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
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, . . . , hn]δpq, (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, . . . , hn]δpq
=
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?.
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!
d˜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
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.
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.
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))
= d∗p. This completes the proof.
The following example shows that the divergence of non-commutative basis is not a constant, in general.
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 d∗1 = 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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