Complexity of Janet basis of a D-module

TL;DR

This paper establishes a double-exponential upper bound on the degree and complexity of Janet bases for D-modules, extending known bounds from polynomial modules to a more general non-commutative setting.

Contribution

It provides the first known double-exponential bound for Janet bases of D-modules, highlighting a significant complexity result in non-commutative algebra.

Findings

Double-exponential upper bound on degree and complexity

Generalization from polynomial modules to D-modules

Bound cannot be derived directly from commutative case

Abstract

We prove a double-exponential upper bound on the degree and on the complexity of constructing a Janet basis of a $D$-module. This generalizes a well known bound on the complexity of a Gr\"obner basis of a module over the algebra of polynomials. We would like to emphasize that the obtained bound can not be immediately deduced from the commutative case.

Equations

$$(dl{)}^{2^{O(n)}}.$$

$$X_v{X}_w=X_w{X}_v,D_v{D}_w=D_w{D}_v,D_v{X}_v-X_v{D}_v=1,X_v{D}_w=D_w{X}_v,v\ne w.$$

$$f=\sum _{i_1,\dots ,i_n,j_1,\dots ,j_n\ge 0}{f}_{i_1,\dots ,i_n,j_1,\dots ,j_n}{X}_1^{i_1}\dots X_n^{i_n}{D}_1^{j_1}\dots D_n^{j_n},$$

$$\begin{array}[]{l}i=(i_{1},\ldots,i_{n}),\quad j=(j_{1},\ldots,j_{n}),\quad f_{i,j}=f_{i_{1},\ldots,i_{n},j_{1},\ldots,j_{n}}\\ X^{i}=X_{1}^{i_{1}}\ldots X_{n}^{i_{n}},\quad D^{j}=D_{1}^{j_{1}}\ldots D_{n}^{j_{n}},\quad f=\sum_{i,j}f_{i,j}X^{i}D^{j},\\ |i|=i_{1}+\ldots+i_{n},\quad i+j=(i_{1}+j_{1},\ldots,i_{n}+j_{n}).\end{array}$$

$$\mathrm{deg}f={\mathrm{deg}}_{X_1,\dots ,X_n,D_1,\dots ,D_n}f=\max \{|i|+|j|:f_{i,j}\ne 0\}.$$

$$\sum _{1\le w\le l}{a}_{v,w}{m}_w,1\le v\le k.$$

$$A^k\stackrel{\iota }{\to }{A}^l\stackrel{\pi }{\to }M\to 0$$

$$(a_{v,1},\dots ,a_{v,l})\in A^l,1\le v\le k.$$

$$A_m=\{a:\mathrm{deg}a\le m\},M_m=\pi (A_m^l),I_m=I\cap A_m^l.$$

$$0\to I_m\to A_m^l\to M_m\to 0$$

$$H(M,m)={\dim }_F{M}_m,m\ge 0.$$

$$f=\sum _{v,i,j}{f}_{v,i,j}{e}_{v,i,j},f_{v,i,j}\in F.$$

$$o(f)=\max \{(v,i,j):f_{v,i,j}\ne 0\},$$

$$o(0)=-\infty <o(f)$$

$$\mathrm{Hdt}(f)=f_{v,i,j}{e}_{v,i,j},$$

$$\mathrm{Hdt}(I)=\sum _{f\in I}A\mathrm{Hdt}(f).$$

$$\mathrm{gr}(A)={\oplus }_{m\ge 0}{A}_m/A_{m-1},\mathrm{gr}(I)={\oplus }_{m\ge 0}{I}_m/I_{m-1},\mathrm{gr}(M)={\oplus }_{m\ge 0}{M}_m/M_{m-1}.$$

$$0\to I_m/I_{m-1}\to (A_m/A_{m-1}{)}^l\to M_m/M_{m-1}\to 0,m\ge 0.$$

$$H(\mathrm{gr}(M),m)={\dim }_F{M}_m/M_{m-1},m\ge 0.$$

$$H(M,m)=\sum _{0\le v\le m}H(\mathrm{gr}(M),v),H(\mathrm{gr}(M),m)=H(M,m)-H(M,m-1).$$

$$I_{m}=\Bigl{\{}\sum_{1\leq v\leq\mu}c_{v}b_{v}\,:\,c_{v}\in A,\quad\deg c_{v}\leq m-\nu_{v},\quad 1\leq i\leq s\Bigr{\}}.$$

$$\begin{array}[]{l}X_{v}X_{w}=X_{w}X_{v},\;D_{v}D_{w}=D_{w}D_{v},\quad\mbox{for all}\quad v,w,\\ D_{v}X_{v}-X_{v}D_{v}=X_{0}^{2},\;1\leq v\leq n,\quad X_{v}D_{w}=D_{w}X_{v}\quad\mbox{for all}\quad v\neq w.\\ \end{array}$$

$$f=\sum _{i_0,i_1,\dots ,i_n,j_1,\dots ,j_n\ge 0}{f}_{i_0,\dots ,i_n,j_1,\dots ,j_n}{X}_0^{i_0}\dots X_n^{i_n}{D}_1^{j_1}\dots D_n^{j_n},$$

$$\begin{array}[]{l}i=(i_{1},\ldots,i_{n}),\quad j=(j_{1},\ldots,j_{n}),\quad f_{i_{0},i,j}=f_{i_{0},\ldots,i_{n},j_{1},\ldots,j_{n}}\\ f=\sum_{i_{0},i,j}f_{i_{0},i,j}X_{0}^{i_{0}}X^{i}D^{j}.\end{array}$$

$$\mathrm{deg}f={\mathrm{deg}}_{X_0,\dots ,X_n,D_1,\dots ,D_n}f=\max \{i_0+|i|+|j|:f_{i_0,i,j}\ne 0\},$$

$${\mathrm{deg}}_{D_1,\dots ,D_n}f=\max \{|j|:f_{i_0,i,j}\ne 0\},$$

$${\mathrm{deg}}_{D_{\alpha }}f=\max \{j_{\alpha }:f_{i_0,i,j}\ne 0\},1\le \alpha \le n$$

$${\mathrm{deg}}_{X_{\alpha }}f=\max \{i_{\alpha }:f_{i_0,i,j}\ne 0\},1\le \alpha \le n$$

$$\mathrm{ord}f={\mathrm{ord}}_{X_0}f=\mu \text{mbox}ifandonlyiff\in X_0^{\mu }({}^{h}A)\setminus X_0^{\mu +1}({}^{h}A),\mu \ge 0.$$

$$\mathrm{ord}z=\underset{1\le i\le l}{\min }\{\mathrm{ord}{z}_i\},\mathrm{deg}z=\underset{1\le i\le l}{\max }\{\mathrm{deg}{z}_i\}.$$

$$({}^{h}A{)}_m=\left\{z\in {}^{h}A:z\text{mbox}ishomogeneous\&\mathrm{deg}z=m\text{mbox}orz=0\right\}$$

$${}^{h}z=\sum _j{z}_j{X}_0^{\mathrm{deg}z-\mathrm{deg}{z}_j}.$$

Full text

Introduction

Let $A$ be the Weyl algebra $F[X_{1},\dots,X_{n},$ ${\partial\over\partial X_{1}},$ $\dots,{\partial\over\partial X_{n}}]$ (or the algebra of differential operators $F(X_{1},$ $\dots,X_{n})[{\partial\over\partial X_{1}},\dots,{\partial\over\partial X_{n}}]$). Denote for brevity $D_{i}={\partial\over\partial X_{i}}$, $1\leq i\leq n$. Any $A$–module is called $D$–module. It is well known that an $A$–module which is a submodule of a free finitely generated $A$-module has a Janet basis. Historically, it was first introduced in [9]. In more recent times of developing computer algebra Janet bases were studied in [5], [13], [10]. Janet bases generalize Gröbner bases which were widely elaborated in the algebra of polynomials (see e. g.[3]). For Gröbner bases a double-exponential complexity bound was obtained in [12], [6] relying on [1] and which was made more precise (with a self–contained proof) in [4].

Surprisingly, no complexity bound on Janet bases was established so far; in the present paper we fill this gap and prove a double-exponential complexity bound. On the other hand, a double-exponential complexity lower bound on Gröbner bases [12], [14] provides by the same token a bound on Janet bases.

There is a folklore opinion that the problem of constructing a Janet basis is easily reduced to the commutative case by considering the associated graded module, and, on the other hand, in the commutative case [6], [12], [4] the double–exponential upper bound is well known. But it turns out to be a fallacy! From a known system of generators of a $D$-module one can not obtain immediately any system of generators (even not necessarily a Gröbner basis) of the associated graded module. The main problem here is to construct such a system of generators of the graded module. It may have the elements of degrees $(dl)^{2^{O(n)}}$, see the notation below. Then, indeed, to the last system of generators of big degrees one can apply the result known in the commutative case and get the bound $((dl)^{2^{O(n)}})^{2^{O(n)}}=(dl)^{2^{O(n)}}$. So new ideas specific to non–commutative case are needed.

We are interested in the estimations for Janet bases of $A$-submodules of $A^{l}$. The Janet basis depends on the choice of the linear order on the monomials (we define them also for $l>1$). In this paper we consider the most general linear orders on the monomials from $A^{l}$. They satisfy conditions (a) and (b) from Section 1 and are called admissible. We prove the following result.

THEOREM 1

For any admissible linear order on the monomials from $A^{l}$ any $A$-submodule $I$ of $A^{l}$ generated by elements of degrees at most $d$ (with respect to the filtration in the corresponding algebra, see Section 1 and Section 9) has a Janet basis with the degrees and the number of its elements less than

[TABLE]

We prove in detail this theorem for the case of the Weyl algebra $A$. The proof for the case of the algebra of differential operators is similar. It is sketched in Section 9. ¿From Theorem 1 we get that the Hilbert function $H(I,m)$, see Section 1, of the $A$-submodule from this theorem is stable for $m\geq(dl)^{2^{O(n)}}$ and the absolute values of all coefficients of the Hilbert polynomial of $I$ are bounded from above by $(dl)^{2^{O(n)}}$, cf. e.g., [12]. This fact follows directly from (10), Lemma 12 from Appendix 1, Lemma 2 and Theorem 2. We mention that in [7] the similar bound was shown on the leading coefficient of the Hilbert polynomial.

Now we outline the plan for the proof of Theorem 1. The main tool in the proof is a homogenized Weyl algebra ${{}^{h}\!A}$ (or respectively, a homogenized algebra of differential operators ${{}^{h}\!B}$). It is introduced in Section 3 (respectively, Section 9). The algebra ${{}^{h}\!A}$ (respectively ${{}^{h}\!B}$) is generated over the ground field $F$ by $X_{0},\ldots,X_{n}$, $D_{1},\ldots,D_{n}$ (respectively over the field $F(X_{1},\ldots,X_{n})$ by $X_{0},D_{1},\ldots,D_{n}$). Here $X_{0}$ is a new homogenizing variable. In the algebra ${{}^{h}\!A}$ (respectively ${{}^{h}\!B}$) relations (12) Section 3 (respectively (50) Section 9) hold for these generators in ${{}^{h}\!A}$.

We define the homogenization ${{}^{h}\!I}$ of the module $I$. It is a ${{}^{h}\!A}$–submodule of ${{}^{h}\!A^{l}}$. The main problem is to estimate the degrees of a system of generators of ${{}^{h}\!I}$. These estimations are central in the paper. They are deduced from Theorem 2 Section 7. This theorem is devoted to the problem of solving systems of linear equations over the ring ${{}^{h}\!A}$; we discuss it below in more detail.

The system of generators of ${{}^{h}\!I}$ gives a system of generators of the graded $\mathop{\rm gr}\nolimits(A)$–module $\mathop{\rm gr}\nolimits(I)$ corresponding to $I$. But $\mathop{\rm gr}\nolimits(A)$ is a polynomial ring. Hence using Lemma 12 Appendix 1 we get a double–exponential bound $(dl)^{2^{O(n)}}$ on the stabilization of the Hilbert function of $\mathop{\rm gr}\nolimits(I)$ and the absolute values of the coefficients of the Hilbert polynomial of $\mathop{\rm gr}\nolimits(I)$. Therefore, the similar bound holds for the stabilization of the Hilbert functions of $I$ and the coefficients of the Hilbert polynomial of $I$, see Section 2.

But the Hilbert functions of the modules $I$ and ${{}^{h}\!I}$ coincide, see Section 3. Hence the last bound holds also for the stabilization of the Hilbert functions of ${{}^{h}\!I}$ and the coefficients of the Hilbert polynomial of ${{}^{h}\!I}$. In Section 5 we introduce the linear order on the monomials from ${{}^{h}\!A^{l}}$ induced by the initial linear order on the monomials from $A^{l}$ (the homogenizing variable $X_{0}$ is the least possible in this ordering). Further, we define the Janet basis of ${{}^{h}\!I}$ with respect to the induced linear order on the monomials. Such a basis can be obtained by the homogenization of the elements of a Janet basis of $I$ with respect to the initial linear order, see Lemma 3.

Let $\mathop{\rm Hdt}\nolimits({{}^{h}\!I})$ be the monomial module (i.e., the module which has a system of generators consisting of monomials) generated by the greatest monomials of all the elements of the module ${{}^{h}\!I}$, see Section 4. Let ${{}^{c}\!I}$, see Section 4, be the module over the polynomial ring ${{}^{c}\!A}=F[X_{0},\ldots,X_{n},D_{1},\ldots,D_{n}]$ generated by all the monomials from $\mathop{\rm Hdt}\nolimits({{}^{h}\!I})$ (they are considered now as elements of ${{}^{c}\!A}$). Then the Hilbert functions of the modules ${{}^{h}\!I}$ and ${{}^{c}\!I}$ coincide. Thus, we have the same as above double–exponential estimation for the stabilization of the Hilbert functions of ${{}^{c}\!I}$ and the coefficients of the Hilbert polynomial of ${{}^{c}\!I}$. Now using Lemma 13 we get the estimation $(dl)^{2^{O(n)}}$ on the monomial system of generators of ${{}^{c}\!I}$, hence also of $\mathop{\rm Hdt}\nolimits({{}^{h}\!I})$. This gives the bound for the degrees of the elements of the Janet bases of ${{}^{h}\!I}$ and hence also for the required Janet basis of $I$, and proves Theorem 1.

The problem of solving systems of linear equations over the homogenized algebra is central in this paper, see Theorem 2. It is studied in Sections 5–7. A similar problem over the Weyl algebra (without a homogenization) was considered in [7]. The principal idea is to try to extend the well known method due to G.Hermann [8] which was elaborated for the algebra of polynomials, to the homogenized Weyl algebra. There are two principal difficulties on this way. The first one is that in the method of G.Hermann the use of determinants is essential which one has to avoid dealing with non-commutative algebras. The second is that one needs a kind of the Noether normalization theorem in the situation under consideration. So it is necessary to choose the leading elements in the analog of the G.Hermann method with the least $\mathop{\rm ord}\nolimits_{X_{0}}$, where $X_{0}$ is a homogenizing variable, see Section 3.

The obtained bound on the degree of a Janet basis implies a similar bound on the complexity of its constructing. Indeed, by Corollary 1 (it is formulated for the case of Weyl algebra but the analogous corollary holds for the case of algebra of differential operators) one can compute the linear space of all the elements $z\in I$ of degrees bounded from above by $(dl)^{2^{O(n)}}$. Further, by Theorem 1 the module $\mathop{\rm Hdt}\nolimits(I)$, see Section 1, is generated by all the elements $\mathop{\rm Hdt}\nolimits(z)$ with $z\in I$ of degrees bounded from above by $(dl)^{2^{O(n)}}$. Hence one can compute a system of generators of $\mathop{\rm Hdt}\nolimits(I)$ and a Janet basis of $I$ solving linear systems over $F$ of size bounded from above $(dl)^{2^{O(n)}}$ (just by the enumeration of all monomials of degrees at most $(dl)^{2^{O(n)}}$ which are possible generators of $\mathop{\rm Hdt}\nolimits(I)$). If one needs to construct the reduced Janet basis it is sufficient to apply additionally Remark 1 Section 4.

For the sake of self–containedness in Appendix 1, see Lemma 12, we give a short proof of the double–exponential estimation for stabilization of the Hilbert function of a graded module over a homogeneous polynomial ring. A conversion of Lemma 12 also holds, see Appendix 1 Lemma 13. It is essential for us. The proof of Lemma 13 uses the classic description of the Hilbert function of a homogeneous ideal in $F[X_{0},\ldots,X_{n}]$ via Macaulay constants $b_{n+2},\ldots,b_{1}$ and the constant $b_{0}$ introduced in [4]. In Appendix 2 we give an independent and instructive proof of Proposition 1 which is similar to Lemma 13. In some sence Proposition 1 is even more strong than Lemma 13 since to apply it one does not need a bound for the stabilization of the Hilbert function. Of course, the reference to Proposition 1 can be used in place of Lemma 13 in our paper.

1 Definition of the Janet basis

Let $A=F[X_{1},\ldots,X_{n},D_{1},\ldots,D_{n}]$, $n\geq 1$, be a Weyl algebra over a field $F$ of zero–characteristic. So $A$ is defined by the following relations

[TABLE]

By (1) any element $f\in A$ can be uniquely represented in the form

[TABLE]

where all $f_{i_{1},\ldots,i_{n},j_{1},\ldots,j_{n}}\in F$ and only a finite number of $f_{i_{1},\ldots,i_{n},j_{1},\ldots,j_{n}}$ are nonzero. Denote for brevity ${\mathbb{Z}}_{+}=\{z\in{\mathbb{Z}}\,:\,z\geq 0\}$ to the set of all nonnegative integers and

[TABLE]

So $i,j\in{\mathbb{Z}}_{+}^{n}$ are multiindices. By definition the degree of $f$

[TABLE]

Let $M$ be a left $A$-module given by its generators $m_{1},\ldots,m_{l}$, $l\geq 0$, and relations

[TABLE]

where $k\geq 0$ and all $a_{v,w}\in A$. We assume that $\deg a_{v,w}\leq d$ for all $v,w$. By (4) we have the exact sequence

[TABLE]

of left $A$-modules. Denote $I=\iota(A^{k})\subset A^{l}$. If $l=1$ then $I$ is a left ideal of $A$ and $M=A/I$. In the general case $I$ is generated by the elements

[TABLE]

For an integer $m\geq 0$ put

[TABLE]

So now $A$, $M$, $I$ are filtered modules with filtrations $A_{m}$, $M_{m}$, $I_{m}$, $m\geq 0$, respectively and the sequence of homomorphisms of vector spaces

[TABLE]

induced by (5) is exact for every $m\geq 0$. The Hilbert function $H(M,m)$ of the module $M$ is defined by the equality

[TABLE]

Each element of $A^{l}$ can be uniquely represented as an $F$-linear combination of elements $e_{v,i,j}=(0,\ldots,0,X^{i}D^{j},0,\ldots,0)$, herewith $i,j\in{\mathbb{Z}}_{+}^{n}$ are multiindices, see (3), and the nonzero monomial $X^{i}D^{j}$ is at the position $v$, $1\leq v\leq l$. So every element $f\in A^{l}$ can be represented in the form

[TABLE]

The elements $e_{v,i,j}$ will be called monomials.

Consider a linear order $<$ on the set of all the monomials $e_{v,i,j}$ or which is the same on the set of triples $(v,i,j)$, $1\leq v\leq l$, $i,j\in{\mathbb{Z}}_{+}^{n}$. If $f\neq 0$ put

[TABLE]

see (7). Set

[TABLE]

for every $0\neq f\in A$. Let us define the leading monomial of the element $0\neq f\in A^{l}$ by the formula

[TABLE]

where $o(f)=(v,i,j)$. Put $\mathop{\rm Hdt}\nolimits(0)=0$. Hence $o(f-\mathop{\rm Hdt}\nolimits(f))<o(f)$ if $f\neq 0$. For $f_{1},f_{2}\in A^{l}$ if $o(f_{1})<o(f_{2})$ we shall write $f_{1}<f_{2}$. We shall require additionally that

1. (a)

for all multiindices $i,j,i^{\prime},j^{\prime}$ for all $1\leq v\leq l$ if $i_{1}\leq i^{\prime}_{1},\ldots,i_{n}\leq i^{\prime}_{n}$ and $j_{1}\leq j^{\prime}_{1},\ldots,j_{n}\leq j^{\prime}_{n}$ then $(v,i,j)\leq(v,i^{\prime},j^{\prime})$. 2. (b)

for all multiindices $i,j,i^{\prime},j^{\prime},i^{\prime\prime},j^{\prime\prime}$ for all $1\leq v,v^{\prime}\leq l$  if $(v,i,j)<(v^{\prime},i^{\prime},j^{\prime})$ then $(v,i+i^{\prime\prime},j+j^{\prime\prime})<(v^{\prime},i^{\prime}+i^{\prime\prime},j^{\prime}+j^{\prime\prime})$.

Conditions (a) and (b) imply that for all $f_{1},f_{2}\in A^{l}$ for every nonzero $a\in A$ if $f_{1}<f_{2}$ then $af_{1}<af_{2}$, i.e., the considered linear order is compatible with the products. Any linear order on monomials $e_{v,i,j}$ satisfying (a) and (b) will be called admissible.

Set

[TABLE]

So $\mathop{\rm Hdt}\nolimits(I)$ is an ideal of $A$. By definition the family $f_{1},\ldots,f_{m}$ of elements of $I$ is a Janet basis of the module $I$ if and only if

$\mathop{\rm Hdt}\nolimits(I)=A\mathop{\rm Hdt}\nolimits(f_{1})+\ldots+A\mathop{\rm Hdt}\nolimits(f_{m})$, i.e., the submodule of $A^{l}$ generated by $\mathop{\rm Hdt}\nolimits(f_{1}),\ldots,\mathop{\rm Hdt}\nolimits(f_{m})$ coincides with $\mathop{\rm Hdt}\nolimits(I)$.

Further, the Janet basis $f_{1},\ldots,f_{m}$ of $I$ is reduced if and only if the following conditions hold.

$f_{1},\ldots,f_{m}$ does not contain a smaller Janet basis of $I$, 2. 3)

$\mathop{\rm Hdt}\nolimits(f_{1})>\ldots>\mathop{\rm Hdt}\nolimits(f_{m})$. 3. 4)

the coefficient from $F$ of every monomial $\mathop{\rm Hdt}\nolimits(f_{v})$, $1\leq v\leq l$, is $1$. 4. 5)

Let $f_{\alpha}=\sum_{v,i,j}f_{\alpha,v,i,j}e_{v,i,j}$ be representation (2) for $f_{\alpha}$, $1\leq\alpha\leq m$. Then for all $1\leq\alpha<\beta\leq m$ for all $1\leq v\leq l$ and multiindices $i,j$ the monomial $f_{\alpha,v,i,j}e_{v,i,j}\not\in\mathop{\rm Hdt}\nolimits(Af_{\beta}\setminus\{0\})$.

Since the ring $A$ is Noetherian for considered $I$ there exists a Janet basis. Further the reduced Janet basis of $I$ is uniquely defined.

2 The graded

module corresponding to a $D$–module

Put $A_{v}=I_{v}=M_{v}=0$ for $v<0$ and

[TABLE]

The structure of the algebra on $A$ induces the structure of a graded algebra on $\mathop{\rm gr}\nolimits(A)$. So we have $\mathop{\rm gr}\nolimits(A)=F[X_{1},\ldots,X_{n},D_{1},\ldots,D_{n}]$ is an algebra of polynomials with respect to the variables $X_{1},\ldots,X_{n}$, $D_{1},\ldots,D_{n}$. Further, $\mathop{\rm gr}\nolimits(I)$ and $\mathop{\rm gr}\nolimits(M)$ are graded $\mathop{\rm gr}\nolimits(A)$-modules. From (6) we get the exact sequences

[TABLE]

The Hilbert function of the module $\mathop{\rm gr}\nolimits(M)$ is defined as follows

[TABLE]

Obviously

[TABLE]

for every $m\geq 0$.

Denote for an arbitrary $a\in M$ by $\mathop{\rm gr}\nolimits(a)\in\mathop{\rm gr}\nolimits(M)$ the image of $a$ in $\mathop{\rm gr}\nolimits(M)$.

LEMMA 1

Assume that $b_{1},\ldots,b_{s}$ is a system of generators of $I$. Let $\nu_{i}=\deg b_{i}$, $1\leq i\leq s$. Suppose that for every $m\geq 0$

[TABLE]

Then $\mathop{\rm gr}\nolimits(b_{1}),\ldots,\mathop{\rm gr}\nolimits(b_{s})$ is a system of generators of the $\mathop{\rm gr}\nolimits(A)$-module $\mathop{\rm gr}\nolimits(I)$.

PROOF This is straightforward.

So it is sufficient to construct a system of generators $b_{1},\ldots,b_{s}$ of $I$ satisfying (11).

3 Homogenization of the Weyl algebra

Let $X_{0}$ be a new variable. Consider the algebra ${{}^{h}\!A}=F[X_{0},X_{1},\ldots,X_{n},D_{1},$ $\ldots,D_{n}]$ given by the relations

[TABLE]

The algebra ${{}^{h}\!A}$ is Noetherian similarly to the Weyl algebra $A$. By (12) an element $f\in{{}^{h}\!A}$ can be uniquely represented in the form

[TABLE]

where all $f_{i_{0},\ldots,i_{n},j_{1},\ldots,j_{n}}\in F$ and only a finite number of $f_{i_{0},\ldots,i_{n},j_{1},\ldots,j_{n}}$ are nonzero. Let $i,j$ be multiindices, see (3). Denote for brevity

[TABLE]

By definition the degrees of $f$

[TABLE]

Set $\mathop{\rm ord}\nolimits 0=\mathop{\rm ord}\nolimits_{X_{0}}0=+\infty$. If $0\neq f\in{{}^{h}\!A}$ then put

[TABLE]

For every $z=(z_{1},\ldots,z_{l})\in{{}^{h}\!A^{l}}$ put

[TABLE]

Similarly one defines $\mathop{\rm ord}\nolimits b$ and $\deg b$ for an arbitrary $(k\times l)$–matrix $b$ with coefficients from ${{}^{h}\!A}$. More precisely, one consider here $b$ as a vector with $kl$ entries.

The element $f\in{{}^{h}\!A}$ is homogeneous if and only if $f_{i_{0},i,j}\neq 0$ implies $i_{0}+|i|+|j|=\deg f$, i.e., if and only if $f$ is a sum of monomials of the same degree $\deg f$. The homogeneous degree of a nonzero homogeneous element $f$ is its degree. The homogeneous degree of [math] is not defined ([math] belongs to all the homogeneous components of ${{}^{h}\!A}$, see below).

The $m$-th homogeneous component of ${{}^{h}\!A}$ is the $F$-linear space

[TABLE]

for every integer $m$. Now ${{}^{h}\!A}$ is a graded ring with respect to the homogeneous degree. By definition the ring ${{}^{h}\!A}$ is a homogenization of the Weyl algebra $A$.

We shall consider the category of finitely generated graded modules $G$ over the ring ${{}^{h}\!A}$. Such a module $G=\oplus_{m\geq m_{0}}G_{m}$ is a direct sum of its homogeneous components $G_{m}$, where $m,m_{0}$. are integers. Every $G_{m}$ is a finite dimensional $F$-linear space and $({{}^{h}\!A})_{p}G_{m}\subset G_{p+m}$ for all integers $p,m$. If $G$ and $G^{\prime}$ are two finitely generated graded ${{}^{h}\!A}$-modules then $\varphi\,:\,G\rightarrow G^{\prime}$ is a morphism (of degree [math]) of the graded modules if and only if $\varphi$ is a morphism of ${{}^{h}\!A}$-modules and $\varphi(G_{m})\subset G^{\prime}_{m}$ for every integer $m$.

The element $z\in{{}^{h}\!A}$ (respectively $z\in A$) is called to be the term if and only if $z=\lambda z_{1}\cdot\ldots\cdot z_{\nu}$ for some $0\neq\lambda\in F$, integer $\nu\geq 0$ and $z_{w}\in\{X_{0},\ldots,X_{n},D_{1},\ldots,D_{n}\}$ (respectively $z_{w}\in\{X_{1},\ldots,X_{n},D_{1},\ldots,D_{n}\}$), $1\leq w\leq\nu$.

Let $z=\sum_{j}z_{j}\in A$ be an arbitrary element of the Weyl algebra $A$ represented as a sum of terms $z_{j}$ and $\deg z=\max_{j}\deg z_{j}$. One can take here, for example, representation (3) for $z$. Then we define the homogenization ${{}^{h}\!z}\in{{}^{h}\!A}$ by the formula

[TABLE]

By (1), (12) the right part of the last equality does not depend on the chosen representation of $z$ as a sum of terms. Hence ${{}^{h}\!z}$ is defined correctly. If $z\in{{}^{h}\!A}$ then ${{}^{a}\!z}\in A$ is obtained by substituting $X_{0}=1$ in $z$. Hence for every $z\in A$ we have ${{}^{a}\!{{}^{h}\!z}}=z$, and for every $z\in{{}^{h}\!A}$ the element $z={{}^{h}\!{{}^{a}\!z}}X_{0}^{\mu}$, where $\mu=\mathop{\rm ord}\nolimits z$.

For an element $z=(z_{1},\ldots,z_{l})\in A^{l}$ put $\deg z=\max_{1\leq i\leq l}\{\deg z_{i}\}$ and

[TABLE]

Similarly one defines $\deg a$ and the homogenization ${{}^{h}\!a}=(a_{v,w})_{1\leq v\leq k,\,1\leq w\leq l}$ for an arbitrary $k\times l$–matrix $a$ with coefficients from $A$. More precisely, one consider here $a$ as a vector with $kl$ entries. Hence if $b=(b_{v,w})_{1\leq v\leq k,\,1\leq w\leq l}={{}^{h}\!a}$ then $b_{v,w}={{}^{h}\!a}_{v,w}X_{0}^{\deg a-\deg a_{v,w}}$ for all $v,w$.

The $m$-th homogeneous component of ${{}^{h}\!A^{l}}$ is

[TABLE]

For an $F$-linear subspace $X\subset A^{l}$ put ${{}^{h}\!X}$ to be the least linear subspace of ${{}^{h}\!A^{l}}$ containing the set $\{{{}^{h}\!z}\,:\,z\in X\}$. If $X$ is a (finitely generated) $A$-submodule of $A^{l}$ then ${{}^{h}\!X}$ is a (finitely generated) graded submodule of ${{}^{h}\!A^{l}}$. The graduation on ${{}^{h}\!X}$ is induced by the one of ${{}^{h}\!A^{l}}$.

For an element $z=(z_{1},\ldots,z_{l})\in{{}^{h}\!A^{l}}$ put ${{}^{a}\!z}=({{}^{a}\!z}_{1},\ldots,{{}^{a}\!z}_{l})\in A^{l}$. For a subset $X\subset{{}^{h}\!A^{l}}$ put ${{}^{a}\!X}=\{{{}^{a}\!z}\,:\,z\in X\}\subset A^{l}$. If $X$ is a $F$-linear space then ${{}^{a}\!X}$ is also a $F$-linear space. If $X$ is a finitely generated graded submodule of ${{}^{h}\!A^{l}}$ then ${{}^{a}\!X}$ is finitely generated submodule of $A^{l}$.

Now ${{}^{h}\!I}$ is a graded submodule of ${{}^{h}\!A^{l}}$. Further, ${{}^{a}\!{{}^{h}\!I}}=I$. Let $({{}^{h}\!I})_{m}$ be the $m$-th homogeneous component of ${{}^{h}\!I}$. Then

[TABLE]

and (17) induces the isomorphism $\iota\,:\,({{}^{h}\!I})_{m}\rightarrow I_{m}$. Set ${{}^{h}\!M}={{}^{h}\!A^{l}}/{{}^{h}\!I}$. Hence ${{}^{h}\!M}$ is a graded ${{}^{h}\!A}$-module and we have the exact sequence

[TABLE]

The $m$-th homogeneous component $({{}^{h}\!M})_{m}$ of ${{}^{h}\!M}$

[TABLE]

by the isomorphism $\iota$. We have the exact sequences

[TABLE]

By definition the Hilbert function of the module ${{}^{h}\!M}$ is

[TABLE]

By (19) we have $H(M,m)=H({{}^{h}\!M},m)$ for every $m\geq 0$, i.e., the Hilbert functions of $M$ and ${{}^{h}\!M}$ coincide.

LEMMA 2

Let $b_{1},\ldots,b_{s}$ be a system of homogeneous generators of the ${{}^{h}\!A}$-module ${{}^{h}\!I}$. Then

[TABLE]

is a system of generators of $\mathop{\rm gr}\nolimits(A)$-module $\mathop{\rm gr}\nolimits(I)$.

PROOF By (17) ${}^{a}\!(({{}^{h}\!I)_{m}})=I_{m}$. Now the required assertion follows from Lemma 1. The lemma is proved.

4 The Janet bases of a module and of

its homogenization

Each element of ${{}^{h}\!A^{l}}$ can be uniquely represented as an $F$-linear combination of elements $e_{v,i_{0},i,j}=(0,\ldots,0,X_{0}^{i_{0}}X^{i}D^{j},0,\ldots,0)$, herewith $0\leq i_{0}\in{\mathbb{Z}}$, $i,j\in{\mathbb{Z}}_{+}^{n}$ are multiindices, see (3), and the nonzero monomial $X_{0}^{i_{0}}X^{i}D^{j}$ is at the position $v$, $1\leq v\leq l$. So every element $f\in{{}^{h}\!A^{l}}$ can be represented in the form

[TABLE]

and only a finite number of $f_{v,i_{0},i,j}$ are nonzero. The elements $e_{v,i_{0},i,j}$ will be called monomials.

Let us replace everywhere in Section 1 after the definition of the Hilbert function the ring $A$, the monomials $e_{v,i,j}$, the multiindices $i$, $i^{\prime}$, $i^{\prime\prime}$, triples $(v,i,j)$, $(v,i^{\prime},j^{\prime})$, the module $I$ and so on by the ring ${{}^{h}\!A}$, monomials $e_{v,i_{0},i,j}$, the pairs $(i_{0},i)$, $(i^{\prime}_{0},i^{\prime})$, $(i^{\prime\prime}_{0},i^{\prime\prime})$ (they are used without parentheses), quadruples $(v,i_{0},i,j)$, $(v,i^{\prime}_{0},i^{\prime},j^{\prime})$, the homogenization ${{}^{h}\!I}$ and so on respectively. Thus, we get the definitions of $o(f)$, $\mathop{\rm Hdt}\nolimits(f)$ for $f\in{{}^{h}\!A^{l}}$, new conditions (a) and (b) which define admissible linear order on the monomials of ${{}^{h}\!A^{l}}$, new conditions 1)–5), the definitions of the Janet basis and reduced Janet basis of ${{}^{h}\!I}$. For example, the new conditions (a) and (b) are

1. (a)

for all indices $i_{0},i^{\prime}_{0}$, all multiindices $i,j,i^{\prime},j^{\prime}$ for all $1\leq v\leq l$ if $i_{0}\leq i^{\prime}_{0}$, $i_{1}\leq i^{\prime}_{1},\ldots,i_{n}\leq i^{\prime}_{n}$ and $j_{1}\leq j^{\prime}_{1},\ldots,j_{n}\leq j^{\prime}_{n}$ then $(v,i_{0},i,j)\leq(v,i^{\prime}_{0},i^{\prime},j^{\prime})$. 2. (b)

for all indices $i_{0},i^{\prime}_{0},i^{\prime\prime}_{0}$, all multiindices $i,j,i^{\prime},j^{\prime},i^{\prime\prime},j^{\prime\prime}$ for all $1\leq v,v^{\prime}\leq l$ if $(v,i_{0},i,j)<(v^{\prime},i^{\prime}_{0},i^{\prime},j^{\prime})$ then $(v,i_{0}+i^{\prime\prime}_{0},i+i^{\prime\prime},j+j^{\prime\prime})<(v^{\prime},i^{\prime}_{0}+i^{\prime\prime}_{0},i^{\prime}+i^{\prime\prime},j^{\prime}+j^{\prime\prime})$.

The Janet basis of ${{}^{h}\!I}$ is homogeneous if and only if it consists of homogeneous elements from ${{}^{h}\!A^{l}}$.

Let $<$ be an admissible linear order on the monomials from $A^{l}$, or which is the same, on the triples $(v,i,j)$, see Section 1. So $<$ satisfies conditions (a) and (b). Let us define the linear order on the monomials $e_{v,i_{0},i,j}$ or, which is the same, on the quadruples $(v,i_{0},i,j)$. This linear order is induced by $<$ on the triples $(v,i,j)$ and will be denoted again by $<$. Namely, for two quadruples $(v,i_{0},i,j)$ and $(v^{\prime},i^{\prime}_{0},i^{\prime},j^{\prime})$ put $(v,i_{0},i,j)<(v^{\prime},i^{\prime}_{0},i^{\prime},j^{\prime})$ if and only if $(v,i,j)<(v^{\prime},i^{\prime},j^{\prime})$, or $(v,i,j)=(v^{\prime},i^{\prime},j^{\prime})$ but $i_{0}<i^{\prime}_{0}$. Notice that this induced linear order satisfies conditions (a) and (b) (in the new sense).

REMARK 1

If $f_{1},\ldots,f_{m}$ is a Janet basis of $I$ (respectively homogeneous Janet basis of ${{}^{h}\!I}$) satisfying 1)–4) then there are the unique $c_{\alpha,\beta}\in A$ (respectively $c_{\alpha,\beta}\in{{}^{h}\!A}$), $1\leq\alpha<\beta\leq m$, such that

[TABLE]

is a reduced Janet basis of $I$ (respectively reduced homogeneous Janet basis of ${{}^{h}\!I}$), cf. [3].

LEMMA 3

Let $f_{1},\ldots,f_{m}$ be a (reduced) Janet basis of $I$ with respect to the linear order $<$. Then ${{}^{h}\!f}_{1},\ldots,{{}^{h}\!f}_{m}$ is a (reduced) homogeneous Janet basis of the module ${{}^{h}\!I}$ with respect to the induced linear order $<$. Conversely, let $g_{1},\ldots,g_{m}$ be a (reduced) homogeneous Janet basis of the module ${{}^{h}\!I}$ with respect to the induced linear order $<$. Then ${{}^{a}\!g}_{1},\ldots,{{}^{a}\!g}_{m}$ is a (reduced) Janet basis of $I$ with respect to the linear order $<$.

PROOF This follows immediately from the definitions.

Let $f\in{{}^{h}\!A^{l}}$ and the module ${{}^{h}\!I}$ be as above. Then there is the unique element $g\in{{}^{h}\!A^{l}}$ such that

[TABLE]

$f-g\in{{}^{h}\!I}$ and if $g_{v,i_{0},i,j}\neq 0$ then $e_{v,i_{0},i,j}\not\in\mathop{\rm Hdt}\nolimits({{}^{h}\!I})$. The element $g$ is called the normal form of $f$ with respect to the module ${{}^{h}\!I}$. We shall denote $g=\mathop{\rm nf}\nolimits({{}^{h}\!I},f)$. Obviously $\mathop{\rm nf}\nolimits({{}^{h}\!I},({{}^{h}\!A^{l}})_{m})\subset({{}^{h}\!A^{l}})_{m}$ is a linear subspace.

Let ${{}^{c}\!A}=F[X_{0},\ldots,X_{n},D_{1},\ldots,D_{n}]$ be the polynomial ring in the variables $X_{0},\ldots,X_{n},D_{1},\ldots,D_{n}$. Each monomial $e_{v,i_{0},i,j}$ can be considered also as an element of ${{}^{c}\!A^{l}}$. Denote by ${{}^{c}\!I}\subset{{}^{c}\!A^{l}}$ the graded submodule of ${{}^{c}\!A^{l}}$ generated by all the monomials $e_{v,i_{0},i,j}$ such that there is $0\neq f\in{{}^{h}\!I}$ with $o(f)=(v,i_{0},i,j)$. The Hilbert functions

[TABLE]

Let us replace in the definition of the normal form above ${{}^{h}\!A},{{}^{h}\!I}$ by ${{}^{c}\!A},{{}^{c}\!I}$ respectively. Thus, for $f\in{{}^{c}\!A^{l}}$ we get the definition of the normal form $\mathop{\rm nf}\nolimits({{}^{c}\!I},f)\in{{}^{c}\!A^{l}}$, cf. [4]. Obviously, $\mathop{\rm nf}\nolimits({{}^{c}\!I},({{}^{c}\!A^{l}})_{m})\subset({{}^{c}\!A^{l}})_{m}$ is a linear subspace. Since the ideals ${{}^{c}\!I}$ and $\mathop{\rm Hdt}\nolimits({{}^{h}\!I})$ are generated by the same monomials we have $\dim\mathop{\rm nf}\nolimits({{}^{c}\!I},({{}^{c}\!A^{l}})_{m})=\dim\mathop{\rm nf}\nolimits({{}^{h}\!I},({{}^{h}\!A^{l}})_{m})$. Hence the Hilbert functions

[TABLE]

coincide. Therefore, see Section 3,

[TABLE]

5 Bound on the kernel of a matrix

over the homogenized Weyl algebra

LEMMA 4

Let $k=l-1$ and $l\geq 1$ be integers. Let $b=(b_{i,j})_{1\leq i\leq k,\,1\leq j\leq l}$ be a matrix where $b_{i,j}\in{{}^{h}\!A}$ are homogeneous elements for all $i,j$. Let $\deg b_{i,j}<d$, $d\geq 1$, for all $i,j$. Assume that there are integers $d_{j}\geq 0$, $1\leq i\leq k$, and $d^{\prime}_{i}\geq 0$, $1\leq j\leq l$, such that

[TABLE]

for all nonzero $b_{i,j}$, and additionally $\min_{1\leq j\leq l}\{d^{\prime}_{j}\}=0$ (hence $d_{i}<d$, $d^{\prime}_{j}<d$ for all $i,j$), $d\geq 1$. Then there are homogeneous elements $z_{1},\ldots,z_{l}\in{{}^{h}\!A}$ such that $(z_{1},\ldots,z_{l})\neq(0,\ldots,0)$,

[TABLE]

all nonzero $b_{i,j}z_{j}$ have the same degree depending only on $i$ and

[TABLE]

Besides that, if all $b_{i,j}$ do not depend on $X_{n}$ (i.e., they can be represented as sums of monomials which do not contain $X_{n}$) then one can choose also $z_{1},\ldots,z_{l}$ satisfying additionally the same property. Finally, dividing by an appropriate power of $X_{0}$ one can assume without loss of generality that $\min\{\mathop{\rm ord}\nolimits z_{i}\,:\,1\leq i\leq l\}=0$.

PROOF We shall assume without loss of generality that $l\geq 2$. At first suppose that that $\deg b_{i,j}=\deg b$ for all nonzero $b_{i,j}$. Consider the linear mapping

[TABLE]

If

[TABLE]

then the kernel of (26) is nonzero. But (27) holds if

[TABLE]

Further, (28) is true if $(1+\deg b/(m-\deg b))^{2n}<l/(l-1)$. The last inequality follows from $m\geq(2n+1)\deg b/\log(l/(l-1))$. Hence also from $m\geq(2n+1)l\deg b$. Notice that $(2n+2)ld\geq 1+(2n+1)l\deg b$. Thus, the existence of $z_{1},\ldots,z_{l}$ is proved, and even more all nonzero $b_{i,j}z_{j}$ have the same degree which does not depend on $i,j$. Notice that in the considered case we prove a more strong inequality $\deg z_{j}\leq(2n+2)ld$ for all $1\leq j\leq l$.

Suppose that $a_{1},\ldots,a_{l}$ do not depend on $X_{n}$. We represent $z_{i}=\sum_{j}z_{i,j}X_{n}^{j}$, $1\leq i\leq l$, where all $z_{i,j}$ do not on $X_{n}$. Let $\alpha=\max_{i}\{\deg_{X_{n}}z_{i}\}$. Obviously in this case one can replace $(z_{1},\ldots,z_{l})$ by $(z_{1,\alpha},\ldots,z_{l,\alpha})$.

Let us return to general case of arbitrary $\deg b_{i,j}$. We shall reduce it to the considered one. Namely, multiplying the $i$-th equation of system (24) to $X_{0}^{\max_{i}\{d_{i}\}-d_{i}}$ we shall suppose without loss of generality that all $d_{i}$ are equal. Let us substitute $z_{j}X_{0}^{d^{\prime}_{j}}$ for $z_{j}$ in (24). Now the degrees of all the nonzero coefficients of the obtained system coincide. Thus, we get the required reduction and estimation (25). The lemma is proved.

REMARK 2

Lemma 4 remains true if one replaces in its statement condition (24) by

[TABLE]

The proof is similar.

REMARK 3

Let the elements $b_{i,j}$ be from Lemma 4. Notice that there are integers $\delta^{\prime}_{i}\geq 0$, $1\leq i\leq k$, and $\delta_{j}\geq 0$, $1\leq j\leq l$, such that

[TABLE]

for all nonzero $b_{i,j}$, and $\min_{1\leq i\leq k}\{\delta^{\prime}_{i}\}=0$. Namely, $\delta^{\prime}_{i}=-d_{i}+\max_{1\leq i\leq k}\{d_{i}\}$, $\delta_{j}=-d^{\prime}_{j}+\max_{1\leq i\leq k}\{d_{i}\}$.

6 Transforming a matrix with coefficients from ${{}^{h}\!A}$ to the

trapezoidal form

Let $b$ be the matrix from Lemma 4 but now $k,l$ are arbitrary. Hence (23) holds. Let $b=(b_{1},\ldots,b_{l})$ where $b_{1},\ldots,b_{l}\in{{}^{h}\!A^{k}}$ be the columns of the matrix $b$ (notice that in Lemma 1 and Lemma 2 $b_{i}$ are rows of size $l$; so now we change the notation). By definition $b_{1},\ldots,b_{l}$ are linearly independent over ${{}^{h}\!A}$ from the right (or just linearly independent if it will not lead to an ambiguity) if and only if for all $z_{1},\ldots,z_{l}\in{{}^{h}\!A}$ the equality $b_{1}z_{1}+\ldots+b_{l}z_{l}=0$ implies $z_{1}=\ldots=z_{l}=0$. By (23) in this definition one can consider only homogeneous $z_{1},\ldots,z_{l}$. For an arbitrary family $b_{1},\ldots,b_{l}$ from Lemma 4 (with arbitrary $k,l$) one can choose a maximal linearly independent from the right subfamily $b_{i_{1}},\ldots,b_{i_{r}}$ of $b_{1},\ldots,b_{l}$. It turns out that $r$ does not depend on the choice of a subfamily. More precisely, we have the following lemma.

LEMMA 5

Let $c_{j}=\sum_{1\leq i\leq l}b_{i}z_{i,j}$, $1\leq j\leq r_{1}$, where $z_{i,j}\in{{}^{h}\!A}$ are homogeneous elements. Suppose that there are integers $d^{\prime\prime}_{j}$, $1\leq j\leq r_{1}$, such that for all $i,j$ the degree $\deg z_{i,j}=d^{\prime}_{i}-d^{\prime\prime}_{j}$. Assume that $c_{j}$, $1\leq j\leq r_{1}$, are linearly independent over ${{}^{h}\!A}$ from the right. Then $r_{1}\leq r$, and if $r_{1}<r$ there are $c_{r_{1}+1},\ldots,c_{r}\in\{b_{i_{1}},\ldots,b_{i_{r}}\}$ such that $c_{j}$, $1\leq j\leq r$, are linearly independent over ${{}^{h}\!A}$ from the right.

PROOF The proof is similar to the case of vector spaces over a field and we leave it to the reader.

We denote $r=\mathop{\rm rankr}\nolimits\{b_{1},\ldots,b_{l}\}$ and call it the rank from the right of $b_{1},\ldots,b_{l}$. In the similar way one can define rank from the left of $b_{1},\ldots,b_{l}$. Denote it by $\mathop{\rm rankl}\nolimits\{b_{1},\ldots,b_{l}\}$. It is not difficult to construct examples when $\mathop{\rm rankr}\nolimits\{b_{1},\ldots,b_{l}\}$ $\neq\mathop{\rm rankl}\nolimits\{b_{1},\ldots,b_{l}\}$. The aim of this section is to prove the following result.

LEMMA 6

Let $b$ be the matrix with homogeneous coefficient from ${{}^{h}\!A}$ satisfying (23), see above. Suppose that $\deg b_{i,j}$ $<d$ for all $i,j$. Assume that $k\geq l\geq 1$. Let $l_{1}=\mathop{\rm rankr}\nolimits\{b_{1},\ldots,b_{l}\}$ and $b_{1},\ldots,b_{l_{1}}$ be linearly independent. Hence $0\leq l_{1}\leq l$. Then there is a matrix $(z_{j,r})_{1\leq j,r\leq l_{1}}$ with homogeneous entries $z_{j,r}\in{{}^{h}\!A}$ and a square permutation matrix $\sigma$ of size $k$ satisfying the following properties.

1. (i)

All the nonzero elements $b_{i,j}z_{j,r}$ for $1\leq j\leq l$ have the same degree depending only on $i,r$ and

[TABLE] 2. (ii)

Set the matrix $e=(e_{i,j})_{1\leq i\leq k,\,1\leq j\leq l_{1}}=\sigma bz$. Then the matrix

[TABLE]

where $e^{\prime}=\mathop{\rm diag}\nolimits(e^{\prime}_{1,1},\ldots,e^{\prime}_{l_{1},l_{1}})$ is a diagonal matrix with $l_{1}$ columns and each $e^{\prime}_{j,j}$, $1\leq j\leq l_{1}$, is nonzero. 3. (iii)

$\mathop{\rm ord}\nolimits e_{i,j}\geq\mathop{\rm ord}\nolimits e^{\prime}_{j,j}$* for all $1\leq i\leq k$, $1\leq j\leq l_{1}$.*

Besides that, if all $a_{i,j}$ (and hence all $b_{i,j}$) do not depend on $X_{n}$ (i.e., they can be represented as sums of monomials which do not contain $X_{n}$) then one can choose also $z_{j,r}$ satisfying additionally the same property. Finally, dividing by an appropriate power of $X_{0}$ one can assume without loss of generality that $\min\{\mathop{\rm ord}\nolimits z_{j,r}\,:\,1\leq j\leq l_{1}\}=0$ for every $1\leq r\leq l_{1}$.

PROOF At first we shall show how to construct $z$ and $e$ such that (ii) and (iii) hold. We shall use a kind of Gauss elimination and Lemma 4. Namely, we transform the matrix $e$. At the beginning we put

[TABLE]

We shall perform some ${{}^{h}\!A}$-linear transformations of columns and permutations of rows of the matrix $e$ and replace each time $e$ by the obtained matrix. These transformation do not change the rank from the right of the family of columns of $e$. At the end we get a matrix $e$ satisfying the required properties (ii), (iii).

We have $\mathop{\rm rankr}\nolimits(e)=l_{1}$. If $l_{1}=0$, i.e, $e$ is an empty matrix, then this is the end of the construction: $z^{\prime}$ is an empty matrix. Suppose that $l_{1}>0$. Let us choose indices $1\leq i_{0}\leq k$, $1\leq j_{0}\leq l_{1}$ such that $\mathop{\rm ord}\nolimits e_{i_{0},j_{0}}=\min_{1\leq j\leq l_{1}}\{\mathop{\rm ord}\nolimits e_{j}\}$. Permuting rows and columns of $e$ we shall assume without loss of generality that $(i_{0},j_{0})=(1,1)$.

By Lemma 4 we get elements $w_{i,1},w_{i,i}\in{{}^{h}\!A}$ of degrees at most $(2n+3)2d$ such that $e_{1,1}w_{1,i}=e_{1,i}w_{i,i}$, $1\leq i\leq l_{1}$, and $\mathop{\rm ord}\nolimits w_{i,i}=0$ for every $1\leq i\leq l_{1}$. Set $w^{\prime}=(-w_{1,2},\ldots,-w_{1,l_{1}})$, and $w^{\prime\prime}=\mathop{\rm diag}\nolimits(w_{2,2},\ldots,w_{l_{1},l_{1}})$ to be the diagonal matrix. Put

[TABLE]

to be the square matrix with $l_{1}$ rows. We replace $e$ by $ew$. Now

[TABLE]

where $E_{2,2}$ has $l_{1}-1$ columns and

[TABLE]

(for the new matrix $e$).

Let us apply recursively the described construction to the matrix $E_{2,2}$ in place of $e$. So using only linear transformations of columns with indices $2,\ldots,l_{1}$ and permutation of rows with indices $2,\ldots,k$ we transform $e$ to the form

[TABLE]

where $\sigma$ is a permutation matrix and $\tau^{\prime}$ is a square matrix with $l_{1}-1$ rows (it transforms $E_{2,2}$), the matrix $E^{\prime}_{2,2}=\mathop{\rm diag}\nolimits(e_{2,2},\ldots,e_{l_{1},l_{1}})$ is a diagonal matrix with $l_{1}-1\geq 0$ columns, and all the elements $e_{2,2},\ldots,e_{l_{1},l_{1}}\in{{}^{h}\!A}$ are nonzero. We shall assume without loss of generality that $\sigma=1$ is the identity matrix. We replace $e$ by $e\tau$. Conditions (ii) and (iii) hold for the obtained $e$ and, more than that, by (iii) applied recursively for $(E_{2,2},E^{\prime}_{2,2},E^{\prime\prime}_{2,2})$ (in place of $(e,e^{\prime},e^{\prime\prime})$), and (31) the same equalities are satisfied for the new obtained matrix $e$.

Let $E^{\prime}_{2,1}=(e_{2,1},\ldots,e_{l_{1},1})^{t}$ where $t$ denotes transposition. By Lemma 4 there are nonzero elements $v_{1,1},\ldots,v_{l_{1},1}\in{{}^{h}\!A}$ of degrees at most

[TABLE]

such that $e_{i,1}v_{1,1}=e_{i,i}v_{i,1}$ and $\min\{\mathop{\rm ord}\nolimits v_{1,1},\mathop{\rm ord}\nolimits v_{1,i}\}=0$ for all $1\leq i\leq l_{1}-1$. Let $v^{\prime}=(-v_{2,1},\ldots,-v_{l_{1},1})^{t}$ and $v^{\prime\prime}$ be the identity matrix of size $l_{1}-1$. Put

[TABLE]

Let us replace $e$ by $ev$. Put $z=w\tau v$, where the matrix $z$ has $l_{1}$ columns. Recall that without loss of generality $\sigma=1$ is the identity permutation. We have $e=(b_{1},\ldots,b_{l_{1}})z$. These Gauss elimination transformations of $e$ do not change the rank from the right of the family of columns of $e$. It can be easily proved using the recursion on $l$, cf. Lemma 8 below. Now the matrix $e$ satisfies required conditions (ii), (iii) and $\sigma=1$.

Let us change the notation. Denote the obtained matrix $z$ by $z^{\prime}$. Let $z^{\prime}=(z^{\prime}_{1},\ldots,z^{\prime}_{l_{1}})$ where $z^{\prime}_{j}$ is the $j$-th column of $z^{\prime}$. Our aim now is to prove the existence of the matrix $z$ satisfying (i)–(iii). By Lemma 4 for every $1\leq r\leq l_{1}$ there are homogeneous elements $z_{j,r}\in{{}^{h}\!A}$, $1\leq j\leq l$, such that $(z_{1,r},\ldots,z_{l,r})\neq(0,\ldots,0)$,

[TABLE]

and estimations for degrees (30) hold. Put the matrix $z=(z_{j,r})_{1\leq j,r\leq l_{1}}$. Let $z=(z_{1},\ldots,z_{l_{1}})$ where $z_{j}$ is the $j$-th column of $z$. Hence $z_{j}=(z_{1,r},\ldots,z_{l,r})^{t}$.

LEMMA 7

For every $1\leq r\leq l_{1}$ we have

[TABLE]

Further, for every $1\leq r\leq l_{1}$ there are nonzero homogeneous elements $g^{\prime}_{r},g_{r}\in{{}^{h}\!A}$ such that $z^{\prime}_{r}g^{\prime}_{r}=z_{r}g_{r}$.

PROOF Consider the matrix $(z^{\prime},z_{r})$ with $l_{1}$ rows and $l_{1}+1$ columns. By Lemma 4 there are homogeneous elements $h_{1},\ldots,h_{l_{1}+1}\in{{}^{h}\!A}$ (they depend on $r$) such that $(h_{1},\ldots,h_{l_{1}+1})\neq(0,\ldots,0)$ and the following property holds. Denote $h=(h_{1},\ldots,h_{l_{1}+1})^{t}$, $h^{\prime}=(h_{1},\ldots,h_{l_{1}})^{t}$. Then

[TABLE]

(we don’t need at present any estimation on degrees from Lemma 4; only the existence of $h$). Denote by $b^{\prime\prime}$ the submatrix consisting of the first $l_{1}$ rows of the matrix $(b_{1},\ldots,b_{l_{1}})$. Multiplying (35) to $b^{\prime\prime}$ from the left we get

[TABLE]

But $b^{\prime\prime}z^{\prime}$ is a diagonal matrix with nonzero elements on the diagonal, see (ii) (for $z^{\prime}$ in place of $z$). Hence by (33) and (36) $h_{j}=0$ for every $j\neq r$. Now $h\neq(0,\ldots,0)^{t}$ implies $h_{r}\neq 0$ and $h_{l_{1}+1}\neq 0$. Therefore, (34) holds. Put $g^{\prime}_{r}=h_{r}$ and $g_{r}=h_{l_{1}+1}$. We have $z^{\prime}_{r}g^{\prime}_{r}=z_{r}g_{r}$ by (36). The lemma is proved.

Let us return to the proof of Lemma 6. Now (i)–(iii) are satisfied by Lemma 7. The last assertions of Lemma 6 are proved similarly to the ones of Lemma 4. Lemma 6 is proved.

7 An algorithm for solving linear systems with

coefficients from ${{}^{h}\!A}$.

Let $u=(u_{1},\ldots,u_{l})^{t}\in{{}^{h}\!A^{l}}$. Let all nonzero $u_{j}$ be homogeneous elements of the degree $-d^{\prime}_{j}+\rho$ for an integer $\rho$. Suppose that $-d^{\prime}_{j}+\rho<d^{\prime}$ for an integer $d^{\prime}>1$. Let $b=(b_{i,j})_{1\leq i\leq k,\,1\leq j\leq l}$ be the matrix with $k$ rows and $l$ columns from the statement of Lemma 6 (but now $k$ and $l$ are arbitrary). So $\deg b_{i,j}=d_{i}-d^{\prime}_{j}<d$ for all $i,j$. Let $Z=(Z_{1},\ldots,Z_{k})$ be unknowns. Consider the linear system

[TABLE]

or, which is the same,

[TABLE]

Denote

[TABLE]

The similar notations will be used for other vectors and matrices. In this section we shall describe an algorithm for solving linear systems over ${{}^{h}\!A}$ and prove the following theorem.

THEOREM 2

Suppose that system (37) has a solution over ${{}^{h}\!A}$. One can represent the set of all solutions of (37) over ${{}^{h}\!A}$ in the form

[TABLE]

where $J\subset{{}^{h}\!A^{l}}$ is a ${{}^{h}\!A}$-submodule of all the solutions of the homogeneous system corresponding to (37) (i.e., system (37) with all $u_{j}=0$) and $z^{*}$ is a particular solution of (37). Moreover, the following assertions hold.

1. (A)

One can choose $z^{*}$ such that $\mathop{\rm ord}\nolimits z^{*}\geq\mathop{\rm ord}\nolimits u-\nu$, where $\nu\geq 0$ is an integer bounded from above by $(dl)^{2^{O(n)}}$ (and depends only on $d$ and $l$). The degree $\deg z^{*}$ is bounded from above by $d^{\prime}(dl)^{2^{O(n)}}$. 2. (B)

There exists a system of generators of $J$ of degrees bounded from above by $(dl)^{2^{O(n)}}$. The number of elements of this system of generators is bounded from above by $k(dl)^{2^{O(n)}}$.

Besides that, if all $b_{i,j}$ and $u_{j}$ do not depend on $X_{n}$ (i.e., they can be represented as sums of monomials which do not contain $X_{n}$) then $z^{*}$ and all the generators of the module $J$ also satisfy this property.

PROOF Let $l_{1}=\mathop{\rm rankr}\nolimits(b_{1},\ldots,b_{l})$. Permuting equations of (37) we shall assume without loss of generality that $(b_{1},\ldots,b_{l_{1}})$ are linearly independent from the right over ${{}^{h}\!A}$. Let $\sigma,z,e,e^{\prime},e^{\prime\prime}$ be the matrices from Lemma 6. Similarly to the proof of Lemma 6 we shall assume without loss of generality that $\sigma=1$. Denote by $b^{\prime}$ the submatrix of $b$ consisting of the first $l_{1}$ columns of $b$, i.e., $b^{\prime}=(b_{1},\ldots,b_{l_{1}})$. By Lemma 4 there are nonzero elements $q_{1,1},\ldots,$ $q_{l_{1},l_{1}}$ of degrees at most (32) such that $e_{1,1}q_{1,1}=e_{i,i}q_{i,i}$ and $\min\{\mathop{\rm ord}\nolimits q_{1,1},\mathop{\rm ord}\nolimits q_{i,i}\}=0$ for all $2\leq i\leq l_{1}$. Set $q=\mathop{\rm diag}\nolimits(q_{1,1},\ldots,q_{l_{1},l_{1}})$ to be the diagonal matrix. Let $\nu_{0}=\mathop{\rm ord}\nolimits e_{1,1}q_{1,1}$. Then by Lemma 6 (iii) $\mathop{\rm ord}\nolimits(b^{\prime}zq)\geq\nu_{0}$. Let $X_{0}^{\nu_{0}}\delta=b^{\prime}zq$. Then $\delta$ is a matrix with coefficients from ${{}^{h}\!A}$ and

[TABLE]

where $\delta^{\prime}=\mathop{\rm diag}\nolimits(\delta_{1,1},\ldots,\delta_{l_{1},l_{1}})$ is a diagonal matrix with homogeneous coefficients from ${{}^{h}\!A}$ and all the elements on the diagonal are nonzero and equal, i.e., $\delta_{j,j}=\delta_{1,1}$ for every $1\leq j\leq l_{1}$. Besides that, $\mathop{\rm ord}\nolimits\delta_{1,1}=0$. Further, $\delta^{\prime\prime}=(\delta_{i,j})_{l_{1}+1\leq i\leq k,\,1\leq j\leq l_{1}}$. We have $\mathop{\rm ord}\nolimits(uzq)\geq\nu_{0}$, since, otherwise, system (37) does not have a solution. Obviously $\mathop{\rm ord}\nolimits u\leq\mathop{\rm ord}\nolimits(uzq)$. Denote $u^{\prime}=(u^{\prime}_{0},\ldots,u^{\prime}_{l})^{t}=X_{0}^{-\nu_{0}}uzq\in{{}^{h}\!A^{l}}$. Hence $\mathop{\rm ord}\nolimits u^{\prime}\geq\mathop{\rm ord}\nolimits(u)-\nu_{0}$. Consider the linear system

[TABLE]

LEMMA 8

Suppose that system (37) has a solution over ${{}^{h}\!A}$. Then linear system (39) is equivalent to (37), i.e., the sets of solutions of systems (39) and (37) over ${{}^{h}\!A}$ coincide.

PROOF The system $Zb^{\prime}z=uz$ is equivalent to (37) by Lemma 5. System (39) is equivalent to $Zb^{\prime}z=uz$ since the ring ${{}^{h}\!A}$ does not have zero–divisors. The lemma is proved.

REMARK 4

Since $\mathop{\rm rankr}\nolimits(b_{1},\ldots,b_{l})=l_{1}$ and by Lemma 6 for every $l_{1}+1\leq j\leq l$ there are homogeneous $z_{j,j},z_{1,j},\ldots,z_{l_{1},j}\in{{}^{h}\!A}$ such that $z_{j,j}\neq 0$ and $b_{j}z_{j,j}+\sum_{1\leq r\leq l_{1}}b_{r}z_{r,j}=0$ and all $\deg z_{j,j}$, $\deg z_{r,j}$ are bounded from above by $(2n+3)(l_{1}+1)d$. Put $u^{\prime}_{j}=u_{j}z_{j,j}+\sum_{1\leq r\leq l_{1}}u_{r}z_{r,j}$, $l_{1}+1\leq j\leq l$. Then system (37) has a solution if and only if system (39) has a solution and $u^{\prime}_{j}=0$ for all $l_{1}+1\leq j\leq l$. This follows from Lemma 8 and Lemma 5. But in what follows for our aims it is sufficient to use only Lemma 8.

REMARK 5

Assume that $\deg_{X_{n}}b_{i,j}\leq 0$ for all $i,j$, i.e., the elements of the matrix $b$ do not depend on $X_{n}$. Then by Lemma 4 and the described construction all the elements of the matrices $b,z,q,\delta,\delta^{\prime},\delta^{\prime\prime}$ also do not depend on $X_{n}$.

By Lemma 4 and Remark 2 for every $l_{1}+1\leq j\leq k$ there are homogeneous elements $g_{j,j},g_{j,i}\in{{}^{h}\!A}$, $1\leq i\leq l_{1}$, such that

[TABLE]

all the degrees $\deg g_{j,j}$, $\deg g_{j,i}$, $1\leq i\leq l_{1}$, are bounded from above by

[TABLE]

and $\min_{1\leq i\leq l_{1}}\{\mathop{\rm ord}\nolimits g_{j,j},\mathop{\rm ord}\nolimits g_{j,i}\}=0$. Hence $\mathop{\rm ord}\nolimits g_{j,j}=0$ for every $l_{1}+1\leq j\leq k$ since $\mathop{\rm ord}\nolimits\delta_{1,1}=0$.

Denote $h=\delta_{1,1}g_{l_{1}+1,l_{1}+1}g_{l_{1}+2,l_{1}+2}\ldots g_{k,k}$. So $h\in{{}^{h}\!A}$ is a nonzero homogeneous element and $\mathop{\rm ord}\nolimits h=0$. Set $\varepsilon=\deg h$. We need an analog of the Noether normalization theorem from commutative algebra, cf. also Lemma 3.1 [7].

LEMMA 9

There is a linear automorphism of the algebra ${{}^{h}\!A}$

[TABLE]

such that all $\alpha_{s,i,j}\in F$, $\deg_{D_{n}}\alpha(h)=\varepsilon$. If $\deg_{X_{n}}h=0$ then one can choose additionally $\alpha(X_{n})=X_{n}$, all $\alpha_{1,n,j}=0$ for $1\leq j\leq n-1$ and $\alpha_{3,n,j}=0$ for $1\leq j\leq n$.

PROOF Recall that $\mathop{\rm ord}\nolimits h=0$. Hence at first it is not difficult to construct a linear automorphism $\beta$ such that $\beta(X_{0})=X_{0}$,

[TABLE]

and $\beta(h)$ contains a monomial $a_{i_{1},\ldots,i_{n}}D_{1}^{i_{1}},\ldots,D_{n}^{i_{n}}$ with $a_{i_{1},\ldots,i_{n}}\neq 0$ and $i_{1}+\ldots+i_{n}=\varepsilon$, i.e., $\varepsilon=\deg_{D_{1},\ldots,D_{n}}\beta(h)$. After that one can find an automorphism $\gamma$ such that $\gamma(X_{0})=X_{0}$,

[TABLE]

and $(\gamma\circ\beta)(h)$ contains a monomial $aD_{n}^{\varepsilon}$ with a coefficient $0\neq a\in F$. Put $\alpha=\gamma\circ\beta$. We leave to prove the last assertion to the reader. The lemma is proved.

We apply the automorphism $\alpha$. In what follows to simplify the notation we shall suppose without loss of generality that $\alpha=1$. So $h$ contains a monomial $aD_{n}^{\varepsilon}$ with a coefficient $0\neq a\in F$, where $\varepsilon=\deg h$. It follows from here that

[TABLE]

Let $z=(z_{1},\ldots,z_{k})\in{{}^{h}\!A^{k}}$ be a solution of (39). Then (42) implies that one can uniquely represent

[TABLE]

where $z^{\prime}_{j},z_{j,s}\in{{}^{h}\!A}$, the degrees $\deg_{D_{n}}z_{j,s}\leq 0$ for all $l_{1}+1\leq j\leq k$, $0\leq s<\deg_{D_{1}}g_{j,j}$. Again by (42) one can uniquely represent

[TABLE]

where $u^{\prime\prime}_{i},u^{\prime}_{i,s}\in{{}^{h}\!A}$, the degrees $\deg_{D_{n}}u^{\prime}_{i,s}\leq 0$ for all $1\leq i\leq l$, $0\leq s<\deg_{D_{1}}g_{j,j}$. Finally, by (42) for all $l_{1}+1\leq j\leq k$, $1\leq i\leq l_{1}$, $0\leq r<\deg_{D_{1}}g_{j,j}$, one can uniquely represent

[TABLE]

where $\delta_{j,r,i},\delta_{j,r,i,s}\in{{}^{h}\!A}$, the degrees $\deg_{D_{n}}\delta_{j,r,i,s}\leq 0$ for all considered $j,r,i,s$. Put

[TABLE]

Therefore,

[TABLE]

Let us introduce new unknowns $Z_{j,r}$, $(j,r)\in{\mathcal{I}}$. By (43)–(45) system (37) is reduced to the linear system

[TABLE]

More precisely, any solution of system (37) is given by (43), (44) where $z^{\prime}_{j}\in{{}^{h}\!A}$ are arbitrary and $z_{j,r}$ is a solution of system (45) over ${{}^{h}\!A}$ (we underline that here this solution $z_{j,r}$ may depend on $D_{n}$ although one can restrict oneself by solutions $z_{j,r}$ which do not depend on $D_{n}$). Note that all $\delta_{j,r,i,s}$ and $u^{\prime}_{i,s}$ are homogeneous elements of ${{}^{h}\!A}$ and there are integers $d_{j,r}$, $(j,r)\in{\mathcal{I}}$, $d^{\prime}_{i,s}$, $(i,s)\in{\mathcal{J}}$, $\widetilde{\rho}$ such that $\deg\delta_{j,r,i,s}=d_{j,r}-d^{\prime}_{i,s}$ and $\deg u^{\prime}_{i,s}=-d^{\prime}_{i,s}+\widetilde{\rho}$ for all $(j,r)\in{\mathcal{I}}$, $(i,s)\in{\mathcal{J}}$. This follows immediately from the described construction.

Now all the coefficients of system (46) do not depend on $D_{n}$. As we have proved if the coefficients of (37) do not depend on $X_{n}$ then the coefficients of (46) also do not depend on $X_{n}$, and hence in the last case they do not depend on $X_{n},D_{n}$.

If the coefficients of (46) depend on $X_{n}$ we perform an automorphism $X_{n}\mapsto D_{n}$ $D_{n}\mapsto-X_{n}$, $X_{i}\mapsto X_{i}$, $D_{i}\mapsto D_{i}$, $1\leq i\leq n-1$. Now the coefficients of system (46) do not depend on $X_{n}$ (but depend on $D_{n}$). After that we apply our construction recursively to system (46).

The final step of the recursion is $n=0$ (although in the statement of theorem $n\geq 1$, see Section 1; we are interested only in Weyl algebras). In this case ${\mathcal{I}}={\mathcal{J}}=\emptyset$. Hence using (44) for $n=0$ we get the required $z^{*}$ and $J$ for $n=0$.

Thus, by the recursive assumption we get a particular solution $Z_{j,r}=z_{j,r}^{*}$, $(j,r)\in{\mathcal{I}}$, of system (46), an integer $\nu_{1}$ (in place of $\nu$ from assertion (A)) such that

[TABLE]

and a system of generators

[TABLE]

of the module $J^{\prime}$ of solutions of the homogeneous system corresponding to (46). Notice that if the coefficients of (37) do not depend on $X_{n}$ then $J^{\prime}$ is a module over the homogenization $F[X_{0},X_{1},\ldots,X_{n-1},D_{1},\ldots,D_{n-1}]$ of the Weyl algebra of $X_{1},\ldots,X_{n-1},D_{1},\ldots,D_{n-1}$. But obviously in the last case (48) gives also a system of generators of the ${{}^{h}\!A}$-module $J^{\prime\prime}={{}^{h}\!A}J^{\prime}$ of solutions of the homogeneous system corresponding to (46). Put

[TABLE]

Then $z^{*}$ is a particular solution of (37). Put

[TABLE]

Then $J=\sum_{1\leq\alpha\leq\beta+k-l_{1}}{{}^{h}\!A}(z_{\alpha,1},\ldots,z_{\alpha,k})$. Hence $(z_{\alpha,1},\ldots,z_{\alpha,k})$, $1\leq\alpha\leq\beta+k-l_{1}$, is a system of generators of the module $J$. By (47) and the definitions of $u^{\prime}$, $u^{\prime\prime}_{i}$ and $u^{\prime}_{i,s}$ we have $\mathop{\rm ord}\nolimits z^{*}\geq\mathop{\rm ord}\nolimits u-\nu_{0}-\nu_{1}$. Put $\nu=\nu_{0}+\nu_{1}$.

LEMMA 10

All the degrees $\deg\delta_{j,i}$, $\deg g_{j,i}$, $\deg\delta_{j,r,i}$, $\deg\delta_{j,r,i,s}$ and $\nu$, see above, are bounded from above by $(nld)^{O(1)}$, the degrees $\deg u^{\prime}_{i}$ are bounded from above $d^{\prime}+(nld)^{O(1)}$, the degrees $\deg u^{\prime\prime}_{i}$, $\deg u^{\prime}_{i,s}$ are bounded from above by $d^{\prime}(nld)^{O(1)}$. Further, all $\mathop{\rm ord}\nolimits u^{\prime\prime}_{i}$, $\mathop{\rm ord}\nolimits u^{\prime}_{i,s}$ are bounded from below by $\mathop{\rm ord}\nolimits u-\nu$. Finally, in system (46) the number of equations $\#{\mathcal{J}}$ is bounded from above by $(nld)^{O(1)}$ and the number of unknowns $\#{\mathcal{I}}$ is bounded from above by $k(nld)^{O(1)}$.

PROOF This follows immediately from the described construction.

Let us return to the proof of Theorem 2. Applying Lemma 10 and recursively assertions (A) and (B) for the formulas giving $z^{*}$ and $J$ we get (A) and (B) from the theorem. The last assertion (related to the case when all $b_{i,j}$ and $u_{j}$ do not depend on $D_{n}$) has been already proved. The theorem is proved.

8 Proof of Theorem 1 for Weyl algebra

Let $a$ be the matrix from Section 1. We shall suppose without loss of generality that the vectors $(a_{i,1},\ldots,a_{i,l})$, $1\leq i\leq k$, are linearly independent over the field $F$. We have $\deg a_{i,j}<d$. This implies $k\leq l{d+2n\choose 2n}$.

Put the matrix $b={{}^{h}\!a}$. Let us define the graded submodules of ${{}^{h}\!I}$

[TABLE]

We have the exact sequence of graded ${{}^{h}\!A}$-modules

[TABLE]

Further, $J_{\nu}\subset J_{\nu+1}\subset{{}^{h}\!I}$ for every $\nu\geq 0$ and ${{}^{h}\!I}=\bigcup_{\nu\geq 0}J_{\nu}$. Since ${{}^{h}\!A}$ is Noetherian there is $N\geq 0$ such that ${{}^{h}\!I}=J_{N}$. So to construct a system of generators of ${{}^{h}\!I}$ it is sufficient to compute the least $N$ such that ${{}^{h}\!I}=J_{N}$ and to find a system of generators of $J_{N}$.

LEMMA 11

${{}^{h}\!I}=J_{N}$* for some $N$ bounded from above by $(dl)^{2^{O(n)}}$. There is a system of generators $b_{1},\ldots,b_{s}$ of the module $J_{N}$ such that $s$ and all the degrees $\deg b_{v}$, $1\leq v\leq s$, are bounded from above by $(dl)^{2^{O(n)}}$.*

PROOF Let us show that the module $J_{N+1}\subset J_{N}$ for $N\geq\nu$. Let $u\in J_{N+1}$. Consider system (37). By assertion (A) of Theorem 2 there is a particular solution $z^{*}$ of (37) such that $\mathop{\rm ord}\nolimits z^{*}\geq 1$. Hence $u\in X_{0}J_{N}\subset J_{N}$. The required assertion is proved. Hence ${{}^{h}\!I}=J_{\nu}$.

Let us replace in (37) $(u_{1},\ldots,u_{l})$ by $(U_{1}X_{0}^{\nu},\ldots,U_{l}X_{0}^{\nu})$, where $U_{1},\ldots,U_{l}$ are new unknowns. Then applying (B) from Theorem 2 to this new homogeneous linear system with respect to all unknowns $U_{1},\ldots,U_{l}$, $Z_{1},\ldots,Z_{k}$ we get the required estimations for the number of generators of $J_{\nu}$ and the degrees of these generators. The lemma is proved.

COROLLARY 1

Let $(a_{i,1},\ldots,a_{i,l})$, $1\leq i\leq l$, be from the beginning of the section and the integer $N$ be from Lemma 3. Then for every integer $m\geq 0$ the $F$–linear space

[TABLE]

PROOF By Lemma 3 we have $(J_{0})_{m+N}\supset X_{0}^{N}(J_{N})_{m}=X_{0}^{N}({{}^{h}\!I})_{m}$. Taking the affine parts we get (49). The corollary is proved.

Now everything is ready for the proof of Theorem 1. By Lemma 11 and Lemma 1 there is a system of generators of the module $\mathop{\rm gr}\nolimits(I)$ with degrees bounded from above by $(dl)^{2^{O(n)}}$. By Lemma 12 from Appendix 1 the Hilbert function $H(\mathop{\rm gr}\nolimits(I),m)$ is stable for $m\geq(dl)^{2^{O(n)}}$. By (10) Section 2 the Hilbert function $H(I,m)$ is stable for all $m\geq(dl)^{2^{O(n)}}$.

Consider the linear order $<$ on the monomials from ${{}^{h}\!A^{l}}$ which is induced by the linear order $<$ on the monomials from $A^{l}$, see Section 4. Then the monomial submodule ${{}^{c}\!I}\subset{{}^{c}\!A^{l}}$ is defined, see Section 4, where ${{}^{c}\!A}=F[X_{0},\ldots,X_{n},D_{1},\ldots,$ $D_{n}]$ is the polynomial ring. By (22) Section 4 the Hilbert function $H({{}^{c}\!I},m)$ is stable for all $m\geq(dl)^{2^{O(n)}}$. Hence all the coefficients of the Hilbert polynomial of ${{}^{c}\!I}$ are bounded from above $(dl)^{2^{O(n)}}$. Therefore, according to (31) the module ${{}^{c}\!I}$ has a system of generators with degrees $(dl)^{2^{O(n)}}$. This means, see Section 4, that the module $\mathop{\rm Hdt}\nolimits({{}^{h}\!I})$ has a system of generators with degrees $(dl)^{2^{O(n)}}$. Therefore, the degrees of all the elements of the Janet basis of ${{}^{h}\!I}$ with respect to the induced linear order $<$ are bounded from above by $(dl)^{2^{O(n)}}$. Hence by Lemma 3 Section 4 the same is true for the Janet basis of the module $I$ with respect to the linear order $<$ on the monomials from $A^{l}$. Theorem 1 is proved for Weyl algebra.

9 The case of algebra of differential operators

Denote by $B=F(X_{1},\ldots,X_{n})[D_{1},\ldots,D_{n}]$ the algebra of differential operators. Recall that $A\subset B$ and hence relations (1) are satisfied. Further, each element $f\in B$ can be uniquely represented in the form

[TABLE]

where all $f_{j_{1},\ldots,j_{n}}=f_{j}\in F(X_{1},\ldots,X_{n})$ and $F(X_{1},\ldots,X_{n})$ is a field of rational functions over $F$. Let us replace everywhere in Section 1 and Section 2 $A$, $X^{i}D^{j}$, $\deg f=\deg_{X_{1},\ldots,X_{n},D_{1},\ldots,D_{n}}f$, $\dim_{F}M$, $e_{v,i,j}$, $f_{v,i,j}\in F$, $(v,i,j)$, $(i,j)$, $(i^{\prime},j^{\prime})$, $(i^{\prime\prime},j^{\prime\prime})$ by $B$, $D^{j}$, $\deg f=\deg_{D_{1},\ldots,D_{n}}f$, $\dim_{F(X_{1},\ldots,X_{n})}M$, $e_{v,j}$, $f_{v,j}\in F(X_{1},\ldots,X_{n})$, $(v,j)$, $j$, $j^{\prime}$, $j^{\prime\prime}$ respectively. Thus, we get the definition of the Janet basis and all other objects from Section 1 for the case of the algebra of differential operators.

We define the homogenization ${{}^{h}\!B}$ of $B$ similarly to ${{}^{h}\!A}$, see Section 3. Namely, ${{}^{h}\!B}=F(X_{1},\ldots,X_{n})[X_{0},D_{1},\ldots,D_{n}]$ given by the relations

[TABLE]

Further, the considerations are similar to the case of the Weyl algebra $A$ with minor changes. We leave them to the reader. For example, Theorem 2 for the case of the algebra of differential operators is the same. One need only to replace everywhere in its statement $A$, ${{}^{h}\!A}$ and $X_{n}$ by $B$, ${{}^{h}\!B}$ and $D_{n}$ respectively. Thus, one can prove Theorem 1 for the case when $A$ is an algebra of differential operators (but now it is $B$). Theorem 1 is proved completely.

One can consider more general algebra of differential operators. Let $\cal F$ be a field with $n$ derivatives $D_{1},\dots,D_{n}$. Then $K_{n}={\cal F}[D_{1},\dots,D_{n}]$ is the algebra of differential operators and similarly one can define its homogenization ${}^{h}K_{n}$ by means of adding the variable $X_{0}$ satisfying the relations

[TABLE]

for all $i,j$ and any element $f\in\cal F$ where $f_{D_{i}}\in\cal F$ denotes the result of the application of $D_{i}$ to $f$. Following the proof of Theorem 1 one can deduce the following statement.

REMARK 6

A similar bound to Theorem 1 holds for $K_{n}$.

Appendix 1: Degrees of generators

of a graded module over a polynomial ring and its Hilbert function.

We give a short proof of the following result, cf. [1], [12], [6], [4].

LEMMA 12

Let $I\subset{\mathcal{A}}^{l}$ be a graded submodule over the graded polynomial ring ${\mathcal{A}}=F[X_{0},\ldots,X_{n}]$, and $I$ is given by a system of generators $f_{1},\ldots,f_{m}$ of degrees less than $d$. Then the Hilbert function $H({\mathcal{A}}^{l}/I,m)=\dim_{F}({\mathcal{A}}^{l}/I)_{m}$ is stable for $m\geq(dl)^{2^{O(n+1)}}$. Further, all the coefficients of the Hilbert polynomial of ${\mathcal{A}}^{l}/I$ are bounded from above by $(dl)^{2^{O(n+1)}}$.

PROOF Denote $M={\mathcal{A}}^{l}/I$. Let $L\in F[X_{0},\ldots,X_{n}]$ be a linear form in general position. Denote by $K$ the kernel of the morphism $M\rightarrow M$ of multiplication to $L$. We have $K=\{z\in{\mathcal{A}}^{l}\,:\,Lz=\sum_{1\leq i\leq m}f_{i}z_{i},\&\,z_{i}\in{\mathcal{A}}\}$. Hence solving a linear system over ${\mathcal{A}}$, we get that $K$ has a system of generators $g_{1},\ldots,g_{\mu}$ with degrees bounded from above by $(dl)^{2^{O(n+1)}}$. Let ${\mathfrak{P}}$ be an arbitrary associated prime ideal of the module $M$ such that ${\mathfrak{P}}\neq(X_{0},\ldots,X_{n})$. Since $L$ is in general position we have $L\not\in{\mathfrak{P}}$. Hence ${\mathfrak{P}}$ is not an associated prime ideal of $K$. Therefore, $K_{N}=0$ for all sufficiently big $N$. So $X_{i}^{N}g_{j}\in I$ for sufficiently big $N$ and all $i,j$. Hence $g_{j}=\sum_{1\leq i\leq m}y_{j,i}f_{i}$ where $y_{j,i}\in F(X_{i})[X_{0},\ldots,X_{n}]$. Solving a linear system over the ring $F(X_{i})[X_{0},\ldots,X_{n}]$ we get an estimation for denominators from $F[X_{i}]$ of all $y_{j,i}$. Since all $g_{j}$ and $f_{i}$ are homogeneous we can suppose without loss of generality that all the denominators are $X_{i}^{N}$. Thus, we get an upper bound for $N$. Namely, $N$ is bounded from above by $(dl)^{2^{O(n+1)}}$.

Therefore, the sequence

[TABLE]

is exact for $m\geq(dl)^{2^{O(n+1)}}$. But $M/LM={\mathcal{A}}^{l}/(I+L{\mathcal{A}}^{l})$ is a module over a polynomial ring of $F[X_{0},\ldots,X_{n}]/(L)\simeq F[X_{0},\ldots,X_{n-1}]$. Hence by the inductive assumption the Hilbert function $H({\mathcal{A}}^{l}/(I+L{\mathcal{A}}^{l}),m)$ is stable for $m\geq(dl)^{2^{O(n)}}$. Therefore, (51) implies that the Hilbert function $H({\mathcal{A}}^{l}/I,m)$ is stable for $m\geq(dl)^{2^{O(n+1)}}$.

Obviously for $m<(dl)^{2^{O(n+1)}}$ the values $H({\mathcal{A}}^{l}/I,m)$ are bounded from above by $(dl)^{2^{O(n+1)}}$. Hence by the Newton interpolation all the coefficients of the Hilbert polynomial of ${\mathcal{A}}^{l}/I$ are bounded from above by $(dl)^{2^{O(n+1)}}$. The lemma is proved.

We need also a conversion of Lemma 12.

LEMMA 13

Let $I\subset{\mathcal{A}}^{l}$ be a graded submodule over the graded polynomial ring ${\mathcal{A}}=F[X_{0},\ldots,X_{n}]$. Assume that the Hilbert function $H({\mathcal{A}}^{l}/I,m)=\dim_{F}({\mathcal{A}}^{l}/I)_{m}$ is stable for $m\geq D$ and all absolute values of the coefficients of the Hilbert polynomial of the module ${\mathcal{A}}^{l}/I$ are bounded from above by $D$ for some integer $D>1$. Then $I$ has a system of generators $f_{1},\ldots,f_{m}$ with degrees $D^{2^{O(n+1)}}$.

PROOF Let us choose $f_{1},\ldots,f_{m}$ to be the reduced Gröbner basis of $I$ with respect to an admissible linear order $<$ on the monomials from ${\mathcal{A}}^{l}$, cf. the definitions from Section 1 and Section 4. The degree of a monomial from ${\mathcal{A}}^{l}$ is defined similarly to Section 1 and Section 4. We shall suppose additionally that the considered linear order is degree compatible, i.e., for any two monomials $z_{1},z_{2}$ if $\deg z_{1}<\deg z_{2}$ then $z_{1}<z_{2}$. For every $z\in{\mathcal{A}}$ the greatest monomial $\mathop{\rm Hdt}\nolimits(z)$ is defined. Further the monomial ideal $\mathop{\rm Hdt}\nolimits(I)$ is generated by all $\mathop{\rm Hdt}\nolimits(z)$, $z\in I$. Now $\mathop{\rm Hdt}\nolimits(f_{1}),\ldots,\mathop{\rm Hdt}\nolimits(f_{m})$ is a minimal system of generators of $\mathop{\rm Hdt}\nolimits(I)$ and $\deg f_{i}=\deg\mathop{\rm Hdt}\nolimits(f_{i})$ for every $1\leq i\leq m$. The values of Hilbert functions $H({\mathcal{A}}^{l}/\mathop{\rm Hdt}\nolimits(I),m)=H({\mathcal{A}}^{l}/I,m)$ coincide for all $m\geq 0$. Thus, replacing $I$ by $\mathop{\rm Hdt}\nolimits(I)$ we shall assume in what follows in the proof that $I$ is a monomial module.

For every $1\leq i\leq l$ denote by ${\mathcal{A}}_{i}\subset{\mathcal{A}}^{l}$ the $i$-th direct summand of ${\mathcal{A}}^{l}$. Put $I_{i}=I\cap{\mathcal{A}}_{i}$, $1\leq i\leq l$. Then $I\simeq\oplus_{1\leq i\leq l}I_{i}$ since $I$ is a monomial module. Further, for every $1\leq\alpha\leq m$ there is $1\leq i\leq l$ such that $f_{\alpha}\in I_{i}$. Let us identify ${\mathcal{A}}_{i}={\mathcal{A}}$. Then $I_{i}\subset{\mathcal{A}}$ is a homogeneous monomial ideal. The case $I_{i}={\mathcal{A}}$ is not excluded for some $i$. For the Hilbert functions we have

[TABLE]

If $({\mathcal{A}}/I_{i})_{D}=0$ for some $i$ then $({\mathcal{A}}/I_{i})_{m}=0$ for every $m\geq D$. In this case the ideal $I_{i}$ is generated by $\sum_{0\leq m\leq D}(I_{i})_{m}$. Hence in (52) for the values $m\geq D$ one can omit this index $i$ in the sum from the right part. Therefore, in this case the proof is reduced to a smaller $l$. So we shall assume without loss of generality that $({\mathcal{A}}/I_{i})_{D}\neq 0$, $1\leq i\leq l$.

Further, we use the exact description of the Hilbert function of a homogeneous ideal, see [4] Section 7. Namely there are the unique integers $b_{i,0}\geq b_{i,1}\geq\ldots\geq b_{i,n+2}=0$ such that

[TABLE]

for all sufficiently big $m$ and

[TABLE]

This description (without constants $b_{i,0}$) is originated from the classical paper [11]. The integers $b_{i,0},\ldots,b_{i,n+2}$ are called the Macaulay constants of the ideal $I_{i}$. Besides that,

[TABLE]

for every $m\geq b_{i,1}$, see [4] Section 7. By Lemma 7.2 [4] for all $1\leq\alpha\leq m$ if $f_{\alpha}\in I_{i}$ then $\deg f_{\alpha}\leq b_{i,0}$. Hence it is sufficient to prove that all $b_{i,0}$, $1\leq i\leq l$, are bounded from above by $D^{2^{O(n+1)}}$.

By (52) and (53) the coefficient at $m^{n-j}$, $0\leq j\leq n$, of the Hilbert polynomial of ${\mathcal{A}}^{l}/I$ is

[TABLE]

where $0\neq\mu_{j}$ is an integer and $\mu_{j,v}\in{\mathbb{Z}}[Z]$, $0\leq v\leq j-1$, is a polynomial with integer coefficients with $\deg\mu_{j,v}=j-v+1$. Moreover, $|\mu_{j}|$ and absolute values of all the coefficients of all the polynomials $\mu_{j,v}$ are bounded from above by, say, $2^{O(n^{2})}$. Denote $b_{j}=\sum_{1\leq i\leq l}b_{i,j}$, $0\leq j\leq n+2$. By the condition of the lemma all the coefficients of the Hilbert polynomial of ${\mathcal{A}}^{l}/I$ are bounded from above by $D$. Hence from (56) one can recursively estimate $b_{n+1},b_{n},\ldots,b_{1}$. Namely, $b_{n+1-j}=(2^{n^{2}}lD)^{2^{O(j+1)}}$, $0\leq j\leq n$. Hence $b_{1}=(lD)^{2^{O(n+1)}}$. Notice that $b_{i,1}\leq\max_{1\leq i\leq l}{b_{i,1}}\leq b_{1}$ for every $1\leq i\leq m$.

Now let $m\geq\max_{1\leq i\leq l}{b_{i,1}}$. By (55) if $h(i,m)\neq 0$ for some $1\leq i\leq l$ then $m<D$, i.e., $m$ is less than the bound $D$ for the stabilization of the Hilbert function of ${\mathcal{A}}^{l}/I$. Thus, $b_{i,0}\leq\max\{b_{i,1},D\}$ by (54). Hence $b_{i,0}$ is bounded from above by $(lD)^{2^{O(n+1)}}$.

We have $({\mathcal{A}}/I_{i})_{D}\neq 0$ for every $1\leq i\leq l$. This implies $H({\mathcal{A}}^{l}/I,D)\geq l$. Denote by $c_{j}$ the $j$-th coefficient of the Hilbert polynomial of the module ${\mathcal{A}}^{l}/I$. Now $|c_{j}|D^{j}\geq l/(n+1)$ for at least one $j$. Hence $D^{n+1}(n+1)\geq l$ by the condition of the lemma. This implies that $l^{2^{O(n+1)}}$ is bounded from above by $D^{2^{O(n+1)}}$. Therefore, $b_{i,0}$ is bounded from above by $D^{2^{O(n+1)}}$. The lemma is proved.

Appendix 2: Bound on the Gröbner basis of a monomial module via

the coefficients of its Hilbert polynomial

Denote by $C_{l}={\mathbb{Z}}_{+}^{n}\cup\cdots\cup{\mathbb{Z}}_{+}^{n}$ the disjoint union of $l$ copies of the semigrid ${\mathbb{Z}}_{+}^{n}=\{(i_{1},\dots,i_{n}):i_{j}\geq 0,1\leq j\leq n\}$. A subset of $C_{l}$ which intersects each disjoint copy of ${\mathbb{Z}}_{+}^{n}$ by a semigroup closed with respect to addition of elements from ${\mathbb{Z}}_{+}^{n}$ is called an ideal of $C_{l}$. Any ideal $I$ in $C_{l}$ has a unique finite Gröbner basis $V=V_{I}$, denote $T=C_{l}\setminus I$. Clearly, $I$ corresponds to a monomial submodule in the free module $(F[X_{1},\dots,X_{n}])^{l}$. The degree of an element $u=(k;i_{1},\dots,i_{n})\in C_{l},1\leq k\leq l$ is defined as $|u|=i_{1}+\cdots+i_{n}$. The degree of a subset in $C_{l}$ is defined as the maximum of the degrees of its elements. The Hilbert function $H_{T}(z)$ equals to the number of vectors $u\in T$ such that $|u|\leq z$. Then $H_{T}(z)=\sum_{0\leq s\leq m}c_{s}z^{s},\quad z\geq z_{0}$ for suitable $z_{0}$, integers $c_{0},\dots,c_{m}$ where the degree $m\leq n$. Denote $c=\max_{0\leq s\leq m}|c_{s}|s!+1$.

PROPOSITION 1

(cf. [6], [12], [4]). The degree of $V$ does not exceed $(cn)^{2^{O(m)}}$.

PROOF An $s$-cone we call a subset of a $k$-th copy of ${\mathbb{Z}}_{+}^{n}$ in $C_{l}$ for a certain $1\leq k\leq l$ of the form

[TABLE]

for suitable $1\leq j_{1},\dots,j_{n-s}\leq n$. The degree of (57) we define as $|P|=i_{1}+\cdots+i_{n-s}$ (note that this definition is different from the one in [4]). By a predessesor of (57) we mean each $s$-cone in the same $k$-th copy of ${\mathbb{Z}}_{+}^{n}$ of the type

[TABLE]

for some $1\leq p\leq n-s$, provided that $i_{p}\geq 1$. Fix an arbitrary linear order on $s$-cones compatible with the relation of predessesors.

By inverse recursion on $s$ we fill gradually $T$ (as a union) by $s$-cones. For the base we start with $s=m$. Assume that a current union $T_{0}\subset T$ of $m$-cones is already constructed (at the very beginning we put $T_{0}=\emptyset$) and an $m$-cone of the form (57) with $s=m$ is the least one (with respect to the fixed linear order on $m$-cones) which is contained in $T$ not being a subset of $T_{0}$. Observe that each predessesor of this $m$-cone was added to $T_{0}$ at earlier steps of its construction. Since the total number of $m$-cones added to $T_{0}$ does not exceed $c_{m}m!<c$ we deduce that the degree of every such $m$-cone is less than $c_{m}m!$ (taking into account that the very first $m$-cone added to $T_{0}$ has the degree [math]).

For the recursive step assume that the current $T_{0}$ is a union of all possible $m$-cones, $(m-1)$-cones,…,$(s+1)$-cones and perhaps, some $s$-cones. This can be expressed as $\deg(H_{T}-H_{T_{0}})\leq s$. Again as in the base take the least $s$-cone of the form (57) which is contained in $T$ not being a subset of $T_{0}$. Observe that each predessesor of the type (58) of this $s$-cone is contained in an appropriate $r$-cone $Q$, $r\geq s$, such that $Q$ was added to $T_{0}$ at earlier steps of its constructing and $Q\subset\{X_{j_{p}}=i_{p}-1\}$. Hence

[TABLE]

The described construction terminates when $T_{0}=T$. Denote by $t_{s}$ the number of $s$-cones added to $T_{0}$ and by $k_{s}$ the maximum of their degrees. We have seen already that $t_{m},k_{m}<c$.

Now by inverse induction on $s$ we prove that $t_{s},k_{s}\leq(cn)^{2^{O(m-s)}}$. To this end we introduce a relevant semilattice on cones. Let ${\cal C}=\{C_{\alpha,\beta}\}_{\alpha,\beta},\quad 0\leq\beta\leq\gamma_{\alpha}$ be a family of cones of the form (57) where $\dim C_{\alpha,\beta}=\alpha$. By an $\alpha$-piece we call an $\alpha$-cone being the intersection of a few cones from ${\cal C}$. All the pieces constitute a semilattice $\cal L$ with respect to the intersection and with maximal elements from ${\cal C}$. We treat $\cal L$ also as a partially ordered set with respect to the inclusion relation. Clearly, the depth of $\cal L$ is less than $n$. Our nearest purpose is to bound from above the size of $\cal L$. For the sake of simplifying the bound we assume (and this will suffice for our goal in the sequel) that $\gamma_{\alpha}\leq(cn)^{2^{O(m-\alpha)}}$ for $s\leq\alpha\leq m$ and $\gamma_{\alpha}=0$ when $\alpha<s$, although one could write a bound in general in the same way. Besides that we assume that th