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The divergence of polynomial derivations

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Andrzej Nowicki

N. Copernicus University, Faculty of Mathematics and Computer Science, ul. Chopina 12–18, 87–100 Toru´ n, Poland

(e-mail: anow@mat.uni.torun.pl) December 29, 1993

Abstract

Let k be a ring k containing Q. A k-derivation d of k[X] = k[x1, . . . , xn] is called special if the divergence of d is equal to zero. We prove the following three theorems (up to some additional assumption on k):

(1) Locally nilpotent k-derivations of k[X] are special.

(2) Every derivation of a commutative basis of Derk(k[X]) is special.

(3) If n = 2 and d 6= 0 is a primitive k-derivation of k[X] then the ring of constants of d is nontrivial if and only if there exists 0 6= h ∈ k[X] such that the derivation hd is special.

Let k be a commutative ring containing the field Q of rational numbers, let k[X] = k[x1, . . . , xn] be the polynomial ring over k, and let d : k[X] −→ k[X] is a k-derivation of k[X]. Denote by d? the divergence of d, that is,

d? = ∂d(x1)

∂x1

+ · · · + ∂d(xn)

∂xn

. The derivation d is said to be special if d? = 0.

It is well known (see [1], [2], [9]) that if d is locally finite and k is reduced (i. e., k has no nonzero nilpotent elements), then d? is an element from k. In this note we shall show that if k is reduced and d is locally nilpotent then d is special.

Consider the k[X]-module Derk(k[X]) of all k-derivations of k[X]. This module is free and the set {∂x

i, . . . ,∂x

n} is a one of its bases. We say that a basis {d1, . . . , dn} of Derk(k[X]) is locally nilpotent if each di is locally nilpotent. In [5] we proved that the Jacobian Conjecture is true in the n-variable case if and only if every commutative basis of Derk(k[X]) is locally nilpotent. We shall show that if {d1, . . . , dn} is a commutative basis of Derk(k[X]), then each di is special.

Assume now that n = 2 and k is a field (of characteristic zero). Denote by k[X]d the ring of constants with respect to d, that is, k[X]d = Ker d. It is obvious that if h is a nonzero element of k[X] then k[X]d= k[X]hd. In Section 5 we prove that if d is nonzero and the polynomials d(x1) and d(x2) are relatively prime, then k[X]d 6= k if and only if there exists a nonzero element h ∈ k[X] such that the k- derivation hd is special.

1

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1 Some properties of the divergence.

Let us start from the following easy

Proposition 1.1. If d, δ ∈ Derk(k[X]) and r ∈ k[X], then:

(1) (d + δ)? = d?+ δ?. (2) (rd)? = rd?+ d(r).

(3) [d, δ]? = d(δ?) − δ(d?). 

The partial derivatives are special derivations. The above proposition implies that the set of all special derivations of k[X] is closed under the sum and the Lie product.

Let us denote by [h1, . . . , hn] the jacobian of h1, . . . , hn ∈ k[X], that is, [h1, . . . , hn] = det ∂hi

∂xj

 .

Proposition 1.2. Let d : k[X] −→ k[X] be a k-derivation and let h1, . . . , hn ∈ k[X].

Then

d([h1, . . . , hn]) = −[h1, . . . , hn]d?+

n

X

p=1

[h1, . . . , d(hp), . . . , hn].

Proof. Put fi = d(xi), fij = ∂x∂fi

j, hij = ∂h∂xi

j, for all i, j ∈ {1, . . . , n}, and let Sn denote the group of all permutations of {1, . . . , n}. First observe that

d(hσ(p)p) = ∂

∂xpd(hσ(p)) −

n

X

q=1

hσ(p)qfqp, (1.1)

for all σ ∈ Sn and p ∈ {1, . . . , n}. Observe also that X

σ∈Sn

(−1)|σ|hσ(1)1· · · hσ(p−1)(p−1)hσ(p)qhσ(p+1)(p+1)· · · hσ(n)n = [h1, . . . , hnpq, (1.2) for all p, q ∈ {1, . . . , n}, where |σ| is the sign of σ, and δpq is the Kronecker delta. Now, using these observations, we get:

d([h1, . . . , hn]) =

n

X

p=1

X

σ∈Sn

(−1)|σ|hσ(1)1· · · d(hσ(p)p) · · · hσ(n)n

(1.1)

=

n

X

p=1

X

σ∈Sn

(−1)|σ|hσ(1)1· · ·

"

∂xpd(hσ(p)) −

n

X

q=1

hσ(p)qfpq

#

· · · hσ(n)n

(1.2)

=

n

X

p=1

[h1, . . . , d(hp), . . . , hn] −X

p=1

n

n

X

q=1

fpq[h1, . . . , hnpq

=

n

X

p=1

[h1, . . . , d(hp), . . . , hn] −

n

X

p=1

fpp[h1, . . . , hn]

=

n

X

p=1

[h1, . . . , d(hp), . . . , hn] − [h1, . . . , hn]d?.

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This completes the proof. 

Corollary 1.3. If d is a special k-derivation of k[X] and h1, . . . , hn∈ k[X], then d([h1, . . . , hn]) =

n

X

p=1

[h1, . . . , d(hp), . . . , hn]. 

2 Automorphism E

d

.

Let A be a k-algebra (commutative with 1), let A[[t]] be the power series ring over A, and let d be a k-derivation of A. Denote by ˜d the k[[t]]-derivation of A[[t]] defined by

d(˜

X

p=0

aptp) =

X

p=0

d(ap)tp, and set

Ed(ϕ) =

X

p=0

1 p!

p(ϕ)tp for all ϕ ∈ A[[t]].

It is well known that Ed is a k[[t]]-automorphism of A[[t]], which is very useful in the differential algebra (see, for example: [7], [8], [4], [3]).

Now let A = k[X] = k[x1, . . . , xn] and let J be the jacobian of Ed(x1), . . . , Ed(xn), that is,

J = [Ed(x1), . . . , Ed(xn)] = det ∂Ed(xi)

∂xj

 . The jacobian J is an element of k[X][[t]]. Let us set J =P

p=0 1

p!Bptp, where each Bp is in k[X].

Assume that n = 2. Put x = x1, y = x2 and let fx, fy denote ∂f∂x, ∂f∂y, respectively, for any f ∈ k[x, y]. In such a case we have:

J = Ed(x)xEd(y)y− Ed(x)yEd(y)x

=

X

p=0

1

p!dp(x)xtp

! X

p=0

1

p!dp(y)ytp

!

X

p=0

1

p!dp(x)ytp

! X

p=0

1

p!dp(y)xtp

!

=

X

p=0

X

i+j=p

1

i!j!(di(x)xdj(y)y− di(x)ydj(y)x)

! tp

=

X

p=0

X

i+j=p

1

i!j![di(x), dj(y)]

! tp

Thus, we see that if n = 2, then

Bp = X

i+j=p

hi, ji[di(x), dj(y)],

for every p > 0. If n is arbitrary then, repeating the same argument, we get

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Proposition 2.1. For every p > 0,

Bp = X

i1+···+in=p

hi1, . . . , ini[di1(x1), . . . , din(xn)], where

hi1, . . . , ini = (i1+ · · · + in)!

i1! · · · in! . 

Let us continue are calculations for n = 2. From Propositions 2.1 and 1.2 we obtain the following equalities:

d(Bp) = X

i+j=p

hi, jid([di(x), dj(y)])

= X

i+j=p

hi, ji([di+1(x), dj(y)] + [di(x), dj+1(y)] − [di(x), dj(y)]d?)

= −Bpd? + X

i+j=p

hi, ji([di+1(x), dj(y)] + [di(x), dj+1(y)]).

Now, by standard formulas and Proposition 2.1, we get:

d(Bp) + Bpd? = [dp+1(x), y] + [x, dp+1(y)]

+

p−1

X

i=0

hi, p − ii[di+1(x), dp−i(y)] +

p

X

i=1

hi, p − ii[di(x), dp+1−i(y)]

= [dp+1(x), y] + [x, dp+1(y)]

+

p

X

i=1

(hi − 1, p + 1 − ii + hi, p − ii)[di(x), dp+1−i(y)]

= X

i+j=p+1

hi, ji[di(x), dj(y)]

= Bp+1.

We used the well known equality: hi − 1, ji + hi, ji = hi, j + 1i (where i > 1). There exists the following generalization of this equality:

hi1, . . . , ini +

n−1

X

j=1

hi1, . . . , ij − 1, . . . , in−1, ini = hi1, . . . , in−1, in+ 1i, (2.1) for i1, . . . , in−1> 1.

Now, using (2.1), Propositions 2.1, 1.2 and repeating the same arguments, one can easily deduce the following

Theorem 2.2. Let d be a k-derivation of k[X] = k[x1, . . . , xn] and let [Ed(x1), . . . , Ed(xn)] =

X

p=0

1 p!Bptp,

where Bp ∈ k[X]. Then B0 = 1, B1 = d? and Bp+1= Bpd?+ d(Bp) for all p > 0. 

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Corollary 2.3. Let d be a k-derivation of k[X]. If d? ∈ k then [Ed(x1), . . . , Ed(xn)] =

X

p=0

1

p!bptp = ebt, where b = d?. 

Corollary 2.4. If d is a special k-derivation of k[X], then [Ed(x1), . . . , Ed(xn)] = 1. 

3 The divergence of locally finite derivations.

Let us recall that if A is a commutative k-algebra and d is a k-derivation of A, then d is called locally nilpotent if for each r ∈ A there exists a natural number s such that ds(r) = 0, and is called locally finite if for any r ∈ A there exists a finite generated k-module M ⊆ A such that r ∈ M and d(M ) ⊆ M .

It is easy to check the following two propositions

Proposition 3.1. If d is a k-derivation of k[X] then, d is locally nilpotent if and only if Ed(k[t][X]) ⊆ k[t][X]. 

Proposition 3.2. If d is a k-derivation of k[X] then the following conditions are equiv- alent:

(1) d is locally finite;

(2) there exists a natural s such that deg dp(xi) 6 s, for all p > 0 and i = 1, . . . , n;

(3) Ed(k[[t]][X]) ⊆ k[[t]][X]. 

The following result is due to H. Bass, G. Meisters [1] and B. Coomes, V. Zurkowski [2]. An another proof of this fact is given in [9].

Theorem 3.3. Let k be a reduced ring containing Q and let k[X] = k[x1, . . . , xn] be a polynomial ring over k. If d is a locally finite k-derivation of k[X] then d?, the divergence of d, is an element of k.

Proof ([1]). Since d is locally finite, the series Ed(x1), . . . , Ed(xn) are elements of k[[t]][X] (Proposition 3.2). By the same way, since the derivation −d is locally finite, the series E−d(x1), . . . , E−d(xn) are also elements of k[[t]][X]. Hence, the mapping Ed is a k[[t]]-automorphism of the polynomial ring k[[t]][X], and hence J = [Ed(x1), . . . , Ed(xn)], the jacobian of Ed, is an invertible element of k[[t]][X]. But k is reduced, so J ∈ k[[t]].

We know, by Theorem 2.2, that J = 1 + d?t1+ · · · . Therefore d? ∈ k.  If k is non-reduced then the above property does not hold, in general.

Example 3.4. Let k = Q[y]/(y2) and let d be the k derivation of k[x] (a polynomial ring in a one variable) defined by d(x) = ax2, where a = y + (y2). Since d2(x) = 2a2x3 = 0, d is locally finite. But d? = 2ax 6∈ k. 

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Now we are ready to prove the main result of this section.

Theorem 3.5. Let k be a reduced ring containing Q. If d is a locally nilpotent k-derivation of k[X] then d? = 0.

Proof. Put b = d? and let J be the jacobian of Ed(x1), . . . , Ed(xn). We know, by Theorem 3.3 and Corollary 2.3, that b ∈ k and J = etb = P

p=0 1

p!bptp. Since d is locally nilpotent, the series Ed(x1), . . . , Ed(xn) are polynomials from k[X][t] and hence J ∈ k[X][t]. Thus bp = 0, for some p, and consequently (since k is reduced), b = 0. 

The derivation d from Example 3.4 is locally nilpotent. This means that if k is non- reduced then there exist locally nilpotent k-derivations of k[X] with a nonzero divergence.

4 Commutative bases of derivations.

Theorem 4.1. Let k be a reduced ring containing Q and let k[X] = k[x1, . . . , xn] be the polynomial ring over k. If {d1, . . . , dn} is a commutative basis of Derk(k[X]) then d?i = 0, for all i = 1, . . . , n.

Proof. Denote by A the matrix [di(xj)] and let w = det(A). Then w is an invertible element of k[X] ([5] Proposition 1) and so, since k is reduced, w is an invertible element of k.

Let [bij] be the matrix such that∂x

i =Pn

j=1bijdj, for i = 1, . . . , n. ThenPn

j=1bijdj(xi) = δij. Thus, [bij] = A−1 and we have: bij = (−1)i+jAjiw−1, where Apq is the determinant of the (n − 1) × (n − 1) matrix obtained from A by removing in A the p-th row and the q-th column. Set

D =

 d1

... dn

 and D(r) =

 d1(r)

... dn(r)

, for r ∈ R.

Then we have A = [D(x1), . . . , D(xn)] and

∂xi

= w−1det[D(x1), . . . , D(xi−1), D, D(xi+1), . . . , D(xn)], for all i = 1, . . . , n. Therefore, if p ∈ {1, . . . , n} then

0 = w−1dp(w)

= w−1dp(det[D(x1), . . . , D(xn)])

= w−1Pn

i=1det[D(x1), . . . , dpD(xi), . . . , D(xn)]

= w−1Pn

i=1det[D(x1), . . . , D(dp(xi)), . . . , D(xn)]

= Pn i=1

∂xi(dp(xi))

= dp. This completes the proof. 

The following example shows that the divergence of non-commutative basis is not a constant, in general.

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Example 4.2. Let

d1 = ∂

∂x1 + x22

∂x2, d2 = ∂

∂x2, . . . , dn = ∂

∂xn.

Then {d1, . . . , dn} is a basis of Derk(k[x1, . . . , xn]) (since det[di(xj)] = 1), and d1 = 2x2 6∈

k. 

5 The ring of constants in two variables

Let d be a k-derivation of the polynomial ring k[x, y]. Put d(x) = P , d(y) = Q and consider k[x, y]d, the ring of constants of k[x, y] with respect to d. It would be of considerable interest to find necessary and sufficient conditions on P and Q for k[x, y]d to be nontrivial (that is, k[x, y]d 6= k). The problem seems to be difficult. We may of course assume that the polynomials P and Q are relatively prime. Moreover, it is clear that k[x, y]d= k[x, y]hd for any nonzero h ∈ k[x, y].

The following two theorems give certain information concerning to our problem Theorem 5.1. Let k be a field of characteristic zero and let d be a nonzero k-derivation of k[x, y]. Assume that the polynomials d(x) and d(y) are relatively prime. Then the following two conditions are equivalent:

(1) k[x, y]d6= k;

(2) there exists a nonzero polynomial h ∈ k[x, y] such that (hd)? = 0.

Proof. (1) ⇒ (2). Let F ∈ k[x, y]dr k. Put d(x) = P , d(y) = Q and h = gcd(Fx, Fy).

Then P Fx+ QFy = 0, h 6= 0 and there exist relatively prime polynomials A, B ∈ k[x, y]

such that Fx = Ah and Fy = Bh. Hence AP = −BQ and hence, A | Q, Q | A, B | P and P | B. This implies that there exists an element a ∈ k r {0} such that A = aQ and B = −aP . Thus, we have:

(hd)? = (hP )x+ (hQ)y = −(a−1hB)x+ (a−1hA)y = −a−1Fyx+ a−1Fxy = 0.

(2) ⇒ (1). Let 0 6= h ∈ k[x, y] with (hd)? = 0 and let δ = hd, δ(x) = P , δ(y) = Q.

We may assume that P 6= 0 and Q 6= 0 (if P = 0 or Q = 0 then k[x, y]d 6= k). Put f = Q and g = −P . Then fy = qx and hence, by [5] Lemma 3, there exists H ∈ k[x, y] such that Hx = f and Hy = g. It is clear that H 6∈ k. Now observe that δ(H) = 0. Indeed,

δ(H) = Hxδ(x) + Hyδ(y) = f P + gQ = QP − P Q = 0.

Thus, H ∈ k[x, y]δ = k[x, y]d and H 6∈ k. 

Using the above theorem and the results of [6] it is not difficult to prove the following Theorem 5.2. Let k be a field of characteristic zero and let d, δ be k-derivations of k[x, y]

such that k[x, y]d6= k and k[x, y]δ 6= k. Then the following two conditions are equivalent:

(1) k[x, y]d= k[x, y]δ;

(2) there exist nonzero polynomials f, g ∈ k[x, y] such that f d = gδ. 

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References

[1] H. Bass, G.H. Meisters, Polynomial flows in the plane, J. Algebra, 55(1985), 173 – 208.

[2] B.A. Coomes, V. Zurkowski, Linearization of polynomial flows and spectra of deriva- tions, J. Dynamics and Diff. Equations, 3(1991), 29 – 66.

[3] A. van den Essen, Locally nilpotent derivations and their applications III, Catholic University, Nijmegen, Report 9330(1993).

[4] A. Nowicki, Differential ideals and rings, Ph.D.thesis (Polish), Toru´n UMK, 1978.

[5] A. Nowicki, Commutative basis of derivations in polynomial and power series rings, J. Pure Appl. Algebra,40(1986), 279–283.

[6] A. Nowicki, On the jacobian equation J (f, g) = 0 for polynomials in two variables, Nagoya J.Math.,109(1988), 151–157.

[7] A. Seidenberg, Differential ideals and rings of finitely generated type, Amer. J. Math., 89(1967), 22 – 42.

[8] W.V. Vasconcelos, Derivations of commutative noetherian rings, Math.Zeitschr.

112(1969), 229 – 233.

[9] V. D. Zurkowski, Locally finite derivations in three dimensions, Preprint 1993.

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