 − 17t2 − 366t+ 233
The second example starts from the extension Q ⊆ Q(β), where β is such
that β4+3β+1 = 0. Here, the unit defining U is u =
(1+β−β2)t+1+β3
t+1+β2−β3
and the
parametrization induced by u(t) is
t4 + 11t3 + 47t2 + 95t+ 72
t4 + 13t3 + 62t2 + 126t+ 81
, ψ1 =
t4 + 7t3 + 15t2 + 17t+ 9
t4 + 13t3 + 62t2 + 126t+ 81
−t4 − 10t3 − 31t2 − 23t
t4 + 13t3 + 62t2 + 126t+ 81
, ψ3 =
t3 + 13t2 + 42t+ 36
t4 + 13t3 + 62t2 + 126t+ 81
The running times for computing the implicit ideal (using a Mac Xserver
with 2 processors G5 2.3 GHz, 2 Gb RAM Maple 10) are
Example 1 Example 2
Gröbner basis method 0.411 0.332
Proposition 5.2 2.094 2.142
Theorem 5.4 0.059 0.021
We refer the interested reader to [11] for a brief discussion and compari-
son of the running times of these algorithms.
6 Characterization of Hypercircles
In the introduction, we defined algebraically a circle as the conic such that its
homogeneous part is x2 + y2 and contains an infinite number of real points.
The condition on the homogeneous part is equivalent to impose that the curve
passes through the points at infinity [±i : 1 : 0]. Analogously, hypercircles
are regular curves of degree n with infinite points over the base field passing
through the points at infinity described in Theorem 4.2. The following result
shows that this is a characterization of these curves.
Theorem 6.1. Let U ⊆ Fn be an algebraic set of degree n such that all whose
components are of dimension 1. Then, it is a primitive α-hypercircle if and
only if it has an infinite number of points with coordinates in K and passes
through the set of points at infinity characterized in Theorem 4.2.
Proof. The only if implication is trivial. For the other one, let U ⊆ Fn be
an algebraic set of pure dimension 1 and degree n passing through P =
{P1, . . . , Pn}, the n points at infinity of a primitive α-hypercircle. Suposse
that U has infinite points with coordinates in K. Then, we are going to prove
that U is irreducible. Let W be an irreducible component of U with infinite
points in K. Note that, since W is irreducible and contains infinitely many
points over K, the ideal I(W) over F is generated by polynomials over K
(see Lemma 2 in [3]). Let q be any point at infinity of W; then q ∈ P . As W
is K-definable it follows that W also contains all conjugates of q. Thus, P is
contained in the set of points at infinity of W. It follows that W is of degree
at least n; since W ⊆ U , U = W. Therefore, U is irreducible and I(U) is
generated by polynomials with coefficients over K. Now, consider the pencil
of hyperplanes Ht ≡ X0+X1α+· · ·+Xn−1α
n−1−t, where t takes values in F.
Notice that Ht ∩ P = {P2, . . . , Pn}. Thus, P1 ∈ U \Ht so, for all t, U 6⊆ Ht.
Moreover, for every point p = (p0, . . . , pn−1) ∈ U , t(p) =
i=0 piα
i ∈ F is
such thatH t(p)∩U = {p, P2, . . . , Pn}. The cardinal of {t(p) | t ∈ U} is infinite,
since otherwise, by the irreducibility of U , it would imply that there is a t0
such that U ⊆ Ht0 , which is impossible. So, for generic t, the intersection
is H t ∩ U = {p(t), P2, . . . , Pn}. Let us check that the coordinates of p(t) are
rational functions in K(α)(t). Take the ideal I(U) of U . The ideal of p(t)
(as a point in F(t)n) is I +Ht, defined over K(α)(t). The reduced Gröbner
basis of the radical I +Ht is of this kind (X0 − ψ0, . . . , Xn−1 − ψn−1) and it
is also defined over K(α)(t)[X0, . . . , Xn−1]. Hence, (ψ0, . . . , ψn−1) is a K(α)-
parametrization of U . Thus, since U is irreducible, it is rational. Moreover∑n−1
i=0 (ψi(t))α
i = t and the parametrization is proper. As the curve is rational
and has an infinite number of points over K, it is parametrizable over K (it
follows, for example from the results in [14]). Let u(t) be a unit such that
Ψ◦u(t) = (φ0(t), . . . , φn−1(t)) is a parametrization over K, where φi(t) ∈ K(t)
i=0 φi(t)α
i = u(t). We conclude that U is the hypercircle associated
to the unit u(t).
Remark that a parametric curve, definable over K and with a regular
point over K, is parametrizable over the same field; for this, it is enough to
K-birationally project the curve over a plane, such that the K-regular point
stays regular on the projection, and then apply the results in [14]. Then, a
small modification of the proof above, yields the following:
Theorem 6.2. Let U ⊆ Fn be a 1-dimensional irreducible algebraic set of
degree n, definable over K . Then, it is a primitive α-hypercircle if and only
if it has a regular point with coordinates in K and passes through the set of
points at infinity characterized in Theorem 4.2.
7 An Application
As mentioned in the introduction, hypercircles play an important role in the
problem of the optimal-algebraic reparametrization of a rational curve (see
[3], [4], [10] [12], [13] for further details). Roughly speaking, the problem
is as follows. Given a rational K–definable curve C by means of a proper
rational parametrization over K(α), decide whether C can be parametrized
over K and, in the affirmative case, find a change of parameter transforming
the original parametrization into a parametrization over K. In [4], a K–
definable algebraic variety in Fn, where n = [K(α) : K], is associated to
C. This variety is called the associated Weil (descente) parametric variety.
In [4], it is proved that this Weil variety has exactly one one-dimensional
component iff C is K–definable (which is our case) and, in this case, C can
be parametrized over K iff this one-dimensional component is a hypercircle.
Moreover, if it is a hypercircle a proper rational parametrization over K of
the hypercircle generates the change of parameter one is looking for; namely
its generating unit.
In the following example, we illustrate how to use the knowledge of the
geometry of hypercircles to help solving the problem. Suppose given the
parametric curve
C ≃ (η1(t), η2(t)) =
(−2t4 − 2t3)α− 2t4
6α2t2 + (4t3 − 2)α+ t4 − 8t
−2t4α
6α2t2 + (4t3 − 2)α + t4 − 8t
where α is algebraic over Q with minimal polynomial x3 + 2. We follow
Weil’s descente method presented in [4] to associate a hypercircle to C. The
method consists in writing ηi(
j=0 tjα
qij(t0,t1,t2)
N(t0,t1,t2)
. In this situation
C is Q−definable if and only if
V = V (q11, q12, q21, q22) \ V (N)
is of dimension 1. Moreover, C is Q-parametrizable if and only if the one-
dimensional component of V is an α-hypercircle. For this example, the equa-
tions of V are:
V = V (2t30t2 − 4t
2 + 3t
1 + 2t
1t2 + 2t0t
2 + 2t
1t2 − t
0t1 + 6t0t1t
2,−6t
0t1t2 +
t40+2t0t
1−8t0t
2−2t0t
0t2−4t1t
2−12t
2, 12t
1−9t0t1t
2−4t0t
2t20t1t2 + 4t
2 − 4t0t
2, 9t0t
2 − 9t
2 − 2t
0t2 − 2t
1t2 + 6t0t1t
2 − 2t
2 + t
0t1 −
2t21t2−2t0t
2, 6t
2+12t
0t1−2t0t
1t2−2t
2+8t1t
2, 6t
2+9t0t1t
0t1t2+4t
2+8t0t
2, 18t2t
1+36t
2t1+14t
0t2+32t
1t2+12t0t1t
7t20t1+14t
1t2+14t0t
2, 6t0t
1t2+2t0t
1t2+ t
0t1+2t
2− 8t1t
2+12t
2t0, 9t
0t2t1−
36t42t1 − 4t
0t2 − 4t
1t2 + 12t0t1t
2 − 4t
2 + 2t
0t1 − 4t
1t2 − 4t0t
2, 6t
1 + 48t
36t42t0−11t
0t1+6t
1+14t0t
1t2−22t
2+64t1t
2, 3t
1t0+6t0t1t
2+2t0t
0t1t2−
2t21t
2 + 2t0t
2, 27t
1 − 27t0t
2 − 9t
2 + 9t
2t1 − 2t
0t2 − 2t
1t2 + 6t0t1t
2 − 2t
t20t1 − 2t
1t2 − 2t0t
2, 6t
0 + 12t
2t1 − 5t0t1t
2 + 2t
2, t0t
2t1 + 2t
Thus the main point is to verify that this curve is a hypercircle. If V is a
hypercircle, then its points at infinity must be as in Theorem 6.1. So, let us
first of all check whether this is the case. The set of generators of the defining
ideal form a Gröbner basis with respect to a graded order, thus to compute
the points at infinity we take the set of leading forms of these polynomials.
Leading forms= {t40 − 2t0t
1 − 6t
0t1t2 − 12t
2 − 8t0t
2, 2t
0t2 − 4t
2 +3t
2t31t2 + 6t0t1t
2, 9t0t
2 − 9t
2, 12t
1 − 9t0t1t
2 + 6t
2, 6t
2 + 12t
2, 6t
9t0t1t
2, 18t2t
1+36t
2t1, t0t
2t1+2t
2, 6t0t
1t2+12t
2t0, 9t
0t2t1−36t
2t1, 6t
48t21t
2 − 36t
2t0, 3t
1t0 + 6t0t1t
2, 27t
1 − 27t0t
2, 6t
0 + 12t
The solutions of this system, after dehomogenizing {t2 = 1}, are t0 =
t21, t
1 + 2 = 0. That is, the points at infinity are of the form [α
i : αi : 1 : 0],
= x2 + αx + α2. Thus, by Proposition 4.4, the points at infinity of V
remind those of an α-hypercircle.
Now, following Proposition 5.1, we may try to parametrize V by the pencil
of hyperplanes t0 + αt1 + α
2t2 − t. Doing so, we obtain the parametrization
(α2 + 2αt+ t2)t
3αt+ α2 + 3t2
−1/2α2t3
3αt+ α2 + 3t2
−1/2αt2(t+ α)
3αt+ α2 + 3t2
Remark that this parametrization can also be computed by means of inverse
computation techniques as described in [13]. Then, by direct computation, we
observe that the parametric irreducible curve defined by this parametrization
is of degree 3, passes through the point (0, 0, 0) and this point is regular.
Moreover, it is Q-definable, since it is the only 1-dimensional component of
V (see [4]), which is, by construction, a Q-definable variety. It follows from
6.2 that it is a hypercircle.
Then, from this parametrization, the algorithm presented in [10] com-
putes a unit u(t) = 2
2t+α2
associated to V. So, V is the hypercircle associated
to u(t) and C is parametrizable over Q. In particular, the parametrization
of V associated to u(t) is
2t3+1
2t3+1
2t3+1
. Moreover, the unit u(t) gives
the change of parameter we need to compute a parametrization of C over the
base field (see [4]), namely:
η (u(t)) =
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	Introduction
	Definition and First Properties
	Main Geometric Properties.
	Properties at Infinity of a Hypercircle
	Parametrization and Implicitation of a Hypercircle
	Characterization of Hypercircles
	An Application
