A CONJECTURE OF BAVULA ON HOMOMORPHISMS OF THE WEYL ALGEBRA Leonid Makar-Limanov

OF THE WEYL ALGEBRA

Leonid Makar-Limanov¹

Department of Mathematics & Computer Science, the Weizmann Institute of Science, Rehovot 76100, Israel and

Department of Mathematics, Wayne State University, Detroit, MI 48202, USA

e-mail: lml@math.wayne.edu

Abstract. In the paperThe inversion formulae for automorphisms of polynomial algebras and differential operators in prime charac- teristic, J. Pure Appl. Algebra 212 (2008), no. 10, 2320–2337, see also Arxiv:math.RA/0604477, Vladimir Bavula states the following Conjecture:

(BC) Any endomorphism of a Weyl algebra (in a finite character- istic case) is a monomorphism.

The purpose of this preprint is to prove BC for A1, show that BC is wrong forAnwhenn >1, and prove an analogue ofBC for symplectic Poisson algebras.

The Weyl algebraAn is an algebra over a field F generated by 2n ele- mentsx₁, . . . x_n; y₁, . . . , y_n which satisfy relations [x_i, y_j](=x_iy_j−y_jx_i) = δ_ij, [x_i, x_j] = 0, [y_i, y_j] = 0, where δ_ij is the Kronecker symbol and 1 ≤ i, j ≤ n. Weyl algebras appeared quite some time ago and initially were considered only over fields of characteristic zero. Arguably the most famous problem related to these algebras is the Dixmier conjecture (see [D]) that any homomorphism of A_n is an automorphism if char(F) = 0. This problem is still open even forn= 1.

The finite characteristic case is certainly less popular but lately appears to attract more attention because it helps to connect questions related to the Weyl algebras and to polynomial rings, e. g. to connect the famous Ja- cobian Conjecture with the Dixmier conjecture (see [T1], [BK], and [AE]).

There is a striking difference between the zero characteristic and the finite characteristic cases. While for characteristic zero the center of An is just

1Supported by an NSA grant H98230-09-1-0008, by an NSF grant DMS-0904713, and a Fulbright fellowship awarded by the United States–Israel Educational Foundation.

For the reader’s convenience the proofs of all necessary lemmas are included, eliminating the need to look for them elsewhere.

1

the ground field and An is infinite-dimensional over the center, when the characteristic is not zero the center ofAn is a polynomial ring in 2ngener- ators andAn is a finite-dimensional free module over the center.

A vector spaceBwith two bilinear operationsx·y(a multiplication) and {x, y}(a Poisson bracket) is calleda Poisson algebraifB is a commutative associative algebra under x·y,B is a Lie algebra under{x, y}, and B sat- isfies the Leibniz identity: {x, y·z}={x, y} ·z+y· {x, z}.

Here we will be concerned with symplectic (Poisson) algebrasP S_n. For eachnthe algebraP S_nis a polynomial algebraF[x₁, . . . x_n; y₁, . . . , y_n] with the Poisson bracket defined by{xi, yj} =δij, {xi, xj}= 0, {yi, yj}= 0, where δij is the Kronecker symbol and 1 ≤ i, j ≤ n. Hence {f, g} = P

i(_∂x^∂f

i

∂g

∂y_i −_∂y^∂f

i

∂g

∂x_i).

To distinguish P Sn and An we will writeP Sn as F{x1, y1, . . . , xn, yn}.

One can think aboutP Sn as a commutative approximation ofAn (and of An as a quantization ofP Sn).

It is clear thatP S_n is a polynomial algebra with some additional struc- ture. It is less clear what is A_n. Of course, we can think about a Weyl algebra as a factor algebra of a free associative algebra by the idealIwhich corresponds to the relations, but it is not obvious even that 1 6∈I so the resulting factor algebra may be the zero algebra.

Lemma on basis. The monomialsy^j₁¹xⁱ₁¹. . . y^j_nⁿxⁱ_nⁿ form a basis P of An overF.

Proof. Any monomial µ in An (which in this consideration may be the zero algebra) can be written as a product µ = µ1µ2. . . µn where µi is a monomial of xi, yi since different pairs (xj, yj) commute. Furthermore, since xiyi = yixi+ 1 any monomial in xi, yi can be written as a linear combination of monomials y^k_ix^l_i with coefficients in Z or inZp depending on the characteristic of F.

It remains to show that the monomials inPare linearly independent over F. This can be done by finding a homomorphic image of A_n in which the images of monomials fromP are linearly independent.

If charF = 0 there is a natural representation of A_n. Consider ho- momorphisms X_i and Y_j of R = F[y₁, . . . , y_n] defined by X_i(r) = _{∂ y}^{∂ r}

i

and Yj(r) = yjr. A straightforward computation shows that α(xi) = Xi, α(yj) =Yj defines a homomorphism ofAn into the ring of homomorphisms of R and that the images of monomials from P are linearly independent.

Unfortunately this representation is not satisfactory when charF =p6= 0 because thenX_i^p= 0.

Here is a way to remedy the problem. ConsiderRn=R[z1, . . . , zn], and homomorphismsXi, Yj, andZk ofRndefined byXi(r) = _{∂ y}^{∂ r}

i,Yj(r) =yjr and Zk(r) = zkr for r ∈ Rn. Since Zk commute withXi, Yj, and Zl, if we replace Xi by Xci =Xi+Zi then [cXi, Yj] = [Xi, Yj] and [cXi,Xcj] = 0.

Now, forσ=PfijY₁^j1Xb₁ⁱ¹. . . Y_n^jnXb_nⁱⁿwe haveσ(1) =Pfijy^j₁¹zⁱ₁¹. . . y_n^jnzⁱ_nⁿ, which is zero only if allfij= 0. (Hereiandjare multi-indices as usual.) Let us call the presentation of an element a∈An through the basisP standard. Further we will use only the standard presentations of elements ofAn. SoAn is isomorphic to a corresponding polynomial ring as a vector space.

Remark 1. An is a domain (does not have zero divisors). To see this consider a degree-lexicographic ordering of monomials inP first by the total degree and then byy₁ >> x₁ >> y₂ >> x₂. . . >> y_n >> x_n. Then the commutation relations of A_n give |f g| = |f||g|| where |h| for h ∈ A_n\0 denotes the largest monomial appearing inh.

If char(F) = 0 then BC is very easy to prove (and is well-known) both for An andP Sn. Suppose thatϕhas a non-zero kernel. Let us take a non-zero element in the kernel ofϕ of the smallest total degree deg possible. It is clear that deg(ab) = deg(a) + deg(b) fora, b∈Anbecause of the commuta- tion relations. ConsiderP Sn first. If Λ is a “minimal” element of the kernel then{xi,Λ} = _∂y^∂Λ

i and {yi,Λ} =−_∂x^∂Λ

i should be identically zero because of the minimality of Λ. So _∂y^∂Λ

i = 0 and _∂x^∂Λ

i = 0 for alli. If char(F) = 0 this means that Λ∈ F. But our homomorphism is overF, so Λ = 0. A similar proof works forAn where the elements [xi,Λ] and [yi,Λ] should be identically zero which again shows that Λ = 0.

From now on we assume that char(F) =p6= 0.

Let us start with purely computational observations.

A straightforward computation shows that [ab, c] = [a, c]b+a[b, c]. There- fore [x^k+1₁ , y1] = [x^k₁, y1]x1+x^k₁[x1, y1] and since [x1, y1] = 1 induction onk gives [x^k₁, y₁] =kx^k−1₁ . Similarly, [x₁, y^k₁] =ky₁^k−1 and the index 1 can be replaced by anyi∈ {1,2, . . . , n}.

Denote [a, b] by ada(b). We will use that ad^p_a(B) = ada^p(b). In order to prove it observe that ada(b) = (al−ar)(b) wherealandarare the operators

of left and right multiplication bya. It is clear thatalandarcommute. So ad^p_a(b) = (al−ar)^p(b) = (a^p_l −a^p_r)(b) = ada^p(b).

Lemma on center. (a) The centerZ(An) ofAn=F[x1, . . . xn; y1, . . . , yn] is the polynomial ringF[x^p₁, . . . x^p_n; y₁^p, . . . , y_n^p].

(b) The Poisson center ofP Sn =F{x1, . . . xn; y1, . . . , yn}is the polynomial ringF[x^p₁, . . . x^p_n; y₁^p, . . . , y_n^p].

Proof. (a) Considerx^p₁. It is clear from the definition ofA_n thatx^p₁ com- mutes with all generators with a possible exception of y₁. As we observed above, [x1, y1] = 1 implies [x^k₁, y1] = kx^k−1₁ . So [x^p₁, y1] = px^p−1₁ = 0 and x^p₁ is in the center of An. Similarly all x^p_j and y_j^p are in the center and Z(An)⊇E whereE =F[x^p₁, . . . x^p_n; y^p₁, . . . , y^p_n].

Any elementa∈Ancan be written asa=P

ci,jy₁^j1xⁱ₁¹. . . y_n^jnxⁱ_nⁿ=a0+σ where 0≤i_s< pand 0≤j_s< p,c_i,j∈E,a₀∈E, andσis the sum of all monomials of a which do not belong toE. If a∈Z(A_n) then [x₁, a] = 0.

Now, [x1, y₁^j1xⁱ₁¹. . . y_n^jnxⁱ_nⁿ] = [x1, y^j₁¹]xⁱ₁¹. . . y_n^jnxⁱ_nⁿ = j1y₁^j1−1xⁱ₁¹. . . y^j_nⁿxⁱ_nⁿ and we have similar formulae whenx1is replaced by anyxioryj. So if one of the monomials in σ is not zero we can take the commutator ofa with an appropriatexi oryj and obtain a non-trivial linear dependence between monomials ofP contrary to the Lemma on basis.

(b) An element abelongs to the Poisson centerZ(A) of a Poisson algebra A if {a, b} = 0 for all b ∈ A. If f ∈Z(P Sn) then _∂x^∂f

i ={f, yi} = _∂y^∂f

i = {xi, f}= 0 for alliwhich is possible only iff ∈F[x^p₁, . . . x^p_n; y^p₁, . . . , y^p_n].

Nousiainen Lemma. Letϕbe a homomorphism of An orP Sn corre- spondingly. Then (a) An=Z(An)[ϕ(x1), . . . ϕ(xn); ϕ(y1), . . . , ϕ(yn)];

(b)P Sn=Z(P Sn)[ϕ(x1), . . . ϕ(xn); ϕ(y1), . . . , ϕ(yn)].

Proof. (a) Let E =F[x^p₁, . . . x^p_n; y₁^p, . . . , y_n^p]. From the Lemma on center Z(An) = E. The algebra An is a finite-dimensional module over E since any element a ∈ An can be written as a = P

ci,jy^j₁¹xⁱ₁¹. . . y^j_nⁿxⁱ_nⁿ where 0 ≤is < p, 0 ≤js < p, andci,j ∈ E. Let K =F(x^p₁, . . . x^p_n; y₁^p, . . . , y_n^p) be the field of fractions of E and letDn =K[x1, . . . xn; y1, . . . , yn]. Alge- braDn is a skew-field since Dn is a finite-dimensional vector space overK and Dn does not have zero divisors according to Remark 1. (Recall that x^p_i, y_j^p ∈ K, so any f ∈ D_n satisfies a non-zero relation PN

i=0k_ifⁱ = 0 wherek_i∈K andN ≤p²ⁿ.)

The monomials y^j₁¹xⁱ₁¹. . . y_n^jnxⁱ_nⁿ, where 0 ≤ is < p, 0 ≤ js < p, are linearly independent over K. Indeed, if Λ = Pc_i,jy^j₁¹xⁱ₁¹. . . y^j_nⁿxⁱ_nⁿ = 0

whereci,j ∈ K then [xm,Λ] = [ym,Λ] = 0 and we can obtain from a non- trivial dependence Λ a “smaller” one. So we can invoke induction on, e. g.

the sum of the total degrees of monomials in Λ.

Since any elementa ∈D_n can be written asa =Pc_i,jy₁^j1xⁱ₁¹. . . y_n^jnxⁱ_nⁿ where 0≤i_s< p, 0≤j_s< p, andc_i,j∈K, the dimension ofD_n over K is p²ⁿ .

Consider now monomialsv^j₁¹uⁱ₁¹. . . v^j_nⁿuⁱ_nⁿ whereui=ϕ(xi), vj=ϕ(yj), 0 ≤ is < p, and 0 ≤ js < p. Let us check that they are also linearly independent over K. If Λ = Pci,jv^j₁¹uⁱ₁¹. . . v^j_nⁿuⁱ_nⁿ = 0 then [um,Λ] = [vm,Λ] = 0 and, since the commutation relations are the same as above, we obtain from a non-trivial dependence Λ a “smaller” one.

Since there are exactly p²ⁿ of these monomials, they also form a basis of Dn over K and any element a ∈ An ⊂ Dn can be written as a = Pc_i,jv^j₁¹uⁱ₁¹. . . v_n^jnuⁱ_nⁿ where c_i,j∈K.

It remains to show that allc_i,j∈E. Order the monomialsv₁^j1uⁱ₁¹. . . v_n^jnuⁱ_nⁿ degree-lexicographically as in Remark 1. Letµ = v₁^j1uⁱ₁¹. . . v_n^jnuⁱ_nⁿ be the largest monomial. Then adⁱ_vⁿ_nad^j_uⁿ_n. . .adⁱ_v¹₁ad^j_u¹₁(a) = (−1)^IQn

m=1(im)!(jm)!ci,j, where I =Pn

m=1i_m, belongs toA_n. Since (−1)^IQn

m=1(i_m)!(j_m)! 6= 0 we conclude that c_i,j ∈A_nTK =E, replaceaby a−c_i,jµ, and finish by in- duction on the number of monomials ofa.

(b) Let T_n be the field of rational functions F(x₁, . . . x_n; y₁, . . . , y_n) en- dowed with the same bracket asP S_n: {f, g}=P

i(_∂x^∂f

i

∂g

∂yi−_∂y^∂f

i

∂g

∂xi). Then T_n becomes a Poisson algebra and we can think of P S_n as a subalgebra of T_n. It is clear that Z(T_n) = F(x^p₁, . . . x^p_n; y^p₁, . . . , y^p_n) and that T_n is a p²ⁿ-dimensional vector space overZ(T_n).

Denoteui=ϕ(xi), vi=ϕ(yi). To show that the monomialsv₁^j1uⁱ₁¹. . . v^j_nⁿuⁱ_nⁿ where 0≤is< p and 0≤js< p are linearly independent overZ(Tn) we, as above, can consider a “minimal” relation Λ =P

ci,jv^j₁¹uⁱ₁¹. . . v^j_nⁿuⁱ_nⁿ= 0 and get “smaller” relations by taking{um,Λ} and{vm,Λ}.

Ifa∈P Sn is presented as a=Pci,jv₁^j1uⁱ₁¹. . . v_n^jnuⁱ_nⁿ where 0≤is< p, 0≤js< p, andci,j∈Z(Tn) then allci,j∈Z(P Sn); to see this just replace, in the considerations above, adz by Adzdefined by Adz(b) ={z, b}.

Corollary. There are no homomorphisms fromAn into A_n−1.

Proof. Assume that we have a homomorphismφ:An →A_n−1. Consider imagesui =φ(xi) andvi=φ(yi). According to Nousiainen LemmaAn−1is a vector space over the center ofAn−1with a basis consisting of monomials v^j₁¹uⁱ₁¹. . . v_n−1^jn−1uⁱ_n−1ⁿ⁻¹, 0≤i_s < p, 0 ≤j_s < p. Therefore u_n and v_n are in the center ofA_n−1and hence commute with each other.

Remark 2. This Lemma is similar to a lemma from Pekka Nousiainen’s PhD thesis (Pennsylvania State University, 1981) which was never published (see [BCW]). Nousiainen proved his Lemma in a commutative setting for a Jacobian set of polynomials, i. e. he proved that ifz1, . . . , zn∈F[y1, . . . , yn] and the Jacobian J(z1, . . . , zn) = 1 thenF[y1, . . . , yn] =F[y^p₁, . . . , y^p_n;z1, . . . , zn].

HenceF[y1, . . . , yn] =F[y^P₁, . . . , y^P_n;z1, . . . , zn] whereP =p^mandmis any natural number. Indeed, if say y₁=Pc_izⁱ₁¹. . . zⁱ_nⁿ where c_i∈F[y₁^p, . . . , y^p_n] theny^p₁=Pc^p_i(z₁ⁱ¹. . . z_nⁱⁿ)^p andc^p_i ∈F[y^p₁², . . . , y^p_n²].

For the same reasonP Sn=Z(P Sn)^P[ϕ(x1), . . . ϕ(xn); ϕ(y1), . . . , ϕ(yn)].

But even forA1 the situation is different. Take e. g. u=x, v =y²x−y when p= 2. Then A1 6= F[x⁴, y⁴;u, v]. Indeed, u² =x², v² =y⁴x² and D16=F(x⁴, y⁴)[u, v] sinceu² andv² are linearly dependent overF(x⁴, y⁴).

This difference between An andP Sn is the reason for BC to be correct forP Sn and wrong forAn.

The Nousiainen Lemma for Weyl algebras is proved in [T2] and [AE].

Theorem 1. BC is true for Poisson algebrasP S_n.

Proof. In the Nousiainen Lemma we proved thatP S_n=Z(P S_n)[u₁, . . . u_n; v₁, . . . , v_n] where ui =ϕ(xi), vi =ϕ(yi). So a= P

ci,jv₁^j1uⁱ₁¹. . . v_n^jnuⁱ_nⁿ where ci,j ∈ Z(P Sn) =F[x^p₁, . . . x^p_n; y^p₁, . . . , y_n^p] for anya∈P Sn. Therefore

a^p=X

c^p_i,jv₁^pj1u^pi₁¹. . . v^pj_nⁿu^pi_nⁿ wherec^p_i,j∈F[x^p₁², . . . , x^p_n²; y₁^p2, . . . , y_n^p2].

Hence

F[x^p₁, . . . , x^p_n; y^p₁, . . . , y^p_n]⊂F[x^p₁², . . . , x^p_n²;y^p₁², . . . , y^p_n²][u^p₁, . . . , u^p_n; v^p₁, . . . , v^p_n] anda=Pd_i,jv^j₁¹uⁱ₁¹. . . v^j_nⁿuⁱ_nⁿ where

di,j∈F[x^p₁², . . . x^p_n²; y₁^p2, . . . , y_n^p2][u^p₁, . . . u^p_n; v₁^p, . . . , v_n^p].

SoP Sn=F[x^p₁², . . . , x^p_n²; y₁^p2, . . . , y^p_n²][u1, . . . , un; v1, . . . , vn]. By iterat-

ing this process we will get thatP Sn=F[x^P₁, . . . x^P_n; y^P₁, . . . , y^P_n][u1, . . . , un; v1, . . . , vn] whereP =p^mfor any positive integerm.

It is clear that u^P_i , v_j^P ∈ F[x^P₁, . . . , x^P_n; y₁^P, . . . , y_n^P] so P Sn is spanned over F[x^P₁, . . . , x^P_n; y₁^P, . . . , y_n^P] by monomialsv^j₁¹uⁱ₁¹. . . , v_n^jnuⁱ_nⁿ where 0 ≤ i_s< P, 0≤j_s< P. Of course,P S_nis spanned overF[x^P₁, . . . x^P_n;y^P₁, . . . , y^P_n] by monomials y₁^j1xⁱ₁¹. . . y^j_nⁿxⁱ_nⁿ where 0 ≤ is < P, 0 ≤ js < P and these monomials are linearly independent overF[x^P₁, . . . x^P_n; y^P₁, . . . , y^P_n].

SoF[x1, . . . xn; y1, . . . , yn] is a free module overF[x^P₁, . . . x^P_n; y^P₁, . . . , y^P_n] of dimensionP²ⁿ.

Ifϕis not an injection then there is a linear dependence overF between monomials v₁^j1uⁱ₁¹. . . v_n^jnuⁱ_nⁿ where 0≤is < P, 0 ≤js < P provided P is sufficiently large. But this is impossible since these monomials are linearly

independent overF[x^P₁, . . . x^P_n; y^P₁, . . . , y^P_n].

We prove now using Gelfand-Kirillov dimension that a homomorphism ofA1 intoAn is an embedding. Here is a definition of the Gelfand-Kirillov dimension (GKdim) suitable for our purpose. LetRbe a finitely-generated associative algebra overF: R=F[r1, . . . , rm]. Consider a free associative algebra F_m =Fhz1, . . . , z_mi and linear subspaces F_m,N of all elements of F_m with total degree at most N. Let α be a homomorphism of F_m onto R defined by α(z_i) = r_i and let R_N = α(F_m,N). Each R_N is a finite- dimensional vector space (overF); denoted_N = dim(R_N).

GKdim(R) =lim_N→∞ln(d_N) ln(N).

Though this definition uses a particular system of generators, it is pos- sible to prove that GKdim(R) does not depend on such a choice (see [GK]

or [KL]). It is not difficult to show that in the commutative case Gelfand- Kirillov dimension coincides with the transcendence degree.

Lemma on GK-dimension. LetR=F[z1, . . . , zm] be a ring of poly- nomials. If a, b ∈ R are algebraically dependent then GKdim(S) ≤ 1 for any finitely generated subalgebraS⊂A=F[a, b].

Proof. If a, b ∈ F then F[a, b] = F and any subalgebra of A is F. So in this case GKdim(S) = 0. Assume now that a 6∈ F, i. e. deg(a) > 0.

If deg(b) = 0 then A = F[a] and d_N = N + 1 where d_N = dim(A_N), so GKdim(A) = 1. Let S be a subalgebra of A generated by a₁, . . . , a_m. Thenai=qi(a) whereqi are polynomials. Assume that degrees of all these polynomials do not exceedd. Therefore a polynomialg(a1, . . . , am) of the total degreeN can be rewritten as a polynomial inaof degree at mostdN.

HenceN < dN ≤dN+ 1 if any ofai is not inF and GKdim(S) = 1. If all ai∈F then GKdim(S) = 0.

Now, let deg(b)>0. Sinceaandbare algebraically dependent,Q(a, b) = 0 for a non-zero polynomialQ. Order monomialsaⁱb^j by total degreei+j and then lexicographically bya >> b. Ifµ=a^kb^l is the largest monomial in Q we can write µ = Q1(a, b) where all monomials of Q1 are less than µ. So we can replace any monomial ν = aⁱb^j where i ≥ k, j ≥ l by a linear combination of monomials which are less thanν. Hence anyc ∈A of the total degree at most N can be written as a linear combination of monomialsaⁱb^j wherei+j ≤N and eitheri < k or j < l. There are less than (k+l)(N+ 1) and more thanN monomials satisfying these properties.

ThereforeN < d_N <(k+l)(N+1) and GKdim(A) = 1. IfSis a subalgebra of A generated by a1, . . . , am then ai = qi(a, b) where qi are polynomials of total degrees bounded by somed. If we take a polynomialg(a1, . . . , am)

of total degreeN theng(a1, . . . , am) can be rewritten as a polynomial ina andbof degree at mostdN. ThereforeN < dN ≤(k+l)(dN+ 1) if any of ai is not inF and GKdim(S) = 1. If allai∈F then GKdim(S) = 0.

Denote by deg the total degree function onAn and bya the element of A which is deg homogeneous and such that deg(a−a)<deg(a). We will refer toa as the leading form ofa. From the commutation relations in A_n it follows that ab=baand that deg([a, b])<deg(ab).

We will think about leading forms not as elements of An but rather as commutative polynomials. Thenab=ab=ba.

Lemma on dependence. Letaandbbe algebraically dependent non- zero homogeneous polynomials andq= deg(a),r= deg(b). Thena^r andb^q are proportional, i. e. there exists anf ∈F so thata^r−f b^q= 0.

Proof. The polynomialsaandbare algebraically dependent, so there is a non-zero polynomialQfor whichQ(a, b) =P

i,jfijaⁱb^j= 0. Since aandb are homogeneous we may assume thatqi+rj is the same for all monomials of Q. Indeed, any Q can be presented as Q = P

iQk where Qk are q, r- homogeneous. Then either Qk(a, b) = 0 or deg(Qk(a, b)) =k and different components cannot cancel out.

Therefore over an algebraic closureFofFwe can writeQ=f₀a^kb^lQ

i(a^r0− f_ib^q0) wheref_i ∈F,r⁰= ^r_d, q⁰= ^q_d, anddis the greatest common divisor of randq. Hencea^r0−fib^q0 = 0 for somefi∈F. But thena^r0b^−q0 ∈F since it is a constant rational function defined overFanda^rb^−q = (a^r0b^−q0)^d ∈F.

Lemma on independence. Letϕbe a homomorphism of A₁into A_n. Then the image ofA₁contains two elements with algebraically independent leading forms.

Proof. Letu=ϕ(x) andv=ϕ(y) wherexandy are generators ofA1and letB=F[u, v] be the image ofA1inAn. Ifuandvare independent, we are done. If not, then by Lemma on dependence there exists a pair of relatively prime natural numbers (q, r) andf ∈F such thatu^q =f v^r. Eitherq orr is not divisible byp. For arguments sake assume that it isq.

We can find k for which kp+ 1 ≡ 0 (modq), f1 ∈ F and a positive integer sso that u^kp+1 =f1v^s. Let us replace the pair (u, v) by the pair (u1=u^kp+1−f1v^s, v1 =v). The commutator [u1, v1] =u^kp is a non-zero element of the centerZ(B) ofB. Ifu₁andv₁are independent we are done, otherwise repeat the step above to get (u₂, v₂), etc.. We claim that after a finite number of steps we produce a pair of elements ofB with independent leading forms.

Consider a function def(a, b) = deg(ab)−deg([a, b]) onAn. Let us check that def(ui+1, vi+1)<def(ui, vi). We will do it for the first step since the computations are the same for every step.

Sinceu^kp+1=f1v^swe see that deg(u1)<(kp+1) deg(u). So def(u1, v1) = deg(u1v1)−deg([u^kp+1−f v^r, v]) = deg(u1)+deg(v)−kpdeg(u)−deg([u, v]))<

deg(u) + deg(v)−deg([u, v] = def(u, v) since [u^kp+1−f v^r, v] =u^kp[u, v] and deg(u1)<(kp+ 1) deg(u).

By definition, def(a, b)>0 if bothaandbare not zero, so after at most def(u, v) steps we either get a pair with zero element or a pair U, V ∈ B with independent U and V. Since [u, v] = 1 the pair we start with does not contain zero. Similarly, since [u_i, v_i]6= 0 we cannot get a pair with zero element.

We can now see that GKdim(B)≥2. Indeed,U and V are “polynomi- als” ofuandv and we may assume that the degrees of these polynomials are at mostd. Then the space of all polynomials in u, v of degree at most Ncontains all polynomials inU, V of degree at most^N_d. SinceU andV are algebraically independent the leading formsUⁱV^j are linearly independent overF and hence UⁱV^j are linearly independent overF. There are about

N²

2d² of these monomials with i+j ≤ ^N_d (exactly ^[Nd₂^]+2

where [^N_d] is the integral part of ^N_d). So the dimensiondN >_2d^N22 and GKdim(B)≥2.

Theorem 2. Letϕ be a homomorphism ofA1 into An. Then ϕis an embedding.

Proof.LetA1=F[x;y] andu=ϕ(x), v=ϕ(y). Ifϕhas a non-zero kernel take an elementain the kernel of smallest total degree possible. Since both [x, a] and [y, a] are also in the kernel of ϕ and have smaller total degrees, a ∈ Z(A1) =F[x^p, y^p]. Therefore u^p and v^p are algebraically dependent and by Lemma on GK-dimension GKdim(F[u^p, v^p]) ≤ 1 (recall that u^p and v^p commute). But GKdim(Z(B)) = GKdim(B) for B = F[u, v]. In- deed, B = Puⁱv^jZ(B) where 0 ≤ i, j < p by Lemma on center. So d_N(B) ≤ p²dN

p(Z(B)) and d_N(B) ≥ dN

p(Z(B)). We showed above that GKdim(B)≥2. Sou^p andv^p are algebraically independent and the kernel ofϕconsists of zero only.

Theorem 2 cannot be extended to A2. Take z = x1+y₁^p−1x2, y1, y2. Then [z, y₁] = 1, [z, y₂] =y^p−1₁ ,and [z^p, y₂] = ad^p_z(y₂) = ad^p−1_z (y₁^p−1) = (p−

1)! =−1. Foru1=z+z^py₁^p−1,v1=y1;u2=y2, v2=z^pthe commutation relations ofA2are satisfied. Soφwhich is defined byφ(xi) =uiandφ(yi) = v_i is a homomorphism of A₂. If B = φ(A₂) then B = F[u₁, u₂;v₁, v₂] =

F[z, y1, y2]. Hence u^p₁ ∈Z(B) and u^p₁ ∈Z(F[z, y1]) =F[z^p, y^p₁] since u1∈ F[z, y1]. Butu^p₁should commute withu2=y2. Thereforeu^p₁∈F[z^p2, y^p₁] = F[v₂^p, v^p₁], u^p₁, v^p₁, v₂^p are algebraically dependent, and φ has a non-zero kernel.

It is an exercise to check that GKdim(B) = 3. On the other hand, GKdim(A2) = 4 sincedN = ^N_2n⁺²ⁿ

forAn which gives GKdim(An) = 2n.

A question about possible GK-dimensions of images of An under homo- morphisms seems reasonable in this setting because clearly the size of the kernel is large when the size of the image is small. Say, GKdim(ϕ(An))≤2n and ifϕis an injection then GKdim(ϕ(An)) = 2n.

Theorem 3. GKdim(ϕ(An)) can be n+i where i = 1,2, . . . , n for a homomorphismϕofA_n intoA_n.

Proof. Denote ϕ(A_n) by B and by u_i = ϕ(x_i), v_i = ϕ(y_i). It is suf- ficient to show that GKdim(B) = n+ 1 is possible for any n because combiningu₁, . . . , u_m;v₁, . . . , v_mof an appropriate map ofA_mto A_mwith x_m+1, . . . , x_n;y_m+1, . . . , y_n we will get an image of A_n of GK-dimension m+ 1 + 2(n−m).

Now we shall findϕsuch that GKdim(B) =n+ 1.

Consider elements z0,0 = 0, zm,0 =xm−y^p−1_m z_m−1,0 for m= 1, . . . , n, and z_m,i = z^p_m,0ⁱ . Then [z_i,0, y_i] = 1, [z_i−k,0, y_i] = 0 and [z_i−k,0, x_i] = 0 if k > 0, and [zi,0, zj,0] = 0. Therefore [zk,i, zm,j] = 0. We can get a relation between zi,j using the equality (yx)^p −yx = y^px^p if [x, y] = 1 (observe that adyx = ad_(yx)p). Take ymzm,0 = ymxm−y_m^pz_m−1,0. Then the summands in the right side commute and (ymzm,0)^p = (ymxm)^p − y^p_m²z^p_m−1,0. So z_m,0^p = y^−p_m [(ymxm)^p−y^p_m²z_m−1,0^p −ymxm+y_m^pzm−1,0] = y^−p_m[y^p_mx^p_m−y_m^p2z_m−1,0^p +y_m^pzm−1,0] =zm−1,0−ym^p(p−1)z_m−1,0^p +x^p_m, i. e.

zm,1 =z_m−1,0−y^p(p−1)m z_m−1,1+x^p_m. Since all summands in the right side of this equality commute

zm,i+1=z_m−1,i−y_m^pi+1(p−1)z_m−1,i+1+x^p_mⁱ⁺¹ and

z_m−1,i=zm,i+1+y_m^pi+1(p−1)z_m−1,i+1−x^p_mⁱ⁺¹. Then by induction we can prove that

zm−1,i=

m−i−2

X

k=0

cm−1,i,kzm,i+1+k+cm−1,i

and that

z_m−j,i=

m−i−j−1

X

k=0

c_m−j,i,kz_m,i+j+k+c_m−j,i

where ci,j, ci,j,k ∈ Z(An). The sums are finite because zm,m ∈ Z(An) (can be proved by induction on m starting with z0,0 = 0 since zm,m = z_m−1,m−1−ym^pm(p−1)z_m−1,m+x^p_m^m).

Now we can show that allci,j,k∈F[y^p₁, . . . , y^p₂].

Since [zm,i, yj] = [z_m−1,i−1, yj]−ym^pi(p−1)[z_m−1,i, yj], we can deduce by induction onmthat [zm,i, yj] is zero ifj > m−i, one ifj=m−i, and belongs toF[y₁^p, . . . , y_n^p] ifj < m−i. Sincez_m−j,i=Pm−i−j−1

k=0 c_m−j,i,kzm,i+j+k+ c_m−j,i we have [z_m−j,i, y_l] = Pm−i−j−1

k=0 c_m−j,i,k[z_m,i+j+k, y_l]; this allows usingl=m−j−i, . . . ,1 to check that all c_m−j,i,k∈F[y₁^p, . . . , y_n^p].

All these computations were done to confirm that

zm,0=

m−1

X

k=0

dm,kz_n,n−m+k+dm

wheredm∈Z(An) anddm,k∈F[y₁^p, . . . , y_n^p].

Recall thatxm=zm,0+y_m^p−1z_m−1,0 and so

xm=

m−1

X

k=0

dm,kz_n,n−m+k+dm+y^p−1_m (

m−2

X

k=0

d_m−1,kz_n,n−m+1+k+d_m−1).

Therefore

um=xm−dm−y^p−1_m d_m−1∈B=F[y1, . . . , yn;zn,0].

Finally,u1, . . . , un;y1, . . . , yn define a homomorphism of An into B and since any monomial inB can be written asy^j₁¹. . . y^j_nⁿz_n,0ⁱ , the Theorem is proved.

By looking atϕ(F[x₁, . . . , x_n]) it is possible to show that GKdim(ϕ(A_n))≥ n. This and Theorem 2 suggest the following

Conjecture. GKdim(ϕ(A_n))> n.

Acknowledgement. The author is grateful to the CRM (Centre de Re- cerca Matem`atica) in Bellaterra, Spain for the hospitality during the work on this project.

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