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© Hindawi Publishing Corp.
ON DEDEKIND’S CRITERION AND MONOGENICITY OVER DEDEKIND RINGS
M. E. CHARKANI and O. LAHLOU Received 29 November 2002
We give a practical criterion characterizing the monogenicity of the integral clo- sure of a Dedekind ringR, based on results on the resultant Res(P , Pi)of the min- imal polynomialPof a primitive integral element and of its irreducible factorsPi modulo prime ideals ofR. We obtain a generalization and an improvement of the Dedekind criterion (Cohen, 1996) and we give some applications in the case where Ris a discrete valuation ring or the ring of integers of a number field, generalizing some well-known classical results.
Mathematics Subject Classification: 11Y40, 13A18, 13F30.
1. Introduction. LetKbe an algebraic number field and letOKbe its ring of integers. IfOK=Z[θ]for some numberθinOK, we say thatOK has a power basis orOKis monogenic. The question of the existence of a power basis was originally examined by Dedekind [5]. Several number theorists were interested in and attracted by this problem (see [7, 8, 9]) and noticed the advantages of working with monogenic number fields. Indeed, for a monogenic number fieldK, in addition to the ease of discriminant computations, the factorization of a primepinK/Qcan be found most easily (see [4, Theorem 4.8.13, page 199]). The main result of this paper isTheorem 2.5which characterizes the monogenicity of the integral closure of a Dedekind ring. More precisely, letR be a Dedekind domain,Kits quotient field,La finite separable extension of degreenofK,αa primitive element ofLintegral overK,P (X)=Irrd(α, K), ma maximal ideal ofR, andOLthe integral closure ofRinL. Assume that P (X)¯ =r
i=1P¯iei(X)in(R/m)[X]withei≥2, and letPi(X)∈R[X]be a monic lifting of ¯Pi(X)for 1≤i≤r. Then we prove thatOL=R[α]if and only if, for every maximal idealmofRandi∈ {1, . . . , r},vm(Res(Pi, P ))=deg(Pi), where vmis them-adic discrete valuation associated tom. This leads to a necessary and sufficient condition for a simple extensionR[α]of a Dedekind ringRto be Dedekind. At the end, we give two illustrations of this criterion. In the second example, we give the converse which was not known yet.
2. Monogenicity over a Dedekind ring. Throughout this paperRis an in- tegral domain,Kits quotient field,Lis a finite separable extension of degree nofK,αis a primitive element ofLintegral overR,P (X)=Irrd(α, K),mis
a maximal ideal ofR, andOLis the integral closure ofRinL. Letf andgbe two polynomials overR; the resultant off andgwill be denoted by Res(f , g) (see [11]).
Definition2.1. IfOL=R[θ]for some numberθ∈OL, thenOLhas a power basis orOLis monogenic.
Proposition2.2. LetRbe an integrally closed ring and letαbe an integral element overR. Then(R[α])p=Rp[α]for every prime idealpofR. In partic- ular,OL=R[α]if and only ifRp[α]is integrally closed for every prime idealp ofRif and only ifR[α]is integrally closed.
Proof. We obtain the result from the isomorphismR[α]R[X]/P (X), the properties of an integrally closed ring and its integral closure, and the properties of a multiplicative closed subset of a ringR, notably,S−1(R[X])= (S−1R)[X](see [1]).
Definition2.3. LetRbe a discrete valuation ring (DVR),p=π Rits maxi- mal ideal, andαan integral element overR. LetPbe the minimal polynomial ofα, and ¯P (X)=r
i=1P¯iei(X)the decomposition of ¯P into irreducible factors in(R/p)[X]. Set
f (X)= r i=1
Pi(X)∈R[X],
h(X)= r i=1
Piei−1(X)∈R[X],
T (X)=P (X)−r
i=1Piei(X)
π ∈R[X],
(2.1)
wherePi(X)∈R[X]is a monic lifting of ¯Pi(X), for 1≤i≤r. We will say that R[α]isp-maximal if(f ,¯T ,¯h)¯ =1 in(R/p)[X](where(·,·)denotes the greatest common divisor (gcd)). IfRis a Dedekind ring andpis a prime ideal ofR, then we say thatR[α]isp-maximal ifRp[α]ispRp-maximal.
Remarks2.4. (1) Ifπ is uniramified inR[α], that is,ei=1 for all i, then h¯=¯1 and thereforeR[α]isp-maximal.
(2) Let π be ramified in R[α], that is, there is at least one i such that ei≥2. LetS = {i∈ {1, . . . , r} |ei≥2}andf1(X)=
i∈SPi(X)∈R[X]. Then (f¯1,T )¯ =(T ,¯f ,¯h)¯ in(R/p)[X]since ¯f1=(f ,¯h). In particular, if every¯ ei≥2, then(f ,¯T )¯ =(T ,¯ f ,¯h), because ¯¯ fdivides ¯hin this case.
(3)Definition 2.3is independent of the choice of the monic lifting of the ¯Pi. More precisely, let
P (X)¯ = r i=1
P¯iei(X)= r i=1
Q¯eii(X) with ¯Pi(X)=Q¯i(X)for 1≤i≤r in(R/p)[X].
(2.2)
Set
g(X)= r i=1
Qi(X)∈R[X], k(X)= r i=1
Qeii−1(X)∈R[X]
U (X)=π−1
P (X)− r i=1
Qeii(X)
∈R[X].
(2.3)
Then(f ,¯T ,¯h)¯ =1 in(R/p)[X]if and only if(g,¯U ,¯ ¯k)=1 in(R/p)[X]. Indeed, we may assume thatR is a DVR and p=π R. LetV1=(g−f )/π andV2= (k−h)/π. Thenπ T=π U+gk−f h. Replacinggbyπ V1+fandkbyπ V2+h, we find that ¯T =U¯+V¯1h¯+V¯2f¯and therefore(T ,¯f ,¯h)¯ =(U ,¯ f ,¯h)¯ =(U ,¯ g,¯k)¯ since ¯f=g¯and ¯h=k.¯
Theorem2.5. LetRbe a Dedekind ring. LetPbe the minimal polynomial of α, and assume that for every prime idealpof R, the decomposition of P¯into irreducible factors in(R/p)[X]verifies:
P (X)¯ = r i=1
P¯iei(X)∈(R/p)[X] (2.4)
withei≥2fori=1, . . . , randPi(X)∈R[X]be a monic lifting of the irreducible factorP¯i for i=1, . . . , r. Then OL=R[α]if only ifvp(Res(Pi, P ))=deg(Pi) for every prime idealpofRand for everyi=1, . . . , r, wherevp is thep-adic discrete valuation associated top.
For the proof we need the following two lemmas.
Lemma2.6. Letp=uR+vRbe a maximal ideal of a commutative ringR.
ThenpRp=vRpif and only if there exista, b∈Rsuch thatu=au2+bv.
Proof. IfpRp=vRp, then there exists∈Randt∈R−psuch thattu=vs.
Sincep is maximal inR, so there existst∈R such thattt−1∈p. Hence u−utt=u−vst∈p2 and there exista, b∈R such that u=au2+bv.
Conversely, u2R+vR⊆vR+p2⊆p. If there exist a, b∈R such that u= au2+bv, thenp=u2R+vRand thereforevR+p2=p. Localizing atp and applying Nakayama’s lemma, we find thatpRp=vRp.
Lemma2.7. LetR be a commutative integral domain, letKbe its quotient field, and consider P , g, h, T ∈ R[X]. If g is monic and P =gh+π T, then Res(g, P )=πdeg(g)Res(g, T ). In particular, ifm=π R is a maximal ideal of Rand ifP (X)¯ =r
i=1P¯iei(X)is the decomposition ofP¯into irreducible factors in(R/m)[X], withPi(X)∈R[X] a monic lifting ofP¯i(X)for 1≤i≤r, and T (X)=π−1(P (X)−r
i=1Piei(X))∈R[X], then Res
Pi, P
=πdeg(Pi)Res Pi, T
(2.5)
and(P¯i,T )¯ =1in(R/m)[X]if and only if
Res Pi, T
=Res Pi, P
πdeg(Pi) ∈R−m. (2.6)
Proof. Letx1, . . . , xmbe the roots ofgin the algebraic closure ¯KofK. It is then easy to see (see [11]) that Res(g, P )=m
i=1P (xi)=πdeg(g)Res(g, T )be- causeP (xi)=π T (xi). The second result follows from Res(P¯i,P )¯ =Res(Pi, P ) and [2, Corollary 2, page 73].
Proof ofTheorem2.5. By Proposition 2.2, we may assume that R is a DVR. Letp be a prime ideal ofRand(OL)(p) the integral closure ofRp inL.
Let ¯P (X)=Πir=1P¯iei(X)in(Rp/pRp)[X]withei≥2 andPi(X)∈Rp[X]a monic lifting of ¯Pi(X)for 1≤i≤r. Let
T (X)=P (X)−Πri=1Piei(X)
π ∈Rp[X] (2.7)
withπ Rp=pRp.
(a) We prove that if(P¯i,T )¯ =1 in(Rp/pRp)[X]for everyi=1, . . . , r, then (OL)(p)=Rp[α]=A. Indeed, ¯P (X)=Πri=1P¯iei(X)in(Rp/pRp)[X]andRpis a lo- cal ring, so by [14, Lemma 4, page 29] (see also [3]) the idealsᏮi=π A+Pi(α)A (i=1, . . . , r) are the only maximal ideals ofA, soAis integrally closed if and only ifᏭᏮi is integrally closed for everyi=1, . . . , r. More generally, we prove that every ᏭᏮi is a DVR. SinceRp is Noetherian, so Rp[α]Rp[X]/P (X) is Noetherian, henceᏭᏮi is Noetherian sinceᏭᏮi is a local integral domain with maximal ideal ᏮiᏭᏮi. It remains to show that ᏮiᏭᏮi is principal. In- deed,(P¯i,T )¯ =1 in(Rp/pRp)[X], hence there exist polynomialsU1, U2, U3∈ Rp[X] such that 1=U1(X)Pi(X)+U2(X)T (X)+π U3(X). Now P (α)= 0= Πrj=1Pjej(α)+π T (α), henceΠrj=1Pjej(α)= −π T (α), so
π=π U1(α)Pi(α)+π2U3(α)−Πrj=1Pjej(α)U2(α)
=π2U3(α)+Pi(α)U4(α) (2.8)
with U4=π U1−Piei−1(r
j=1, j≠iPjej)U2 ∈Rp[X]. It follows fromLemma 2.6 thatᏮiᏭᏮi =Pi(α)ᏭᏮi, in other words,ᏮiᏭᏮi is principal. We conclude that ᏭᏮiis a DVR and therefore an integrally closed ring, and(OL)(p)=Rp[α].
(b) We will now prove that(P¯i,T )¯ =1 in(Rp/pRp)[X]for everyi=1, . . . , rif (OL)(p)=Rp[α]. We first show that the ringᏭᏮiis a DVR, for everyi. Indeed, Rpis a Dedekind ring andLis a finite extension ofK, and it follows from [10, Theorem 6.1, page 23] that(OL)(p)=Rp[α]=Ais a Dedekind ring, soᏭᏮi is a DVR. Let us show next thatT (α)is a unit in everyᏭᏮi. Indeed,ᏭᏮiis a DVR and so its maximal idealᏮiᏭᏮi=πᏭᏮi+Pi(α)ᏭᏮi is principal. Letλ∈ᏭᏮi
be a generator of ᏮiᏭᏮi. Then there existu, v ∈ᏭᏮi such thatλ =π u+ Pi(α)v ∈ᏮiᏭᏮi−(ᏮiᏭᏮi)2. NowRp is a DVR, P =Irrd(α, Rp), ¯P =Πrj=1P¯jej
in(Rp/π Rp)[X],π Rp∈SpecRp, and(OL)(p)=Rp[α]=Ais the integral clo- sure ofRpinL=K(α)withK=Fr(Rp), and we find thatπ A=Πrj=1Ꮾejj. Hence π∈Ꮾi2becauseei≥2. Nowλ ∈(ᏮiᏭᏮi)2, hencePi(α)∉(ᏮiᏭᏮi)2, because λ=uπ+Pi(α)v. It then follows thatPi(α)is a generator ofᏮiᏭᏮi=Pi(α)ᏭᏮi
since πᏭᏮi =(ᏮiᏭᏮi)ei =Piei(α)ᏭᏮi, and π =Piei(α)1 with 1∈U (ᏭᏮi).
We now show thatPj(α)∈U (ᏭᏮi)for everyj≠i. Indeed, ifPj(α)∈ᏮiᏭᏮi, then there exists ai ∈ Ꮾi and bi ∈ A−Ꮾi such that Pj(α)= ai/bi. Then ai=Pj(α)bi∈Ꮾi. Now,Ꮾi is a prime ideal of A, hencePj(α)∈Ꮾi. AsᏮj= π A+Pj(α)A, soᏮj⊆Ꮾi. The idealᏮjis a maximal ideal of A, soᏮi=Ꮾj. This is impossible because theᏮi are distinct, and it follows thatPj(α)∈U (ᏭᏮi) for everyj≠i. Thus there exists2∈U (ᏭᏮi)such thatr
j=1, j≠iPjej(α)=2. Sincer
j=1Pjej(α)= −π T (α), π =Piei(α)1, and r
j=1, j≠iPjej(α)=2, then T (α)= −21−1∈U (ᏭᏮi). SoT (α)∈U (ᏭᏮi)for everyi, and T (α)∈U (A);
otherwise, Krull’s theorem implies the existence of a maximal idealᏮi of A such thatT (α)∈Ꮾi, andT (α)∈ᏮiᏭᏮi=ᏭᏮi−U (ᏭᏮi), which is impossible.
We conclude thatT (α)is a unit inRp[α], and, by [2, Corollary 1, page 73], there existU1, V1∈Rp[X]such that 1=U1(X)P (X)+V1(X)T (X). Consequently ¯1= U¯1(X)P (X)¯ +V¯1(X)T (X)¯ in(Rp/π Rp)[X], which is principal. Hence(P ,¯T )¯ = 1 in (Rp/π Rp)[X] since ¯P = r
i=1P¯iei in (Rp/π Rp)[X] then (P¯i,T )¯ =1 in (Rp/π Rp)[X] for every i. Our result now follows fromProposition 2.2 and Lemma 2.7.
Remarks2.8. (1) Letπ be ramified inR[α],S = {i∈ {1, . . . , r} |ei≥2}, andf1(X)=
i∈SPi(X)∈R[X]. It follows fromLemma 2.7that the following statements are equivalent:
(i) (f¯1,T )¯ =1 in(R/p)[X];
(ii) vp(Res(f1, P ))=deg(f1);
(iii) for everyi∈S, we havevp(Res(Pi, P ))=deg(Pi), wherevpis thep-adic discrete valuation associated top.
(2) It follows from the above equivalence andRemark 2.4(2) and (3) that the condition inTheorem 2.5is independent of the choice of the monic lifting of ¯Pi. More precisely, ifei≥2 for everyi, and if we take another monic liftingQiof ¯Pi, thenvp(Res(Pi, P ))=deg(Pi)for alli=1, . . . , rif and only ifvp(Res(Qi, P ))= deg(Qi)for alli=1, . . . , r.
(3)Theorem 2.5states that, under the assumption thatei≥2 for everyi, OL=R[α]if and only ifR[α]isp-maximal for every prime idealpofR.
Corollary2.9. Under the assumptions ofTheorem 2.5, ifOL=R[α], then, for every prime idealpofR,Rp[α]is principal andᏮi=Pi(α)Rp[α]for everyi.
Proof. Indeed, a Dedekind ring having only a finite number of prime ideals is principal. To prove the second statement, takex∈Asuch thatᏮi=xA.
Then ᏮiᏭᏮi =xᏭᏮi =Pi(α)ᏭᏮi, hence Pi(α)=xε with ε∈ U (ᏭᏮi). Then ε∈U (A), soᏮi=Pi(α)A.
Definition2.10. LetRbe a DVR with maximal idealm=π R, withf , g∈ R[X]monic polynomials. Thenf is called an Eisenstein polynomial relative togif there existsT∈R[X]and an integere≥1 such thatf=ge+π T and (g,¯T )¯ =1 in(R/π R)[X].
Remark 2.11. As in the classical Eisenstein’s criterion, we have a crite- rion for the irreducibility of an Eisenstein polynomial relative to g, called the Schönemann criterion, see [12, page 273]; if f = ge+π T is an Eisen- stein polynomial relative to g such that ¯g ∈ (R/m)[X] is irreducible and deg(T ) < edeg(g), thenfis irreducible inK[X].
Corollary2.12. LetRbe a DVR with maximal idealm=π R. IfP¯=g¯ein (R/m)[X]withe≥2, thenOL=R[α]if and only ifPis an Eisenstein polynomial relative tog.
Proof. We obtain the result usingTheorem 2.5,Definition 2.10, andLemma 2.7.
Remark 2.13. Corollary 2.12 generalizes [14, Propositions 15 and 17]; it integrates the two results in one statement and provides the converse.
3. Monogenicity over the ring of integers. LetK=Q(α)be a number field of degreen,P (X)∈Z[X]a minimal polynomial ofα,OK the ring of integers ofK, andpa prime number.
Proposition3.1. LetK=Q(α)be a number field andPthe minimal poly- nomial ofα. ThenOK=Z[α]if and only if for every prime numberpsuch that p2dividesDisc(P ), the prime numberpdoes not divideInd(α).
Proof. We obtain the result from the fact thatOK=Z[α]if and only if Ind(α)=1, and Disc(P )=(Ind(α))2dK(see [6], [4, page 166]).
Proposition3.2. LetP (X)¯ =r
i=1P¯iei(X)be the factorization ofP (X)mod- ulopinFp[X], and putf (X)=r
i=1Pi(X)withPi(X)∈Z[X]a monic lifting ofP¯i(X)andei≥2for alli. Leth(X)∈Z[X]be a monic lifting ofP (X)/¯ f (X)¯ andT (X)=(f (X)h(X)−P (X))/p∈Z[X]. Then the following statements are equivalent:
(i) pdoes not divideInd(α)=[OK:Z[α]];
(ii) (f ,¯T )¯ =1inFp[X];
(iii) vp(Res(f , P ))=deg(f );
(iv) vp(Res(Pi, P ))=deg(Pi), for everyi∈ {1, . . . , r}.
Proof. (i)(ii). Let(OK)(p)be the integral closure ofZ(p)inK. We first show thatpdoes not divide Ind(α)if and only if(OK)(p)=Z(p)[α]. By the finiteness theorem [13, page 48],(OK)(p)= ⊕ni=0−1Z(p)xi, and, becauseZ(p) is principal, αi= nj=−01aijxjwithaij∈Z(p), and therefore[(OK)(p):Z(p)[α]]= |det(aij)|.
On the other hand, Ind(α)=[OK:Z[α]]=[(OK)(p):(Z[α])(p)]=[(OK)(p) : Z(p)[α]], hence(OK)(p)=Z(p)[α]if and only ifpdoes not divide Ind(α)if and only if Ind(α)∈ ∪(Z(p))=Z(p)−pZ(p). Hence by the proof ofTheorem 2.5,p does not divide Ind(α)if and only if(P¯i,T )¯ =1 inFp[X]for everyi=1,2, . . . , r (in other words, if and only if(f ,¯T )¯ =1 inFp[X]).
(ii)(iii). By [2, Corollary 2, page 73], (f ,¯T )¯ =1 in Fp[X] if and only if Res(f ,¯T )¯ =Res(f , T )≠¯0 inFpif and only if Res(f , T )∈Z−pZ. On the other hand,
Res(f , T )=(−1)deg(f )
pdeg(f ) Res(f , P ). (3.1) (ii)(iv). We have(f ,¯T )¯ =1 inFp[X]if and only if Res(f , T )∈Z−pZ. On the other hand, Res(f , T )=r
i=1Res(Pi, T )and
Res Pi, T
=(−1)deg(Pi) pdeg(Pi) Res
Pi, P
. (3.2)
Theorem3.3. LetK=Q(α)be a number field of degreen,P (X)∈Z[X]
a monic minimal polynomial ofα, and OK the ring of integers ofK. Assume P (X)¯ =r
i=1P¯iei(X)inFp[X], for every prime numberpsuch thatp2divides Disc(P ), withPi(X)∈Z[X]a monic lifting ofP¯i(X)andei≥2for1≤i≤r. ThenOK=Z[α]if and only if for every prime numberp, such thatp2divides Disc(P ),vp(Res(Pi, P ))=deg(Pi)for1≤i≤r.
Proof. It suffices to apply Propositions3.1and3.2, andTheorem 2.5.
Remark3.4. Proposition 3.2provides a complement to the Dedekind cri- terion (see [4, page 305]). Indeed, inFp[X], we have(f ,¯T )¯ =(f ,¯T ,¯ h)¯ since all ei≥2.
We finish this section giving other conditions equivalent top not being a divisor of Ind(α).
Proposition3.5. The following statements are equivalent:
(i) pdoes not divideInd(α)=[OK:Z[α]];
(ii) Z[α]+pOK=OK; (iii) Z[α]∩pOK=pZ[α].
Proof. (ii)(iii). Consider the following map ofFp-vector spaces:
j:Z[α]/pZ[α] →OK/pOK, j
x+pZ[α]
=x+pOK. (3.3)
As OK and Z[α]are two free groups of the same rank n, Z[α]/pZ[α] and OK/pOK are two Fp-vector spaces of the same dimension n and injectivity ofjis equivalent to surjectivity ofj. Moreover,j is one-to-one if and only if Z[α]∩pOk=pZ[α]andjis onto if and only ifZ[α]+pOK=OK.
(i)(iii). Ifpdoes not divide Ind(α)andpZ[α]⊂Z[α]∩pOK, then there existsx∈OKsuch thatx∉ Z[α]andpx∈Z[α], so the order of the subgroup generated byx+Z[α]of the finite groupOK/Z[α]is equal top, and, by La- grange’s theorem,pdivides Ind(α), which is the order of the groupOK/Z[α], and this is impossible.
Conversely, assume thatZ[α]∩pOK=pZ[α]andpdivides Ind(α). Cauchy’s theorem implies that there exists an element of orderpinOK/Z[α]; in other words, there exists x∈OK such thatx∉ Z[α] andpx∈Z[α]. Thenpx∈ Z[α]∩pOK=pZ[α], hencex∈Z[α], which is impossible.
4. Applications
4.1. Monogenicity of cyclotomic fields
Proposition4.1. Letn≥3be an integer,ξna primitiventh root of unity, K=Q(ξn), andφn(X)thenth cyclotomic polynomial overQ. ThenOK=Z[ξn].
Proof. We know from [15] that
φn(X)=
1≤i≤n i∧n=1
X−ξni
=Irrd ξn,Q
,
Disc φn
=(−1)ϕ(n)/2 nϕ(n)
p|npϕ(n)/(p−1) =(−1)ϕ(n)/2 s i=1
pϕ(n)(ri i−1/(pi−1)), (4.1)
whereϕ(n)is the Eulerϕ-function and
n= s i=1
piri=priimi withmi= s j=1, j≠i
prjj. (4.2)
Letqbe a prime number such thatq2divides Disc(φn). Then there existsi∈ {1, . . . , s}such thatq=pi. We have ¯φn(X)=(φ¯mi(X))ϕ(prii ) (modpi), where ϕ(piri)≥2, and
Res
φmi, φn
=(−1)ϕ(mi)ϕ(n)Res
φn, φmi
=Res
φn, φmi
=piϕ(mi), (4.3)
and we obtain thatvpi(Res(φn, φmi))=deg(φmi(X)).
Now the result follows immediately fromTheorem 3.3andProposition 3.2.
4.2. Monogenicity of the fieldK=Q(α), withαa root ofP (X)=Xp−a Proposition4.2. Letαbe a root of the irreducible polynomialP (X)=Xp− a, whereais a squarefree integer andpis a prime number.
(i)Ifpdividesa, thenOK=Z[α]if and only ifais squarefree.
(ii)Ifpdoes not dividea, thenOK=Z[α]if and only ifais squarefree and vp(ap−1−1)=1.
Proof. We haveP (X)=Xp−a=Irrd(α,Q)and
Disc(P )=(−1)p((p−1)/2)NK/Q P(α)
=(−1)(3p2−p−2)/2p(ap)p−1. (4.4)
Ifpis odd, the only prime numbersqsuch thatq2divides Disc(P )arepand the prime divisors ofa. Ifp=2, then 2 is the only prime numberqsuch that q2divides Disc(P ).
Letqbe a prime number such thatq2divides Disc(P ). We have two cases:
(1) ifqdoes not dividea, then ¯P (X)=g(X)pinFp[X], withg(X)=X−a, and then Res(g, P )=P (a)=ap−a;
(2) ifq dividesa, then ¯P (X)=g(X)p in Fq[X], withg(X)=X and then Res(g, P )=P (0)= −a.
In both cases, the result is deduced fromTheorem 3.3.
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M. E. Charkani: Department of Mathematics, Faculty of Sciences Dhar-Mahraz, Uni- versity of Sidi Mohammed Ben Abdellah, BP 1796, Fes, Morocco
E-mail address:[email protected]
O. Lahlou: Department of Mathematics, Faculty of Sciences Dhar-Mahraz, University of Sidi Mohammed Ben Abdellah, BP 1796, Fes, Morocco
E-mail address:[email protected]