Bases for Projective modules in An(k)

BASES FOR PROJECTIVE MODULES IN An(k) JES´ US GAGO-VARGAS Abstract. Let An(k) be the Weyl algebra, with ka field of characteristic zero. It is known that every projective finitely generated left module is free or isomorphic to a left ideal. Let Mbe a left submodule of a free module. In this paper we give an algorithm to compute the projective dimension of M. If M is projective and rank(M)≥2 we give a procedure to find a basis. Introduction. The study of finitely generated projective modules over a ring is an interesting topic. We know that over polynomial rings they are free, as it was shown by Quillen and Suslin. There are several algorithmic versions of this theorem [Logar et al.(1992), Laubenbacher et al.(1997), Gago(2002)] that compute a basis from a system of generators. All of these procedures use Gr¨obner bases in polynomial rings. It is natural to extend these results to the Weyl Algebra An(k), with k a field with characteristic zero. It is known that if a left finitely generated An(k)- module is projective and has rank greater or equal 2 then is free [Stafford(1978)]. Our goal is to give an algorithm to find a basis of these modules. Projective modules in An(k) are stably free [Stafford(1977)], so the first step is to find an isomorphism P⊕An(k)s≃An(k)tfor some s, t. We develop this procedure in Section 1, together with an algorithm to compute the projective dimension of a module, that is valid for a broad class of rings. We note by pdim(M) the projective dimension of a module M. We require the computation of Gr¨obner bases in the ring and that every module has a finite free resolution. If Mis projective we find a matrix that defines an isomorphism M⊕Rs≃Rt. The starting point is a left R-module Mdefined by a system of generators in some Rm. In Section 2 we follow the proof of [Stafford(1978)] with algorithmic tools to find a basis of a projective module. We develop, for completeness, the reference to [Swan(1968)] used in [Stafford(1978), Thm. 3.6(a)], to clarify where these computations are needed . We follow describing the minor changes to [Hillebrand et al.(2002)] to obtain two special generators of a left ideal, according to [Stafford(1978), Theorem 3.1]. Finally, we give an example of this procedure to build a basis of a projective module in A2(Q). For all the computations we need an effective field kin the sense of [Cohen(1999)] to apply the Gr¨obner bases algorithm in An(k). We have used in the examples k=Q. Key words and phrases. Projective modules, non commutative rings, Gr¨obner bases. Partially supported by project BFM2001-3164 and FQM-813. This is a preliminary version of this article. 1 2 JES´ US GAGO-VARGAS 1. Computing projective dimension. Let Rbe a ring where it is possible to compute a finite free resolution of a left module, and we can determine if a right submodule of Rkis equal to Rk. Such a ring may be k[x1, . . . , xn], An(k) or more general rings like PBW algebras [Bueso et al.(1998)]. We make use of a characterization given in [Logar et al.(1992)], based on a finite free resolution of a module. The existence of a finite free resolution for a projective module Mis equivalent for Mto be stably free [McConnell et al.(1987)]. With the algorithm described in this section we test wether Mis projective, and if the answer is yes we compute an isomorphism M⊕Rs≃Rtfor some s, t. The procedure is by induction on the length of the resolution. We identify the homomorphisms with their matrices to simplify the notation. Suppose 0→F1 α1 →F0 α0 →M→0 is a free resolution of M, with rank(Fi) = ri. If Mis a projective module, this sequence splits, so there exists β1:F0→F1such that β1α1=Ir1. We can compute this matrix from the rows of the matrix α1: if we consider them as vectors of F1, the right R-module generated must be equal to F1. We express each vector of the canonical basis of F1as a linear combination of the rows of α1, and with these coefficients we construct the matrix β1. So we can give the isomorphism F1⊕ker(β1)≃F0≃F1⊕Mand a basis of F1⊕ker(β1). Let F: 0 →Ft αt −→ Ft−1 αt−1 −→ Ft−2 αt−2 −→ Ft−3 αt−3 −→ . . . α1 −→ F0 α0 −→ M→0 be a finite free resolution of M with rank(Fi) = riand t≥2 (we take α−1the null homomorphism). Again, if Mis a projective module, then the short exact sequence 0→ker(α0)→F0→M→0 splits, so ker(α0) = im(α1) is projective. By induction, the modules im(αi), i= 1, . . . , t are projective. In particular, im(αt−1) is projective and the exact sequence 0→Ft αt −→ Ft−1 αt−1 −→ im(αt−1)→0 splits. Then there exists βt:Ft−1→Ftsuch that Irt=βtαt. The module ker(βt) is projective, isomorphic to im(αt−1) and we can compute the isomorphism ker(βt)⊕Ft≃Ft−1. We consider the following sequence: 0→Ft eαt −→ Ft−1⊕Ft eαt−1 −→ Ft−2⊕Ft eαt−2 −→ Ft−3 αt−3 −→ . . . α1 −→ F0 α0 −→ M→0 where eαt(vt) = (αt(vt),0),eαt−1(vt−1,vt) = (αt−1(vt−1),vt), eαt−2(vt−2,vt) = αt−2(vt−2) Then it is an exact sequence and again the module im(eαt−1) is projective. As before, the sequence (1) 0 →Ft eαt −→ Ft−1⊕Ft eαt−1 −→ im(eαt−1)→0 splits and there exists e βt:Ft−1⊕Ft→Ftsuch that Irt=e βteαt. In this case, e βt=βtθ BASES FOR PROJECTIVE MODULES IN An(k) 3 where θis the null matrix with order rt×rt. Then e β(vt−1,vt) = βt(vt−1), so ker(e βt) = ker(βt)⊕Ft≃Ft−1. We can compute the isomorphism eνt−1:Ft−1→ker(e βt). Let (2) eγt−1=eαt−1eνt−1:Ft−1→Ft−2⊕Ft. Then the sequence 0→Ft−1 eγt−1 −→ Ft−2⊕Ft eαt−2 −→ Ft−3 αt−3 −→ . . . α1 −→ F0 α0 −→ M→0 is exact. Because the sequence (1) splits, the homomorphism eαt−1is an isomorphism between ker(e βt) and im(eαt−1), so eγt−1is an isomorphism between Ft−1and im(eαt−1) = ker(eαt−2), and we have the exactness of the sequence (2). We apply again the process to eγt−1to check the projectiveness of the module M. We need the following result: Theorem 1. Let Rbe a ring and F:. . . →Fd→Fd−1→. . . →F1→F0→M→0 a projective resolution. Let dbe the smallest number such that {im Fd→Fd−1}is projective. Then ddoes not depend on the resolution and pdim(M) = d. Proof. [Eisenbud(1995)], exercise A.3.13.  Theorem 2. The previous algorithm allows us to compute the projective dimension of a module. Proof. Let 0→Fn αn −→ Fn−1 αn−1 −→ . . . →F1 α1 −→ F0 α0 −→ M→0 be a finite free resolution given by the procedure. Then im(αn−1) is not projective, because the matrix αnhas not left inverse. We can suppose that Mis not projective, otherwise we have had shortened the resolution. Then the sequence 0→ker(α0)→F0→M→0 does not split, so im(α1) = ker(α0) is not projective. In the same way, the short exact sequence 0→ker(α1)→F1→im(α1)→0 does not split and im(α2) = ker(α1) is not projective. Then the modules im(α1),im(α2),...,im(αn−1) are not projective and the module im(αn) is projective. Then the projective dimension of Mis equal to n. Algorithm 1.1. Projective dimension. Input: a left R-module Mdefined by its generators in Rr. Output: Projective dimension of Mand a minimal length free resolution. If pdim(M) = 0, i.e. Mis projective, the algorithm returns an isomorphism M⊕Rs≃Rt. Let Fbe a finite free resolution of M: 0→Ft αt −→ Ft−1 αt−1 −→ Ft−2 αt−2 −→ Ft−3 αt−3 −→ . . . α1 −→ F0 α0 −→ M→0 START: 4 JES´ US GAGO-VARGAS if αthas no left inverse then pdim(M) = t. STOP. else let βtbe a left inverse of αt. end if if t= 1 then pdim(M) = 0 and M⊕F1≃ker(β1)⊕F1≃F0. STOP. else compute the exact sequence 0→Ft eαt −→ Ft−1⊕Ft eαt−1 −→ Ft−2⊕Ft eαt−2 −→ Ft−3 αt−3 −→ . . . α1 −→ F0 α0 −→ M→0 and the matrix eνt−1that gives the isomorphism ker(βt)⊕Ft≃Ft−1. end if Let eγt−1=eαt−1eνt−1. Let Fbe the finite free resolution 0→Ft−1 eγt−1 −→ Ft−2⊕Ft eαt−2 −→ Ft−3 αt−3 −→ . . . α1 −→ F0 α0 −→ M→0. go to START. This algorithm has being programmed with Macaulay 2 [Grayson et al.(1999)], using the routines for D-modules developed by A. Leykin and H. Tsai [Leykin et al.(2002)]. Example 1.Let W=A2(Q) and I=Whx∂x−1, x∂y, ∂2 x, ∂2 yi. We found a resolution of Iof the form 0←I eα0 ←W4 eγ1 ←W3←0 where eγ1=    −∂2 x−x∂x+ 1 0 ∂y0−x 0∂y∂x −∂x−x0    . The rows of the matrix eγ1do not generate W3, because a Gr¨obner basis is given by the columns of the matrix   0 0 ∂y∂x 1 0 0 0 0 1 0 0  . Then the ideal Iis not projective, and its projective dimension is 1. 2. Computing a basis. Let kbe a field of characteristic zero. Given a projective module over An(k) with rank greater than 1, we are going to describe a procedure to compute a basis. We will need the standard Gr¨obner basis theory on An(k) to make the computations. See for example [Castro(1987)] for a description of this algorithm. In [Hillebrand et al.(2002)] we found Theorem 3. Let R=k(x1, . . . , xn)[∂1, . . . , ∂n]and I=Rha, b, ci. Then we can compute ˜a,˜ b∈ R such that I=Rha+ ˜ac, b +˜ bci. BASES FOR PROJECTIVE MODULES IN An(k) 5 As pointed out in [Hillebrand et al.(2002), Remark 3.15], the algorithm can be extended to W=An(k) = k[x1, . . . , xn][∂1, . . . , ∂n]. We need the following stronger result [Stafford(1978), Thm. 3.1]: Theorem 4. Let I=Wha, b, cibe a left W-ideal, and let d1, d2∈W− {0}be arbitrary elements. Then we can find f1, f2∈Wsuch that I=Wha+d1f1c, b +d2f2ci. This can be accomplished with some minor changes to the proof of [Hillebrand et al.(2002), Lemma 3.10]. Following their notation, it is enough to take g1, g2∈Wsuch that h1d1g1+h2d2g2= 0, and to apply [Hillebrand et al.(2002), Lemma 3.9] to v=td2g2. These changes appear in the proof of [Stafford(1978), Theorem 3.1]. The procedure is analogous for right ideals. Definition 1. Let Mbe a left W-module and v∈M. We say that vis unimodular in Mif there exists ϕ∈HomW(M, W ) such that ϕ(v) = 1. Remark 1.If vis a column vector in some Wmthen vis unimodular if and only if the right ideal generated by its entries is equal to W. Through Gr¨obner bases, we can give the homomorphism that apply vin 1. The following Lemma is a direct consequence of Theorem 4, and it will allow a ’cancellation’ in some direct sums. Lemma 1. [Stafford(1978), Lemma 3.5] Let M⊂Wmbe a left W-module with rank(M)≥2and a⊕t∈M⊕Wunimodular. Then there is an algorithm to find Φ∈HomW(W, M)such that a+ Φ(t)is unimodular in M. Proof. Let a1∈M⊂Wmbe a non zero element and consider Φ1:Wm→Wa projection such that Φ1(a1)6= 0. Let M1=M∩ker(Φ1), that we can compute by Gr¨obner bases. Then rank(M1) = rank(M)−1≥1, so there exists a2∈M1−0. Let Φ2:Wm→Wbe a projection such that Φ2(a2)6= 0. If Φ2(a1)6= 0 we can compute syzygies to get r1, r2∈Wsuch that Φ1(a1)r1+ Φ2(a2)r2= 0 and replace Φ2by the homomorphism Φ1r1+ Φ2r2. Then Φ1(a2)=Φ2(a1) = 0. Let d1= Φ1(a1), d2= Φ2(a2) and consider the right ideal I=hΦ1(a),Φ2(a), tiW. Then there exist f1, f2∈Wsuch that I=hΦ1(a) + tf1d1,Φ2(a) + tf2d2iW. Let Φ : W→Mbe the homomorphism defined by Φ(1) = f1a1+f2a2. Then, as shown in [Stafford(1978), Lemma 3.5], a+Φ(t) is unimodular, and we can compute j∈HomW(M, W ) such that j(a+ Φ(t)) = 1.  Remark 2.The case a6=0is of special interest. In this case we can take a1=a and obtain Φ2(a) = 0, d1= Φ1(a). We have to find f1, f2such that I=hd1,0, tiW=hd1+tf1d1, tf2d2iW. Note that the problem is not to find two generators for the ideal I. We are looking for two special generators. Proposition 1. [Swan(1968), Corollary 12.6] Let M⊂Wmbe a left W-module with rank(M)≥2and h:W⊕N→W⊕Mbe an isomorphism with Na left W-module. Then M≃N. 6 JES´ US GAGO-VARGAS Proof. Let h(1,0)=(t0,a0)∈W⊕M. The vector (1,0) is unimodular so (t0,a0) too. Then we compute Φ : W→Msuch that a0 0=a0+ Φ(t0) is unimodular in Mand we get the homomorphism j:M→Wwith j(a0 0) = 1. We consider the following homomorphisms: g:W⊕M→W⊕M, g(t, a) = (t, a+ Φ(t)) k:W→W, k(1) = t0 l:W⊕M→W⊕M, l(t, a)=(t−(k◦j)(a),a), i:W⊕N→W⊕M, i =l◦g◦h Then iis isomorphism and i(1,0) = (0,a0 0). We have M=Wa0 0⊕ker(j) and the following chain of isomorphisms N≃(W⊕N)/We1 i →(W⊕M)/Wa0 0= (W⊕ker(j)⊕Wa0 0)/Wa0 0≃ W⊕ker(j)≃Wa0 0⊕ker(j) = M The isomorphism is defined as follows. Take v1,...,vra set of generators of N. Let i(0,vi) = (αi,ui) where αi∈W, ui∈M. The map (W⊕M)/W a0 0→W⊕ker(j) works taking an element of W⊕M, decomposes the component in Mas a sum v+wwith v∈Wa0 0,w∈ker(j) and takes w. For this step note that if u∈M and λ=j(u) then u= (λa0 0)+(u−λa0 0) is the desired decomposition.  Remark 3.When the module Nis of the form Ws, then Mis isomorphic to a free module, so it has a basis. Such a basis is the image of ei, i = 1, . . . , s. Algorithm 2.1. Computing a basis. Input: an isomorphism Wth ≃Ws⊕M, with t−s≥2. Output: a basis of the module M. START: if s= 0 then {h(e1,), . . . , h(et)}is a basis. STOP. end if Let h(1,0)=(t0,a0), with t0∈W, a0∈Ws−1⊕M. Compute Φ : W→Ws−1⊕Msuch that a0 0=a0+ Φ(t0)is unimodular. Compute j:Ws−1⊕M→Wsuch that j(a0 0)=1. Let i:W⊕Wt−1→W⊕(Ws−1⊕M)as defined in Prop. 1. Let h:Wt−1→Ws−1⊕Mthe isomorphism defined by h(ei) = αia0 0+ui−λia0 0 where i(0,ei) = (αi,ui), αi∈W, ui∈Ws−1⊕M, λi=j(ui). go to START As in the previous section, this algorithm has been programmed with Macaulay2. Example 2.Let W=A2(Q), and f=x∂yxy ∂x. Then P= ker fis a projective module, because fis a unimodular row. Let β=  −y∂x ∂x∂y −x  . BASES FOR PROJECTIVE MODULES IN An(k) 7 Then f·β= 1, and im β⊕P=W3. The isomorphism h:W⊕W2→W⊕Pis given by the matrix h=    x∂yxy ∂x xy∂x∂y+x∂x+ 1 xy2∂xy∂2 x −x∂x∂2 y−xy∂x∂y+ 1 −∂2 x∂y x2∂yx2y x∂x+ 2    . Then t0=x∂y,a0=  xy∂x∂y+x∂x+ 1 −x∂x∂2 y x2∂y  . We must find Φ : W→Psuch that a0 0=a0+ Φ(t0) is unimodular. Let Φ1:P→Wbe the projection over the first component and a2∈P∩ker(Φ1) not null. For example, a2=  0 ∂2 x∂y −xy∂x∂y−x∂x−2y∂y−2   and let Φ2:W→Pbe the projection over the second component. Because Φ2(a0)6= 0, we have to compute r1, r2∈Wsuch that Φ1(a0)r1+ Φ2(a2)r2= 0. In this case, we get r1=−∂2 x∂y, r2=xy∂x∂y−2y∂y+ 1, and following the notation of the proof of Lemma 1, d1=xy∂x∂y+x∂x+ 1, d2=xy∂3 x∂2 y+x∂3 x∂y+∂2 xdy. We have to find f1, f2∈Wsuch that hd1, t0iW=hd1+t0f1d1, t0f2d2iW. Applying the modified procedure of [Hillebrand et al.(2002)], we find f1= 0, f2=x+y. Let Φ : W→Pbe the morphism defined by Φ(1) = (x+y)a2. Then a0 0=a0+Φ(t0) is unimodular and we can compute the morphism j:P→Wsuch that j(a0 0) = 1. The output is too large to be included here, but has the form j= (−2 63 x2y7∂4 x∂5 y−2 63 xy8∂4 x∂5 y+5 126 x3y6∂3 x∂6 y+. . . −433 9x∂x+ 17x∂y+ 1, 2 63 xy8∂3 x∂4 y−5 126 x2y7∂2 x∂5 y+10 63 xy8∂2 x∂5 y+ + . . . +5 3xy −137 6y2,0) Also we can build the matrices associated to the other morphisms g=10 ΦI3, k = (x∂y), l =1−k·j 0I3, i=l·g·h=0α2α3 a0 0u2u3 where u2= (xy2∂x, x2y∂2 x∂y+xy2∂2 x∂y−xy∂x∂y+ 1, −x3y2∂x∂y−x2y3∂x∂y−x3y∂x−x2y2∂x−2x2y2∂y−2xy3∂y−x2y−2xy2)t, u3= (y∂2 x, x∂3 x∂y+y∂3 x∂y, −x2y∂2 x∂y−xy2∂2 x∂y−x2∂2 x−xy∂2 x−4xy∂x∂y−3y2∂x∂y−3x∂x−3y∂x−2y∂y)t Then w1= (α2−λ2)a0 0+u2,w2= (α3−λ3)a0 0+u3 is a basis of P, where λi=j(ui), i = 2,3. 8 JES´ US GAGO-VARGAS References [Bueso et al.(1998)] J.L. Bueso, J. G´omez-Torrecillas, F.J. Lobillo, F.J. Castro, An introduction to effective calculus in quantum groups in Rings, Hopf algebras, and Brauer groups, Lecture Notes in Pure and Appl. Math., 197, Dekker, New York, 1998. [Castro(1987)] F.J. 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Departamento de ´ Algebra, Universidad de Sevilla, Apdo. 1160, 41080 Sevilla, ESPA˜ NA E-mail address:[email protected]