# Relations of the Algebra of Polynomial Integrodifferential Operators

Authors: Andre S. E. Mialebama Bouesso; G. Mobouale Wamba
Journal: Journal of Mathematics
 (2019)

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## Abstract

We revisit the algebra of polynomial integrodifferential operators and we give a generalization of its relations. Generators of prime ideals of height 1, of the maximal ideal, and of the smallest ideal ofIn are discussed. A technique for obtaining a generating set for a given two-sided ideal of In is also discussed.


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## Body

## 1. Introduction

The algebra of polynomial integrodifferential operators was introduced in [1] and was studied extensively in [2–6]. Early in 2009, the author of [1] introduced the Jacobian algebra (see [7]) which has similar properties to the algebra In of polynomial integrodifferential operators and both algebras are ideally equivalent. In 2010, the same author introduced the algebra Sn (see [8]) of one-sided inverses of a polynomial algebra which is a generalized Weyl algebra as well as the Jacobian algebra An and the algebra of polynomial integrodifferential operators In. As applications, the algebra In of polynomial integrodifferential operators has interesting applications in computer algebra as described in [9, 10]. Our goal in this work is to firstly study new relations of the algebra of polynomial integrodifferential operators in order to facilitate the use of this algebra in a computational point of view. Our starting point is to prove some relations introduced in [1] in Proposition 4. What is more, we give a generalization of these relations in Corollary 7 and then we give new other relations in Lemma 10. In the second part of this paper we study generators of the ideals of In. Prime ideals of height 1 listed in [1] are revisited in Theorem 11, sets of generators of the only maximal ideal and of the smallest ideal of In are studied, and a given two-sided ideal of In is also discussed. The relations discussed in this paper can be used to revisit the special case of In, and some results were obtained in the general case of Generalized Weyl algebras especially those described in [11–16].


## 2. Relations ofIn

LetK be a field and Pn=K[x1,…,xn] be a polynomial ring over K. Denote ∂i=∂/∂xi∀i=1,…,n.Definition 1.
By a Weyl algebra we mean the free associative algebraAn=K〈x1,…,xn,∂1,…,∂n〉 satisfying [xi,xj]=[∂i,∂j]=[∂i,xj]=0∀i≠j and ∂ixi=xi∂i+1 where [a,b]=ab-ba.Definition 2.
By the algebra of polynomial integro-differential operators we mean the algebraIn generated by the Weyl algebra An and endomorphisms (1)∫i​:Pn→Pn,∀i=1,…,nxα=x1α1⋯xnαn↦αi+1-1xi·xα i.e (2)In=Kx1,…,xn,∂1,…,∂n,∫1​,…,∫n​.Remark 3.
Observe that fori=1,…,n(3)∂i:Pn→Pnx1α1⋯xnαn↦αix1α1⋯xiαi-1⋯xnαnifαi≠0.∫i​:Pn→Pn,x1α1⋯xnαn↦xiαi+1x1α1⋯xiαi⋯xnαn.
SetHi=∂ixi=xi∂i+1; regarding 1 as the identity map of Pn, we have the endomorphism(4)Hi:Pn→Pn,x1α1⋯xnαn↦αi+1x1α1⋯xiαi⋯xnαnHi-1:Pn→Pn,x1α1⋯xnαn↦1αi+1x1α1⋯xiαi⋯xnαn.
Denotingxα=x1α1⋯xnαn, we see that ∫ixα=xiHi-1(xα); that is, ∫iHi=xi. From here, we recall Proposition 2.2 introduced in [1].Proposition 4 (see Proposition 2.2, [1]).
(1)
The algebraIn is generated by the elements {∂i,∫i​,Hi∣i=1,…,n} that satisfy the following defining relations: ∀i,
(a)
∂ i ∫ i ​ = 1; (b)
[ H i , ∫ i ​ ] = ∫ i ​; (c)
[ H i , ∂ i ] = - ∂ i; (d)
H i ( 1 - ∫ i ∂ i ) = ( 1 - ∫ i ∂ i ) H i = ( 1 - ∫ i ∂ i ); (e)
Fori≠j, we have aiaj=ajai∀ak∈{Hk,∫k,∂k}. (2)
The algebraIn=⨂i=1nD1(i)(σi,1)=Dn((σ1,…,σn),(1,…,1)) is a Generalized Weyl algebra where Dn=⨂i=1nD1(i),D1(i)=K[Hi]⊕(⨁j∈Nejj(i)) and Hiejj(i)=Hiejj(i)=(j+1)ejj(i). Moreover, the algebra In=⨁α∈ZnIn,α is Zn-graded where In,α=vα,+Dnvα,-, where vα,+=∏αi>0vαi and vα,-=∏αi<0vαi such that(5)vji=∫ij​ifj>01ifi=j∂i-jifj<0. (3)
Each element ofIn is a unique finite sum a=∑α∈Znvα,+aαvα,- for unique elements aα∈Dn.Proposition5 proves the first item of Proposition 4 left in [1].Proposition 5.
The algebraIn=K〈∂i,∫i​,Hi∣i=1,…,n〉 satisfies the following relations: ∀i=1,…,n(1)
∂i∫i​=1;(2)
[Hi,∫i​]=∫i​;(3)
[Hi,∂i]=-∂i;(4)
Hi∫i∂i=∫i∂iHi(5)
Hi(1-∫i∂i)=(1-∫i∂i)Hi=(1-∫i∂i);(6)
Fori≠j, we have aiaj=ajai∀ak∈{Hk,∫k​,∂k}.Proof.
(1)
∂ i ∫ i ​ = ∂ i ( x i H i - 1 ) = ( ∂ i x i ) H i - 1 = H i H i - 1 = 1. (2)
By definition(6)Hi,∫i​=Hi∫i​-∫iHi=xi∂i+1∫i​-∫iHi Thus,[Hi,∫i​]=xi(∂i∫i​)+∫i​-Hi∫i​=xi+∫i​-xi=∫i​. (3)
[ H i , ∂ i ] = H i ∂ i - ∂ i H i = ∂ i ( x i ∂ i ) - ∂ i ( x i ∂ i + 1 ) = - ∂ i. (4)
[ H i , ∫ i ∂ i ] = H i ∫ i ∂ i - ∫ i ∂ i H i = ( x i ∂ i + 1 ) ∫ i ∂ i - ∫ i ∂ i ( x i ∂ i + 1 ) = x i ∂ i - ( ∫ i H i ) ∂ i = 0. (5)
H i ( 1 - ∫ i ∂ i ) = H i - H i ∫ i ∂ i = H i - ∫ i ∂ i H i = ( 1 - ∫ i ∂ i ) H i. Observe that (7)1-∫i∂iHi=xi∂i+1-∫i∂ixi∂i-∫i∂i=xi∂i+1-xi∂i-∫i∂i=1-∫i∂i since∫i​=xiHi-1. (6)
Applying Remark3, where Hi=xi∂i+1 and ∂jxi=xi∂j for all i≠j, then we have [Hi,∂j]=Hi∂j-∂jHi=(xi∂i+1)∂j-∂j(xi∂i+1)=xi∂i∂j-(∂jxi)∂i=xi∂i∂j-xi∂j∂i. Using now Definition1, we see that (8)Hi,∂j=Hi∂j-∂jHi=0.Settingeij(k)=∫ki∂kj-∫ki+1∂kj+1, we easily see that eij(k)=∫kie00(k)∂kj where e00(k)=1-∫k∂k.Example 6 (construction ofI2).
Let us consider the polynomial ringP=C[x,y] in two indeterminates. By a monomial in P we mean an element of the form xαyβ where α,β∈N, and by a polynomial we mean a finite linear combination of monomials. Consider the following operators: O1
(9)∫x​:P→Pxαyβ↦α+1-1xα+1yβ;O2
(10)∫y​:P→Pxαyβ↦β+1-1xαyβ+1;O3
(11)∂x:P→Pxαyβ↦α·xα-1yβ;O4
(12)∂y:P→Pxαyβ↦β·xαyβ-1;O5
(13)Hx=x∂x+1:P→Pxαyβ↦xα·xα-1yβ+xαyβ=α+1xαyβ;O6
(14)Hy=y∂y+1:P→Pxαyβ↦yβ·xαyβ-1+xαyβ=β+1xαyβ.The algebra I2 of polynomial integrodifferential operators is the free associative algebra generated by the operators defined in O1,...,O6 with coefficients in C and multiplication given by the composition of applications. The relations between operators are given by Proposition 5.With notations as above we have the following result.Corollary 7.
Fork=1,…,n we have (1)
Hkl∂kj=∂kj(Hk-j)l;(2)
∂kjHkl=(Hk+j)l∂kj;(3)
∫klHkj=(Hk-l)j∫kl​;(4)
Hkj∫kl​=(Hk+l)j∫kl​;(5)
Hklejj(k)=ejj(k)Hkl=(j+1)lejj(k);(6)
Hkleij(k)=(i+1)leij(k);(7)
eij(k)Hkl=(j+1)leij(k).Proof.
(1) Forl=1 we have (15)Hk∂kj=Hk∂k∂kj-1.
By Proposition5 we have Hk∂k=∂kHk-∂k. Replacing this in the previous expression we have(16)Hk∂kj=∂kHk-∂k∂kj-1=∂kHk∂kj-1-∂kj.
Observe that(17)Hk∂kj-1=Hk∂k∂kj-2=∂kHk-∂k∂kj-2=∂kHk∂kj-2-∂kj-1.
Replacing this in the previous expression we have the following.(18)Hk∂kj=∂k2Hk∂kj-2-2∂kj
By induction using the same technique we see that(19)Hk∂kj=∂k3Hk∂kj-3-3∂kj=∂k4Hk∂kj-4-4∂kj=⋯=∂kjHk-j∂kj=∂kjHk-j.For l>1 we have(20)Hkl∂kj=Hkl-1∂kjHk-j=Hkl-2Hk∂kjHk-j.
We have seen thatHk∂kj=∂kj(Hk-j), and using this we have(21)Hkl∂kj=Hkl-1∂kjHk-j=Hkl-2∂kjHk-j2.
By induction we easily see that(22)Hkl∂kj=∂kjHk-jl.(2)(23)∂kjHkl=∂kj-1∂kHkHkl-1
Using Proposition5 we see that ∂kHk=Hk∂k+∂k. Applying the same techniques as in the previous item we see that (24)∂kjHklHk+jl∂kj.(3) ∫klHkj=∫kl-1(∫kHk)Hkj-1. By Proposition 5 we see that (25)∫kHk=Hk∫k​-∫k​=Hk-1∫k​
and then∗∫klHkj=∫kl-1Hk-1∫kHkj-1.
Observe that∗∗∫kHkj-1=∫kHkHkj-2=Hk-1∫kHkj-2=⋯=Hk-1j-1∫k​,
and replacing this in∗ we see that3∗∫klHkj=∫kl-1Hk-1j∫k​.
By∗∗ we see that 4∗∫kl-1Hk-1j=Hk-lj∫kl-1​
and combining3∗ and 4∗ we have(26)∫klHkj=Hk-lj∫kl​.(4) Using the same technique as in the previous item and the fact that (27)Hk∫k​=∫kHk+∫k​=∫kHk+1
we see that(28)Hkj∫kl​=Hk+lj∫kl​.(5)∗′Hkejjk=Hk∫kj​e00k∂kj=∫kjHk+je00k∂kj
since we have shown above thatHk∫kj​=∫kj(Hk+j). Observe that (29)Hk+je00k=e00kHk+j,Hk+j∂kj=j∂kj+Hk∂kj=j∂kj+∂kjHk-j=∂kjHk.
Replacing all this in∗′ we have the following result:(30)Hkejjk=ejjkHk.
Now let us prove thatejj(k)Hk=(j+1)ejj(k). Observe that EHkejjk=Hk∫kj​e00k∂kj=∫kjHk+j∫kj​e00k∂kj=∫kjHke00k∂kj+jejjk.
Observe thatHke00(k)=e00(k), and using this in E we have the following result: (31)Hkejjk=j+1ejjk.
Now it becomes easy to see that(32)Hklejjk=Hkl-1Hkejjk=j+1Hkl-1ejjk=j+1lejjk.(6) The same techniques as in the previous item can be easily applied.Example 8 (application).
LetP=C[x] be a univariate polynomial ring and let us consider the algebra I1 constructed from P; we have I1=〈Hx,∫x​,∂x〉. Setting Hx>∫x​>∂x, one can study the noncommutative polynomial ring I1. (1)
Monomials inI1: in our case monomials are of the form Hxα∫xβ∂xγ where α,β,γ∈N. Note that ∫xβHxα,∂xγHxα are not monomials but polynomials for α,β,γ∈N∖{0} since by Corollary 7 we have
(i)
∂xγHxα=(Hx+γ)α∂kγ;(ii)
∫xβHxα=(Hx-β)α∫xβ​.(2)
Polynomials inI1: a polynomial in I1 is a linear combination of monomials. As a specific case, let us consider the polynomial f=∂x4Hx2∫x2​-(Hx+4)2∫x∂x3. According to the order we have imposed to our variables, this polynomial is not well-organized. Observe that (33)∂x4Hx2∫x2​=Hx+42∂x2after considering (1) in Proposition5 and (1) in Corollary 7. Replacing in f we get (34)f=Hx+421-∫x∂x∂x2=Hx+42e00∂x2=Hx2e00+8Hxe00+16e00∂x2.Using (5) in Corollary7 we get (35)f=25e00∂x2.We have seen in [1] that In=I1(1)⊗I1(2)⊗⋯⊗I1(n) where I1(k)=K〈Hk,∫k​,∂k〉 and the only proper ideal of I1(k) is F=⨁i,j∈NKeij(k).Corollary 9.
(1) The algebraI1(k) is principal and its only proper ideal is F=〈e00(k)〉;
(2)F=〈e00(k)〉 is an K[Hk]-module.Proof.
Consider the notation above.
(1) We have seen in [1] that the only proper ideal of I1(k) is F=⨁i,j∈NKeij(k)=⨁i,j∈NK∫kie00(k)∂kj. Let us now prove that F=〈e00(k)〉.
Letf∈F; then (36)f=a00e00k+a10∫k1e00k+a00e00k∂k1+a00∫k1e00k∂k1+⋯+ass∫kse00k∂ks.
Observe that each term off belongs in 〈e00(k)〉 as a two-sided ideal; therefore f∈〈e00(k)〉.
The converse is straightforward sincee00(k)∈F.
(2)F can be regarded as an I1-module. Moreover, (37)I1k=KHk⨁⨁j∈NK∫kje00k∂kj.
We easily see that the external multiplication is only concern withK[Hk].Lemma 10.
Forn>0 and k=1,…,n, (1)
∂kne00(k)=0≠e00(k)∂kn;(2)
e00(k)∫kn​=0≠∫kne00(k);(3)
e00(k)ejj(k)=0;(4)
ejj(k)e00(k)=0;(5)
eij(k)2=0.Proof.
(1)
Seeing that algebraI1(k) is associative and the fact that ∂k∫k​=1 (by Proposition 4), then (38)∂kne00k=∂kn1-∫k∂k=∂kn-1∂k-∂k∫k​∂k=0. Furthermoree00(k)∂kn=(1-∫k∂k)∂kn=∂kn-∫k∂kn+1=e0n(k). (2)
e 00 ( k ) ∫ k n ​ = [ e 00 ( k ) ∫ k ​ ] ∫ k n - 1 ​. Observe that (39)e00k∫k​=∫k​-∫k∂k∫k​=∫k​-∫k​=0. But(40)∫kne00k=∫kn​-∫kn+1∂. (3)
e 00 ( k ) e j j ( k ) = e 00 ( k ) ∫ k j e 00 ( k ) ∂ k j = [ e 00 ( k ) ∫ k j ​ ] e 00 ( k ) ∂ k j = 0. (4)
e j j ( k ) e 00 ( k ) = ∫ k j e 00 ( k ) [ ∂ k j e 00 ( k ) ] = 0. (5)
e i j ( k ) 2 = ( ∫ k i e 00 ( k ) ∂ k j ) ( ∫ k i e 00 ( k ) ∂ k j ) = ( ∫ k i e 00 ( k ) ( ∂ k j ∫ k i ​ ) e 00 ( k ) ∂ k j ) = 0.
(a)
Ifj>i then j=i+s and ∂kj∫ki​=∂ks, since ∂kse00(k)=0 then eij(k)2=0. (b)
Ifj<i, then i=j+s and ∂kj∫ki​=∫ks​, since e00(k)∫ks​=0.


## 3. Two-Sided Ideals ofIn

The theory of ideals ofIn is already studied in [1]. The goal of this section is to study generating sets of those ideals and to study the behavior of a given ideal.


### 3.1. Generators of Primes Ideals of Height 1

Corollary 3.3 in [1] proved that the only prime ideals of In of height 1 are (41)P1=F1⊗In-1,P2=I11⊗F2⊗In-2,…,Pn=In-1⊗FnwhereF(k)=⨁i,j∈NKeij(k).With notations as above we have the following result.Theorem 11.
Fork=1,…,n we have (42)Pk=e00kg1…gk-1gk+1…gn for some gj∈K[Hj].Proof.
By definitionPk=Ik-1⊗F⊗In-k where (43)Ik-1=I11⊗I12⊗⋯⊗I1k-1,F=⨁i,j∈NKeijk,In-k=I1k+1⊗I1k+2⊗⋯⊗I1n-k.If f∈Pk, then f=f1·f2·f3 where f1∈Ik-1,f2∈F and f3∈In-k.
Observe thatf1∈Ik-1⇒f1∈I1(1)⊗I1(2)⊗⋯⊗I1(k-1); that is, (44)f1=g11⋯gk-11 where gj1∈I1(j) for j=1,…,k-1. By Proposition 2.2 in [1], we see that (45)I1j=∫jαjKHj⨁⨁i,j∈NKeikj∂jβj.
g j 1 = ∫ j α j ( p j 1 + q j 1 ) ∂ j β j where pj1∈K[Hj] and qj1∈⨁i,k∈NKeik(j).
It is easy to seegj1∈〈pj1,qj1〉 as a two-sided ideal of I1(j). Observe that (46)qj1=a001e00j+a101∫j1e00+a011e00∂j1+a111∫j1e00∂j1+⋯+ass1∫jse00∂js∈e00j. We then have
〈 p j 1 , q j 1 〉 = 〈 p j 1 , e 00 ( j ) 〉 = 〈 p j 1 〉 since e00(j)=e00(j)/b01+⋯+br11pj1; i.e., if (47)pj1=br11Hjr1+br1-11Hjr1-1+⋯+b11Hj1+b01we have(48)e00jpj1=br11Hjr1e00j+br1-11Hjr1-1e00j+⋯+b11Hj1e00j+b01e00j=b01+⋯+br11e00j.
This means thatgj1∈〈pj1〉 for j=1,…,k-1; thus (49)f1∈p11⋯pk-11.
f 2 ∈ F implies straightforwardly (50)f2∈e00k.
Using the same technique as above, we have(51)f3∈pk+13⋯pn3.
Thus(52)f∈p11⋯pk-11·e00k·pk+13⋯pn3=e00kp11⋯pk-11pk+13⋯pn3.Corollary 12.
(1)
The only maximal ideal ofIn is (53)M=e001g21⋯gn1,e002g12g32⋯gn2,…,e00ng1ng2n⋯gn-1n wheregik∈K[Hi]∀i,k. (2)
The smallest nonzero ideal ofIn is Fn=⋂i=1nPi=〈e00(1)⋯e00(n)〉.Proof.
It is well known (see [1]) that (1)
the only maximal ideal ofIn is (54)M=P1+⋯+Pn,that is,(55)M=e001g21…gn1+e002g12g32…gn2+…+e00ng1n…gn-1nM=e001g21…gn1,e002g12g32…gn2,…,e00ng1n…gn-1n;(2)
the smallest nonzero ideal ofIn is (56)Fn=⋂i=1nPi.
It is straightforward to see that〈e00(1)⋯e00(n)〉⊂Fn. We prove the converse for n=2 since the general case can be obtained by induction. Let f∈F2; then f∈Pk∀k=1,2. That is, ∗′′f=q11e001g2kq12=q21e002g1kq22
for someq11,q12,q21,q22∈In≅I1(1)⊗I1(2). That is,(57)q11=∫1α11∫2α21u11+v11r11+w11∂1β11∂2β21q12=∫1α12∫2α22u12+v12r12+w12∂1β12∂2β22q21=∫1σ11∫2σ21u21+v21r21+w21∂1δ11∂2δ21,q22=∫1σ12∫2σ22u22+v22r22+w22∂1δ12∂2δ22
foruij∈K[H1],rij∈K[H2],j=1,2, and vij∈⨁l∈NKell(1),wij∈⨁l∈NKell(2)i,j=1,2.
We have the following.(58)f=∫1α11∫2α21u11+v11r11+w11∂1β11∂2β21e001g2k∫1α12∫2α22u12+v12r12+w12∂1β12∂2β22,(59)f=∫1σ11∫2σ21u21+v21r21+w21∂1δ11∂2δ21e002g1k∫1σ12∫2σ22u22+v22r22+w22∂1δ12∂2δ22
Let us expand (58):(60)f=∫1α11∫2α21u11+v11r11+w11∂1β11e001∂2β21g2k∫1α12∫2α22u12+v12r12+w12∂1β12∂2β22
since different variable commutes. Observe that
∂ 1 β 1 1 e 00 ( 1 ) = 0 for β11≠0; this leads to f=0. To avoid this, let us take β11=0; in this case we have the following.(61)f=∫1α11∫2α21u11+v11e001r11+w11∂2β21g2k∫1α12∫2α22u12+v12r12+w12∂1β12∂2β22Observe that
( u 1 1 + v 1 1 ) e 00 ( 1 ) = d 1 e 00 ( 1 ) for some d1∈K. This is because(i)
u11e00(1)=d1e00(1) and;(ii)
v11e00(1)=(1∑j∈Najejj(k))e00(1)=a0e00(k) since ejj(k)e00(k)=0∀j>0.
Observe also that
∂ 2 β 2 1 g 2 k = ϕ 2 ∂ 2 β 2 1 for some ϕ2∈K[H2] (using Corollary 7).
Replacing this inf we get(62)f=∫1α11∫2α21d1e001ϕ2r11+ϕ2w11∂2β21∫1α12∫2α22u12+v12r12+w12∂1β12∂2β22.
Since different variable commutes, we havee00(1)∫1α12​=0 by Proposition 5(6), and this leads directly to f=0. To avoid this case, we take α21=0; in this case f becomes(63)f=∫1α11∫2α21d1e001ϕ2r11+ϕ2w11∂2β21∫2α22u12+v12r12+w12∂1β12∂2β22.
Again since different variable commutes, we have(64)e001u12+v12=d2e001for  somed2∈K.
Replacing this inf we get(65)f=∫1α11∫2α21d1+d2e001ϕ2r11+ϕ2w11∂2β21∫2α22r12+w12∂1β12∂2β22.
Observe the following.(66)∂2β21∫2α22​=∂2rfor  somerifβ21>α221ifβ21=α22∫2l​for  somelifβ21<α22
In any of these cases we finish with an expression of the form(67)f=∫1α11∫2λ1d1+d2e001ρ1+ρ2∂1β12∂2λ2 for some λ1,λ2∈N and ρ1∈K[H2],ρ2∈⨁j∈NKejj(2); thus(68)f=∫1α11∫2λ1d1e001ρ1∂1β12∂2λ2+∫1α11∫2λ1d1e001ρ2∂1β12∂2λ2.
Let us expand (59). Applying the same technique as in (58) we get(69)f=∫1c11∫2c21d2e002θ1∂1c12∂2c22+∫1b11∫2b21d2e002θ2∂1b12∂2b22
wherecij,bij∈N∀i,j=1,2 and θ1∈K[H1],θ2∈⨁j∈NKejj(1).
In (68) we see that f∈〈e00(1)ρ1,e00(1)e00(2)〉 and in (69) f∈〈e00(2)θ1,e00(1)e00(2)〉; thus(70)f∈e001ρ1,e001e002⋂e002θ1,e001e002=e001e002.Example 13.
Let us consider again the algebraI2=K〈Hx,Hy,∫x​,∫y​,∂x,∂y〉. The maximal ideal of I2 is M=P1+P2 where P1=〈e00(x)g2〉 and P2=〈e00(y)g1〉 for some polynomials g1∈K[Hy] and g2∈K[Hx]. Thus M=〈e00(x)g2,e00(y)g1〉. The smallest ideal is F2=〈e00(x)e00(y)〉.


### 3.2. Generators of Ideals ofIn

LetI=〈f1,…,fs〉 be a two-sided ideal of In; then I can be written as I=〈f1〉+⋯+〈fn〉.Fori=1,…,n, let fi∈In≅I1(1)⊗⋯⊗I1(n); then (71)fi=h1i⋯hniwherehji∈I1j.It is proven in [1], Proposition 2.2, that I1(j)=vαj,+D1(j)vβj,- where vαj,+=∫jαj​,vβj,-=∂jβj and (72)D1j=KHj⊕⨁i,k∈NKeikj. For j=1,…,n we have(73)hji=∫jαjgji+qji∂jβj where gji∈K[Hj] and qji∈⨁i,k∈NKeik(j). Thus we have the following.(74)f1=∫1α1​⋯∫nαng11+q11⋯g1n+q1n∂1β1⋯∂nβn∗′′′f1=∫1α1​⋯∫nαng11⋯g1n∂1β1⋯∂nβn+∫1α1​⋯∫nαnq11⋯q1n∂1β1⋯∂nβn(1) If eachgik≠0≠qik, then (75)f1=g11⋯g1n,q11⋯q1n.Observe that(76)qjk=a00ke00j+a10k∫je00j+a01ke00j∂j+a11k∫je00j∂j+⋯+arrk∫jre00j∂jrand〈qjk〉=〈e00(j)〉; this implies the following.(77)f1=g11⋯g1n,e00j⋯e00nIf(78)gj1=br11Hjr1+br1-11Hjr1-1+…+b11Hj1+b01note that(79)e00j=e00jb0l+⋯+br1lgjland then(80)fk=gk1⋯gkn,e00j⋯e00n=gk1⋯gkn.(2) Assume without loss of generalities that there existk1,…,n such that (81)gkk1=⋯=gkn=0,and in this case(82)fk=gk1⋯gkk1-1e00k1⋯e00n,e001⋯e00n=gk1⋯gkk1-1e00k1⋯e00n.Remark 14.
Obtaining a set of generators of a given two-sided ideal ofIn is very important. According to the theory of ideals studied in [1], we see that any two-sided ideal of In can be written as a product or intersection of primes ideals of In. A very interesting result will be to propose an algorithm for finding these prime ideals from a given ideal. This work proposes the first step of tools necessary to deal with this problem, but to propose a decomposition algorithm, we will need to study the theory of two-sided Groebner bases (see [17]) on In as well as some applications such as the two-sided module of syzygies, since it is very important for the computation of generating sets of intersection of ideals as described in [18]. Our next project will develop computer algebra in In and solve this problem.


## 3.1. Generators of Primes Ideals of Height 1

Corollary 3.3 in [1] proved that the only prime ideals of In of height 1 are (41)P1=F1⊗In-1,P2=I11⊗F2⊗In-2,…,Pn=In-1⊗FnwhereF(k)=⨁i,j∈NKeij(k).With notations as above we have the following result.Theorem 11.
Fork=1,…,n we have (42)Pk=e00kg1…gk-1gk+1…gn for some gj∈K[Hj].Proof.
By definitionPk=Ik-1⊗F⊗In-k where (43)Ik-1=I11⊗I12⊗⋯⊗I1k-1,F=⨁i,j∈NKeijk,In-k=I1k+1⊗I1k+2⊗⋯⊗I1n-k.If f∈Pk, then f=f1·f2·f3 where f1∈Ik-1,f2∈F and f3∈In-k.
Observe thatf1∈Ik-1⇒f1∈I1(1)⊗I1(2)⊗⋯⊗I1(k-1); that is, (44)f1=g11⋯gk-11 where gj1∈I1(j) for j=1,…,k-1. By Proposition 2.2 in [1], we see that (45)I1j=∫jαjKHj⨁⨁i,j∈NKeikj∂jβj.
g j 1 = ∫ j α j ( p j 1 + q j 1 ) ∂ j β j where pj1∈K[Hj] and qj1∈⨁i,k∈NKeik(j).
It is easy to seegj1∈〈pj1,qj1〉 as a two-sided ideal of I1(j). Observe that (46)qj1=a001e00j+a101∫j1e00+a011e00∂j1+a111∫j1e00∂j1+⋯+ass1∫jse00∂js∈e00j. We then have
〈 p j 1 , q j 1 〉 = 〈 p j 1 , e 00 ( j ) 〉 = 〈 p j 1 〉 since e00(j)=e00(j)/b01+⋯+br11pj1; i.e., if (47)pj1=br11Hjr1+br1-11Hjr1-1+⋯+b11Hj1+b01we have(48)e00jpj1=br11Hjr1e00j+br1-11Hjr1-1e00j+⋯+b11Hj1e00j+b01e00j=b01+⋯+br11e00j.
This means thatgj1∈〈pj1〉 for j=1,…,k-1; thus (49)f1∈p11⋯pk-11.
f 2 ∈ F implies straightforwardly (50)f2∈e00k.
Using the same technique as above, we have(51)f3∈pk+13⋯pn3.
Thus(52)f∈p11⋯pk-11·e00k·pk+13⋯pn3=e00kp11⋯pk-11pk+13⋯pn3.Corollary 12.
(1)
The only maximal ideal ofIn is (53)M=e001g21⋯gn1,e002g12g32⋯gn2,…,e00ng1ng2n⋯gn-1n wheregik∈K[Hi]∀i,k. (2)
The smallest nonzero ideal ofIn is Fn=⋂i=1nPi=〈e00(1)⋯e00(n)〉.Proof.
It is well known (see [1]) that (1)
the only maximal ideal ofIn is (54)M=P1+⋯+Pn,that is,(55)M=e001g21…gn1+e002g12g32…gn2+…+e00ng1n…gn-1nM=e001g21…gn1,e002g12g32…gn2,…,e00ng1n…gn-1n;(2)
the smallest nonzero ideal ofIn is (56)Fn=⋂i=1nPi.
It is straightforward to see that〈e00(1)⋯e00(n)〉⊂Fn. We prove the converse for n=2 since the general case can be obtained by induction. Let f∈F2; then f∈Pk∀k=1,2. That is, ∗′′f=q11e001g2kq12=q21e002g1kq22
for someq11,q12,q21,q22∈In≅I1(1)⊗I1(2). That is,(57)q11=∫1α11∫2α21u11+v11r11+w11∂1β11∂2β21q12=∫1α12∫2α22u12+v12r12+w12∂1β12∂2β22q21=∫1σ11∫2σ21u21+v21r21+w21∂1δ11∂2δ21,q22=∫1σ12∫2σ22u22+v22r22+w22∂1δ12∂2δ22
foruij∈K[H1],rij∈K[H2],j=1,2, and vij∈⨁l∈NKell(1),wij∈⨁l∈NKell(2)i,j=1,2.
We have the following.(58)f=∫1α11∫2α21u11+v11r11+w11∂1β11∂2β21e001g2k∫1α12∫2α22u12+v12r12+w12∂1β12∂2β22,(59)f=∫1σ11∫2σ21u21+v21r21+w21∂1δ11∂2δ21e002g1k∫1σ12∫2σ22u22+v22r22+w22∂1δ12∂2δ22
Let us expand (58):(60)f=∫1α11∫2α21u11+v11r11+w11∂1β11e001∂2β21g2k∫1α12∫2α22u12+v12r12+w12∂1β12∂2β22
since different variable commutes. Observe that
∂ 1 β 1 1 e 00 ( 1 ) = 0 for β11≠0; this leads to f=0. To avoid this, let us take β11=0; in this case we have the following.(61)f=∫1α11∫2α21u11+v11e001r11+w11∂2β21g2k∫1α12∫2α22u12+v12r12+w12∂1β12∂2β22Observe that
( u 1 1 + v 1 1 ) e 00 ( 1 ) = d 1 e 00 ( 1 ) for some d1∈K. This is because(i)
u11e00(1)=d1e00(1) and;(ii)
v11e00(1)=(1∑j∈Najejj(k))e00(1)=a0e00(k) since ejj(k)e00(k)=0∀j>0.
Observe also that
∂ 2 β 2 1 g 2 k = ϕ 2 ∂ 2 β 2 1 for some ϕ2∈K[H2] (using Corollary 7).
Replacing this inf we get(62)f=∫1α11∫2α21d1e001ϕ2r11+ϕ2w11∂2β21∫1α12∫2α22u12+v12r12+w12∂1β12∂2β22.
Since different variable commutes, we havee00(1)∫1α12​=0 by Proposition 5(6), and this leads directly to f=0. To avoid this case, we take α21=0; in this case f becomes(63)f=∫1α11∫2α21d1e001ϕ2r11+ϕ2w11∂2β21∫2α22u12+v12r12+w12∂1β12∂2β22.
Again since different variable commutes, we have(64)e001u12+v12=d2e001for  somed2∈K.
Replacing this inf we get(65)f=∫1α11∫2α21d1+d2e001ϕ2r11+ϕ2w11∂2β21∫2α22r12+w12∂1β12∂2β22.
Observe the following.(66)∂2β21∫2α22​=∂2rfor  somerifβ21>α221ifβ21=α22∫2l​for  somelifβ21<α22
In any of these cases we finish with an expression of the form(67)f=∫1α11∫2λ1d1+d2e001ρ1+ρ2∂1β12∂2λ2 for some λ1,λ2∈N and ρ1∈K[H2],ρ2∈⨁j∈NKejj(2); thus(68)f=∫1α11∫2λ1d1e001ρ1∂1β12∂2λ2+∫1α11∫2λ1d1e001ρ2∂1β12∂2λ2.
Let us expand (59). Applying the same technique as in (58) we get(69)f=∫1c11∫2c21d2e002θ1∂1c12∂2c22+∫1b11∫2b21d2e002θ2∂1b12∂2b22
wherecij,bij∈N∀i,j=1,2 and θ1∈K[H1],θ2∈⨁j∈NKejj(1).
In (68) we see that f∈〈e00(1)ρ1,e00(1)e00(2)〉 and in (69) f∈〈e00(2)θ1,e00(1)e00(2)〉; thus(70)f∈e001ρ1,e001e002⋂e002θ1,e001e002=e001e002.Example 13.
Let us consider again the algebraI2=K〈Hx,Hy,∫x​,∫y​,∂x,∂y〉. The maximal ideal of I2 is M=P1+P2 where P1=〈e00(x)g2〉 and P2=〈e00(y)g1〉 for some polynomials g1∈K[Hy] and g2∈K[Hx]. Thus M=〈e00(x)g2,e00(y)g1〉. The smallest ideal is F2=〈e00(x)e00(y)〉.


## 3.2. Generators of Ideals ofIn

LetI=〈f1,…,fs〉 be a two-sided ideal of In; then I can be written as I=〈f1〉+⋯+〈fn〉.Fori=1,…,n, let fi∈In≅I1(1)⊗⋯⊗I1(n); then (71)fi=h1i⋯hniwherehji∈I1j.It is proven in [1], Proposition 2.2, that I1(j)=vαj,+D1(j)vβj,- where vαj,+=∫jαj​,vβj,-=∂jβj and (72)D1j=KHj⊕⨁i,k∈NKeikj. For j=1,…,n we have(73)hji=∫jαjgji+qji∂jβj where gji∈K[Hj] and qji∈⨁i,k∈NKeik(j). Thus we have the following.(74)f1=∫1α1​⋯∫nαng11+q11⋯g1n+q1n∂1β1⋯∂nβn∗′′′f1=∫1α1​⋯∫nαng11⋯g1n∂1β1⋯∂nβn+∫1α1​⋯∫nαnq11⋯q1n∂1β1⋯∂nβn(1) If eachgik≠0≠qik, then (75)f1=g11⋯g1n,q11⋯q1n.Observe that(76)qjk=a00ke00j+a10k∫je00j+a01ke00j∂j+a11k∫je00j∂j+⋯+arrk∫jre00j∂jrand〈qjk〉=〈e00(j)〉; this implies the following.(77)f1=g11⋯g1n,e00j⋯e00nIf(78)gj1=br11Hjr1+br1-11Hjr1-1+…+b11Hj1+b01note that(79)e00j=e00jb0l+⋯+br1lgjland then(80)fk=gk1⋯gkn,e00j⋯e00n=gk1⋯gkn.(2) Assume without loss of generalities that there existk1,…,n such that (81)gkk1=⋯=gkn=0,and in this case(82)fk=gk1⋯gkk1-1e00k1⋯e00n,e001⋯e00n=gk1⋯gkk1-1e00k1⋯e00n.Remark 14.
Obtaining a set of generators of a given two-sided ideal ofIn is very important. According to the theory of ideals studied in [1], we see that any two-sided ideal of In can be written as a product or intersection of primes ideals of In. A very interesting result will be to propose an algorithm for finding these prime ideals from a given ideal. This work proposes the first step of tools necessary to deal with this problem, but to propose a decomposition algorithm, we will need to study the theory of two-sided Groebner bases (see [17]) on In as well as some applications such as the two-sided module of syzygies, since it is very important for the computation of generating sets of intersection of ideals as described in [18]. Our next project will develop computer algebra in In and solve this problem.



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*Source: 2409193-2019-03-18.xml*