CUBO, A Mathematical Journal

Vol. 24, no. 01, pp. 95–103, April 2022

DOI: 10.4067/S0719-06462022000100095

A characterization of Fq-linear subsets of affine
spaces Fn

q2

Edoardo Ballico
1

1Department of Mathematics,

University of Trento,

38123 Povo (TN), Italy.

edoardo.ballico@unitn.it

ABSTRACT

Let q be an odd prime power. We discuss possible definitions

over F
q2 (using the Hermitian form) of circles, unit segments

and half-lines. If we use our unit segments to define the

convex hulls of a set S ⊂ Fn
q2

for q /∈ {3, 5, 9} we just get the

Fq-affine span of S.

RESUMEN

Sea q una potencia de primo impar. Discutimos posi-

bles definiciones sobre F
q2 (usando la forma Hermitiana)

de ćırculos, segmentos unitarios y semi-ĺıneas. Si usamos

nuestros segmentos unitarios para definir las cápsulas con-

vexas de un conjunto S ⊂ Fn
q2

para q /∈ {3, 5, 9} simplemente

obtenemos el Fq-generado af́ın de S.

Keywords and Phrases: Finite field, Hermitian form.

2020 AMS Mathematics Subject Classification: 15A33; 15A60; 12E20.

Accepted: 27 December, 2021

Received: 17 May, 2021

c©2022 E. Ballico. This open access article is licensed under a Creative Commons

Attribution-NonCommercial 4.0 International License.

http://cubo.ufro.cl/
http://dx.doi.org/10.4067/S0719-06462022000100095
https://orcid.org/0000-0002-1432-7413
mailto:edoardo.ballico@unitn.it


96 E. Ballico CUBO
24, 1 (2022)

1 Introduction

Fix a prime p and a p-power q. There is a unique (up to isomorphism) field Fq with #Fq = q.

The field Fq2 is a degree 2 Galois extension of Fq and the Frobenius map t 7→ t
q is a generator

of the Galois group of this extension. This map allows the definition of the Hermitian product

〈 , 〉 : Fn
q2

× Fn
q2

−→ Fq2 in the following way: if u = (u1, . . . , un) ∈ F
n
q2

and v = (v1, . . . , vn) ∈ F
n
q2
,

then set 〈u, v〉 =
∑n

i=1 u
q
i vi. The degree q + 1 hypersurface {〈(x1, . . . , xn), (x1, . . . , xn)〉 = 0} is

the famous full rank Hermitian hypersurface ([11, Ch. 23]).

In the quantum world the classical Hermitian product over the complex numbers is fundamental.

The Hermitian product 〈 , 〉 is one of the tools used to pass from a classical code over a finite field

to a quantum code ([17, pp. 430–431], [14, Introduction], [20, §2.2]).

The Hermitian product was used to define the numerical range of a matrix over a finite field ([1,

2, 3, 4, 8]) by analogy with the definition of numerical range for complex matrices ([9, 12, 13, 21]).

Over C a different, but equivalent, definition of numerical range is obtained as the intersection of

certain disks ([5, §15, Lemma 1]). It is an important definition, because it was used to extend the

use of numerical ranges to rectangular matrices ([7]) and to tensors ([16]). This different definition

immediately gives the convexity of the numerical range of complex matrices. Motivated by that

definition we look at possible definitions of the unit disk of Fq2. It should be a union of circles

with center at 0 and with squared-radius in the unit interval [0, 1] ⊂ Fq.

For any c ∈ Fq and any a ∈ Fq2 set

C(0, c) := {z ∈ Fq2 | z
q+1 = c}, C(a, c) := a + C(0, c).

We say that C(a, c) is the circle of Fq2 with center a and squared-radius c. Note that C(a, 0) = {a}

and #C(a, c) = q + 1 for all c ∈ Fq \ {0}.

Circles occur in the description of the numerical range of many 2×2 matrices over Fq2 ([8, Lemmas

3.4 and 3.5]). Other subsets of Fq2 (seen as a 2-dimensional vector space of Fq) appear in [6] and are

called ellipses, hyperbolas and parabolas, because they are affine conics whose projective closure

have 0, 2 or 1 points in the line at infinity.

All these constructions are inside Fq2 seen as a plane over Fq. Restricting to planes we get the

following definition for Fnq2.

Definition 1.1. A set E ⊂ Fnq2 is said to be a circle with center 0 ∈ F
n
q2 and squared-radius c

if there is an Fq-linear embedding f : Fq2 −→ F
n
q2

such that E = f(C(0, c)). A set E ⊂ Fn
q2

is

said to be a circle with center a ∈ Fnq2 and squared-radius c if E − a is a circle with center 0 and

squared-radius c. A set S ⊆ Fn
q2
, S 6= ∅, is said to be circular with respect to a ∈ Fn

q2
if it contains

all circles with center a which meet S.



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97

In the classical theory of numerical range over C the numerical range of a square matrix which

is the orthogonal direct sum of the square matrices A and B is obtained taking the union of all

segments [a, b] ⊂ C with a in the numerical range of A and b in the numerical range of B ([21,

p. 3]). For the numerical range of matrices over Fq2 instead of segments [a, b] one has to use the

affine Fq-span of {a, b} ([1, Lemma 1], [8, Proposition 3.1]). We wonder if in other linear algebra

constructions something smaller than Fq-linear span occurs. A key statement for square matrices

over C (due to Toeplitz and Hausdorff) is that their numerical range is convex ([9, Th. 1.1-2], [21,

§3]). Convexity is a property over R and to define it one only needs the unit interval [0, 1] ⊂ R.

Obviously [0, 1] = [0, +∞) ∩ (−∞, 1] and (−∞, 1] = 1 − [0, +∞). As a substitute for the unit

interval [0, 1] ⊂ R (resp. the half-line [0, +∞) ⊂ R) we propose the following sets Iq and I
′
q (resp.

Eq).

Definition 1.2. Assume q odd. Set Eq := {a
2}a∈Fq ⊂ Fq, Iq := Eq ∩(1−Eq), I

′′
q := Eq ∩(1+xEq)

with x ∈ Fq \ Eq, and I
′
q := I

′′
q ∪ {0}.

Note that I′q = {0, 1} ∪(Eq ∩(1 + (Fq \ Eq)). In the first version of this note we only used Iq, but a

referee suggested that it is more natural to consider I′′q . We use Iq and I
′
q because {0, 1} ⊆ Iq ∩ I

′
q,

while 0 ∈ I′′q if and only if −1 is not a square in Fq, i. e. if and only if q ≡ 3 (mod 4) ([10, (ix)

and (x) at p. 5], [22, p. 22]). In all statements for odd q we handle both Iq and I
′
q.

In the case q even we propose to use {a(a + 1)}{a∈Fq} as Eq, i. e. Eq := Tr
−1
Fq/F2

(0). Thus Eq is a

subgroup of (Fq, +) of index 2. If q is even we do not have a useful definition of Iq.

Thus we restrict to odd prime powers, except for Propositions 1.8, 2.9 and Remarks 2.1 and 2.2.

We see Iq or I
′
q (resp. Eq) as the unit segment [0, 1] (resp. positive half-line starting at 0) of Fq ⊂

Fq2. In most of the proofs we only use that {0, 1} ⊆ Iq and that #Iq is large, say #Iq > (q − 1)/4.

Remark 1.3. Note that #Eq = (q + 1)/2 for all odd prime powers q.

We prove that #Iq = #I
′
q − 1 = (q + 3)/4 if q ≡ 1 mod 4 and #Iq = #I

′
q = (q + 5)/4 if q ≡ 3

(mod 4) (Proposition 2.3).

We only use the case A = Eq, A = Iq and A = I
′
q of the following definition.

Definition 1.4. Fix S ⊆ Fn
q2
, S 6= ∅, and A ⊆ Fq such that 0 ∈ A. We say that S is A-closed if

a + (b − a)A ⊆ S for all a, b ∈ S.

In the set-up of Definition 1.4 for any a, b ∈ Fnq2 the A-segment [a, b]A of {a, b} is the set a+(b−a)A.

Note that [a, a]A = {a} and that if b 6= a then b ∈ [a, b]A if and only if 1 ∈ A. If S is a subset of

a real vector space and A is the unit interval [0, 1] ⊂ R, Definition 1.4 gives the usual notion of

convexity, because a + (b − a)t = (1 − t)a + tb for all t ∈ [0, 1].



98 E. Ballico CUBO
24, 1 (2022)

Remark 1.5. Take any A ⊆ Fq such that 0 ∈ A. Any translate by an element of F
n
q2 of an

Fq-linear subspace of F
n
q2

is A-closed. In particular Fnq and F
n
q2

are A-closed. The intersection of

A-closed sets is A-closed, if non-empty. Hence we may define the A-closure of any S ⊆ Fnq2, S 6= ∅,

as the intersection of all A-closed subsets of Fn
q2

containing S.

In most cases Iq is not Iq-closed. We prove the following result.

Theorem 1.6. Assume q odd. Then:

(a) If q /∈ {3, 5, 9} (resp. q 6= 3), then Fq is the Iq-closure of Iq (resp. the I
′
q-closure of I

′
q).

(b) If q /∈ {3, 5, 9} (resp. q 6= 3), then the Iq-closed (resp. I
′
q-closed) subsets of F

n
q2

are the

translations of the Fq-linear subspaces.

Remark 1.7. Fix A ⊆ Fq such that 0 ∈ A. Assume that Fq is the A-closure of Fq. Then S ⊆ F
n
q2,

S 6= ∅, is A-closed if and only if it is the translation of an Fq-linear subspace by an element of F
n
q2
.

Thus part (b) of Theorem 1.6 follows at once from part (a) and similar statements are true for the

A-closures for any A whose A-closure is Fq.

As suggested by one of the referees a key part of one of our proofs may be stated in the following

general way.

Proposition 1.8. Let A, B be subsets of Fq containing 0. Assume A 6= {0} and let G be the

subgroup of the multiplicative group Fq \ {0} generated by A \ {0}. Assume that B is A-closed.

Then B \ {0} is a union of cosets of G.

Fix S ⊂ Fn
q2

and a set A ⊂ Fq such that {0, 1} ⊆ A. Instead of the A-closure of S the following

sets Si,A, i ≥ 1, seem to be better. In particular both circles and S1,A appear in some proofs

on the numerical range. Let S1,A be the set of all a + (b − a)A, a, b ∈ S. For all i ≥ 1 set

Si+1,A := (S1,A)1,A. Obviously Si,A is A-closed for i ≫ 0. Note that {0, 1}A = A and hence if we

start with S = {0, 1} we obtain the A-closure of A after finitely many steps.

We thank the referees for an exceptional job, making key corrections and suggestions.

2 The proofs and related observations

We assume q odd, except in Remarks 2.1 and 2.2, Proposition 2.9 and the proof of Proposition

1.8.

Remark 2.1. The notions of Eq-closed, Iq-closed and I
′
q-closed subsets of F

n
q2

are invariant by

translations of elements of Fnq2 and by the action of GL(n, Fq).



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A characterization of Fq-linear subsets of affine spaces F
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99

Remark 2.2. Fix any A ⊆ Fq such that 0 ∈ A. Any translate by an element of F
n
q2 of an A-closed

set is A-closed. The Fq-closed subsets of F
n
q2

are the translates by an element of Fn
q2

of the Fq-linear

subspaces. If A ⊆ {0, 1}, then any nonempty subset of Fnq2 is A-closed.

Proof of Proposition 1.8: Since Fq \ {0} is cyclic, G is cyclic. Let a ∈ A \ {0} be a generator of G.

Fix c ∈ B \ {0} and take t ∈ Fq \ {0} such that c = ta
z for some positive integer z. We need to

prove that B \ {0} contains all tak, k ∈ Z. Since b ∈ B, B is A-closed, a ∈ A and a = 0 + (a − 0),

we get taz+1 ∈ B. Iterating this trick we get that B contains all tak for large k and hence the

coset tG, because G is cyclic.

Proposition 2.3. We have #Iq = #I
′
q −1 = (q+3)/4 if q ≡ 1 (mod 4) and #Iq = #I

′
q = (q+5)/4

if q ≡ 3 (mod 4).

Proof. Since A := {x2 +y2 = 1} ⊂ F2q is a smooth affine conic, its projectivization B := {x
2 +y2 =

z2} ⊂ P2(Fq) has cardinality q + 1 ([10, th. 5.1.8]). Note that the line z = 0 is not tangent to

B and hence B ∩ {z = 0} has 2 points over Fq2. It has 2 points over Fq if and only if −1 is a

square in Fq, i. e. if and only if q ≡ 1 (mod 4) ([10, (ix) and (x) at p. 5], [22, p. 22]). Hence

#A = q + 1 if q ≡ 3 (mod 4) and #A = q − 1 if q ≡ 1 (mod 4). Note that a ∈ Iq if and only if

there is (e, f) ∈ F2q such that e
2 + f2 = 1 and a = e2. Note that (e, f) ∈ A and that conversely for

each (e, f) ∈ A, e2 ∈ Iq. Obviously 0 ∈ Iq and (0, f) ∈ A if and only if either f = 1 or f = −1.

Thus 0 ∈ Iq comes from 2 points of A. Obviously 1 ∈ Iq. If either e = 1 or e = −1, then (e, f) ∈ A

if and only if f = 0. Thus 1 ∈ Iq comes from 2 points of A. If e
2 /∈ {0, 1} and e2 ∈ Iq, then e

2

comes from 4 points of A.

Fix a non-square c ∈ Fq and set A
′ := {x2 − cy2 = 1} ⊂ F2q. Let B

′ := {x2 − cy2 = z2} ⊂ P2(Fq)

be the smooth conic which is the projectivization of A′. The line {z = 0} is not tangent to B′ and

{z = 0} ∩ A′ = ∅. Thus #A′ = q + 1. Note that a ∈ I′′q if and only if there is (e, f) ∈ F
2
q such

that a = e2 and e2 − cf2 = 1. The element 1 ∈ I′′q comes from two elements of A
′. If 0 ∈ I′′q , then

it comes from two elements of A′. If 0 /∈ I′′q , i. e. if q ≡ 3 (mod 4), we get #I
′′
q = (q + 1)/4 and

#I′q = (q + 5)/4. If 0 ∈ I
′′
q we get #I

′′
q = #I

′
q = (q + 7)/4.

Remark 2.4. If q ∈ {3, 5}, then Iq = {0, 1} and hence each non-empty subset of F
n
q2 is Iq-closed if

q ∈ {3, 5}. Since {0, 1} ⊆ I′q, Proposition 2.3 gives I
′
3 = I3. We have I

′
5 = {0, 1, 4} = E5, because

3 is not a square in F5.

Remark 2.5. Fix any t ∈ Fq \ Eq. Then Fq \ Eq = t(Eq \ {0}). Obviously EqEq = Eq.

The following result characterizes Eq2 and hence characterizes all Er with r a square odd prime

power.

Proposition 2.6. The set of Eq2 \ {0} of all squares of Fq2 \ {0} is the set of all ab such that

a ∈ Fq \ {0} and b
q+1 = 1. We have ab = a1b1 if and only if (a1, b1) ∈ {(a, b), (−a, −b)}.



100 E. Ballico CUBO
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Proof. Fix z ∈ Fq2 \ {0}. Hence z
q2−1 = 1. Thus z(q−1)

q+1
= 1 (and so z(1−q)

q+1
= 1) and

z(q+1)
q−1

= 1, i. e. zq+1 ∈ Fq \ {0}. Note that z
2 = zq+1z1−q. Assume ab = a1b1 with a,

a1 ∈ Fq \ {0} (i.e., with a
q−1 = a1

q−1 = 1) and bq+1 = b1
q+1 = 1. Taking aa1

−1 and bb1
−1 instead

of a and b we reduce to the case a1 = b1 = 1 and hence ab = 1. Thus a
q+1bq+1 = 1. Hence a2 = 1.

Since q is odd and a 6= 1, then a = −1. Thus b = −1.

Proposition 2.7. Take S ⊆ Fnq2. The set S is Eq-closed if and only if it is a translation of an

Fq-linear subspace.

Proof. Remark 2.2 gives the “if” part. Assume that S is not a translation of an Fq-linear subspace

and fix a, b ∈ S such that a 6= b and the affine Fq-line L spanned by {a, b} is not contained in S.

By Remark 2.1 it is sufficient to find a contradiction in the case n = 1 and L = Fq with a = 0 and

b = 1. Thus Eq ⊆ S. Since S is Eq-closed and 0 ∈ S, c + (−c)Eq ⊆ S for all c ∈ Eq. First assume

−1 ∈ Eq. In this case −cEq = Eq. Thus S contains all sums of two squares. Thus S = Fq. Now

assume −1 /∈ Eq. In this case we obtained that S contains all differences of two squares. Thus

−Eq ⊂ S. Since −1 /∈ Eq, −Eq = {0} ∪ (Fq \ Eq) (Remark 2.5). Thus S ⊇ L.

The cases of Iq-closures and I
′
q-closures are more complicated, because Iq = I

′
q = {0, 1} if q = 3, 5

and hence all subsets of Fnq2 are Iq-closed if q = 3, 5. The following observation shows that the

I9-closed subsets of F
n
81 are exactly the translations of the F3-linear subspaces and gives many

examples with Iq * I
′
q.

Remark 2.8. We always have 2 /∈ 1 + cEq, c a non-square, because 1 is a square. If q is a square,

say q = s2, then obviously Fs ⊆ Eq ∩ (1 − Eq) = Iq and hence 2 ∈ Iq. Take q = 9. We get F3 ⊆ I9.

Since #I9 = 3 (Proposition 2.3), we get Iq = F3. Thus the I9-closed subsets of F
n
81 are exactly the

translations of the F3-linear subspaces. Now assume that q is not a square. We have 2 ∈ 1 − Eq if

and only if −1 is a square, i. e. if and only if q ≡ 1 (mod 4). Since q is not a square, we have

2 ∈ Eq if and only if 2 is a square in Fp, i. e. if and only if p ≡ −1, 1 (mod 8) ([15, Proposition

5.1.3]). Thus for a non-square q holds: 2 ∈ Iq if and only if p ≡ 1 (mod 8).

Proof of Theorem 1.6: Let Y be the Iq-closure of Iq. By Proposition 1.8, Y
′ := Y \ {0} is a union

of the cosets of H := 〈Iq \ {0}〉. Since #(Iq \ {0}) ≥ (q − 1)/4 with equality if and only if q ≡ 1

(mod 4), H is either F∗q, the set of non-zero squares, the set of non-zero cubes or (only if q ≡ 1

mod 4), the set of all non-zero 4-powers. Since Iq ⊆ Eq, H 6= F
∗
q. If H is the set of cubes, then,

as all elements of Iq are squares, it would be the set of 6-th powers, contradicting the inequality

#Iq > (q − 1)/4.

(a) Assume that H = Eq \{0}. It suffices to show that the Iq-closure of the set of squares contains

a non-square. Suppose otherwise. Take an element a ∈ Iq with a /∈ {0, 1}. Then we obtain

that for all squares x, y, x + (y − x)a is also a square. Since a is a non-zero square, this is the



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A characterization of Fq-linear subsets of affine spaces F
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101

same as the statement that for all squares x, z the element z + (1 − a)x is a square. If 1 − a

is a square we deduce that the set of all squares is closed under addition, a contradiction. If

1 − a is not a square we may take x = 1, z = 0 to obtain a contradiction.

(b) Assume q ≡ 1 (mod 4), q 6= 9, and that H is the set of all non-zero 4-powers. We also saw

that H = Iq \ {0}. The proof of step (a) works using the word “4-power” instead of “square”

with a a 4-power. We get that the set of all 4-powers is closed under taking differences. Thus

Iq is closed under taking differences and, since it contains 0, under the multiplication by −1.

H is obviously closed under taking products. Thus Iq is a subfield of order (q + 3)/4, which is

absurd if q 6= 9.

(c) Now we consider I′q and set H
′ := 〈Iq \ {0}〉. The cases in which H

′ is the set of all squares

or all cubes are excluded as above. Since #(I′q \ {0}) > (q − 1)/4, Y is not the set of all 4-th

powers.

Proposition 2.9. Assume q even and set Eq := {a(a + 1)}a∈Fq.

(1) If q = 2, 4, then Eq is the Eq-closure of itself.

(2) If q ≥ 8, then Fq is the Eq-closure of Eq.

Proof. We have E2 = {0} and E4 = {0, 1}.

Now assume q ≥ 8 and call B the Eq-closure of Eq. Let G be the subgroup of the multiplicative

group Fq \ {0} generated by Eq \ {0}. By Proposition 1.8 it is sufficient to prove that G = Fq \ {0}.

Since #Eq = q/2, Eq \ {0} 6= ∅. Fix a ∈ Eq \ {0} and a positive integer k. The Eq-closure

of {0, ak} contains ak+1. Thus B contains the multiplicative subgroup of Fq \ {0} generated by

Eq \ {0}. Since q ≥ 8, #(Fq \ {0}) = q − 1 is odd and q − 1 < 3(q/2 − 1) = 3#(Fq \ {0}), we get

G = Fq \ {0}.



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	Introduction
	The proofs and related observations