A characterization of linearized polynomials with maximum kernel

Contents lists available atScienceDirect

Finite

Fields

and

Their

Applications

www.elsevier.com/locate/ffa

A

characterization

of

linearized

polynomials

with

maximum

kernel

Bence Csajbóka_{, Giuseppe Marino}b,c_{, Olga Polverino}b_,

Ferdinando Zullob,∗

a_{MTA–ELTEGeometricandAlgebraicCombinatoricsResearchGroup,ELTE}

EötvösLorándUniversity,Budapest,Hungary,DepartmentofGeometry,1117

Budapest,Pázmány P.stny.1/C,Hungary

b

DipartimentodiMatematicaeFisica,UniversitàdegliStudidellaCampania “LuigiVanvitelli”,VialeLincoln5,I-81100Caserta,Italy

c_{DipartimentodiMatematicaeApplicazioni“RenatoCaccioppoli”,Universitàdegli}

StudidiNapoli“FedericoII”,ViaCintia,MonteS.Angelo,I-80126Napoli,Italy

a r t i c l e i n f o a b s t r a c t

Articlehistory:

Received15August2018 Receivedinrevisedform26 November2018

Accepted29November2018 Availableonline6December2018 CommunicatedbyDieterJungnickel

MSC:

11T06 15A04

Keywords:

Linearizedpolynomials

Weprovidesufficient andnecessaryconditions forthe coef-ficientsofa q-polynomial f overFqn whichensure that the

numberofdistinctrootsof f inFqn equals thedegreeof f .

Wesaythatthesepolynomialshavemaximumkernel.Asan applicationwestudyindetail q-polynomialsofdegree qn−2 overFqn whichhave maximumkernelandforn≤ 6 welist

all q-polynomialswithmaximum kernel.Wealso obtain in-formationonthesplittingfieldofanarbitraryq-polynomial. Analogousresultsareprovedforqs_{-polynomialsaswell,where}

gcd(s,n)= 1.

©2018ElsevierInc.Allrightsreserved.

✩ _{TheresearchwassupportedbyMinistryforEducation,UniversityandResearchofItalyMIUR(Project} PRIN2012“GeometriediGaloisestrutturediincidenza”)andbytheItalianNationalGroupforAlgebraic andGeometricStructuresandtheirApplications(GNSAGA- INdAM).Thefirstauthorwassupportedby theJánosBolyai ResearchScholarshipoftheHungarian AcademyofSciencesandbyOTKAGrantNo. K 124950.

* Correspondingauthor.

E-mailaddresses:csajbokb@cs.elte.hu(B. Csajbók),giuseppe.marino@unicampania.it, giuseppe.marino@unina.it(G. Marino),olga.polverino@unicampania.it(O. Polverino), ferdinando.zullo@unicampania.it(F. Zullo).

https://doi.org/10.1016/j.ffa.2018.11.009

Lineartransformations Semilineartransformations

1. Introduction

Aq-polynomialoverFqnisapolynomialoftheformf (x)=

iaixq i

,whereai∈ Fqn.

Wewill denotethesetofthesepolynomials byLn,q.LetK denotethealgebraicclosure

ofFqn.ThenforeveryF_qn≤ L≤ K,f definesanF_q-lineartransformationofL,whenL

isviewedasanFq-vectorspace.IfL isafinitefieldofsizeqmthenthepolynomialsofLn,q

consideredmodulo(xqm_{−x) formanF}

q-subalgebraoftheFq-lineartransformationsofL.

OncethisfieldL isfixed,wecandefinethekernel off asthekernelofthecorresponding Fq-linear transformation ofL, which isthe sameas theset of roots of f inL; and the

rank off astherankofthecorrespondingFq-lineartransformation ofL.Notethatthe

kernelandtherankoff dependonthisfieldL andfromnowonwewillconsiderthecase L=Fqn.Inthis caseL_n,q consideredmodulo(xq

n

− x) isisomorphictotheFq-algebra

of Fq-lineartransformationsof then-dimensionalFq-vectorspace Fqn.The elementsof

this factor algebraare representedby L˜n,q :={

n−1 i=0 aixq

i

: ai ∈ Fqn}.For f ∈ ˜L_n,q if

deg f = qk then wecall k theq-degree of f .It is clearthatinthis casethekernel off hasdimensionat mostk and therankoff isat leastn− k.

Let U = u1,u2,. . . ,ukFq be ak-dimensional Fq-subspace of Fqn. It is well known

that, up to a scalar factor, there is a unique q-polynomial of q-degree k, which has kernel U . Wecangetsuchapolynomialasthedeterminantofthematrix

⎛ ⎜ ⎜ ⎜ ⎜ ⎝ x xq _{· · · x}qk u1 uq1 · · · u qk 1 .. . uk uq_k · · · uq k k ⎞ ⎟ ⎟ ⎟ ⎟ ⎠.

Theaimofthispaperistostudytheotherdirection,i.e.whenagivenf ∈ ˜Ln,q with

q-degreek haskernelofdimensionk.Ifthishappensthenwesaythatf isaq-polynomial

with maximumkernel.

If f (x) ≡ a0x+ a1xσ+· · · + akxσ k

(mod xqn

− x), with σ = qs _{for some s with}

gcd(s,n)= 1,thenwesaythatf (x) isaσ-polynomial(orqs_{-polynomial)withσ-degree}

(or qs_{-degree)k.Regardingσ-polynomialsthefollowingisknown.}

Result 1.1.[7, Theorem 5] Let L be a cyclic extension of a field F of degree n, and suppose thatσ generates the Galois group of L over F. Let k bean integer satisfying

1 ≤ k ≤ n, and let a0,a1,. . . ,ak be elements of L, not all them are zero. Then the

f (x) = a0x + a1xσ+· · · + akxσ k

haskernelwithdimensionatmostk inL.

Similarlyto thes= 1 case wewill saythataσ-polynomialis ofmaximum kernel if thedimensionofitskernelequalsitsσ-degree.

Linearized polynomials have been used to describe families of Fq-linear maximum

rank distancecodes (MRD-codes), i.e. Fq-subspacesof L˜n,q of order qnk inwhich each

element has kernel of dimension at most k. The first examples of MRD-codes found were the generalized Gabidulin codes [3,5], thatis Gk,s = x,xq

s

,. . . ,xqs(k−1)Fqn with

gcd(s,n)= 1;thefactthatGk,sisanMRD-codecanbeshownsimplybyusingResult1.1.

Itisimportant tohaveexplicitconditionsonthecoefficientsofalinearizedpolynomial characterizingthenumberofitsroots.Furtherconnectionswithprojective polynomials canbe foundin[8].

Ourmain resultprovides sufficient and necessary conditions onthe coefficients ofa

σ-polynomialwith maximumkernel.

Theorem1.2. Consider

f (x) = a0x + a1xσ+· · · + ak−1xσ

k−1 − xσk_,

with σ = qs_{,gcd(s,n)= 1 and a}

0,. . . ,ak−1 ∈ Fqn. Thenf (x) is of maximumkernel if

andonly ifthematrix

A = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 0 · · · 0 a0 1 0 · · · 0 a1 0 1 · · · 0 a2 .. . ... ... ... 0 0 · · · 1 ak−1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ (1) satisfies AAσ· · · Aσn−1 = Ik,

whereAσi _{isthematrixobtainedfrom A byapplying toeachof itsentriesthe}

automor-phism x→ xσi _andI

k istheidentity matrixoforder k.

Animmediateconsequenceofthisresultgivesinformationonthesplittingfieldofan arbitraryσ-polynomial,cf. Theorem4.1.

InSection3.1westudyindetailstheσ-polynomialsofσ-degreen− 2 for eachn.For

n≤ 6 wealsoprovidealistofallσ-polynomialswithmaximumkernelcf.Sections3.2,3.3

and3.4.TheseresultsmightyieldfurtherclassificationresultsandexamplesofFq-linear

2. Preliminaryresults

In this section we recall some results of Dempwolff, Fisher and Herman from [4], adaptingthemto ourneedsinordertomake thispaperself-contained.

LetV beak-dimensionalvectorspaceoverthefieldF andletT beasemilinear

trans-formationofV .AT -cyclicsubspaceofV isanF-subspaceofV spannedby{v,T (v),. . .} overF forsomev∈ V ,whichwillbedenotedby[v].Wefirstrecallthefollowinglemma. Lemma2.1.[4,Theorem1] LetV beann-dimensionalvectorspaceoverthefieldF,σ an

automorphism ofF and T an invertible σ-semilineartransformation onV .Then

V = [u1]⊕ . . . ⊕ [ur]

forT -cyclicsubspaces satisfying dim[u1]≥ dim[u2]≥ . . . ≥ dim[ur]≥ 1.

Theorem 2.2.Let T be an invertible semilinear transformation of V = V (k,qn) of

or-der n,withcompanionautomorphismσ∈ Aut(Fqn) suchthatFix(σ)=F_q.ThenFix(T )

is ak-dimensionalFq-subspaceof V andFix(T )Fqn = V .

Proof. First assumethatthe companionautomorphism of T is x→ xq _{and thatthere}

exists v∈ V suchthat

V =v, T (v), . . . , Tk−1(v)Fqn.

Followingtheproofof[4,MainTheorem],considertheorderedbasisBT = (v,T (v),. . . , Tk−1(v)) andletA bethematrixassociatedwithT withrespect tothebasisBT,i.e.

A = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 0 · · · 0 α0 1 0 · · · 0 α1 0 1 · · · 0 α2 .. . ... ... ... 0 0 · · · 1 α_k−1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ∈ Fk×k qn , (2) where Tk_{(v) =} k

i=1αi−1Ti(v) with α0,. . . ,αk−1 ∈ Fqn and, since T is invertible,

we have α0 = 0. Denote by T the semilinear transformation of Fkqn having A as the

associated matrix withrespect to thecanonical orderedbasis BC = (e1,. . . ,ek) of Fkqn

and companion automorphism x→ xq_{.Note thatc}

BT(Fix(T ))= Fix(T ),where cBT is

thecoordinatization withrespectto thebasisBT. Also,sinceT hasordern,we have

AAq· · · Aqn−1 = Ik, (3)

where Aqi,fori∈ {1,. . . ,n− 1},is thematrixobtainedfrom A byapplying toeachof itsentriestheautomorphismx→ xqi.Avectorz= (z0,. . . ,zk−1)∈ Fkqn isfixedbyT if

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ α0z_k−1q = z0 zq₀+ α1zq_k−1= z1 .. . zq_k−2+ αk−1zk−1q = zk−1

Eliminatingz0,. . . ,zk−2, weobtain theequation

αq₀k−1zq_k−1k + αq₁k−2zq_k−1k−1+ . . . + αk−1zk−1q − zk−1= 0,

whichhasqk distinct solutionsinthealgebraic closureK ofFqn bythederivativetest.

Each solution determines a unique vector of Fix(T ) in Kk_{. Also, the set Fix(T ) is an}

Fq-subspace of Kk and hence dimFqFix(T ) = k. Let {w1,. . . ,wk} be an Fq-basis of

Fix(T ) andnote thatsince|Fix(T )|= qk_{, avector} k

i=1

aiwi isfixed byT ifand onlyif ai∈ Fq.Thisimpliesthatw1,. . . ,wkarealsoK-independent.ThusFix(T )K=Kk and {w1,. . . ,wk} isalsoaK-basisofKk.Denotebyφ theK-lineartransformationsuchthat

φ(wi)= ei and by P theassociated matrix withrespect to the canonicalbasis BC, so

P ∈ GL(k,K).Thesemilineartransformation φ◦ T ◦ φ−1 hascompanionautomorphism

x→ xq_{, ordern andassociated matrixwithrespect to thecanonicalbasisP · A· P}−q_,

whereP−q istheinverseofP in whichtheautomorphismx→ xq _{isappliedentrywise.}

Notethatφ◦ T ◦ φ−1(ei)= φ(T (wi))= φ(wi)= ei,hence

P· A · P−q= Ik, (4)

i.e.

Pq= P · A. (5)

ByEquations(3) and(5) andusinginductionweget Pqn = P· A · Aq· . . . · Aqn−1 = P,

i.e. P ∈ Fk×k_qn . This implies thatFix(T ) is an F_q-subspace of Fk_qn of dimension k and

henceFix(T )= c−1_BT(Fix(T )) isak-dimensionalsubspaceofV (k,qn_{) withtheproperty}

thatFix(T )Fqn = V .

Considernowthegeneralcase,i.e.supposeT asinthestatement,thatisT isan invert-iblesemilinearmapofordern withcompanionautomorphismx→ xqs_{andgcd(s,n)= 1.}

Sincegcd(s,n)= 1 thereexistl,m∈ N suchthat1= sl + mn, andhencegcd(l,n)= 1. Then thesemilinear transformation Tl _{hasorder n, companion automorphism x→ x}q

andFix(T )= Fix(Tl).ByLemma2.1,we maywrite V = [u1]⊕ . . . ⊕ [ur],

where [ui] is a Tl-cyclic subspace of V of dimension mi ≥ 1, for each i ∈ {1,. . . ,r},

and r_i=1mi = k.Then we can restrictTl to eachsubspace [ui] and byapplying the

previousargumentswe getthatUi= Fix(Tl|[ui]) isanFq-subspaceof[ui] ofdimension mi withthepropertythatUiFqn = [ui].Thus

Fix(T ) = Fix(Tl) = U1⊕ . . . ⊕ Ur

is anFq-subspaceofdimensionk ofV withtheproperty thatFix(T )Fqn = V . 2

TheexistenceofamatrixP ∈ GL(k,K),withK thealgebraicclosureofafinitefield of order q,satisfying(4) isalso aconsequenceofthecelebratedLang’s Theorem[9] on connectedlinearalgebraic groups.Moreprecisely,byLang’sTheorem,sinceGL(k,K) is a connected linear algebraic group, the map M ∈ GL(k,K) → M−1· Mq _{∈ GL(k,K)}

is onto. InTheorem 2.2 it is proved that, if the semilinear transformation of V (k,qn₎

having A asassociated matrixhasorder n,thenP ∈ GL(k,Fqn).

Remark 2.3.Let T be an invertible semilinear transformation of V = V (k,qn) with companion automorphism x → xq and let K be the algebraic closure of Fqn. Denote

by T the semilinear transformation of Kk _{associated with T as in the proof of}

The-orem 2.2. If λ ∈ K, then the set E(λ) := {v ∈ Kk_{: T (v) = λv} is an F}

q-subspace

of Kk_{. By [4, page 293], it follows that E(λ) = λ}q−11 _{Fix(T ) and by [4, Main}

The-orem] E(λ) is a k-dimensional Fq-subspace of Kk. Also, when T has order n and λq−11 ∈ F_qn, by Theorem 2.2, E(λ) is a k-dimensional F_q-subspace contained in Fk_qn

suchthatE(λ)_Fqn =F

k qn.

3. Mainresults

Now weareableto proveourmain result:

Proof of Theorem1.2. First suppose dim_Fqker f = k. Then there exist u0,u1, . . . , uk−1∈ Fqn whichform anF_q-basisofker f .

Put u := (u0,u1,. . . ,uk−1)∈ Fkqn.Sinceu₀,u₁,. . . ,uk−1 areF_q-linearlyindependent,

by [10, Lemma 3.51],we get thatB := (u,uqs_{,. . . ,u}qs(k−1)_{) is anordered F}

qn-basisof Fk qn.Also, uq sk = a0u + a1uq s

+· · · + a_k−1uqs(k−1)_{. It canbe seenthatthe matrix(1)}

representstheFqn-linearpartoftheF_qn-semilinearmap σ : v∈ Fk_qn→ vq s

∈ Fk qn w.r.t.

thebasisB. Sincegcd(s,n)= 1,σ hasordern and hencetheassertionfollows. Viceversa, letτ bedefinedasfollows

τ : ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ x0 x1 .. . x_k−1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠∈ F k qn → A ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ x0 x1 .. . x_k−1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ qs ∈ Fk qn, (6)

whereA isas in(1) with thepropertyAAqs· · · Aqs(n−1) = Ik. Thenτ has ordern and,

byTheorem 2.2, it fixesa k-dimensionalFq-subspace S ofFkqn with the property that SFqn =F

k qn.

LetB_S = (s0,. . . ,sk−1) beanFq-basisofS and notethat,sinceSFqn =F k

qn,BS is

alsoanFqn-basisofFk_qn, thendenotingbyBC thecanonicalordered basisofFkqn,there

existsauniqueisomorphismφ ofFk

qn suchthatφ(s_i)= e_i foreachi∈ {1,. . . ,k}.Then σ = φ◦ τ ◦ φ−1,whereσ : v∈ Fk qn→ vq s ∈ Fk qn.Also, σi= φ◦ τi◦ φ−1, (7)

foreachi∈ {1,. . . ,n− 1}.Also,by(6) τ (e0) = e1, τ (e1) = τ2(e0) = e2, .. . τ (e_k−1) = τk_(e 0) = (a0, . . . , ak−1) = a0e0+· · · + ak−1ek−1.

So,wegetthat

τk(e0) = a0e0+ a1τ (e0) +· · · + ak−1τk−1(e0),

andapplyingφ itfollowsthat

φ(τk(e0)) = a0φ(e0) + a1φ(τ (e0)) +· · · + ak−1φ(τk−1(e0)).

By(7) thepreviousequationbecomes

σk(φ(e0)) = a0φ(e0) + a1σ(φ(e0)) +· · · + ak−1σk−1(φ(e0)).

Putu = φ(e0),then

uqsk= a0u + a1uq

s

+· · · + a_k−1uqs(k−1).

Thisimpliesthatu0,u1,. . . ,uk−1 areelementsofker f ,whereu = (u0,. . . ,uk−1).Also,

theyareFq-independentsinceB = (u,. . . ,uq s(k−1)

)= (φ(e0),. . . ,φ(ek−1)) isanordered

Fqn-basisofFk

qn.Thiscompletes theproof. 2

As a corollary we get the second part of [6, Theorem 10], see also [12, Lemma 3] for the case s = 1 and [11] for the case when q is a prime. Indeed, by evaluating the determinantsinAAqs· · · Aqs(n−1) = Ik weobtainthefollowing corollary.1

1 _{Forx∈ Fqn andforasubfieldF}

qm ofF_qn wewilldenotebyN_qn_/qm(x) thenormofx overF_qm andby

Trqn_/qm(x) wewilldenotethetraceofx overF_qm.Ifn isclearfromthecontextandm= 1 thenwewill

Corollary3.1. Ifthekernelofaqs-polynomialf (x)= a0x+a1xq

s

+· · ·+ak−1xqs(k−1)−xqsk

has dimensionk, thenN(a0)= (−1)n(k+1).

Corollary 3.2.LetA beamatrix asinTheorem1.2.The condition AAqs· · · Aqs(n−1) = Ik

is satisfiedifandonly if AAqs

· · · Aqs(n−1) _{fixes e}

0= (1,0,. . . ,0).

Proof. The only if part is trivial, we prove the if part by induction on 0 ≤ i ≤ k − 1. Suppose AAqs

· · · Aqs(n−1)_eT

i = eTi for some 0 ≤ i ≤ k − 1. Then by

tak-ing qs-th powers of each entry we get AqsAq2s· · · AeT_i = eT_i . Since AeT_i = eT_i+1 this yields Aqs_Aq2s_{· · · A}qs(n−1)_eT

i+1 = eTi . Then multiplying both sides by A yields AAqs_Aq2s_{· · · A}qs(n−1)_eT i+1 = eTi+1. 2 Consideraqs_{-polynomialf (x)= a} 0x+ a1xq s +· · · + a_k−1xqs(k−1)_{− x}qsk_{, thematrix} A∈ Fk_qn×k asinTheorem1.2andthesemilinearmapτ definedin(6).

Note that eτ₀ = (0, 1, 0, . . . , 0) = e1 eτ₀2 = (0, 0, 1, . . . , 0) = e2 .. . eτ0k−1 = (0, 0, 0, . . . , 1) = ek−1 eτ₀k= (a0, a1, a2, . . . , ak−1) eτ₀k+1= (a0aq s k−1, aq s 0 + a1aq s k−1, aq s 1 + a2aq s k−1, . . . , aq s k−2+ aq s₊₁ k−1 ). (8) Hence,if

eτ0i= (Q0,i, Q1,i, . . . , Qk−1,i)

where Qj,icanbeseenaspolynomials ina0,a1,. . . ,ak−1,fori≥ 0,then eτ₀i+1 = (a0Qq s k−1,i, Qq s 0,i+ a1Qq s k−1,i, . . . , Qq s k−2,i+ ak−1Qq s k−1,i),

i.e. the polynomials Qj,i for 0 ≤ j ≤ k − 1 can be defined by the following recursive

relationsfor0≤ i≤ k − 1:

Qj,i=



1 if j = i, 0 otherwise, and bythefollowing relationsfori≥ k:

Q0,i+1= a0Qq s k−1,i Qj,i+1= Qq s j−1,i+ ajQq s k−1,i. (9) Now,weareabletoprovethefollowing.

Theorem3.3. The kernel of a qs-polynomial f (x)= a0x+ a1xq

s

+· · · + ak−1xqs(k−1)− xqsk ∈ Fqn[x],wheregcd(s,n)= 1, has dimensionk ifandonly if

Qj,n(a0, a1, . . . , ak−1) =



1 if j = 0,

0 otherwise. (10)

Proof. Relations(9) canbewrittenasfollows ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ Q0,i+1 Q1,i+1 .. . Qk−1,i+1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠= ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 0 · · · 0 a0 1 0 · · · 0 a1 0 1 · · · 0 a2 .. . ... ... ... 0 0 · · · 1 ak−1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ Qq_0,is Qq_1,is .. . Qq_k−1,is ⎞ ⎟ ⎟ ⎟ ⎟ ⎠,

with i ∈ {0,. . . ,n − 1}. Also, (Q0,0,Q1,0,. . . ,Qk−1,0) = (1,0,. . . ,0) and eτ

t

0 =

(Q0,t,. . . ,Qk−1,t) for t ∈ {0,. . . ,n}. By Theorem 1.2 and by Corollary 3.2, the

ker-nel of f (x) has dimension k if and only if e0 = (Q0,0,Q1,0,. . . ,Qk−1,0) is fixed by

AAqs· · · Aqs(n−1),sothis happensifandonlyif

eτ₀n= (Q0,n, Q1,n, . . . , Qk−1,n) = (1, 0, . . . , 0). 2

Theorem 3.3 with k = n− 1 and s = 1 givesthe following well-known result as a corollary.

Corollary3.4.[10,Theorem2.24] Thedimensionof thekernel ofaq-polynomialf (x)∈ Fqn[x] isn− 1 ifandonly ifthere existα,β ∈ F∗_qn suchthat

f (x) = α Tr(βx). AgainfromTheorem3.3 wecandeducethefollowing. Corollary3.5. [10,Ex. 2.14] Theqs-polynomiala0x− xq

sk

∈ Fqn[x], withgcd(s,n)= 1

and1≤ k ≤ n− 1, admitsqk _{roots ifand onlyif k| n and N}

qn_/qk(a₀)= 1.

3.1. Whentheqs_{-degreeequalsn− 2}

Inthissectionweinvestigateqs-polynomials f (x) = a0x + a1xq

s

+· · · + an−3xq s(n−3)

with gcd(s,n)= 1.ByTheorem 3.3,dim ker f (x) = n− 2 if andonlyifa0,a1,. . . ,an−3

satisfythefollowingsystemofequations ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ Q0,n= a0(aq 2s n−4+ aq 2s_+qs n−3 ) = 1, Q1,n= aq s 0 a q2s n−3+ a1(aq 2s n−4+ aq 2s +qs n−3 ) = 0, Q2,n= aq 2s 0 + a q2s n−3aq s 1 + a2(aq 2s n−4+ aq 2s_+qs n−3 ) = 0, Q3,n= aq 2s 1 + a q2s n−3aq s 2 + a3(aq 2s n−4+ aq 2s +qs n−3 ) = 0, .. . Qn−3,n = aq_n−52s + aq_n−32s a_n−4qs + an−3(aq_n−42s + aq_n−32s+qs) = 0, (11) whichisequivalent to ⎧ ⎪ ⎨ ⎪ ⎩ a0(aq 2s n−4+ aq 2s +qs n−3 ) = 1, a1=−aq s₊₁ 0 a q2s n−3=: g1(a0, an−3), aj=−aq 2s j−2a0− aq 2s n−3aq s j−1a0=: gj(a0, an−3), for 2≤ j ≤ n − 3. (12)

So,dim_Fqker f (x) = n− 2 if andonlyifa0 andan−3 satisfytheequations

 a0(gn−4(a0, an−3)q 2s + aq_n−32s+qs) = 1, an−3 = gn−3(a0, an−3), and aj= gj(a0,an−3) forj∈ {1,. . . ,n− 4}.

Theorem 3.6.Suppose thatf (x)= a0x+ a1xq+· · · + an−3xqn−3− xqn−2 has maximum

kernel.Thenfort≥ 2 withgcd(t− 1,n)= 1 thecoefficients at−2 andan−t arenon-zero

and, with s= n− t+ 1,

a_n−2t+1aq_t−22s+qs=−a_n−tqs+1aq_2t2s₋₃. (13)

Also, itholdsthat

−an−t(−aq s t−2aq 2s 3t−4+ aq 2s_+qs 2t−3 ) = aq 2s_+qs₊₁ t−2 . (14)

In particular,fort≥ 2 withgcd(t− 1,n)= 1 weget

N(a_n−t) = (−1)nN(a_t−2) (15)

and

N(a_n−2t+1) = (−1)nN(a2t−3), (16)

Proof. Lett≥ 2 withgcd(t− 1,n)= 1 andconsiderthepolynomialF (x)= f (xqt),that is, F (x) = a0xq t + a1xq t+1 +· · · + a_n−3xqn+t−3− xqn+t−2.

Clearly dim_Fqker F = dimFqker f = n− 2. Byrenaming the coefficients, F (x) can be

writtenas F (x) = α0x + α1xq n−t+1 + α2xq 2(n−t+1) +· · · + αn−3xq (n−t+1)(n−3) + αn−2xq (n−t+1)(n−2) = α0x + α1xq n−t+1 +· · · + α_n−3xq3t−3+ α_n−2xq2t−2.

SinceF (x) hasmaximumkernel,bythesecondequationof(12) wegetα0= 0,αn−2= 0

andthefollowingrelation

− α1 α_n−2 =−  − α0 α_n−2 qs₊₁ −αn−3 α_n−2 q2s . (17)

The coefficient αj of F (x) equals the coefficient ai of f (x) with i ≡ n− t+ j(1− t)

(mod n),inparticular ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ α0= an−t, α1= an−2t+1, αn−3= a2t−3, αn−2= at−2, αn−4= a3t−4, (18)

andby(17),wegetthatat−2 andan−t arenonzero,and an−2t+1aq 2s_+qs t−2 =−aq s₊₁ n−t aq 2s 2t−3,

whichgives(13).Thefirstequationof(12) gives

− α0 αn−2  −αn−4 αn−2 q2s +  −αn−3 αn−2 q2s+qs = 1, thatis, −α0(−αq s n−2αq 2s n−4+ αq 2s_+qs n−3 ) = αq 2s_+qs₊₁ n−2 .

Then(18) andα_n−4= a3t−4 imply

−an−t(−aq s t−2aq 2s 3t−4+ a q2s+qs 2t−3 ) = a q2s+qs+1 t−2 ,

whichgives(14).ByCorollary3.1withs= n− t+ 1 weobtain N  − α0 α_n−2  = 1, and taking(18) intoaccountwe get

N(an−t) = (−1)nN(at−2).

Then (13) andthepreviousrelationyield

N(an−2t+1) = (−1)nN(a2t−3). 2

Proposition 3.7. Let f (x) be a qs_{-polynomial with q}s_{-degree n− 2 and with maximum} kernel.Ifthecoefficientofxqs iszero,thenn isevenandf (x)= α Trqn_/q2(βx) forsome

α,β∈ F∗_qn.

Proof. We mayassumef (x)= a0x+ a1xq

s

+· · · + a_n−3xqs(n−3)_{− x}qs(n−2) _{with a}

1= 0.

Bythesecondequation of(12),itfollowsthatan−3= 0.Bythethirdequationof (12), we getthataj= 0 for everyoddintegerj∈ {3,. . . ,n− 3}.Ifj iseventhenwehave

aj = (−1) j

2aq

sj_+qs(j−2)_+···+q2s₊₁

0 . (19)

Ifn−3 iseven,thenthisgivesusacontradictionwithj = n−3.Itfollowsthatn−3 isodd andhencen iseven.ByN(a0)= (−1)n,thereexistsλ∈ F∗qn suchthata₀=−λ1−q

s(n−2) . So,by(19) wegetaj = λq js_−qs(n−2) ,and hence f (x) = Trqn/q2(λx) λqs(n−2) . 2

In the next sections we list all the qs-polynomials of Fqn with maximum kernel for n≤ 6. ByCorollaries 3.4and 3.5 then≤ 3 case canbe easily describedhence wewill consider onlythen∈ {4,5,6} cases.

For f (x) = n−1_i=0 aixq i ∈ ˜Ln,q we denote by f (x)ˆ := n−1_i=0 aq n−i i xq n−i the adjoint (w.r.t.thesymmetricnon-degeneratebilinear formdefinedbyx,y= Tr(xy))off .

By [1, Lemma 2.6],see also [2, pages 407–408],the kernel of f and f hasˆ the same dimensionand hencethefollowingresultholds.

Proposition 3.8. Iff (x)∈ ˜Ln,q is aqs-polynomialwith maximumkernel, thenf (x) isˆ a

qn−s-polynomialwithmaximum kernel.

3.2. The n= 4 case

InthissectionwedeterminethelinearizedpolynomialsoverFq4withmaximumkernel.

Withoutlossofgenerality,wecansupposethattheleadingcoefficientofthepolynomial is−1.

Because of Proposition 3.8, we canassume s = 1. Corollaries 3.4 and 3.5 cover the caseswhentheq-degreeoff is1 or3 sofromnowonwesupposef (x)= a0x+a1xq−xq

2

. Ifa1= 0 thenwecanuseagainCorollary3.5andwegeta0x− xq

2

,withNq4_/q2(a₀)= 1.

Supposea1= 0.ByEquation(12),wegettheconditions

 a0(aq 2 0 + a q2_+q 1 ) = 1, a1=−aq+10 a q2 1 , whichisequivalentto  Nq4_/q(a₀) = 1, aq+1₁ = aq₀2+q+1− aq₀, see(A1) ofSection5.

InTable1we listtheq-polynomialsof L4,q withmaximum kernel,upto anon-zero

scalarin F∗_q4. Applying theadjoint operation we can obtain thelist of q3-polynomials

overFq4 withmaximumkernel.Inthetabletheq-degreewillbe denotedbyk.

Table 1

LinearizedpolynomialsofFq4 withmaximumkernelwiths= 1.

k polynomial form conditions

3 Tr(λx) λ∈ F∗q4 2 a0x− xq 2 Nq4_/q2(a0) = 1 2 a0x + a1xq− xq 2 N_q4_/q(a0) = 1 aq+11 = a q2_+q+1 0 − a q 0 1 a0x− xq Nq4_/q(a0) = 1 3.3. The n= 5 case

InthissectionwedeterminethelinearizedpolynomialsoverFq5withmaximumkernel.

Withoutlossofgenerality,wecansupposethattheleadingcoefficientofthepolynomial is−1.BecauseofProposition3.8,wecanassumes∈ {1,2}.Corollaries3.4and3.5cover thecaseswhentheqs_{-degreeoff is1 or4.Firstwesupposethatf hasq}s_{-degree3,i.e.}

f (x) = a0x + a1xq

s

+ a2xq 2s

From (12), f (x) hasmaximum kernel ifand only ifa0, a1 and a2 satisfy thefollowing system: ⎧ ⎪ ⎨ ⎪ ⎩ a1=−aq s₊₁ 0 a q2s 2 , −aq3s+q2s+1 0 a q4s 2 + a q2s+qs 2 a0= 1, a2=−aq 2s₊₁ 0 + a q3s_+q2s 2 a q2s_+qs₊₁ 0 , whichis equivalentto ⎧ ⎪ ⎨ ⎪ ⎩ N(a0) = 1, a1=−aq s₊₁ 0 a q2s 2 , −aq3s_+q2s₊₁ 0 a q4s 2 + a0aq 2s_+qs 2 = 1,

see(A2)ofSection5.

Supposenow thattheqs_{-degreeis2,i.e.} f (x) = a0x + a1xq

s − xq2s_.

ByTheorem3.3thepolynomialf (x) hasmaximumkernelifandonlyifitscoefficients satisfy  a0(aq 2s 0 a q3s 1 + a qs 1 (a q3s 0 + a q3s_+q2s 1 )) = 1, aq₀s+1(aq₀3s+ aq₁3s+q2s) + a1= 0, whichis equivalentto  N(a0) =−1, a₀qs+ aq₁s+1= a₀q2s+qs+1aq₁3s, see(A3)ofSection5.

In Table 2we list the qs_{-polynomials,s ∈ {1,2} of L}

5,q with maximum kernel, up

to a non-zero scalar in F∗_q5. Applying the adjoint operation we can obtain the list of

qt_{-polynomials, t ∈ {3,4}, over F}

q5 with maximum kernel. As before, the qs-degree is

denotedbyk.

Table 2

LinearizedpolynomialsofFq5withmaximumkernelwiths∈ {1,2}.

k polynomial form conditions

4 Tr(λx) λ∈ F∗ q5 3 a0x + a1xq s + a2xq 2s − xq3s ⎧ ⎪ ⎨ ⎪ ⎩ N(a0) = 1 a1=−aq s₊₁ 0 a q2s 2 −aq3s_+q2s₊₁ 0 a q4s 2 + a0aq 2s_+qs 2 = 1 2 a0x + a1xq s − xq2s N(a0) =−1 aq1s+1+ a qs 0 = a q3s 1 a q2s_+qs₊₁ 0 1 a0x− xq s N(a0) = 1

3.4. The n= 6 case

InthissectionwedeterminethelinearizedpolynomialsoverFq6withmaximumkernel.

Without loss of generality, we can suppose that the leading coefficient of the polyno-mial is −1. Because of Proposition 3.8, we can assume s = 1. Corollaries 3.4 and 3.5

coverthe cases whenthe q-degree of f is1 or 5.As before, denote by k theqs_-degree

of f .

Wefirstconsiderthecasek = 2,i.e.f (x)= a0x+ a1xq

s − xq2s

.ByTheorem3.3,f (x) hasmaximumkernelifandonlyifthecoefficientssatisfy

⎧ ⎪ ⎨ ⎪ ⎩ N(a0) = 1, (aq₀+ aq+1₁ )q3 = aq₀5+q4+q3(aq₀+ aq+1₁ ), a₁q4aq₀3+ aq₁2(a₀q4+ aq₁4+q3) =− a1 aq+10 ,

see(A4) ofSection5.

Ifk = 3,thenf (x)= a0x+a1xq

s

+a2xq 2s

−xq3s

,andbyTheorem3.3ithasmaximum kernelifandonlyifthe coefficientsfulfill

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ N(a0) = 1, aq₀3+q+1+ aq₂3a₁q2aq+1₀ − aq₂a1= aq0, a₂q+1=−a₀q3+q2+q+1aq₁4− aq₁, aq+1₁ = a2aq0+ a q2_+q+1 0 a q3 2 ,

see(A5) ofSection5.Notethata1= 0 ifandonlyifa2= 0 andinthiscasewegetthe

traceover Fq3.

Finally,letk = 4.Thenthepolynomialf (x)= a0x+ a1xq

s + a2xq 2s + a3xq 3s − xq4s

hasmaximumkernelifandonlyifthecoefficientssatisfy ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ N(a0) = 1, a0(−aq 4_+q2 0 + a q5_+q4 3 a q4_+q3_+q2 0 + a q2_+q 3 ) = 1, a1=−aq+10 a q2 3 , a2=−aq 2₊₁ 0 + a q3_+q2 3 a q2_+q+1 0 , a3= aq 4 3 a q3+q2+1 0 + a q2 3 a q3+q+1 0 − a q3+q2+q+1 0 a q4+q3+q2 3 ,

see(A6) ofSection5.

InTable3we listtheq-polynomialsof L6,q withmaximum kernel,upto anon-zero

scalarin F∗_q6. Applying theadjoint operation we can obtain thelist of q5-polynomials

Table 3

LinearizedpolynomialsofFq6withmaximumkernelwiths= 1.

k polynomial form conditions 5 Trq6_/q(λx) λ∈ F∗_q6 4 a0x + a1xq+ a2xq 2 + a3xq 3 − xq4 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ a1= 0 N(a0) = 1 a0(−aq 4_+q2 0 + a q5_+q4 3 a q4_+q3_+q2 0 + a q2_+q 3 ) = 1 a1=−aq+10 a q2 3 a2=−aq 2₊₁ 0 + a q3_+q2 3 a q2_+q+1 0 a3= aq 4 3a q3_+q2₊₁ 0 + a q2 3 a q3_+q+1 0 − a q3_+q2_+q+1 0 a q4_+q3_+q2 3 4 Trq6_/q2(λx) λ∈ F∗_q6 3 a0x + a1xq+ a2xq 2 − xq3 ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ N(a0) = 1 aq03+q+1+ a q3 2a q2 1a q+1 0 − a q 2a1= aq0 aq+12 =−a q3_+q2_+q+1 0 a q4 1 − a q 1 aq+11 = a2aq0+ a q2_+q+1 0 a q3 2 3 Trq6_/q3(λx) λ∈ F∗_q6 2 a0x + a1xq− xq 2 ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ a1= 0 N(a0) = 1 (aq0+ a q+1 1 )q 3 = aq05+q4+q3(a q 0+ a q+1 1 ) aq14a q3 0 + a q2 1(a q4 0 + a q4_+q3 1 ) =− a1 aq+1 0 2 a0x− xq 2 Nq6_/q2(a0) = 1 1 a0x− xq Nq6_/q(a0) = 1 4. Application

As an application of Theorem 1.2 we are able to prove the following result on the splitting fieldofq-polynomials.

Theorem 4.1.Let f (x) = a0x+ a1xq+· · · + ak−1xq

k−1 − xqk

∈ Fqn[x] with a₀ = 0

and let A be defined as in (1). Then the splitting field of f (x) is Fqnm where m is the

(multiplicative) order ofthematrixB := AAq_{· · · A}qn−1_.

Proof. The derivativeof f (x) isnon-zeroandhencef (x) has qk _{distinct rootsinsome}

algebraic extension of Fqn. Suppose that F_qnm is the splitting field of f (x) and let t

denote the order of B. Then the kernel of the Fq-linear Fqnm → F_qnm map defined as

x→ f(x) hasdimensionk over Fqand hencebyTheorem 1.2wehave

AAq· · · Aqnm−1 = Ik.

Since thecoefficientsofA areinFqn,thisisequivalenttoBm= I_k andhencet| m.On

theother hand

andhenceagainbyTheorem1.2thekerneloftheFq-linearFqnt→ F_qnt mapdefinedas x→ f(x) hasdimensionk over Fq. ItfollowsthatFqnm isasubfieldofF_qnt from which

m| t. 2

AfurtherapplicationofTheorem1.2 isthefollowing.

Theorem4.2.Letn,m,s andt bepositiveintegerssuchthatgcd(s,nm)= gcd(t,nm)= 1

and s ≡ t (mod m). Let f (x) = a0x+ a1xq

s +· · · + ak−1xq s(k−1) − xqsk _{and g(x) =} a0x+ a1xq t +· · · + a_k−1xqt(k−1)_{− x}qtk_,wherea 0,a1,. . . ,ak−1∈ Fqm.The kernelof f (x)

consideredas alineartransformation ofFqnm has dimensionk if andonly ifthekernel

ofg(x) considered asalinear transformationof Fqnm has dimensionk.

Proof. DenotebyA thematrixassociatedwithf (x) asin(1).Byhypothesis,A∈ Fk×k_qm

and it is the same as the matrix associated with g(x). By Theorem 1.2 the kernel of

f (x),consideredasalineartransformation ofFqnm,hasdimensionk ifandonlyif

AAqs· · · Aqs(nm−1) = Ik.

Sinces≡ t (mod m),wehave

AAqs· · · Aqs(nm−1) = AAqt· · · Aqt(nm−1) = Ik,

and, againbyTheorem 1.2, thisholds ifand onlyifthekernel ofg(x), consideredasa lineartransformationofFqnm,hasdimensionk. 2

Addendum

Duringthe“Combinatorics2018”conference,thefourthauthor presentedtheresults ofthispaperinthetalkentitled“Onq-polynomialswithmaximumkernel”.Inthesame conferenceJohnSheekeypresentedajointworkwithGaryMcGuire[8] inhistalkentitled “Ranksof LinearizedPolynomials and Roots of ProjectivePolynomials”.It turned out that, independently from the authors of the present paper, they also obtained similar results.

5. Appendix

Inthissectionwedevelopsomecalculationsregardingtherelationsonthecoefficients ofalinearizedpolynomialswithmaximumkernelpresentedinSections3.2,3.3and3.4, seealso[13].

(A1) ByEquation(11) withn= 4,s= 1 andk = 2,wegettheconditions

Σ :  a0(aq 2 0 + a q2_+q 1 ) = 1, a1=−aq+10 a q2 1 .

ByCorollary3.1, thesystemΣ isequivalentto thefollowingsystem Σ: ⎧ ⎪ ⎨ ⎪ ⎩ Nq4_/q(a₀) = 1, a0(aq 2 0 + a q2_+q 1 ) = 1, a1=−aq+10 a q2 1 ,

whichcanberewrittenas follows

Σ: ⎧ ⎪ ⎨ ⎪ ⎩ Nq4_/q(a₀) = 1, aq₁2−1=− 1 aq+10 , aq+1₁ = aq₀2+q+1− aq₀. Now considerthesystem

Σ∗: 

Nq4_/q(a₀) = 1,

aq+1₁ = aq₀2+q+1− aq₀.

Clearly, S(Σ)⊆ S(Σ∗),where S(Σ) andS(Σ∗) denotetheset of solutionsofΣ andΣ∗,respectively.Let(a0,a1)∈ S(Σ∗),thenbyusingthenormconditionona0

aq₁2−1=  1 aq₀3 − a q 0 q−1 =  1− aq+q₀ 3 aq₀3 q−1 = = 1− a 1+q2 0 1− aq+q₀ 3a q3₋₁ 0 = 1− 1 aq+q30 1− aq+q₀ 3a q3₋₁ 0 =− 1 aq+1₀ , i.e. (a0,a1)∈ S(Σ) and henceS(Σ∗)= S(Σ)= S(Σ).

(A2) From (11) withn= 5,gcd(s,5)= 1 andk = 3,wegetthefollowing conditions:

Σ : ⎧ ⎪ ⎨ ⎪ ⎩ a0(aq 2s 1 + a q2s_+qs 2 ) = 1, a1=−aq s +1 0 a q2s 2 , a2=−aq 2s₊₁ 0 − a q2s 2 a qs 1 a0. ByCorollary3.1, Σ isequivalentto Σ: ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ Nq5_/q(a₀) = 1, a0(aq 2s 1 + a q2s_+qs 2 ) = 1, a1=−aq s₊₁ 0 a q2s 2 , a2=−aq 2s₊₁ 0 − a q2s 2 a qs 1 a0,

Σ: ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ Nq5_/q(a₀) = 1, a1=−aq s₊₁ 0 a q2s 2 , −aq3s_+q2s₊₁ 0 a q4s 2 + a q2s_+qs 2 a0= 1, a2=−aq 2s₊₁ 0 + a q3s_+q2s 2 a q2s_+qs₊₁ 0 .

Byraisingthethirdequationtoqs_{andmultiplyingbya}q2s₊₁

0 ,sinceN(a0)= 1,we

getthefourthequation.Therefore Σ, andhenceΣ,isequivalentto ⎧ ⎪ ⎨ ⎪ ⎩ Nq5_/q(a₀) = 1, a1=−aq s +1 0 a q2s 2 , −aq3s_+q2s₊₁ 0 a q4s 2 + a0aq 2s_+qs 2 = 1.

(A3) ApplyingTheorem3.3withn= 5 andk = 2,wegetthatthepolynomialf (x) has maximumkernelifandonlyifitscoefficientssatisfy

Σ :  Q0,5= a0(aq 2s 0 a q3s 1 + a qs 1 (a q3s 0 + a q3s+q2s 1 )) = 1, Q1,5= aq s 0 (a q3s 0 + a q3s_+q2s 1 ) + a1(aq 2s 0 a q3s 1 + a qs 1 (a q3s 0 + a q3s_+q2s 1 )) = 0, whichisequivalentto ⎧ ⎪ ⎨ ⎪ ⎩ Nq5_/q(a₀) =−1, a0(aq 2s 0 a q3s 1 + a qs 1 (a q3s 0 + a q3s+q2s 1 )) = 1, aq₀s+1(aq₀3s+ aq₁3s+q2s) + a1= 0,

becauseofCorollary3.1andsinceaq₀2saq₁3s+ a₁qs(aq₀3s+ aq₁3s+q2s)= 1 a0

.Theabove systemcanberewrittenas follows

⎧ ⎪ ⎨ ⎪ ⎩ Nq5_/q(a₀) =−1, aq₀3s+ aq₁3s+q2s =− a1 aqs +10 , aq₁3sa₀q2s+qs+1− aq₁s+1= aq₀s, whichisequivalentto ⎧ ⎪ ⎨ ⎪ ⎩ Nq5_/q(a₀) =−1, a₁qs+1+ aq₀s= a₁q3saq₀2s+qs+1, a1aq 4s_+q3s_+q2s 0 =−_aqs +1a1 0 .

Ifthefirstandthesecondequationsaresatisfied,clearlyalsothelastoneisfulfilled, henceΣ isequivalent tothefollowingsystem



Nq5_/q(a₀) =−1,

(A4) By Theorem3.3, withn= 6,s= 1 andk = 2,weget  Q0,6= a0(aq 2 0 (a q4 0 + a q4 +q3 1 ) + a q 1(a q3 0 a q4 1 + a q2 1 (a q4 0 + a q4 +q3 1 ))) = 1, Q1,6= aq 2 0 a1(aq 4 0 + a q4 +q3 1 ) + (a q+1 1 + a q 0)(a q3 0 a q4 1 + a q2 1 (a q4 0 + a q4 +q3 1 )) = 0, whichisequivalent to  a₀q2(aq₀4+ a₁q4+q3) + aq₁(aq₀3aq₁4+ aq₁2(a₀q4+ aq₁4+q3)) = _a1 0, a1 a0 + a q 0(a q3 0 a q4 1 + a q2 1 (a q4 0 + a q4_+q3 1 )) = 0, i.e.  a₀q2(aq₀4+ a₁q4+q3) + aq₁(aq₀3aq₁4+ aq₁2(a₀q4+ aq₁4+q3)) = _a1 0, a₀q3aq₁4+ aq₁2(a₀q4+ aq₁4+q3) =− a1 aq+10 . ByCorollary3.1, theprevioussystemisequivalentto

⎧ ⎪ ⎨ ⎪ ⎩ N(a0) = 1, a₀q2(aq₀4+ a₁q4+q3) + aq₁(aq₀3aq₁4+ aq₁2(a₀q4+ aq₁4+q3)) = _a1 0, a₀q3aq₁4+ aq₁2(a₀q4+ aq₁4+q3) =− a1 aq+10 , whichisequivalent to ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ N(a0) = 1, aq₀2(aq₀+ aq+1₁ )q3 −aq+11 aq+1 0 = 1 a0, a₀q3aq₁4+ aq₁2(a₀q4+ aq₁4+q3) =− a1 aq+10 , henceitisequivalentto ⎧ ⎪ ⎨ ⎪ ⎩ N(a0) = 1, (aq₀+ aq+1₁ )q3 = aq₀5+q4+q3(aq₀+ aq+1₁ ), a₁q4aq₀3+ aq₁2(a₀q4+ aq₁4+q3) =− a1 aq+10 . (A5) By Theorem3.3with n= 6,s= 1 andk = 3,weget

⎧ ⎪ ⎨ ⎪ ⎩ a0Qq2,5 = 1, Qq_0,5+ a1Qq2,5= 0, Qq_1,5+ a2Qq2,5= 0, where Q0,5= a0(aq 2 1 + a q2+q 2 ), Q1,5= aq0a q2 2 + a1(aq 2 1 + a q2_+q 2 ), Q2,5= aq 2 0 + a q2 2 a q 1+ a2(aq 2 1 + a q2+q 2 ),

hencewe obtainthefollowingsystem ⎧ ⎪ ⎨ ⎪ ⎩ a0(aq 3 0 + a q3 2 a q2 1 + a q 2(a q3 1 + a q3+q2 2 )) = 1, a1 a0 + a q 0(a q3 1 + a q3_+q2 2 ) = 0, a2 a0 + a q3 2 a q2 0 + a q 1(a q3 1 + a q3_+q2 2 ) = 0. ByCorollary 3.1itisequivalentto ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ N(a0) = 1, a0(aq 3 0 + a q3 2 a q2 1 + a q 2(a q3 1 + a q3_+q2 2 )) = 1, aq₁3+ aq₂3+q2 =− a1 a1+q0 , a2 a0 + a q3 2 a q2 0 + a q 1(a q3 1 + a q3+q2 2 ) = 0,

bysubstitutingthethirdequation intotheothersweget ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ N(a0) = 1, a0(aq 3 0 + a q3 2 a q2 1 − aq2a1 a1+q0 ) = 1, aq₁3+ aq₂3+q2=− a1 a1+q0 , a2 a0 + a q3 2 a q2 0 − aq+11 a1+q0 = 0, i.e. ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ N(a0) = 1, aq₀3+q+1+ aq₂3a₁q2aq+1₀ − aq₂a1= aq0, a₂q+1=−a₀q3+q2+q+1aq₁4− aq₁, aq+1₁ = a2aq0+ a q2+q+1 0 a q3 2 .

(A6) Equations(11) withn= 6,s= 1 andk = 4 are ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ a0(aq 2 2 + a q2_+q 3 ) = 1, aq₀aq₃2+ a1(aq 2 2 + a q2_+q 3 ) = 0, aq₀2+ aq₃2aq₁+ a2(aq 2 2 + a q2_+q 3 ) = 0, aq₁2+ aq₃2aq₂+ a3(aq 2 2 + a q2_+q 3 ) = 0,

which,byCorollary3.1,isequivalent to ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ N(a0) = 1, a0(aq 2 2 + a q2_+q 3 ) = 1, aq₀aq₃2+ a1(aq 2 2 + a q2+q 3 ) = 0, aq₀2+ aq₃2aq₁+ a2(aq 2 2 + a q2_+q 3 ) = 0, aq₁2+ aq₃2aq₂+ a3(aq 2 2 + a q2+q 3 ) = 0,

thusitcanberewrittenas follows ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ N(a0) = 1, aq₂2+ aq₃2+q=_a1 0, aq₀aq₃2+a1 a0 = 0, aq₀2+ aq₃2aq₁+a2 a0 = 0, aq₁2+ aq₃2aq₂+a3 a0 = 0, and hence ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ N(a0) = 1, a0(aq 2 2 + a q2_+q 3 ) = 1, a1=−aq+10 a q2 3 , a2=−aq 2₊₁ 0 − a q2 3 a q 1a0, a3=−aq 2 1 a0− aq 2 3 a q 2a0, i.e. ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ N(a0) = 1, a0(−aq 4_+q2 0 + a q5_+q4 3 a q4_+q3_+q2 0 + a q2_+q 3 ) = 1, a1=−aq+10 a q2 3 , a2=−aq 2₊₁ 0 + a q3_+q2 3 a q2_+q+1 0 , a3= aq 4 3 a q3_+q2₊₁ 0 + a q2 3 a q3_+q+1 0 − a q3_+q2_+q+1 0 a q4_+q3_+q2 3 . References

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