Locally nilpotent derivations of affine

domains

L. Makar-Limanov ^{∗}

Abstract

Any locally nilpotent derivation of an affine domain is equivalent to a restriction of a Jacobian type derivation of a polynomial ring.

Introduction.

Let us recall that a locally nilpotent derivation is a generalization of a partial derivative of a polynomial ring. It is a derivation of a ring which when applied sufficiently many times to a given element of a ring sends it to the zero.

∗The author is supported by an NSA grant. It is also a pleasure to acknowledge the hospitality of several institutions the author enjoyed while working on this project, to wit the Faculty of Mathematics and Computer Science of Nicolaus Copernicus University in Torun, Poland, the Department of Mathematics of University of Hong Kong, China, and the Max Plank Institute of Mathematics in Bonn, Germany.

Locally nilpotent derivations (let us call them lnd) are useful though rather elusive objects. Though on “majority” of rings we do not have them at all, when we have them it is rather hard to find them and even harder to find all of them or to give some qualitative statements. Even for polynomial rings we do not know much. Of course it is rather easy to understand the situation for a polynomial ring with one variable. For two variable case it is not easy and requires a considerable effort (see [Re] or [ML2]). For three variables the kernel of a non-zero lnd is a polynomial ring in two variables ([Mi]) and there is a description of all weighted homogeneous derivations, i. e. derivations which send forms which are homogeneous relative to a given set of variables supplied with given positive weights to homogeneous forms ([Fr1], [Da2], [DR]). There are examples of lnd on polynomial rings with more than four variables with non-finitely generated kernels ([Ro], [Fr2], [DF1]) and though it is not known whether such derivations exist in the case of four generators, it is know that here there is no uniform bound on the number of generators of the kernel of an lnd ([DF2]). There are also several classes of surfaces for which we know everything as far as lnds are concerned ([ML2], [ML3], [ML4], [Da3], [Wi]) and a class of threefolds for which we know a lot ([ML1], [KML]), and that is all.

The purpose of this paper is to give a standard form for an lnd on the affine domains. This form is somewhat analogous to a matrix representation of a linear operator. Hopefully this representation of lnds will be helpful in the future research.

Definitions and claim.

Let ℂ be the field of complex numbers and let A be a domain over ℂ. A ℂ-linear mapping ∂ : A → A is called a derivation if it distributes a product of elements of A according to the Leibnitz law: ∂(ab) = ∂(a)b + a∂(b).

Constants A^{∂} = {a ∈ A|∂(a) = 0} of ∂ form a subalgebra of A.

A derivation is locally nilpotent if for any a ∈ A there exists a natural n = n(a) for which ∂^{n}(a) = 0. Let us denote by LND(A) the set of all lnd of A.

Let ℂ_{n} = ℂ[x_{1}, …, x_{n}] be the polynomial ring with n generators. A class of derivations on ℂ_{n} can be obtained as follows. Take f_{1}, …, f_{n−1} ∈ ℂ_{n}. Define ∂(g) = J(f_{1}, …, f_{n−1}, g) where J stands for the Jacobian of its arguments, that is for the determinant of the corresponding Jacobi matrix. Sometimes these derivations are called Jacobian derivations. A Jacobian derivation is also a derivation in every parameter f_{i}, which helps to make computations.

Now we are ready to formulate the main result.

Theorem. Let I be a prime ideal of ℂ_{n}, R be the factor ring ℂ_{n}/I, and π : ℂ_{n} → R be the corresponding projection. Let ∂ ∈ LND(R). Then there exist elements f_{1}, …, f_{n−1} ∈ ℂ_{n} and non-zero elements a, b ∈ R^{∂} for which a∂(π(f)) = bπ(J(f_{1}, …, f_{n−1}, f)) for any f ∈ C_{n}.

The proof consists of eight lemmas. To make it less boring for the reader, here is an outline of what is going to happen.

Let us assume that the transcendence degree of R is n − k.

In the first five Lemmas we will see that it is possible to find k elements i_{1}, …, i_{k} ∈ I so that J(i_{1}, …, i_{k}, f_{k+1}, …, f_{n}) ∈ I if and only if the elements f_{k+1}, …, f_{n} are algebraically dependent over I (this is probably obvious to a geometer). Only in the last three Lemmas the derivation ∂ appears.

Exposition is made as elementary as possible to make this technique as popular as possible, possibly a misguided approach!

First five Lemmas.

Let I be an ideal of ℂ_{n}. Choose a non-zero element p ∈ I of the minimal possible total degree. We are not concerned, of course, with the case p ∈ ℂ. Up to renumbering of generators x_{1}, …, x_{n} we may assume that p contains x_{1}. If we consider p as a polynomial in x_{1} with coefficients in ℂ[x_{2}, …, x_{n}] then the coefficients of p with positive degrees of x_{1} do not belong to I since their degrees are too small. By looking at all elements of I as polynomials of x_{1} we can find an element p_{1} ∈ I satisfying the following two conditions:

The coefficient q_{1} of the highest degree of x_{1} of p_{1} is not in I.

The degree d of p_{1} relative to x_{1} is the smallest degree possible provided the first condition is satisfied.

The degree d is positive because otherwise p_{1} ∈ I is the coefficient of x_{1}^{0} and the first condition is not satisfied.

Let I_{1} = I ∩ ℂ[x_{2}, …, x_{n}]. Find now p_{2} ∈ I_{1} with the same property relative to x_{2} (again up to a possible renumbering) and so on until we reach k for which I_{k} = 0. In the process we will obtain polynomials p_{1}, p_{2}, …, p_{k}.

Lemma 1. Any elements f_{k}, …, f_{n} ∈ ℂ_{n} are algebraically dependent over I.

Proof. We may assume that f_{k}, …, f_{n} are algebraically independent since otherwise the statement is clear. Let us consider the linear subspace V_{N} of ℂ_{n} of all polynomials of f_{k}, …, f_{n} with total degree relative to f_{k}, …, f_{n} not exceeding N. The linear dimension of V_{N} over ℂ is at least (\frac{N}{n})^{n−k+1}.

Denote by deg the total degree of elements of ℂ_{n} relative to x_{1}, …, x_{n} and by deg_{j} the degree relative to x_{j}. If D = max(deg(f_{k}), …, deg(f_{n})) then deg(f) ≤ DN for any f ∈ V_{N}. Also let e_{i} = deg(p_{i}).

Take p = ax_{1}^{m} + … with deg(p) ≤ DN. Recall p_{1} = q_{1}x_{1}^{d} + . . .. (Here … are the terms of the smaller degree in x_{1}.)

Let r_{1,1} = q_{1}p − ax_{1}^{m−d}p_{1} if m ≥ d. Then deg_{1}(r_{1,1}) < deg_{1}(p). If m < d take r_{1,1} = q_{1}p. Since deg(ax_{1}^{m−d}) ≤ deg(p) and deg(q_{1}) < deg(p_{1}) in both cases deg(r_{1,1}) ≤ DN + e_{1}. Since DN ≥ deg_{1}(p) after DN steps like this we obtain r_{1} with deg_{1}(r_{1}) < deg_{1}(p_{1}) and deg(r_{1}) ≤ DN(1 + e_{1}).

Now apply the same procedure to r_{1} and p_{2}. After DN(1 + e_{1}) steps we obtain r_{2} with deg_{2}(r_{2}) < deg_{2}(p_{2}) and deg(r_{2}) ≤ DN(1 + e_{1})(1 + e_{2}). Also deg_{1}(r_{2}) ≤ deg_{1}(r_{1}) since deg_{1}(q_{2}) = deg_{1}(p_{2}) = 0 .

Considering further on the pairs r_{2}, p_{3}, etc., we will get r_{k} with deg(r_{k}) ≤ DN(1 + e_{1}) … (1 + e_{k}) and deg_{i}(r_{k}) < deg_{i}(p_{i}).

Take m = q_{1}q_{2}^{(1+e_{1})} … q_{k}^{(1+e_{1})...(1+e_{k−1})}. We observed that m^{DN}p = ∑ α_{i}p_{i}+ r_{k} where deg_{j}(r_{k}) < deg_{j}(p_{j}) for j = 1, …, k and deg(r_{k}) ≤ DN(1 + e_{1}) … (1 + e_{k}) if deg(p) ≤ DN. Denote r_{k} by r_{k}(p). It is clear that when N is fixed the restriction of p → r_{k}(p) on V_{N} is a linear mapping of V_{N} into ℂ_{n}.

Since the degrees of r_{k}(p) relative to x_{1}, …, x_{k} are bounded and deg(r_{k}(p)) ≤ DN(1 + e_{1}) … (1 + e_{k}), all remainders belong to a subspace R_{N} of ℂ_{n} with dimension not exceeding cN^{n−k} where c is a constant which does not depend on N. So dim(V_{N}) > dim(R_{N}) for a sufficiently large N and a linear mapping from V_{N} to R_{N} should have a kernel. It implies that there is a non-zero f ∈ V_{N} with r_{k}(f) = 0. So m^{DN}f = ∑ α_{i}p_{i} ∈ I and since I is prime and all q_{i} ∉ I we conclude that f ∈ I.

Remark. If f ∈ I then r_{k}(f) = 0 for any N. Indeed, in this case r = r_{k}(f) ∈ I. Since deg_{1}(r) < deg_{1}(p_{1}), from the definition of p_{1} all coefficients of r as a monomial in x_{1} must be in I_{1}. Looking at these coefficients as monomials in x_{2} and using deg_{2}(r) < deg_{2}(p_{2}) and definition of p_{2} we see that they in turn have all coefficients in I_{2}, etc.. Since I_{k} = 0 we see that r = 0.

Lemma 2. J(p_{1}, p_{2}, …, p_{k}, x_{k+1}, …, x_{n}) ∉ I.

Proof. Indeed, J(p_{1}, p_{2}, …, p_{k}, x_{k+1}, …, x_{n}) = \frac{∂p_{1}}{∂x_{1}} \frac{∂p_{2}}{∂x_{2}} … \frac{∂p_{k}}{∂x_{k}} since only p_{1} depends on x_{1}, only p_{2} depends on x_{2}, etc.. But \frac{∂p_{i}}{∂x_{i}} ∉ I for all i by definition of p_{i}, so \frac{∂p_{1}}{∂x_{1}} \frac{∂p_{2}}{∂x_{2}} … \frac{∂p_{k}}{∂x_{k}} ∉ I.

Lemma 3. J(f_{1}, …, f_{k+1}, …, f_{n}) ∈ I if f_{1}, …, f_{k+1} ∈ I.

Proof. By Lemma 1 and Remark to Lemma 1 we can find a monomial M such that Mf_{i} = ∑ α_{i,j}p_{j} for i ≤ k+1. Therefore J(Mf_{1}, …, Mf_{k+1}, f_{k+2}, …, f_{n}) = M^{k+1}J(f_{1}, …, f_{k+1}, …, f_{n}) + Δ = ∑ J(α_{1,j_{1}}p_{j_{1}}, …, α_{k+1,j_{k+1}}p_{j_{k+1}}, …, f_{n}). Each summand J(α_{1,j_{1}}p_{j_{1}}, …, α_{k+1,j_{k+1}}p_{j_{k+1}}, …, f_{n}) ∈ I since there is only k different p_{i}’s. Also Δ ∈ I because it is a linear combination of Jacobians with coefficients containing f_{i}’s where i ≤ k + 1.

So M^{k+1}J(f_{1}, …, f_{k+1}, …, f_{n}) = ∑ J(α_{1,j_{1}}p_{j_{1}}, …, α_{k+1,j_{k+1}}p_{j_{k+1}}, …, f_{n})− Δ ∈ I. Since I is prime we see that J(f_{1}, …, f_{k+1}, …, f_{n}) ∈ I.

Lemma 4. J(f_{1}, f_{2}, …, f_{k}, f_{k+1}, …, f_{n}) ∈ I if f_{1}, f_{2}, …, f_{k} ∈ I and f_{k+1}, …, f_{n} are dependent over I.

Proof. Let P be a non-zero polynomial of f_{k+1}, …, f_{n} with value in I of minimal total degree possible. We may assume (up to renumbering) that P depends on f_{n}. So J(f_{1}, f_{2}, …, f_{k}, f_{k+1}, …, f_{n−1}, P) ∈ I by Lemma 3. On the other hand J(f_{1}, f_{2}, …, f_{k}, f_{k+1}, …, f_{n−1}, P) = \frac{∂P}{∂f_{n}} J(f_{1}, f_{2}, …, f_{k}, f_{k+1}, …, f_{n}). Since \frac{∂P}{∂f_{n}} ∉ I it implies that J(f_{1}, f_{2}, …, f_{k}, f_{k+1}, …, f_{n}) ∈ I.

Lemma 5. J(p_{1}, p_{2}, …, p_{k}, f_{k+1}, …, f_{n}) ∉ I if f_{k+1}, …, f_{n} are independent over I.

Proof. Assume that J(p_{1}, p_{2}, …, p_{k}, x_{k+1}, …, x_{k+i}, f_{k+i+1}, …, f_{n}) ∉ I and that i is minimal possible. From Lemma 2 we know that i ≤ n − k. Elements x_{k+1}, …, x_{k+i}, f_{k+i+1}, …, f_{n} are algebraically independent over I by Lemma 4. Let us consider now x_{k+1}, …, x_{k+i}, f_{k+i}, f_{k+i+1}, …, f_{n}. By Lemma 1 they are algebraically dependent over I. So let us take a nonzero polynomial P of x_{k+1}, …, x_{k+i}, f_{k+i}, f_{k+i+1}, …, f_{n} with value in I of minimal total degree possible. Its degree relative to f_{k+i} is positive as we noticed above. Since by assumption of the Lemma f_{k+i}, f_{k+i+1}, …, f_{n} are algebraically independent over I, polynomial P (up to renumbering) depends on x_{k+i}.

From Lemma 3 follows that I ∋ J(p_{1}, p_{2}, …, p_{k}, x_{k+1}, …, x_{k+i−1}, P, f_{k+1+i}, …, f_{n}) = \frac{∂P}{∂x_{k+i}} J(p_{1}, p_{2}, …, p_{k}, x_{k+1}, …, x_{k+i}, f_{k+i+1}, …, f_{n}) + \frac{∂P}{∂f_{k+i}} J(p_{1}, p_{2}, …, p_{k}, x_{k+1}, …, x_{k+i−1}, f_{k+i}, …, f_{n}).

Since \frac{∂P}{∂x_{k+i}} ∉ I, \frac{∂P}{∂f_{k+i}} ∉ I, and J(p_{1}, p_{2}, …, p_{k}, x_{k+1}, …, x_{k+i}, f_{k+i+1}, …, f_{n}) ∉ I, we see that J(p_{1}, p_{2}, …, p_{k}, x_{k+1}, …, x_{k+i−1}, f_{k+i}, …, f_{n}) ∉ I.

To avoid a contradiction we must assume that J(p_{1}, p_{2}, …, p_{k}, x_{k+1}, …, x_{k+i}, f_{k+i+1}, …, f_{n}) does not contain x_{j}’s at all which proves the Lemma.

Additional definitions and facts about lnd.

Let A be a domain of characteristic zero, F = Frac(A), and ∂ ∈ LND(A). Derivation ∂ can be extended on F by ∂(ab^{−1}) = ∂(a)b^{−1} − a∂(b)b^{−2}.

Denote this extension also by ∂.

Field F contains an element s for which ∂(s) = 1, s is transcendental over F^{∂}, A ⊂ F^{∂}[s], and F = F^{∂}(s). So every element of A can be looked at as a polynomial in s and we can define deg_{∂}(a) for a ∈ A as the degree of this polynomial. As usual, deg_{∂}(0) = −∞.

Element s = \frac{t}{∂(t)} for some t ∈ A with ∂(t) ∈ A^{∂}. Also F^{∂} = Frac(A^{∂}) and since the transcendence degree of F over F^{∂} is one we have trdeg(A^{∂}) = trdeg(A) − 1 if trdeg(A) < ∞.

All these facts can be found in [Es].

Two lnds of A are equivalent if they determine the same degree function on A.

Remaining Lemmas.

Recall that R = ℂ_{n}/I, π is the projection of ℂ_{n} on R, and ∂ ∈ LND(R). So all of the above is applicable to our setting.

We will be looking at elements of R also as polynomials from F^{∂}[s] where F is the field of fractions of R, s ∈ F, and ∂(s) = 1 and denote by deg the corresponding degree.

Let i_{1}, …, i_{k} ∈ I and f_{k+1}, …, f_{n−1} ∈ ℂ_{n}. From Lemma 3 it follows that ϵ(π(f)) = π(J(i_{1}, …, i_{k}, f_{k+1}, …, f_{n−1}, f) is a well-defined derivation on R since replacement of f by f + i in the right side does not change the left side if i ∈ I.

Let us chose now some algebraically independent r_{k+1}, …, r_{n−1} ∈ R^{∂}. It is possible since trdeg(R^{∂}) = n − k − 1. Let p_{k+1}, …, p_{n−1} ∈ ℂ_{n} be any elements for which π(p_{i}) = r_{i}. Denote the derivation π(J(i_{1}, …, i_{k}, p_{k+1}, …, p_{n−1}, f)) of R by ϵ(i, r)(g) where g = π(f). Again from Lemma 3 it is clear that ϵ(i, r)(g) does not depend on a particular choice of p_{k+1}, …, p_{n−1}. As always we will think about these derivations as derivations of F.

From Lemma 5 we know that for some choices of i these derivations are non-zero.

Lemma 6. Non-zero derivations ϵ(i, r) and ϵ(i, s) are linearly dependent over F^{∂}.

Proof. The claim is obvious if r and s are the same. Let us assume that ϵ(i, r) and ϵ(i, s) do not satisfy the claim and that the number of common elements of r and s is maximal possible under this condition. Up to renumbering we may assume that s_{n−1} ∉ {r_{k+1}, …, r_{n−1}}. Elements r_{k+1}, …, r_{n−1}, s_{n−1} are algebraically dependent. Let us take a non-zero polynomial P of r_{k+1}, …, r_{n−1}, s_{n−1} with value zero of minimal total degree possible. Its degree relative to s_{n−1} is positive. P also must depend on at least one r_{j} ∉ {s_{k+1}, …, s_{n−1}} since the elements {s_{k+1}, …, s_{n−1}} are algebraically independent. Again up to renumbering we may assume that it is r_{n−1}.

Let p_{k+1}, …, p_{n−1} ∈ ℂ_{n} be co-images of r_{k+1}, …, r_{n−1} and let q be a coimage of s_{n−1}. So P (p_{k+1}, …, p_{n−1}, q) ∈ I and I ∋ J(i_{1}, …, i_{k}, p_{k+1}, …, p_{n−2}, P, f) = \frac{∂P}{∂p_{n−1}} J(i_{1}, …, i_{k}, p_{k+1}, …, p_{n−1}, f)+\frac{∂P}{∂q} J(i_{1}, …, i_{k}, p_{k+1}, …, p_{n−2}, q, f) for any f ∈ ℂ_{n} by Lemma 3. Therefore π(\frac{∂P}{∂p_{n−1}} J(i_{1}, …, i_{k}, p_{k+1}, …, p_{n−1}, f) + \frac{∂P}{∂q} J(i_{1}, …, i_{k}, p_{k+1}, …, p_{n−2}, q, f)) = 0. Both π(\frac{∂P}{∂p_{n−1}} ) and π(\frac{∂P}{∂q}) are polynomials of ∂-constants r_{k+1}, …, r_{n−1}, s_{n−1}, and so π(\frac{∂P}{∂p_{n−1}} ), π(\frac{∂P}{∂q}) ∈ R^{∂} \ 0 by the definition of P. So the corresponding derivations are linearly dependent over F^{∂}, are both non-zero, and are linearly independent with ϵ(i, s).

Since {s_{k+1}, …, s_{n−1}} has more common elements with {π(p_{k+1}), …, π(p_{n−2}), π(q)} = {r_{k+1}, …, r_{n−2}, s_{n−1}} than with {r_{k+1}, …, r_{n−1}} we reached a contradiction which proves the Lemma.

Remark. Since r is a transcendence basis of R^{∂}, the kernel of ϵ(i, r) is F^{∂} by Lemma 4.

This Lemma shows that the derivations ϵ(i, r) essentially do not depend on r. So let us fix a basis r and omit it from the notation for ϵ: ϵ(i, r) = ϵ(i).

From Lemma 5 we know that ϵ(i) is a non-zero derivation for some choices of i. Recall s ∈ F satisfying ∂(s) = 1. If ϵ(i) ≠ 0 then ϵ(i)(s) ≠ 0 since otherwise the kernel of ϵ(i) is too large. So there are choices of i_{1}, …, i_{k} ∈ I for which q = ϵ(i)(s) ≠ 0 and deg(q) is minimal possible. Let us fix such a q.

Lemma 7. For any j_{1}, …, j_{k} ∈ I the element ϵ(j)(s) is divisible by q as an element of F^{∂}[s].

Proof. If the claim is wrong choose a pair i, j so that q = ϵ(i)(s), p = ϵ(j)(s) is not divisible by q, and the number d of common elements of i and j is maximal possible.

For any h_{1}, …, h_{k} ∈ I denote ϵ(h)(s) by Δ(h_{1}, …, h_{k}). Since Δ(h_{1}, …, h_{k}) = π(J(h_{1}, …, h_{k}, p_{k+1}, …, p_{n−1}, s) it is skew-symmetric. Also J(fh_{1}, h_{2}, …, h_{k}, p_{k+1}, …, p_{n−1}, s) = h_{1}J(f, …, h_{k}, p_{k+1}, …, p_{n−1}, s)+fJ(h_{1}, …, h_{k}, p_{k+1}, …, p_{n−1}, s) for any f ∈ ℂ_{n}. So π(J(fh_{1}, …, h_{k}, p_{k+1}, …, p_{n−1}, s)) = π(f)π(J(h_{1}, …, h_{k}, p_{k+1}, …, p_{n−1}, s)) since π(h_{1}) = 0. This can be written with a slight abuse of notations as Δ(gh_{1}, h_{2}, …, h_{k}) = gΔ(h_{1}, …, h_{k}) where g = π(f) because any co-image of g in the left side gives the same right side. Since Δ is skewsymmetric Δ(h_{1}, h_{2}, …, gh_{j}, …, h_{k}) = gΔ(h_{1}, …, h_{k}) for any j.

It is easy to see now that δ(i) = Δ(i_{1}, …, i_{k−1}, i) is divisible by q for any i ∈ I. Consider q and δ(i) as elements of F^{∂}[s] and choose u, v ∈ F^{∂}[s] so that w = uq + vδ(i) is the greatest common divisor of q and δ(i). If q does not divide δ(i) then deg(w) < deg(q). Since F^{∂} = Frac(R^{∂}) and s = \frac{t}{∂(t)} where t ∈ R and ∂(t) ∈ R^{∂} we may assume that u, v ∈ R. But Δ(i_{1}, …, i_{k−1}, ui_{k}+vi) = w which leads to a contradiction with the definition of q.

Without loss of generality we may assume that j_{k} ∉ {i_{1}, …, i_{k}} and that i_{k} ∉ {j_{1}, …, j_{k}}. As we already know Δ(i_{1}, …, i_{k−1}, j_{k}) = rq for some r ∈ F^{∂}[s]. If r ∉ R find f ∈ R^{∂} so that fr ∈ R. For j = fj_{k} − fri_{k} we have Δ(i_{1}, …, i_{k−1}, j) = 0. Now, Δ(j_{1}, …, j_{k−1}, j + i_{k}) = fΔ(j_{1}, …, j_{k}) + (1 − fr)Δ(j_{1}, …, j_{k−1}, i_{k}). The element Δ(j_{1}, …, j_{k−1}, i_{k}) is divisible by q since the number of common elements of {i_{1}, …, i_{k}} and {j_{1}, …, j_{k−1}, i_{k}} is d + 1. On the other hand Δ(i_{1}, …, i_{k−1}, j + i_{k}) = q since Δ(i_{1}, …, i_{k−1}, j) = 0. The sets {j_{1}, …, j_{k−1}, j + i_{k}} and {i_{1}, …, i_{k−1}, j + i_{k}} also have d + 1 common elements, so q divides Δ(j_{1}, …, j_{k−1}, j +i_{k}). Therefore fΔ(j_{1}, …, j_{k}) = Δ(j_{1}, …, j_{k−1}, j+i_{k})−(1−r)Δ(j_{1}, …, j_{k−1}, i_{k}) is divisible by q. Since f ∈ R^{∂} we can conclude that Δ(j_{1}, …, j_{k}) is divisible by q in F^{∂}[s].

Lemma 8. q ∈ F^{∂} \ 0.

Proof. Let us show that 1 = π(J(x_{1}, …, x_{n})) is divisible by q in F^{∂}[s]. Write π(x_{a}) = ∑ f_{a,b}s^{b} where f_{a,b} ∈ F^{∂}.

As we know it is possible to find a v ∈ R^{∂} \ 0 so that vπ(x_{a}) = ∑ g_{a,b}t^{b} where g_{a,b} ∈ R^{∂} and t ∈ R. We can lift these expressions into ℂ_{n}: V x_{a} = ∑ G_{a,b}T^{b} + i_{a} where V, G_{a,b}, and T are corresponding co-images and i_{a} ∈ I.

Let A = ℂ_{n}[V^{−1}] i. e. A is a subring of the field of fractions of ℂ_{n} with denominators which are powers of V. Projection π can be defined on A and it will take it to R[v^{−1}], which is a subring of F^{∂}[s].

So J(x_{1}, …, x_{n}) = ∑ J(y_{1}, …, y_{n}) where y_{a} is either \frac{G_{a,b}}{V} T^{b} or \frac{i_{a}}{V}. Since J(…, HT^{b}, …) = T^{b}J(…, H, …) + bT^{b−1}HJ(…, T, …) and J(…, \frac{G}{V}, …) = V^{−1}J(…, G, …) − GV^{−2}J(…, V, …) we can write J(x_{1}, …, x_{n}) = ∑ u_{m}J(z_{1}, …, z_{n}) where u_{m} ∈ ℂ[T, G_{1,0}, …, G_{n,N}, i_{1}, …, i_{n}, V^{−1}] and z_{j} is either G_{a,b}, or T, or i_{a}.

Since π(u_{m}) ∈ F^{∂}[s] it suffices to show that q divides π(J(z_{1}, …, z_{n})). If T appears in this Jacobian in two positions then it is zero, so we may assume that there is at most one T in it. Also by Lemma 3 at most k positions have elements from I. If exactly k positions have elements from I then q divides π(J(z_{1}, …, z_{n})) by Lemmas 6 and 7. So we may assume that at least n − k positions have elements G_{a,b}. Since π(G_{a,b}) ∈ F^{∂} these elements are algebraically dependent over I.

Assume that π(J(z_{1}, …, z_{n})) is not divisible by q and rewrite it so that the elements from I are in the beginning and T is in the last position if at all. So J(z_{1}, …, z_{n}) = J(i_{1}, …, i_{m}, h_{m+1}, …, h_{n−1}, U) where i_{j} ∈ I, π(h_{j}) ∈ F^{∂}, π(U) ∈ F^{∂} ⋃ t and m < k. Assume also that the number of elements from I is maximal possible under this condition.

Let P be a non-zero polynomial of h_{m+1}, …, h_{n−1} with value in I and of minimal total degree possible. We may assume (up to renumbering) that P depends on h_{m+1}. Therefore J(i_{1}, …, i_{m}, P, h_{m+2}, …, h_{n−1}, U) = \frac{∂P}{∂h_{m+1}} J(i_{1}, …, i_{m}, h_{m+1}, …, h_{n−1}, U) is divisible by q. Since π(\frac{∂P}{∂h_{m+1}} ) ∈ R^{∂}\0 it implies that J(i_{1}, …, i_{m}, h_{m+1}, …, h_{n−1}, U) is divisible by q. So q is a unit in F^{∂}[s], i. e. q ∈ F^{∂}.

Find now any i_{1}, …, i_{k} ∈ I for which ϵ(i)(s) = q and take ϵ = ϵ(i). It is clear that ϵ = ϵ(s)∂. Since q = \frac{a}{b} where a, b ∈ R^{∂} \0 the Theorem is proved.

Remarks. Some forms of this Theorem appeared before. In [ML2] it was proved for any lnd of ℂ_{n}, and in [Da1] a more precise result was found for n = 3. Also in [KML] a somewhat confusing form of it was proved for hypersurfaces. It should be mentioned that for hypersurfaces and complete intersections the choice of elements from the corresponding ideal is essentially unique.

Example. The following example shows that in general we can give with a Jacobian only an equivalent lnd. Let ∂ be an lnd on ℂ_{4} given by ∂(z) = y, ∂(y) = x, ∂(x) = a, and ∂(a) = 0. Since ∂ is a homogeneous operator any invariant of ∂ is the sum of homogeneous invariants. It is clear that there is just one invariant of degree 1. So J(f_{1}, f_{2}, f_{3}, x) ≠ a if f_{i} ∈ ker(∂). Indeed f_{i} may be assumed homogeneous and then only one of them can have degree 1. Therefore deg(J(x, f_{1}, f_{2}, f_{3})) > 1.

Acknowledgments. Many colleagues are familiar with preliminary versions of the proof and helped to improve the exposition. I want to single out Daniel Daigle and Shulim Kaliman for very helpful suggestions.

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Department of Mathematics & Computer Science, Bar-Ilan University, 52900 Ramat-Gan, Israel

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

E-mail address: lml@math.wayne.edu