PRELIMINARIES ON DIXMIER CONJECTURE

YOSHIFUMI TSUCHIMOTO

Abstract. We gather some basic facts concerning algebra endomorphisms of Weyl algebras A_{n}(k) for fields k with positive characteristic p. These facts could lead some preliminary results on Dixmier conjecture.

Contents

1. Introduction 1 2. Notations 2 2.1. A presentation of the full matrix algebra 3 2.2. light exponential function 3 3. First properties of Weyl algebras 4 3.1. reduction to characteristic p 4 3.2. Weyl algebras over fields of positive characteristics and their centers 5 3.3. Algebra endomorphisms and centers of Weyl algebras 6 3.4. A splitting algebra of A_{n} 7 3.5. The quotient field of the Weyl algebra 7 3.6. definition of reduced trace and reduced norm 8 3.7. Algebra endomorphism and splitting of Weyl algebra 8 3.8. the geometric degree 9 3.9. uniqueness of an operator p-th root for generators of A_{n}(k) 9 3.10. Algebra endomorphisms of Weyl algebras are determined by its

restriction to the center 9

4. Reduced trace and reduced norm 10 4.1. calculation of reduced trace 10 4.2. invariance of trace and inversion formula 10 5. injectivity (n = 1) 11 5.1. comments on this section 11 5.2. injectivity(n = 1) 11 6. multidegree monoids and lattices 12 6.1. injectivity for almost all primes 13 References 14

1. Introduction

Our aim in this paper is to gather some basic facts concerning algebra endomorphisms of Weyl algebras A_{n}(k) for fields k with positive characteristic p.

1991 Mathematics Subject Classification. 14R15(16W20).

Key words and phrases. Jacobian conjecture, Dixmier conjecture, Noncommutative algebra.

A study of such object may lead (Lemma 2) to some progress in the Dixmier conjecture (conjecture 1) which states that any algebra endomorphism of Weyl algebra over a field of characteristic zero is actually invertible.

It turns out that A_{n}(k) is a free module of rank p^{2n} over its center Z(A_{n}(k)) (Lemma 3) and that any k-algebra endomorphism ϕ of A_{n}(k) sends central elements to central elements (Lemma4).

Thus the study of ϕ may be deeply related to the study of sheaves of matrix algebras over polynomial algebras Z(A_{n}(k)).

In particular, we may use traces to obtain a nice formula for candidate of inverse of ϕ (Proposition 1).

On the other hand, suppose we are given an algebra endomorphism ϕ of the Weyl algebra A_{n}(𝔎) over an algebraic number field 𝔎. Then we have, for almost all (that is, for all except finite number of) prime ideal 𝔭 of 𝔎, an algebra endomorphism of A_{n}(k(𝔭) over the residue field k(𝔭). We prove that almost all such maps are injective and that geometric degree of these maps are bounded by a constant which is independent of 𝔭 (Proposition 2).

2. Notations

All rings are assumed to be unital, associative. All homomorphisms are assumed to be unital. ℕ = {0, 1, 2, 3, … }

For any ring R and for any positive integer l, we put

M_{l}(R) = (the set of l × l-matrix with coefficients in R.)

For any ring k, we denote by A_{n}(k) the Weyl algebra.

A_{n}(k) = k⟨ξ_{1}, ξ_{2}, …, x_{n}, η_{1}, η_{2}, …, η_{n}⟩/(η_{j}ξ_{i} − ξ_{i}η_{j} − δ_{ij}; 1 ≤ i, j ≤ n)

where δ_{ij} is the Kronecker’s delta.

If k is a field with characteristic p, then we further employ the following notations. Z_{n}(k) = Z(A_{n}(k)) the center of A_{n}(k) (Later we prove that Z_{n}(k) is actually

equal to k[ξ_{1}^{p}, ξ_{2}^{p}, …, ξ_{n}^{p}, η_{1}^{p}, η_{2}^{p}, …, η_{i}^{p}]).

T_{i} = (ξ_{i}^{p})^{1/p} U_{i} = (η_{i}^{p})^{1/p} S_{n}(k) = Z_{n}(k)^{1/p} = k^{1/p}[T_{1}, T_{2}, …, T_{n}, U_{1}, U_{2}, …, U_{n}]. (p-th roots are taken

in the usual sense of commutative algebra. S_{n}(k) is a commutative algebra over Z_{n}(k).)

K_{n}(k) = Q(Z_{n}(k)), the quotient field of Z_{n}(k). D_{n}(k) = A_{n}(k) ⊗_{Z_{n}(k)} K_{n}(k) L_{n}(k) = Q(S_{n}(k)). B_{n}(k) = A_{n}(k) ⊗_{Z_{n}(k)} S_{n}(k) (≅ M_{p^{n}}(S_{n}(k))) V_{n}(k) = ⊕_{i=1}^{p^{n}}S_{n}(k) E_{n}(k) = A_{n}(k) ⊗_{Z_{n}(k)} L_{n}(k) (≅ M_{p^{n}}(L_{n}(k))) α_{i} = ξ_{i} − T_{i} β_{i} = η_{i} − U_{i} \overline{A_{n}} : a copy of A_{n} (Identified with the image of ϕ when ϕ is injective). \overline{Z_{n}} = Z(\overline{A_{n}}) : a copy of Z_{n} \overline{S_{n}} : a copy of S_{n} (extension of \overline{Z_{n}})

Similarly we use a notation \overline{ξ_{i}}, \overline{η_{i}}, \overline{T_{i}}, \overline{U_{i}} to indicate a copy of non-bar counterparts.

For a matrix x, we denote by λ(x), ρ(x), ad(x) the left action, the right action, and the adjoint action by x, respectively.

λ(x)y = xy, ρ(x)y = yx, ad(x)y = xy − yx

2.1. A presentation of the full matrix algebra.

Lemma 1. Let k be a field of characteristic p. Then a k-algebra 𝔐 which is generated by α_{1}, α_{2}, …, α_{n}, β_{1}, β_{2}, …, β_{n} with the relations

[β_{j}, α_{i}] = δ_{ji}, β_{j}^{p} = 0, α_{i}^{p} = 0 (i, j = 1, 2, …, n)

(where δ_{ij} is the Kronecker’s delta) is isomorphic to the full matrix algebra M_{p^{n}}(k).

Proof. Since M_{p^{n}}(k) is isomorphic to a tensor product of n copies of the matrix algebra M_{p}(k), we may assume that n = 1. We define elements μ, ν ∈ M_{p}(k) as follows.

(1)  μ = (δ_{i+1,j}),  ν = ((p − j)δ_{i,j+1})

Then μ, ν satisfies the same relation as α_{1}, β_{1}. In other words, we have a kalgebra homomorphism ϕ from 𝔐 to M_{p}(k) with ϕ(α_{1}) = μ, ϕ(β_{1}) = ν.

On the other hand, it is easy to see that the algebra 𝔐 is linearly generated by {α^{i}β^{j}; 0 ≤ i, j ≤ p − 1} and hence that its dimension is not greater than p^{2}.

By a dimension argument we see that the algebra homomorphism ϕ is an isomorphism.

□

2.2. light exponential function. For an element L of an algebra over a field k of characteristic p > 0, we may define light exponential of L by

ex(L) = ∑_{i=0}^{p−1} \frac{1}{i!}L^{i}.

(The ex is obtained by cut exponential function off the tail after p).

If L_{1}, L_{2} commutes and \underline{L_{1}^{i}L_{2}^{j} = 0 whenever i + j ≥ p}, then we have

ex(L_{1} + L_{2}) = ex(L_{1}) ex(L_{2})

In particular, if L^{p} = 0 then we have

ex(L) ex(−L) = 1

There is a differential equation for the light exponential function.

d/dT (ex(T)) = ex(T) + T^{p−1}

Furthermore, the following equations holds for any constant c_{1}, c_{2}.

ex(c_{1}μ) ex(c_{2}ν) = ex(c_{2}(ν − c_{1}E − c_{1}^{p}μ^{p−1})) ex(c_{1}μ)
ex(c_{1}μ) ex(c_{2}ν) ex(−c_{1}μ) = ex(c_{2}(ν − c_{1}E − c_{1}^{p}μ^{p−1}))
[ex(c_{1}ν), μ] = c_{1} ex(c_{1}ν) + c_{1}^{p}ν^{p−1}
[ex(c_{1}μ), ν] = −c_{1} ex(c_{1}μ) − c_{1}^{p}μ^{p−1}
ex(c_{1}μ)ν ex(−c_{1}μ) = ν − c_{1}1_{p} − c_{1}^{p}μ^{p−1}

ex(c_{1}ν)μ ex(−c_{1}ν) = μ + c_{1}1_{p} + c_{1}^{p}ν^{p−1} ex(−c_{1}μ)ν ex(c_{1}μ) = ν + c_{1}1_{p} + c_{1}^{p}μ^{p−1} ex(−c_{1}ν)μ ex(c_{1}ν) = μ − c_{1}1_{p} − c_{1}^{p}ν^{p−1}

3. First properties of Weyl algebras

For any ring k we denote by A_{n}(k) the Weyl algebra:

A_{n}(k) = k⟨ξ_{1}, ξ_{2}, …, ξ_{n}, η_{1}, η_{2}, …, η_{n}⟩/(η_{j}ξ_{i} − ξ_{i}η_{j} − δ_{ij})

where δ_{ij} is the Kronecker’s delta.

One of the good ways to compute multiplications of elements in A_{n}(k) appears in [2, formula (11,4)]. For any variable η, ξ with the canonical commutation relation ηξ−ξη = 1,and for any pair of “normally ordered” polynomials f(ξ, η) = ∑ f_{i,j}ξ^{i}η^{j} and g(ξ, η) = ∑ g_{i,j}ξ^{i}η^{j}, one has the following formula

f(ξ, η)g(ξ, η) = ∑_{k=0}^{∞} \frac{1}{k!}∂_{η}^{k}f(ξ, η) ∗ ∂_{ξ}^{k}g(ξ, η)

where ∗ stands for an ‘multiplication as though ξ, η commutes’. This formula is valid and proved in the book cited above only if characteristic of the coefficient field is 0. If the characteristic p of the coefficient field is positive, then we replace ∞ in the above formula by p − 1 and obtain a valid formula in this case. The proof is almost the same. It is well suited for computer calculation using commutative polynomials and differentiations.

Dixmier conjectures that

Conjecture 1. Every ℂ-algebra endomorphism ϕ of A_{n}(ℂ) is invertible (that is, it is an automorphism).

Since A_{n}(ℂ) is simple (has no nontrivial both-sided ideal), we know that ϕ above is injective. The question therefore is the surjectivity of ϕ.

3.1. reduction to characteristic p. Suppose we are given a ℂ-algebra endomorphism ϕ of A_{n}(ℂ). Since the algebra A_{n}(ℂ) is finitely generated over ℂ, the endomorphism ϕ is actually defined over a ring R which is finitely generated algebra over ℚ.

By a specialization argument we may assume R = 𝔒(𝔎)[1/f], where 𝔎 is a finite extension field of ℚ, 𝔒(𝔎) is the ring of all algebraic integers in 𝔎, f is a non zero element of 𝔒(𝔎).

For almost all (that is, all except finite number of) prime ideals 𝔭 of 𝔒(𝔎), we obtain an algebra endomorphism ϕ_{𝔭} of an algebra A_{n}(k(𝔭)) over k(𝔭) where k(𝔭) = 𝔒(𝔎)/𝔭 is a field of a positive characteristic p.

Lemma 2. ϕ is invertible if and only if homomorphisms ϕ_{𝔭} are invertible for all except finite number of primes 𝔭 ∈ Spec(O(𝔎)).

Proof. The “only if” part is clear. To prove “if”, we suppose on the contrary that ϕ is not invertible. This means, as we mentioned, that ϕ is not surjective. Then there exists a nonzero linear functional ^{1} χ such that χ ∘ ϕ = 0. It is easy to see that χ defines a non zero linear functional χ_{𝔭} on A_{n}(k(𝔭)) for all except finite number of primes 𝔭, and that χ_{p}e ∘ ϕ_{𝔭} = 0. this is a contradiction, and the lemma is proved. □

1Note added in proof: The functional χ should be chosen more carefully. We will correct this in a forthcoming paper.

It is thus worthwhile to study A_{n}(k) and its automorphism when k has a positive characteristic.

3.2. Weyl algebras over fields of positive characteristics and their centers.

Lemma 3. Let k be a field of characteristic p. We have the following facts. (1) ξ_{i}^{p}, η_{j}^{p} belongs to the center of A_{n}(k).

(2) More precisely, the center Z_{n}(k) = Z(A_{n}(k)) of the ring A_{n}(k) is given by

Z_{n}(k) = k[ξ_{1}^{p}, …, ξ_{n}^{p}, η_{1}^{p}, …, η_{n}^{p}].

(3) A_{n}(k) is a free Z_{n}(k)-module of rank p^{2n}.

(4) Let a = (a_{1}, …, a_{n}), b = (b_{1}, … b_{n}) be elements of k^{n}. Let I_{a,b} be an ideal of A_{n}(k) generated by (ξ_{1} − a_{1})^{p}, …, (ξ_{n} − a_{n})^{p}, (η_{1} − b_{1})^{p}, …, (η_{n} − b_{n})^{p}. Then we have

A_{n}(k)/I_{a,b} ≅^{π_{a,b}} M_{p^{n}}(k).

(5) If the field k is algebraically closed, then any maximal (both-sided) ideal of A_{n}(k) is of the form I_{a,b} above.

Proof. (1) easy.

(2) Write

f = ∑_{I,J} f_{IJ}ξ^{I}η^{J}

and compute [ξ_{i}, f] and [η_{j}, f] term by term. (3)

A_{n}(k) = ⨁_{0≤i_{1},...,i_{n},j_{1},...,j_{n}≤p−1} Z_{n}(k)ξ_{1}^{i_{1}}ξ_{2}^{i_{2}}ξ_{3}^{i_{3}} … ξ_{n}^{i_{n}}η_{1}^{i_{1}}η_{2}^{i_{2}}η_{3}^{i_{3}} … η_{n}^{i_{n}}.

(4) is a direct consequence of Lemma 1. For the reference purposes, we record here another explanation. By considering a translation in ξ- and η-directions, we may well assume a = b = 0. Consider an action of A_{n}(k) on A = k[ξ_{1}, …, ξ_{n}]/(ξ_{1}^{p}, …, ξ_{n}^{p}) defined by the following formula.

(2)  P.f = P (ξ_{1}, …, ξ_{n}, ∂_{1}, …, ∂_{n}).f

We may then verify that this gives a well-defined action of A_{n}(k)/I_{0,0} on the p^{n}- dimensional vector space A.

If an element P in A_{n}(k)/I_{0,0} satisfy

P.1 = 0, P.ξ = 0, …, P.ξ^{p−1} = 0

then we may easily see that P = 0. This means the homomorphism

π_{0,0} : A_{n}(k)/I_{0,0} → End_{k_{-module}}(A)

is injective. A dimension argument now shows that π_{0,0} is a bijection.

(5) Let M be an ideal of A_{n}(k). We consider a sheaf { \atop (A_{n}(k)/M)} of algebras on Spec(Z_{n}(k)) which corresponds to the Z_{n}(k)-module A_{n}(k)/M. Since A_{n}(k) is a finitely generated module over a polynomial ring Z_{n}(k), There exists a closed point 𝔪 of Z_{n}(k) such that fiber ({ \atop (A_{n}(k)/M)}|𝔪) = A_{n}(k)/(M + 𝔪A_{n}(k)) is nonzero. Since M is maximal, this implies that 𝔪 ⊂ M. The rest of the proof is easy (Use Nullstellensatz.)

3.3. Algebra endomorphisms and centers of Weyl algebras.

Lemma 4. Let k be a field of characteristic p > 0. For any k-algebra endomorphism ϕ of A_{n}(k),

(1) π_{a,b} ∘ ϕ is a surjective homomorphism for all (a, b) ∈ k^{2n}.

(2) ϕ(Z_{n}(k)) ⊂ Z_{n}(k).

Proof. We may assume that k is an algebraically closed field.

(1) The kernel of π_{a,b}∘ϕ is an ideal of A_{n}(k) and therefore, by a dimension argument, is one of the maximal ideals of A_{n}(k). It also follows that π_{a,b} ∘ ϕ is surjective.

(2) The result (1) implies that for any (a, b) ∈ k^{2n} we have

ϕ(Z_{n}(k)) ⊂ I_{a,b} + k

Thus the claim deduces to the following sublemma.

Sublemma 1.

⋂_{(a,b)∈k^{2n}}(I_{a,b} + k) = Z_{n}(k)

[proof of sublemma]

The left hand side may be identified with a section of { \atop A_{n}(k)} which reduces to 0 at each k-valued point when we regard it as a section of { \atop (A_{n}(k)/Z_{n}(k))}. Since k is an infinite field, this implies that f ∈ Z_{n}(k).

Corollary 1. A_{n}(k) is generated by ϕ(A_{n}(k)) and Z(A_{n}(k)).

[proof] We may assume k is algebraically closed. Let B be the algebra generated by ϕ(A_{n}(k)) and Z(A_{n}(k)). Then the claim (1) of the previous lemma shows that for any maximal ideal 𝔪 of Z(A_{n}(k)), we have an isomorphism

B/𝔪B ≅ A_{n}(k)/𝔪A_{n}(k).

as Z(A_{n}(k))-modules. Since A_{n}(k) is finitely generated module over Z(A_{n}(k)), we see immediately that M = A_{n}(k) as required.

Corollary 2. Let \overline{A_{n}(k)} be a copy of A_{n}(k). Let ϕ : \overline{A_{n}(k)} → A_{n}(k) be a khomomorphism. Let Z_{n}(k) = Z(A_{n}(k)), \overline{Z_{n}(k)} = Z(\overline{A_{n}(k))} be the center of the algebras A_{n}(k) \overline{A_{n}(k)}, respectively. Then the natural homomorphism

\overline{A_{n}(k)} ⊗\overline{Z_{n}(k)} Z_{n}(k) → A_{n}(k)

is an isomorphism.

Proof. By the corollary above we already know that it is surjective. Since both hand sides are free Z_{n}(k)-modules of rank p^{2n}, and since Z_{n}(k) is an integral domain, the map is a bijection. [The surjection admits a splitting. Then we consider determinants.]

□

Corollary 3. An algebra homomorphism ϕ : A_{n}(k) → A_{n}(k) is surjective if and only if its restriction to the center

ϕ|_{Z(A_{n}(k))}^{Z(A_{n}(k))} : Z(A_{n}(k)) → Z(A_{n}(k))

is surjective.

By the birational case of the Jacobian conjecture (which is already known to be true), we conclude that

Corollary 4. ϕ : A_{n}(k) → A_{n}(k) is bijective if and only if the following three conditions hold.

(1) ψ = ϕ|_{Z(A_{n}(k))}^{Z(A_{n}(k))} : Z(A_{n}(k)) → Z(A_{n}(k)) is injective.

(2) ψ is birational.

(3) The Jacobian det D(ψ) is a nonzero constant.

3.4. A splitting algebra of A_{n}. In this subsection we assume that k is a field of characteristic p > 0. Let S_{n}(k) = Z_{n}(k)^{1/p} = k^{1/p}[T_{1}, …, T_{n}, U_{1}, …, U_{n}] where T_{i} = (ξ_{i}^{p})^{1/p}, U_{i} = ((η_{i})^{p})^{1/p}). It is a splitting algebra of A_{n}(k), as the following lemma tells.

Lemma 5. The algebra A_{n}(k) acts on V = ⊕_{i=1}^{p^{n}}S_{n}(k). In other words, there exists a representation Φ of A_{n}(k) on V.

Φ(ξ_{i}) = μ_{i} + T_{i}, Φ(η_{i}) = ν_{i} + U_{i},

where elements μ_{i}, ν_{i} of M_{p^{n}}(k) are defined (using notation in Lemma 1) as follows.

μ_{i} = 1_{p^{i−1}} ⊗ μ ⊗ 1_{p^{n−i}}, ν_{i} = 1_{p^{i−1}} ⊗ ν ⊗ 1_{p^{n−i}}.

The representation Φ may be extended to the following isomorphism.

Φ : A_{n}(k) ⊗_{Z_{n}(k)} S_{n}(k) ≅ M_{p^{n}}(S_{n}(k))

Proof. Put α_{i} = ξ_{i} − T_{i}, β_{i} = η_{i} − U_{i}(i = 1, …, n). Then it is easy to show that elements α_{1}, …, α_{n}, β_{1}, …, β_{n} satisfy the relation of generators of the algebra 𝔐 as in Lemma 1.

□

3.5. The quotient field of the Weyl algebra. Let k be a field of characteristic p > 0. There is a nice “quotient field of the Weyl algebra A_{n}(k).

Lemma 6. Let K_{n}(k) be the quotient field of the center Z_{n}(k) of the Weyl algebra Z_{n}(k). Then the following statements hold.

(1) D_{n}(k) = A_{n}(k) ⊗_{Z_{n}(k)} K_{n}(k) is a skew field.

(2) dim_{K_{n}(k)} D = p^{2n}.

(3) M_{p^{n}}(L_{n}(k)) may be regarded as a p^{2n}-dimensional vector space over D with a basis {T^{I}U^{J}; ||I||_{ℓ^{∞}} ≤ p − 1, ||J||_{ℓ^{∞}} ≤ p − 1}.

Proof. (1) D_{n}(k) is a simple algebra with no zero-divisor except for 0. Then we use Wedderburn’s structure theorem.

(2),(3): we may easily see that

dim_{K_{n}(k)} D_{n}(k) ≤ p^{2n}, dim_{D_{n}(k)} M_{p}(L_{n}(k)) ≤ p^{2n}.

On the other hand we have

dim_{K_{n}(k)} M_{p}(L_{n}(k)) = p^{4n}
□

3.6. definition of reduced trace and reduced norm. From a general theory of central simple algebra, we have a notion of reduced traces and reduced norms. (See for example [1] for a theory of reduced norms and reduced traces.)

It may be defined as follows. D_{n}(k) is a central simple algebra over K_{n}(k). Let Ω be a splitting field Ω of D_{n}(k). That means, we have an isomorphism of k-algebras

ι : D_{n} ⊗_{K_{n}(k)} Ω ≅ M_{p^{n}}(Ω).

Then for each element x of D_{n}, it is known that the trace tr(ι(x)) and the determinant det(ι(x)) actually belongs to K_{n} and that they do not actually depend on the choice of the splitting field Ω.

We call them reduced trace and reduced norm of x respectively.

As we already saw in the previous section, S_{n}(k) is a splitting algebra of A_{n}(k). Thus we see that the quotient field L_{n}(k) of S_{n}(k) is one of the splitting field of D_{n}(k). We have thus proved that reduced norm and reduced determinant of A_{n}(k) actually lie in Z_{n}(k).

3.7. Algebra endomorphism and splitting of Weyl algebra.

Lemma 7. Let k be a algebraically closed field with characteristic p > 0. Let ϕ : A_{n}(k) → A_{n}(k) be a k-homomorphism, ψ : Z_{n}(k) → Z_{n}(k) its restriction to the center. Then we have the following.

(1) ψ extends uniquely to a homomorphism

ψ̂ : S_{n}(k) → S_{n}(k)

(2) ϕ extends uniquely to a homomorphism

ϕ̂: A_{n}(k) ⊗_{Z_{n}(k)} S_{n}(k) → A_{n}(k) ⊗_{Z_{n}(k)} S_{n}(k)

(3) Under the isomorphism Φ of lemma 5, ϕ̂ may be identified with a map

M_{p}(S_{n}(k)) ∋ M(T, U) ↦ G(T, U)M(f(T, U))G(T, U)^{−1}

where G(T, U) is an element of GL_{p}(S_{n}(k)) and f = ^{a}ψ : 𝔸^{2n} → 𝔸^{2n} is a polynomial map associated to the algebra homomorphism ψ̂. In other words, we have

Φ(ϕ(x)) = Gf^{∗}(Φ(x))G^{−1}

Proof. (1): In any commutative integral domain of characteristic p, p-th root of an element is unique.

(2): For each k-valued point (t, u) of 𝔸^{2n} we obtain a homomorphism ψ_{t,u} : M_{p^{n}}(k) → M_{p^{n}}(k). In other words, we obtain a morphism

Ǧ : 𝔸^{2n} → PGL_{p} .

But since the sheaf cohomology H^{1}(𝔸^{2n}, 𝔾_{m}) (in Zariski topology) is trivial, we have a lift G of Ǧ and the claim is proved.

Corollary 5 (invariance of trace under algebra endomorphism).

trd(ϕ(a)) = ϕ(trd(a))

Lemma 8. The polynomial map f of the lemma above is determined by G(T, U) uniquely up to an addition of an element of k[T^{p}, U^{p}]. That means, if we have two f’s, namely f_{1} and f_{2}, with the same G(T, U), then we have

f_{1}^{∗}(T) − f_{2}^{∗}(T) ∈ k[T^{p}, U^{p}], f_{1}^{∗}(U) − f_{2}^{∗}(U) ∈ k[T^{p}, U^{p}]

Proof. Put α = ξ − T, β = η − U. Then

ϕ(ξ) − ϕ(T) = ϕ(α) = G(T, U)αG(T, U)^{−1}, (3)

ϕ(η) − ϕ(U) = ϕ(β) = G(T, U)βG(T, U)^{−1}. (4)

ϕ(ξ), ϕ(η) ∈ k[ξ, η], ϕ(T), ϕ(U) ∈ k[T, U].

3.8. the geometric degree.

Definition 1. Let k be a field of positive characteristic. Then for any k-algebra homomorphism. ϕ : \overline{A_{n}(k)} → A_{n}(k), the geometric degree geomdeg(ϕ) of ϕ is defined to be the index [Q(Z_{n}(k)) : Q(ϕ(Z_{n}(k))] of the corresponding field extension.

Note that if the geometric degree is finite, then by comparing the transcendent degree we see that ϕ is actually injective and the geometric degree is equal to [K_{n}(k) : \overline{K_{n}(k)}].

3.9. uniqueness of an operator p-th root for generators of A_{n}(k).

Lemma 9. If F ∈ A_{n}(k) satisfies an identity F^{p} = ξ_{1}^{p}, then F = ξ_{1}.

Proof. See the principal symbol (highest degree part) of F. Then it should be ξ_{1}. Thus there exists a constant c such that F = ξ_{1} + c. On the other hand, we have (ξ_{1} + c)^{p} = ξ_{1}^{p} + c^{p}. □

3.10. Algebra endomorphisms of Weyl algebras are determined by its restriction to the center. In this subsection we assume that k is a field of positive char.

Lemma 10. The natural group homomorphism Aut_{k_{-alg}}(A_{n}(k)) → Aut_{k_{-alg}}(Z_{n}(k)) (restriction map) is injective.

Proof. The uniqueness of operator p-th root for generators of Z_{n}. □ Note.

It is not surjective. For example, let k be a field of odd characteristic and consider an algebra automorphism ψ of Z_{1}(k) given by

ψ(ξ^{p}) = ξ^{p}, ψ(η^{p}) = −η^{p}.

Then by the uniqueness of the operator p-th root we see that the lift ϕ of ψ should satisfy

ϕ(ξ) = ξ, ϕ(η) = −η.

But this ϕ is not an algebra homomorphism.

Lemma 11. The restriction map

End_{k_{-alg}}(A_{n}(k)) → End_{k_{-alg}}(Z_{n}(k))

is injective.

Proof. Let \overline{A_{n}(k)} be a copy of A_{n}(k) and consider two k-algebra homomorphisms

ϕ^{(1)}, ϕ^{(2)} : \overline{A_{n}(k)} → A_{n}(k).

Then we have k-algebra isomorphisms

ϕ^{(1)} ⊗ id_{Z_{n}(k)} : \overline{A_{n}(k)} ⊗\overline{Z_{n}(k)} Z_{n}(k) → A_{n}(k)

where \overline{Z_{n}(k)} is the center of \overline{A_{n}(k)}. Since these two isomorphisms coincide on the center, we deduce from the previous lemma that both maps coincide. □

4. Reduced trace and reduced norm

In this section we study some properties of reduced trace and reduced norm defined in subsection3.6.

4.1. calculation of reduced trace.

Lemma 12. The trace of a differential operator ξ^{k}(d/dξ)^{l} on a vector space k[ξ]/(ξ^{p}) is non zero if and only if k = l = p − 1.

[proof] If k ≠ l then ξ^{i}(d/dξ)^{j} is represented by a strictly triangular matrix and the trace is 0. If k = l, we have

(ξ^{l}(d/dξ)^{l}).ξ^{s} = s(s − 1) … (s − k + 1)ξ^{s}.

Summing up this we obtain the result.

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Corollary 6.

tr((μ + T)^{k}(ν + U)^{l}) = \begin{cases}−T^{k_{0}p}U^{l_{0}p} & if k = k_{0}p + (p − 1), l = l_{0}p + (p − 1) \\ 0 & otherwise \end{cases}

Lemma 13. (1) If 0 ≤ i, j ≤ p − 1, then trd(ξ^{i}η^{j}) is an element of k. (2)

trd(ξ^{i}η^{j}) = \begin{cases}−1 & (if i = j = p − 1) \\ 0 & otherwise \end{cases}

A formula for the reduced trace when n > 1 is easily obtained by noting that A_{n}(k) is isomorphic to a tensor product A_{1}(k) ⊗_{k} A_{1}(k) ⊗_{k} ⋯ ⊗_{k} A_{1}(k) of A_{1}(k) ’s.

Lemma 14. Let k be a field of characteristic p. Let ϕ be a k-endomorphism of A_{n}(k). Let I, J be an element of {0, 1, 2, …, p − 1}^{n} (multi-index). Then for any f ∈ Z(A_{n}(k)), we have

trd(ξ^{I}η^{J}f) = \begin{cases}−f & if I = J = (p − 1, p − 1, …, p − 1) \\ 0 & otherwise \end{cases}

4.2. invariance of trace and inversion formula.

Lemma 15. Let k be a field of characteristic p. Let ϕ be a k-endomorphism of A_{n}(k). Let I, J be an element of {0, 1, 2, …, p − 1}^{n} (multi-index). Then for any f ∈ Z(A_{n}(k)), we have

trd(ϕ(ξ^{I}η^{J}f)) = \begin{cases}−ϕ(f) & if I = J = (p − 1, p − 1, …, p − 1) \\ 0 & otherwise \end{cases}

Proof. This is the same as Corollary 5. One may also prove this by using Corollary 2.

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For any index set I ∈ ℕ^{2n}, we denote by I^{c} the index set (p−1, p−1, …, p−1)−I.

Corollary 7. Let k be a field of characteristic p. Let ϕ be a k-endomorphism of A_{n}(k). Assume we are given a set of p^{2n} elements {f_{IJ} ∈ Z(A_{n}(k)); I, J ⊂ {0, 1, 2, …, p − 1}^{n}}. Let M be the maximum of total degree of ϕ(ξ_{1}), …, ϕ(ξ_{n}), ϕ(η_{1}), …, ϕ(η_{n}). Then for any set {f_{IJ}} of p^{2n} elements of Z(A_{n}(k)), we have

totaldeg(ϕ(∑_{I,J} f_{IJ}ξ^{I}η^{J})) ≥ max_{I,J}(totaldeg ϕ(f_{IJ})) − Mp^{2n}

Proof. Put F = ϕ(∑_{I,J} f_{IJ}ξ^{I}η^{J}). Then

trd F ϕ(η)^{J^{c}}ϕ(ξ)^{I^{c}} = f_{IJ}

Thus totaldeg(F ϕ(η)^{J^{c}}ϕ(ξ)^{I^{c}}) ≥ totaldeg(f_{IJ}). Noting that total degree is additive (totaldeg(F G) = totaldeg(F) + totaldeg(G)), we complete the proof.

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Proposition 1 (inversion formula). Let k be a field of characteristic p. Assume we have an injective algebra endomorphism ϕ of A_{n}(k). We use the notation \overline{ξ_{i}} = ϕ(ξ_{i}), \overline{η_{j}} = ϕ(η_{j}). Then for any element x ∈ A_{n}(k), we have

x = − ∑_{I,J} trd(x\overline{η}^{J^{c}} \overline{ξ}^{I^{c}})\overline{ξ}^{I} \overline{η}^{J}

Corollary 8. Under the assumption of the Lemma above, if elements trd(ξ_{i}\overline{η}^{J^{c}} \overline{ξ}^{I^{c}}), trd(η_{i}\overline{η}^{J^{c}} \overline{ξ}^{I^{c}}) (i = 1, …, n) are all constants, then ϕ is invertible.

5. injectivity (n = 1)

5.1. comments on this section. When characteristic of a field k is 0, then we know that A_{n}(k) is simple and that any algebra endomorphism is injective. Even if the characteristic of k is nonzero, all examples of algebra endomorphisms of A_{n}(k) author knows are injective.

In this section we prove this is true for n = 1 case.

5.2. injectivity(n = 1).

Lemma 16. ϕ : A_{n}(k) → A_{n}(k) be a k-homomorphism. Then trans.deg(ϕ(ξ_{1}^{p}), …, ϕ(ξ_{n}^{p})) is equal to n.

[proof] We may assume that k is algebraically closed. The field k(ξ_{1}, …, ξ_{n}) has n linearly independent k-derivations {ad(η_{i})}_{i=1}^{n}. Since k(ξ_{1}, …, ξ_{n}) is separable over k, its transcendent degree is equal to the number of linear independent derivations [3, Theorem 4.4.2.]. Thus we conclude that transcendent degree of k(ξ_{1}, …, ξ_{n}) is no less than (hence is equal to) n. That means, ξ_{1}, …, ξ_{n} are algebraically independent over k.

Lemma 17 (n = 1). Any k-algebra endomorphism ϕ : A_{1}(k) → A_{1}(k) is injective.

[proof] We may assume that the base field k is algebraically closed. ϕ(A_{1}(k)) has no zero-divisor except for 0. Thus

ϕ(A_{1}(k)) ⊗_{ϕ(Z(A_{1}(k))} Q(ϕ(Z(A_{1}(k)))

is a skew field which is of finite rank over Q(ϕ(Z(A_{1}(k)))). If the transcendent degree of the field K is 1, then it contradicts with Tsen’s theorem. Thus ϕ(ξ^{p}), ϕ(η^{p}) are algebraically independent over k.

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6. multidegree monoids and lattices

For any element f = ∑_{I,J} f_{IJ}ξ^{I}η^{J} ∈ A_{n}(k), we denote by multideg(f) the multidegree of f. That means,

multideg(f) = max{(I, J) ∈ ℕ^{2n}; f_{IJ} ≠ 0},

where we employ the lexicographic order on ℕ^{2n}. For any subalgebra A of A_{n}(k), we define its multidegree monoid multideg(A) as follows.

multideg(A) = {multideg(f); f ∈ A}.

It is a sub monoid of ℕ^{2n}. We use several norms for indices in ℕ^{2n}. Among them is the “ℓ^{1}-norm” |t_{1}| + ⋯ + |t_{2n}|. of (t_{1}, …, t_{2n}) The total degree totaldeg(f) of an element f of A_{n}(k) is then defined to be the ℓ^{1}-norm of multideg(f). By a total degree of a derivation or an algebra endomorphism of A_{n}(k) we mean the maximum of the total degree of the image of the standard generators ξ_{1}, …, ξ_{n}, η_{1}, …, η_{n}.

Definition 2. For any subset S of ℕ^{2n}, we denote by S_{≤d} the set of all elements of S whose ℓ^{1}-norms are less than or equal to d. We denote by ^{a}HF ”the affine Hilbert function” of S, that is,

aHF_{S}(i) = #(S_{≤i}) (i ∈ ℕ)

For any subalgebra A of A_{n}(k), we define its affine Hilbert function ^{a}HF_{A} as the affine Hilbert function ^{a}HF_{multideg A} of multidegree of A.

Lemma 18. Let m be a positive integer. Let S be a sub monoid of ℕ^{m}. Let L = ℤS be the submodule of ℤ^{n} generated by S. Then the following conditions are equivalent.

(1) There exists a positive real number ϵ such that ^{a}HF_{S}(i) ≥ ϵi^{m} for all i >> 0.

(2) rank(L) = m

Proof. (2) ⟹ (1): Take I_{1}, I_{2}, …, I_{m} ∈ S which are linearly independent over ℚ. Put c = max(||I_{1}||_{ℓ^{1}}, ||I_{2}||_{ℓ^{1}}, …, ||I_{n}||_{ℓ^{1}}). Then a map α defined by

α : ℕ^{m} ∋ (a_{1}, …, a_{m}) ↦ a_{1}I_{1} + a_{2}I_{2} + ⋯ + a_{m}I_{m} ∈ S

is injective and satisfies α(ℕ_{≤d}^{m}) ⊂ S_{≤cd} for every positive integer d. Thus we have

aHF_{S}(d) ≥ \binom{m + [d/c]}{m} ≥ \frac{d^{m}}{(c + 1)^{m}m!}

when d is large enough. ([•] denotes the Gaussian symbol.)

(1) ⟹ (2): Assume on the contrary that r = rank(L) < m. Then the module L, being torsion free, is isomorphic to ℤ^{r}. Let I_{1}, …, I_{r} be elements of S which forms a ℤ-basis of L. Then a map

β : ℝ^{r} ∋ (t_{1}, t_{2}, …, t_{r}) ↦ (t_{1}I_{1} + t_{2}I_{2} + ⋯ + t_{r}I_{r}) ∈ ℝ^{m}

is an injective linear map from ℝ^{r} to ℝ^{m}. Thus we may easily see that there exists a real number M such that

|| ∑_{i=1}^{r} t_{i}I_{i}||_{ℓ^{1}} < 1 ⟹ ||(t_{1}, …, t_{r})||_{ℓ^{1}} < M

This implies that #S_{<d} is smaller than the number of elements of ℤ^{r} which are shorter (in ℓ^{1}-norm) than Md. □

6.1. injectivity for almost all primes.

Lemma 19. Let k be a field. Assume ϕ : A_{n}(k) → A_{n}(k) is injective (which is always the case if char k = 0). Then rank(ℤ multideg(ϕ(A_{n}(k))) = 2n.

Proof. Let N be the maximum of total degrees of elements ϕ(ξ_{1}), …, ϕ(ξ_{n}), ϕ(η_{1}), …, ϕ(η_{n}). Then we see that

ϕ(A_{n}(k)_{≤i}) ⊂ A_{n}(k)_{≤Ni}

holds for any i > 0. Thus for any i > 0, we have

aHF_{ϕ(A_{n}(k))}(Ni) ≥ ^{a}HF_{A_{n}(k)}(i) = \binom{i + 2n}{i}.

This together with Lemma 18 gives the result. □

Lemma 20. Let A be a subalgebra of a polynomial algebra P = k[X_{1}, X_{2}, …, X_{m}] over a field k. If there exists an positive integer c such that

dim(A_{≤d}) ≥ \frac{1}{cm!}d^{m}

holds for all integers d > 0, then we have

[Q(P) : Q(A)] ≤ c

Proof. Suppose on the contrary that [Q(P) : Q(A)] > c. Take elements f_{1}, …, f_{c+1} ∈ Q(P) which are linearly independent over Q(A). By multiplying a “common denominator”, we may assume that f_{i} are elements of P. Then it follows that the sum

f_{1}A + f_{2}A + ⋯ + f_{c+1}A

is direct in P. Let M be the maximum of total degrees of f_{1}, …, f_{c+1}. Then the directness above implies that an inequality

dim ((f_{1}A + f_{2}A + ⋯ + f_{c+1}A)_{≤d}) ≥ (c + 1) dim A_{≤(d−M)}

holds for any integer d > M. Since the left hand side is not greater than dim P_{≤d}, we obtain the following inequation

\frac{d^{m}}{m!} + O(d^{m−1}) ≥ (c + 1)\frac{1}{cm!}(d − M)^{m}

which leads to a contradiction when d is large enough. □

Proposition 2 (injectivity for almost all primes). Let 𝔎 be an algebraic number field, 𝔒 = 𝔒(𝔎) be the ring of integers in 𝔎. suppose we are given an 𝔎-algebra endomorphism ϕ of A_{n}(𝔎).

Then the multidegree monoid of the image Image(ϕ) has rank 2n. Furthermore, for almost all prime ideals 𝔭 of 𝔒, we have the following facts.

(1) ϕ induces an k(𝔭)-algebra endomorphism ϕ_{𝔭} of A_{n}(k(𝔭)) (where k(𝔭) = 𝔒/𝔭 is the residue field of 𝔭).

(2) The multidegree monoid of the image Image(ϕ_{𝔭}) has rank 2n.

(3) ϕ_{𝔭} is injective.

(4) There exists a constant C such that geomdeg(ϕ_{𝔭}) ≤ C for all 𝔭.

Proof. The first statement is an easy consequence of Lemma 19. This in turn implies (2) (except for finite primes). In precise, let x_{1}, x_{2}, …, x_{n} be elements in A_{n}(𝔎) such that multi degrees of ϕ(x_{1}), ϕ(x_{2}), …, ϕ(x_{2n}) are linearly independent. Then for almost all primes, their reductions x_{1}, x_{2}, …, x_{n} are defined as elements of A_{n}(k(𝔭)) and their multi degrees stays invariant under the reduction.

We apply Lemma 18 to see that there exists a positive real number ϵ which is independent of 𝔭 such that an inequality

dim(ϕ_{𝔭}(A_{n}(k))_{≤s}) ≥ ϵs^{2n}

holds for any large integer s.

(1) is already proved in subsection 3.1. To prove (4), we denote by k = k(𝔭) the quotient field with characteristic p(> 0). Since A_{n}(k) is a free Z_{n}(k)-module of rank 2n with generators {ξ^{I}η^{J}; ||I||_{ℓ^{∞}}, ||J||_{ℓ^{∞}} ≤ p − 1}, we have

ϕ_{𝔭}(Z_{n}(k)) · ( ∑_{||I||_{ℓ^{∞}} ,||J||_{ℓ^{∞}} ≤p−1} kϕ_{𝔭}(ξ)^{I}ϕ_{𝔭}(η)^{J}) = ϕ_{𝔭}(A_{n}(k)).

Then Corollary 7 gives us an relationship of total degrees of both hands sides. Namely, there exists a positive number N such that

ϕ_{𝔭}(Z_{n}(k))_{≤s+N} · ( ∑_{||I||_{ℓ^{∞}} ,||J||_{ℓ^{∞}} ≤p−1} kϕ_{𝔭}(ξ)^{I}ϕ_{𝔭}(η)^{J}) ⊃ ϕ_{𝔭}(A_{n}(k))_{≤s}

holds for any positive integer s.

Combining these, we obtain

dim(ϕ_{𝔭}(Z_{n}(k))_{≤s+N})p^{2n} ≥ ϵs^{2n}

Then we use Lemma20 to see that (4) is true. (One should be very careful about grading of Z_{n}(k) here. degrees of generators ξ_{i}^{p}, η_{i}^{p} of Z_{n}(k) are all p. not 1.) (3) follows from (4) and a consideration on transcendence degrees.

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References

[1] P. K. Draxl, Skew fields (London mathematical society lecture note series, Cambridge Univ. press, 1983.

[2] D. Mumford, Tata lectures on theta II, Birkhäuser, 1984.

[3] M. Nagata, Field theory, Marcel Dekker, 1977.

Kochi University Department of Mathematics Faculty of Science Kochi, 780 Japan

                                   E-mail: docky@math.kochi-u.ac.jp