On the resolvent of an ideal and some applications

PII. S0161171203205378 http://ijmms.hindawi.com © Hindawi Publishing Corp.

ON THE RESOLVENT OF AN IDEAL

AND SOME APPLICATIONS

DRISS BOUZIANE and ABDELILAH KANDRI RODY

Received 5 May 2002

We give an algorithm to compute a resolvent of an algebraic variety without com-puting its irreducible components; we decompose the radical of an ideal into prime ideals and we test the primality of a regular ideal.

2000 Mathematics Subject Classification: 68W30.

1. Introduction. A fundamental construction in algebraic geometry is the decomposition of a variety into irreducible components; this is connected from commutative algebra viewpoint with the primary decomposition of ideals. The purpose of this paper is to study the structure of an affine varietyVdefined by a zero-set of a finite set of the polynomial ringK[x1, . . . , xn]. We characterize

the associated irreducible varieties ofVby a resolvent.

The concept of resolvent was introduced by Ritt [14] in his work on differ-ential algebra. He showed that generic zeros of a prime differdiffer-ential ideal are birationally equivalent to general zeros of one differential polynomial.

Based on Ritt-Wu’s algorithm to decompose a variety into irreducible vari-eties, Gao and Chou [8] extended Ritt’s concept of resolvent to an ideal, not necessarily prime, with respect to a parametric set. They use a factorization over a tower of algebraic extensions of the field of coefficients. In the ordinary differential case, Cluzeau and Hubert [6] extended also Ritt’s concept of resol-vent to regular differential ideals. We exploit the interplay between both results to compute a resolvent of an idealᏵwith respect to a parametric set. We use a decomposition of√Ᏽas an intersection of regular ideals, we compute a basis of√Ᏽ, and then we deduce the resolvent of the ideal√Ᏽ. The approach taken in this paper is interesting. Avoiding factorization, we compute irreducible va-rieties associated to a given affine variety and we check whether a regular ideal is prime.

2. Preliminaries and notation

2.1. Definitions and notation. LetK[x]=K[x1, . . . , xn]be the ring of

alge-braic polynomials innindeterminates with coefficients in a fieldKof charac-teristic zero. We fix an order on the indeterminates such thatx1≺ ··· ≺xn.

Letf be a polynomial not inK. The leading variable off is the highest inde-terminatexiappearing inf; it is denoted by lv(f ). Theinitialoff, init(f ), is

the coefficient of the highest power of lv(f )inf. Therank off, rank(f ), is the monomial lv(f )d_{, wheredis the degree offin ld(f ). Thetailoff, tail(f ),}

is the polynomialf−init(f )·lv(f )d_{. Theseparant off, sep(f ), is equal to}

∂f /∂v withv=lv(f ). We also definehf to be the product of theinitial and

theseparantoff.

LetΣbe a subset ofK[x]. We denote, respectively, by(Σ)and(Σ)the ideal and the radical ideal generated byΣ. An idealᏵis said to be radical if√Ᏽ=Ᏽ. A polynomialgis said to bereducedwith respect tofif the degree ofgin lv(f )is strictly less than the degree offin lv(f ).

Letf andgbe two elements ofK[x]. With a finite number of pseudodivi-sions, we can compute a polynomial rem(g;f )reduced with respect tof such that there existsα∈Nsatisfying

init(f )α_{·g≡rem(g;f )mod(f ). (2.1)}

Any order≺onxcan be extended to a partial order onK[x]as follows: forf andginK[x], we say thatfis less thang, and we writef≺gif either

(i) f∈Kandg∉K; (ii) lv(f )≺lv(g); or

(iii) lv(f )=lv(g)=vand degree(f , v) <degree(g, v).

If neitherf≺gnorg≺f, we say thatfandgare equivalent, we writef≡g. 2.2. Autoreduced sets. A subsetᏭofK[x]is called anautoreduced setif every element ofᏭis reduced with respect to the others. An autoreduced set is finite (see [13, page 77]). An autoreduced setᏭ= {A1, . . . , Ap}is denoted by A1, . . . , Ap ifA1≺ ··· ≺Ap. If Ꮽ=A1, . . . , Ap and Ꮾ=B1, . . . , Bq are two

autoreduced sets, we say thatᏭis less thanᏮand we writeᏭ≺Ꮾif either (i) there existsk≤min(p, q)such thatAi≡Bifori < kandAk≺Bk; or

(ii) p > qandAi≡Bifor 1≤i≤q.

If neitherᏭ≺Ꮾnor Ꮾ≺Ꮽ, we say that Ꮽ and Ꮾare equivalent, we write Ꮽ≡Ꮾ.

Remark _2.1. _{The order on the set of autoreduced sets is Artinian (well} ordering) (see [14, page 4] and [13, page 81]).

Proposition_2.2. _{LetᏭ=A1, . . . , Ap be an autoreduced set. Then for any}

polynomialf, there exist nonnegative integerαand a polynomialgreduced with respect toᏭsuch thatIα

Ꮽ·f≡gmod(Ꮽ), whereIᏭ is the product of the initials of elements inᏭ.

LetSbe a nonempty subset ofK[x]and letᏵbe an ideal ofK[x]. We define the saturation ofᏵ byS asᏵ:S∞= {f∈K[x]:h·f ∈Ᏽforha product of elements ofS}. It is also an ideal ofK[x]. WhenS is finite,Ᏽ:S∞is in fact equal to {f ∈K[x]| ∃α∈N, sα_{·f ∈Ᏽ}that is usually denoted byᏵ:s}∞_, wheresis the product of elements ofS.

Proposition_2.3. _{LetΣbe a nonempty subset ofK[x]. Letf1, . . . , fr∈K[x]}

and letSbe a finite subset ofK[x]. Then the following properties hold true:

(i) (Σ,ri=1fi)=ri=1

(Σ, fi);

(ii) (Σ):S∞_{=(Σ):s, wheresis the product of elements ofS.} Proof. _{See [1,9].}

2.3. Regular and characterizable ideal. LetᏭbe an autoreduced set with respect to some given order onx. LetHᏭ=IᏭSᏭ, whereIᏭandSᏭare, respec-tively, the product of initials and the product of separants of elements inᏭ. The autoreduced setᏭis said to be consistent if 1∉(Ꮽ):H_Ꮽ∞.

Definition_2.4. _{LetᏵbe an ideal ofK[x].}

(i) The idealᏵis said to be aregular ideal with respect to some order≺

on the variablesxif it is of the form(Ꮽ):H_Ꮽ∞, whereᏭis an autoreduced set with respect to the same order≺.

(ii) The idealᏵis said to be acharacterizable ideal with respect to some order≺onxif there is an autoreduced setᏭwith respect to the same order

≺such thatᏭis a characteristic set ofᏵandᏵ=(Ꮽ):H_Ꮽ∞. In this case we say thatᏭis a characteristic set.

Remark_2.5. _{(1) Every characterizable ideal is regular. The converse is false} (see [10, Example 2.5]).

(2) Every prime ideal is characterizable for any order on the variables. But a characterizable ideal is not necessarily prime.

(3) There exists some ideal that is characterizable with respect to some order onxbut not with respect to another order (see [10, Example 3.6]).

(4) There is an algorithm to decompose a radical of an ideal as intersection of characterizable ideals (see [5,10]).

(5) There is an algorithm to decompose a radical of an ideal as intersection of regular ideals (see [3]).

Proof. _{See [4,10].}

The following theorem gives a necessary and sufficient condition for a reg-ular ideal to be characterizable.

Theorem_2.7. _{LetᏭ=A1, . . . , Ap be an autoreduced set ofK[x]. The}

au-toreduced setᏭis a characteristic set if and only if

(i) init(Ai)is not a zero divisor modulo(Ꮽi−1):IᏭ∞i−1; (ii) sep(Ai)is not a zero divisor modulo(Ꮽi):I_Ꮽ∞_i, whereᏭi=A1, . . . , Ai.

Proof. _{See [1,5].}

2.4. Irreducible autoreduced set. LetᏭ=A1, . . . , Apbe an autoreduced set

ofK[x]andy=y1, . . . , ypare the leading variables of elements inᏭ, and let

u=u1, . . . , uqbe the other indeterminates that are present in the elements of

Ꮽ. The autoreduced setᏭbecomes an autoreduced set in the ringK[u, y].

Definition_2.8. _{An autoreduced setᏭis said to be irreducible if either} (i) p=1 andA1is irreducible inK(u)[y1]; or

(ii) A1, . . . , Ap−1isirreducibleandApis irreducible as a polynomial inyp

with coefficients considered in the quotient field ofK(u)[y1, . . . , yp−1]/ Ᏽp−1, whereᏵp−1=(Ꮽp−1):HᏭ∞p−1andᏭp−1=A1, . . . , Ap−1.

Proposition _2.9. _{LetᏭ=A1, . . . , Ap be an autoreduced set inK[x] such}

thatA1is irreducible inK(u)[y1]and for alli=2, . . . , p,Aihas degree one in its leading variable. ThenᏭis an irreducible autoreduced set.

Proposition_2.10. _{An irreducible autoreduced setᏭis a characteristic set}

of a prime ideal. This ideal is exactly the regular one associated toᏭ, that is,

(Ꮽ):H_Ꮽ∞.

2.5. The link between Gröbner basis and characteristic set. In this section, we recall an interesting result cited in [11,12], which says that we can extract a characteristic set from a lexicographical Gröbner basis.

Lemma _2.11. _{Let Ꮽbe an autoreduced set in K[x] and let Ᏽ be an ideal}

containingᏭ. ThenᏭis a characteristic set ofᏵif and only if for all nonzero polynomialfinᏵ,f is not reduced with respect toᏭ.

Proof. _{See [14, page 5].}

LetᏮ=B1≺ ··· ≺Brbe the reduced Gröbner basis (see [2] for more details

about this notion) ofᏵan ideal in K[x]with respect to the lexicographical term order such thatx1≺ ··· ≺xn.

LetC1=B1,C2=rem(Bi1;C1)withBi1 the first polynomial that contains a new variable not appearing inB1.

We continue this finite processes; we obtain a family of polynomials Ꮿ=

C1, . . . , Cs which is an autoreduced set with respect to the orderx1≺ ··· ≺xn.

Proposition_2.12. _{With the same notation as above,Ꮿis a characteristic}

set ofᏵ with respect to the orderx1≺ ··· ≺xn and is called the extracted characteristic set fromᏮ.

Proof. _{Letf∈Ᏽandf≠0; byLemma 2.11it is sufficient to show thatfis} not reduced with respect toᏯ. SinceᏮis a Gröbner basis ofᏵ, then there isBiin

Ꮾsuch that the leading lexicographical monomial ofBidivides some monomial

inf. So there existsCjinᏯsuch that deg(f ,lv(Cj))≥deg(Cj,lv(Cj)). Hence,

fis not reduced with respect toᏯ.

In what follows, autoreduced sets are supposed to be consistent.

3. Resolvent of an ideal. Gao and Chou have introduced in [8] the notion of resolvent for an arbitrary ideal with respect to a parametric set as a gener-alization of the one introduced by Ritt for a prime ideal; they have given an algorithm to compute a resolvent using the decomposition of a radical ideal into prime ideals. In this section, we recall some definitions and properties about this notion. Then we give an algorithm to compute a resolvent of an idealᏵusing a decomposition of√Ᏽinto regular ideals without using factor-ization over a tower of algebraic extensions.

Definition_3.1. _{LetᏵbe an ideal ofK[x]. A subsetu=u1, . . . , uqof{x1, . . . ,} xn}is said to be a parametric set ofᏵif K[u]∩Ᏽ=(0)and for every y∈ {x1, . . . , xn}\{u1, . . . , uq},K[u, y]∩Ᏽ≠(0).

The set of nonleading variables of elements in an autoreduced setᏭis called the parametric set ofᏭ.

Remark_3.2. _{LetᏭbe an irreducible autoreduced set with the parametric} setu. Thenuis a parametric set of the prime ideal(Ꮽ):H_Ꮽ∞.

Lemma_3.3. _{Letᏼbe a prime ideal with a parametric setuand letf∉ᏼ.}

Then(ᏼ, f )∩K[u]≠(0).

Proof. _{See [14].}

Lemma_3.4. _{LetᏭbe an autoreduced set inK[x]withuthe parametric set}

ofᏭ. Thenuis a parametric set of(Ꮽ):H_Ꮽ∞.

Proof. _{By Lazard’s lemma, each minimal prime component of(Ꮽ):H}∞ Ꮽ hasuas a parametric set. Then(Ꮽ):H_Ꮽ∞is a decomposition of prime ideals such that each one has u as a parametric set. So u is a parametric set of (Ꮽ):H_Ꮽ∞.

Lemma_3.5. _{LetᏵbe an ideal ofK[x]=K[u, y]havingu=u1, . . . , uqas a}

by replacing eachyiby a new variablezi. Consider᏶the ideal generated byᏵ andᏵinK[u, y, z]. Then᏶hasuas a parametric set.

Proof. _{Leth∈᏶∩K[u], sincehis independent of y and z, then if we} replacezi’s by theyi’s, we obtain thathis inᏵ, hence᏶∩K[u]=(0). Since

Ᏽ⊆᏶, then for alli=1, . . . , p, we have᏶∩K[u, yi]≠(0)and᏶∩K[u, zi]≠(0).

In the following theorem proved in [7] for a prime ideal, we extend the same result for an arbitrary ideal.

Theorem_{3.6(Ritt’s theorem)}. _{LetᏵbe an ideal ofK[u, y]withu=u1, . . . ,} uqas a parametric set. Then there existG∈K[u]\{0}and integersM1, . . . , Mp such that two distinct zeros ofᏵwith theutaking the same values for whichG

does not vanish give different values forQ=M1y1+···+Mpyp.

Proof. _{Let᏶=(Ᏽ,Ᏽ)be the ideal defined inLemma 3.5.} (a) Let᏶=᏶1∩···∩᏶tbe the decomposition of

᏶into prime ideals. (A) If, for somej∈ {1, . . . , t},uis not a parametric set of᏶j, then there

ishj(u)∈᏶j∩K[u]andhj≠0.

(B) If, for somej, the ideal(y1−z1, . . . , yp−zp)⊆᏶j anduis a

para-metric set of᏶j, then we puthj=1.

(C) If, for somej, the ideal(y1−z1, . . . , yp−zp)is not a subset of᏶j

anduis a parametric set of᏶j. Then there existsk∈ {1, . . . , p}such

that yk−zk∉᏶j. Since᏶j is a prime ideal anduis a parametric

set of᏶j, then byLemma 3.3,(᏶j, yk−zk)∩K[u]≠(0), hence there

existsh_j(u)∈(᏶j, yk−zk)∩K[u],hj(u)≠0.

The cases (A), (B), and (C) exhaust all possibilities.

(b) Letj1, . . . , jsbe such thatuis a parametric set of᏶jand(y1−z1, . . . , yp−

zp)is not a subset of᏶jfor allj∈ {j1, . . . , js}. Then there exist integers

M1, . . . , Mp such that ¯c=M1(y1−z1)+ ··· +Mp(yp−zp)∉᏶j for all

j ∈ {j1, . . . , js}. Consequently, by Lemma 3.3,(᏶j,c)¯ ∩K[u]≠(0). We

puth=h1···hs withhj∈(᏶j,c)¯ ∩K[u],hj ≠0.

(c) LetGbe the product ofhj(case (A)), ofhj(cases (B) and (C)), and ofh.

Let(¯u, y)and(u, y¯ )be two distinct zeros ofᏵ, then(¯u, y, y)is a zero of ᏶.

We assume that pi=1Mi(yi−yi)=0; there are three cases to be

distin-guished:

(i) (u, y¯ , y)is a zero of some᏶jsatisfying case (A), thenhj(¯u)=0, and

henceG(u)¯ =0;

(ii) (u, y¯ , y)is a zero of some᏶j satisfying case (B), theny=y and

hence(¯u, y)=(¯u, y). This case is impossible; (iii) (u, y¯ , y)is a zero of some᏶jsatisfying case (C).

for allk∈ {1, . . . , p}, then(u, y¯ , y)will be a zero of(᏶j,c), hence¯ h(¯u)=0,

and thenG(u)¯ =0.

Lemma_3.7. _{LetᏵ1 andᏵ2 be two ideals in K[x], ωa new variable, and} Q=pi=1Miyi, where theMi’s are integers. Then(Ᏽ1∩Ᏽ2, ω−Q)=(Ᏽ1, ω− Q)∩(Ᏽ2, ω−Q).

Proof. _{It is sufficient to prove the indirect inclusion.}

Let f be in (Ᏽ1, ω−Q)∩(Ᏽ2, ω−Q), then f =si=1λifi+h1(ω−Q)= t

j=1µjgj+h2(ω−Q), where h1, h2,λi,µj are inK[x, ω]and fi∈Ᏽ1 and gj∈Ᏽ2fori=1, . . . , sandj=1, . . . , t.

We can consider theλi’s and theµj’s as free fromωbecause otherwise we

reduce these polynomials with respect toω−Q, thensi=1λifi−tj=1µjgj=

(h2−h1)(ω−Q).

Since the left-hand side is free ofω, thensi=1λifi= t

j=1µjgj, andfis in

(Ᏽ1∩Ᏽ2, ω−Q).

Lemma_3.8. _{LetᏵbe a prime ideal inK[u, y]such thatuis a parametric}

set,ωa new variable, andQ=pi=1Miyi, where theMi’s integers. Then the ideal(Ᏽ, w−Q)is prime and hasuas a parametric set.

Proof. _{See [14, page 40].}

Lemma_3.9. _{LetᏵbe an ideal ofK[u, y]havinguas a parametric set. Letω}

be a new variable andQ=pi=1Miyiwith theMi’s integers. Then

(Ᏽ, w−Q)=

(√Ᏽ, w−Q)anduis a parametric set of the ideal᏶=(Ᏽ, w−Q)inK[x, ω].

Proof. _{For the first point, letf∈(Ᏽ, w−Q), then there existsα∈Nsuch} thatfα_{∈(Ᏽ, ω−Q), hencef}α_{=g+h(ω−Q), whereg∈(Ᏽ)inK[x, ω]and}

h∈K[x, ω]. We reducefandgwith respect toω−Q, then we obtainf=f1+

f2(ω−Q)andg=g1+g2(ω−Q)withf1∈K[x],g1∈Ᏽ, andf2, g2∈K[x, ω]. Consequently,fα_=fα

1+F (ω−Q)=g1+g2(ω−Q)for someF∈K[x, ω], then f1α=g1∈Ᏽ; this implies thatf∈(

√

Ᏽ, ω−Q).

For the second property, we show that᏶∩K[u]=(0)and᏶∩K[u, ω]≠(0). Firstly, we assume that᏶∩K[u]≠(0), then there exist a nonzero polyno-mial P (u) in ᏶∩K[u]; this implies that there exist f , f1, . . . , fr ∈ K[x, ω]

and g1, . . . , gr ∈Ᏽ such that P (u) = ri=1figi+f (w−Q); for ω =Q, we

will have P (u)∈ K[u]∩Ᏽ, but this is impossible because u is a paramet-ric set ofᏵ. Secondly, we decompose √Ᏽ into prime ideals such that √Ᏽ=

(si=1mi)∩(tj=1pj), where themi’s are the prime components havinguas a

parametric set and thepj’s are those satisfyingpj∩K[u]≠(0). ByLemma 3.8,

we have(mi, ω−Q)∩K[u, ω]≠(0), then by the first property andLemma 3.7,

we deduce thatuis a parametric set of᏶.

Theorem_3.10. _{LetᏵbe an ideal ofK[u, y]withu a parametric set. Let}

w be a new variable, M1, . . . , Mp integers satisfying the Ritt’s theorem, and

Q=pi=1Miyi. Let᏶=(Ᏽ, w−Q). Then

R(u, w), R1(u, w, y1), . . . , Rp(u, w, yp)with respect to the orderu1<···< uq<

w < y1<···< ypsuch thatdeg(Rj, yj)=1forj=1, . . . , p.

Proof. _{See [8, page 7].}

Corollary_3.11. _{LetᏵbe an ideal ofK[u, y]withua parametric set. Let} w, λ1, . . . , λp be indeterminates and Q=pi=1λiyi. Then᏶=(Ᏽ, w−Q)has

u as parametric set and᏶has a characteristic set of the form R(λ, u, w), R1(λ, u, w, y1), . . . , Rp(λ, u, w, yp) with respect to the order λ1 ≺ ··· ≺λp ≺

u1<···< uq< w < y1<···< ypsuch thatdeg(Rj, yj)=1forj=1, . . . , p.

Proof. _{For showing thatuis a parametric set of᏶, we use the same proof} as the one given forLemma 3.9.

It is clear thatλ1, . . . , λp satisfy the Ritt’s theorem for the idealᏵ, then by

Theorem 3.10theRi’s are linear in their leading variables.

Definition_3.12. _{The polynomialRdefined inTheorem 3.10is said to be} a resolvent ofᏵwith respect to the parametric setu.

4. Computation of the resolvent

4.1. Computation of a basis of the radical of an ideal

Lemma_4.1. _{LetᏭbe an autoreduced set inK[x]andza new indeterminate.}

Then(Ꮽ):H_Ꮽ∞=(Ꮽ, zHᏭ−1)∩K[x].

Proof. _{For the direct inclusion, letf ∈(Ꮽ):H}∞

Ꮽ, then there existsα∈N such thatH_Ꮽαf∈(Ꮽ), hence(zHᏭ)αf∈(Ꮽ)inK[x, z]sincef=f+f (zHᏭ)α− f (zHᏭ)α, thereforef∈(Ꮽ, zHᏭ−1)∩K[x]. For the other inclusion, we take f∈(Ꮽ, zH_Ꮽ−1)∩K[x], thenf will be a linear combination of elements inᏭ and the polynomialzH_Ꮽ−1 with coefficients inK[x, z]sincefis independent ofz, so if we replacezby 1/HᏭ, we will obtain thatf∈(Ꮽ):HᏭ∞.

Remark_4.2. _{ByLemma 4.1, one can compute a basis of the ideal(Ꮽ):H}∞ Ꮽ using Gröbner basis.

Lemma_4.3. _{LetᏵbe an ideal inK[x]. Then there is an effective method to}

compute a basis of√Ᏽ.

Proof. _{We decompose}√_{Ᏽinto regular ideals as}√_Ᏽ=s

i=1((Ꮽi):H_Ꮽ∞_i)(see

Remark 2.5). ByLemma 4.1, we can determine a basis of each(Ꮽi):H_Ꮽ∞_i. So, by

the use of Gröbner-basis computation, we deduce a basis of√Ᏽ.

Lemma_4.4. _{LetᏭbe an autoreduced set inK[x],uthe parametric set ofᏭ,}

ωa new variable,M1, . . . , Mpintegers satisfying Ritt’s theorem, andQ= p

i=1.

Then the ideal(Ꮽ, ω−Q):H_Ꮽ∞ is a characterizable ideal with respect tou≺

ω≺y.

Proof. _{See [6].}

Lemma_4.5. _{LetᏭbe an autoreduced set withuthe parametric set andy}

the other variables. ThenᏭis a characteristic set inK[u, y]if and only ifᏭis a characteristic set inK(u)[y].

Proof. _{See [10].}

Theorem_4.6. _{LetᏭbe an autoreduced set inK[u, y]withuthe parametric}

set. Then there is an algorithm to compute a resolvent of(Ꮽ):H_Ꮽ∞.

Proof. _{Letλ1, . . . , λpbe new variables. Theλi’s satisfy the Ritt’s theorem for} (Ꮽ):H_Ꮽ∞, then, byLemma 4.4,(Ꮽ, w−pi=1λiyi):H∞_Ꮽ is characterizable with

respect toλ≺u≺w≺yand has a characteristic set of the formR, R1, . . . , Rp

such that deg(Ri, yi)=1 (byCorollary 3.11). LetD=II1···IpwithI, I1, . . . , Ip,

respectively, the initials ofR, R1, . . . , Rp. LetM1, . . . , Mp be integers such that

D(M, u, w) ≠0. After substituting the Mi’s in λi’s, we obtain R, R1, . . . , Rp

which will be a characteristic set of ((Ꮽ) : H_Ꮽ∞, ω−pi=1Miyi) verifying

deg(R_i, yi)=1.

The polynomialRis a resolvent of(Ꮽ):H_Ꮽ∞with respect to the parametric setu.

Algorithm_4.7. _{LetᏭ be an autoreduced set inK[u, y], whereuis the} parametric set.

Step₁. _{Letλ1, . . . , λpandzbe new variables. We computeᏮ, the Gröbner} ba-sis inK(λ, u)[w, y, z], of the zero-dimensional ideal(Ꮽ, zHᏭ−1, w−pi=1λiyi)

with respect to a lexicographical order such thatw≺y≺z.

Step₂. _{Let Ꮿ:=C, C1, . . . , Cpbe the extracted characteristic set fromᏮ∩} K(λ, u)[w, y].

Step₃. _{Let᐀:=T , T1, . . . , Tpbe obtained fromᏯby clearing out the} denom-inators.

Step₄. _{Let᏾:=R, R1, . . . , Rp be obtained from᐀ by replacing theλi’s by} Mi’s such thatD(M, u, y)≠0, whereD=II1···IpwithI, I1, . . . , Ip, respectively,

the initials ofT , T1, . . . , Tp.

The polynomialRis a resolvent of(Ꮽ):H_Ꮽ∞with respect to the parametric setu.

Correctness_4.8. _{It is a consequence of Lemmas4.4,4.5andTheorem 4.6.}

4.3. Computation of the resolvent of an ideal

Step_{1(the computation of a parametric set ofᏵ)}. _{We decompose}√_Ᏽinto regular idealssi=1(Ꮽi):HᏭ∞i. The parametric set ofᏭisuch that|Ꮽi|is minimal

is a parametric set ofᏭ.

Step₂. _{We compute a basisGof}√_Ᏽ. Step₃. _{Letω, λ1, . . . , λpbe new variables.}

We compute the Gröbner basisᏮof the ideal(√Ᏽ, ω−pi=1λiyi)=(G, ω− p

i=1λiyi)with respect to the lexicographical order term satisfyingλ≺u≺

ω≺y.

LetᏯbe the extracted characteristic set fromᏮ.

The autoreduced setᏯhas the formR, R1, . . . , Rpsuch that deg(Ri, yi)=1.

Step ₄. _{Let D = II1···Ip with I, I1, . . . , Ip, respectively, the initials of} R, R1, . . . , Rp.

LetM1, . . . , Mpbe integers such thatD(M, u, w)≠0.

LetR, R1, . . . , Rpbe obtained fromR, R1, . . . , Rpafter substituting theMi’s in

λi’s.

The autoreduced setR, R₁, . . . , Rp is a characteristic set of( √

Ᏽ, ω−pi=1Miyi)

verifying deg(R_i, yi)=1.

The polynomialRis a resolvent ofᏵwith respect to the parametric setu. Correctness_4.10. _{It is a consequence ofProposition 2.12,Corollary 3.11,} andLemma 4.3.

5. Applications. The resolvent has a wide range of applications, namely it transforms a set of polynomial equations to a single polynomial equation such that their varieties are birationally equivalent, it permits to compute a primitive element for a finitely generated algebraic extension over a field of characteristic zero, and obviously it has other areas of applications.

In this section, we show how the resolvent can be used to decompose a variety into irreducible varieties and how to test that a variety associated to a regular ideal is irreducible.

5.1. Decomposition of a variety into irreducible varieties. LetᏭ=A1, . . . , Apbe an autoreduced set,M1, . . . , Mpintegers satisfying Ritt’s theorem for the

regular ideal(Ꮽ):H_Ꮽ∞, andwa new indeterminate. Put᏶=(Ꮽ, w−Q):H_Ꮽ∞, whereQ=M1y1+ ··· +Mpyp. We know that᏶is characterizable and has a

characteristic set of the form᏾=R, R1, . . . , Rp, where eachRi is linear inyi.

We can assumeRsquare free because᏶is radical.

Lemma_5.1. _{With the same notations as above, the following properties hold:} (1) ᏶=(Ꮾ):H_Ꮾ∞;

(2) ᏶∩K[u, y]=(Ꮽ):H_Ꮽ∞.

Proof. _{(1) It is a corollary ofLemma 4.4.}

existsh∈(w−Q)such thatfr_{=g+hforr∈N. Let ¯g=rem(g;w−Q); we}

obtainfr_=g¯+h¯_{for ¯h∈(w−Q). Sincef}r _{is free ofw, thenf}r_{=g, hence¯}

fr_∈(Ꮽ):H∞

Ꮽ, thereforef∈(Ꮽ):HᏭ∞because(Ꮽ):HᏭ∞is perfect by Lazard’s lemma.

Theorem_5.2. _{With the same notations as inLemma 5.1. LetR=B1···Bs}

be the factorization of R into irreducible polynomials in K[u][ω] andᏮi =

Bi, R1, . . . , Rpfori=1, . . . , s. Then,(Ꮽ, ω−Q):H_Ꮽ∞=si=1(Ꮾi):H_Ꮾ∞_i and each

(Ꮾi):H_Ꮾ∞_i is a prime ideal withᏮia characteristic set.

Proof. _{Since the Mi’s satisfy The Ritt’s theorem, then, by Lemma 4.4,} (Ꮽ, ω−Q):H_Ꮽ∞is a characterizable ideal with respect tou≺ω≺y and by Lazard’s lemma the ideal(Ꮽ, ω−Q):H_Ꮽ∞is radical and is equal to(᏾᏿):H_᏾᏿∞ with᏾᏿=R, R1, . . . , Rp, then(Ꮽ, ω−Q):H_Ꮽ∞=(᏾᏿):H_᏾᏿∞ =

((᏾᏿):H_᏾᏿∞)=

(᏾᏿):H_᏾᏿ =si=1

(Bi, R1, . . . , Rp):H_᏾᏿∞ (by Proposition 2.3). To finish the

proof, it is sufficient to show that(Bi, R1, . . . , Rp):H᏾᏿=(Ꮾi):HᏮi.

For this letf∈(Bi, R1, . . . , Rp):H᏾᏿, then

f·H_᏾᏿∈Bi, R1, . . . , Rp. (5.1)

We haveH_᏾᏿=hR·H᏾᏿=I·S·H_᏾᏿, where᏾᏿_{=R1, . . . , R}

p,I=init(R)and

S=sep(R). We have sep(R)=∂R/∂ω=∂(B1···Bs)/∂ω=sk=1(Sk·j≠kBj),

whereSk=sep(Bk). Furthermore, init(R)= s

k=1Ik, whereIk=init(Bk). Since

all terms insk=1(Sk·j≠kBj), exceptSi·j≠iBj, are in

(Bi, R1, . . . , Rp), then

f·H_᏾᏿∈Bi, R1, . . . , Rp

⇒f·init(R)·Si·

j≠i

Bj·H᏾᏿∈Bi, R1, . . . , Rp

⇒f· j≠i

IjBj

∈Bi, R1, . . . , Rp

:H_Ꮾi

since init(R)= s

k=1

Ik, HᏮi=IiSiH᏾᏿

⇒f· j≠i

IjBj

∈Bi, R1, . . . , Rp

:H_Ꮾ∞_i

(5.2)

(the last implication follows byProposition 2.3and Lazard’s lemma).

The autoreduced setᏮi is irreducible, then it is a characteristic set of the

prime ideal(Ꮾi):HᏮ∞i. Forj≠i,Bj∉(Ꮾi):H

∞

Ꮾi because otherwiseBiwill be

equal toBj. Sof∈(Ꮾi):H_Ꮾ∞_i.

In the following, we will illustrate how one obtains the decomposition of (Ꮽ):H_Ꮽ∞into prime ideals from the decomposition of the characterizable ideal ᏶=(Ꮽ, w−Q):H_Ꮽ∞. Forj=1, . . . , s, let ¯Ᏸj=D1, . . . , Dp, ¯Djbe a characteristic

Lemma_5.3. _{With the same notations as above, the ideal(Ꮾj):H}∞

Ꮾj∩K[u, y] hasᏰ˜j=D1, . . . , Dpas a characteristic set with respect tou≺yand it is equal to(_Ᏸ˜_j):H_Ᏸ∞_˜

j forj=1, . . . , s.

Proof. _{We have thatᏮjis irreducible byProposition 2.9, then(Ꮾj):H}∞ Ꮾj

is a prime ideal and also(Ꮾj):HᏮ∞j∩K[u, y].

Letf be in(Ꮾj):H∞_Ꮾj∩K[u, y]and ¯f=rem(f; ˜Ᏸj). Sincef is inK[u, y],

then rem(f; ˜Ᏸj)=rem(f; ¯Ᏸj)=0 because ¯Ᏸj is a characteristic set of(Ꮾj):

H_Ꮾ∞_j. It follows that(Ꮾj):HᏮ∞j∩K[u, y]=(Ᏸ˜j):H

∞ ˜

Ᏸjbecause it is a prime ideal

and ˜Ᏸjis its characteristic set.

The following proposition is the aim result; it gives the decomposition of (Ꮽ):H_Ꮽ∞into prime ideals that each one is given by its characteristic set ˜Ᏸj.

Proposition_5.4. _{With the same notations as above,}

(Ꮽ):H_Ꮽ∞= s

j=1

_˜

Ᏸj

:H_Ᏸ∞_˜

j. (5.3)

Proof. _{This result is a corollary ofLemma 5.1andTheorem 5.2.}

Remark_5.5. _{For the radical of an idealᏵ=(Σ), we firstly decompose}√_Ᏽ into regular ideals (see [3,4,5,9]), and afterwards, using the techniques above, decompose each regular ideal into prime ideals.

5.2. Test of the primality of a regular ideal

Proposition_5.6. _{LetᏵ=(Ꮽ):H}∞

Ꮽ withᏭan autoreduced set havinguas the parametric set andRa resolvent ofᏵwith respect tou. ThenRis irreducible overK(u)if and only if√Ᏽis a prime ideal.

Proof. _{LetM1, . . . , Mpbe integers satisfying the Ritt’s theorem for the ideal} Ᏽwith respect to the parametric setu, then there existR1, . . . , Rplinear in their

leading variables such thatR, R1, . . . , Rpis a characteristic set of

(Ᏽ, w−Q), whereQ=M1y1+ ··· +Mpyp, hence, byLemma 4.4,

(Ᏽ, w−Q)=(᏾):H_᏾∞, where᏾=R, R1, . . . , Rp. This implies, byProposition 2.9andLemma 5.1, that

Ris irreducible overK(u)if and only ifᏵis prime.

6. Examples

Example_6.1. _{LetᏵ be the ideal, in the ringK[x, y, z], generated by the} following polynomials:

f1:=y6_−2x5_y3_+x10_, f2:=x2_y3_z−x7_z−y5_+x5_y2_, f3:=y4_z−x5_yz−x3_y3_+x8_, f4:=x4z2−2x2y2z+y4.

Firstly, we compute the decomposition of√Ᏽinto regular ideals; we obtain

Ᏽ:=Ᏽ1∩Ᏽ2, Ᏽ1:=x2z−y2, y3−x5:(x, y)∞,Ᏽ2:=(x, y). (6.2)

We remark thatxis a parametric set ofᏵ.

We verify thatM1=1,M2=1 are integers satisfying Ritt’s theorem for the ideal Ᏽ1 with respect to the parametric set x (that is two distinct zeros of Ᏽ1, with the x taking the same value, give different values for Q=M1y+

M2z=y+z), and so we obtain a characteristic setR, R1, R2of(Ᏽ1, w−y−z), satisfying the definition of the resolvent, that is, deg(R1, y)=deg(R2, z)=1, whereR:= −3x3_w+w3_−x5_−x4_,R1:=(−x+x2_)y+w2_−x2_w−2x3_{, and} R2:=(−x+x2_)z+xw−w2_+2x3_{. SinceRis irreducible, then}√_Ᏽ:=Ᏽ

1∩Ᏽ2 is the decomposition into prime ideals. The polynomialxRis a resolvent ofᏵ with respect to the parametric setx.

Example_6.2. _{LetᏭ:=A1, A2be an autoreduced set inK[x, y, z]withA1:=} y2_{−(1+x)y+xandA2:=z}2_{−(3+x)z+3x.}

The autoreduced setR, R1, R2is a characteristic set of((Ꮽ):H_Ꮽ∞, w−y−z) with respect tox≺w≺y≺z, whereR:=24x+w4_+5w2_x2_−4w3_x+32x2₋ 2x3_w+8x3_−8w3_+19w2_{−54xw−28x}2_w−12w+28xw2_{,R1:=(−4+2x)y−} 8x2_+15xw−6w2_−14x+9w+w3_−3xw2_+2x2_{w,R2:=(−4+2x)z+3xw}2₋ 2x2_w−w3_+6w2_+14x+8x2_{−17xw−5w,Ris a resolvent of(Ꮽ):H}∞

Ꮽ with respect to the parametric setx, andR:=(w−4)(w−2x)(w−x−3)(w−x−1) is the factorization ofRoverK[u].

We put B1:= w−4, B :=w−2x, B3 := w−x−3, and B4:= w−x−1. Then ((Ꮽ): H_Ꮽ∞, w−y−z) := 4

i=1Ꮾi, where Ꮾi := ((Bi, R1, R2): H∞) and

H:=init(R1)init(R2):=(4−2x)2_.

Changing of the order on the variables, we obtain

(Ꮽ):H∞_Ꮽ :=(z−3, y−1)∩(z−x, y−x)∩(z−x, y−1)∩(z−3, y−x). (6.3)

7. Conclusion. We have developed an algorithm to compute a resolvent of an algebraic variety. No factorization is needed. Some of the main problems in polynomial ideal theory can be solved by means of the resolvent. We com-pute the irreducible varieties associated to a given affine variety, we test the primality of a regular ideal.

The algebraic complexity of the resolvent and the computational complex-ity of the associated algorithms have been explicitly explored by Gallo and Mishra [7].

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[11] A. Kandri Rody,Effective methods in the theory of polynomial ideals, Ph.D. thesis, Rensselaer Polytechnic Institute, New York, 1984.

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Driss Bouziane: Département de Mathématiques, Faculté des Sciences Semlalia, Uni-versité Cadi Ayyad, BP 2390, Marrakech, Morocco

E-mail address:[email protected]

Abdelilah Kandri Rody: Département de Mathématiques, Faculté des Sciences Semlalia, Université Cadi Ayyad, BP 2390, Marrakech, Morocco