Families of k-derivations on k-algebras

Journal of Pure and Applied Algebra 199 (2005) 11–26

www.elsevier.com/locate/jpaa

Families ofk-derivations onk-algebras

Philippe BonnetMathematisches Institut, Universität Basel, Rheinsprung 21, 4051 Basel, Switzerland

Received 14 April 2004; received in revised form 19 November 2004Available online 10 March 2005Communicated by C.A. Weibel

Abstract

LetA be an integralk-algebra of finite type over a fieldk of characteristic zero. LetF be a familyof k-derivations onA andMF theA-module spanned byF. In this paper, we generalise a result dueto Nowicki and construct an element� of MF such thatker � = ⋂

d∈Fker d. Such a derivation iscalledF-minimal. Then we establish a density theorem forF-minimal derivations inMF.© 2004 Elsevier B.V. All rights reserved.

MSC:13N15; 13A99; 13A50

1. Introduction

Let A be an integralk-algebra of finite type over a fieldk of characteristic zero. Forconvenience, we will say that a sub-algebraB of A is algebraically closed in Aif everyelementa of A that is algebraic overB belongs toB. LetF be a family ofk-derivations onA. In this paper, we are interested in describing the kernel of this family, i.e. the followingset

ker F =⋂d∈F

ker d.

Let MF be theA-module spanned by the elements ofF. By analogy with the theory offoliations, we say that an elementf of A is afirst integral of a k-derivationd if d(f ) = 0andf /∈ k. Similarly f is afirst integral ofF if d(f )= 0 for everyd ∈ F, andf /∈ k. First,integrals correspond to the notion of constants for a derivation (see[10]), except that theymust not belong to the coefficient fieldk.

E-mail address:[email protected](P. Bonnet).

0022-4049/$ - see front matter © 2004 Elsevier B.V. All rights reserved.doi:10.1016/j.jpaa.2004.12.004

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mailto:[email protected]

12 P. Bonnet / Journal of Pure and Applied Algebra 199 (2005) 11–26

The description of the kernelsker F is usually quite tricky because of their complexity.Indeed, since Nagata’s works (see[8]), it is well-known that the sub-algebraB = ker Fneed not be finitely generated. Nagata’s construction uses locally nilpotent derivations on ak-algebraAof Krull dimensionn�32. This result has been refined by Deveney and Finston,who constructed a locally nilpotentk-derivation onk[x1, . . . , x7] whose kernel is not finitelygenerated (see[2]). Recently this result has been improved by Daigle and Freudenburg (see[1]), with an example of a locally nilpotent derivation onk[x1, . . . , x5] having as kernel anon-finitely generated algebra. In contrast, such behaviours do not occur in low dimensions.For instance, derivations onk[x1, . . . , xn] have as kernel a finitely generatedk-algebra ifn�3 (see[9]).

In what follows, we will choose to expressker F not as ak-algebra, but in terms of thederivations involved in its construction. Our starting point is an article of Nowicki (see[10])where he proved the following two theorems.

Theorem 1.1(Nowicki[10]). Let A be an integral k-algebra of finite type over a field k ofcharacteristic zero. LetF be a family of k-derivations onA. Then there exists a k-derivationd on A such thatker d = ker F.

Theorem 1.2(Nowicki[10]). Let A be an integral k-algebra of finite type over a field k ofcharacteristic zero. Let B be a sub-algebra of A. Then B is algebraically closed in A if andonly if B is the kernel of a k-derivation on A.

The proof ofTheorem 1.1 is very elegant and uses Noether normalisation lemma. Howeverthe construction of the derivationd is independent of the familyF, and it only uses the factthat the ringB = ker F is algebraically closed inA. In this paper we will refine Theorem1.1, and we will express the derivationd in terms of the elements ofF. More precisely.

Theorem 1.3. Let A be an integral k-algebra of finite type over a field k of characteristiczero. LetF = {di}i∈I be a family of k-derivations on A. Then there exists an A-linearcombinationd =∑

aidi such thatker d = ker F.

A k-derivationd in MF isF-minimalif ker d = ker F. HereF-minimality means thatthe kernel ofd is smallest among all kernels ofk-derivations inMF. We denote byMF,minthe set of allF-minimalk-derivations onA. In terms of first integrals, Theorem 1.3 can bereinterpreted as follows:

Corollary 1.4. Let A be an integral k-algebra of finite type over a field k of characteristiczero. LetF = {di}i∈I be a family of k-derivations on A. If every A-linear combinationd =∑

aidi admits a first integral, thenF admits a first integral.

Let d =∑aidi be anA-linear combination of elements ofF. A priori its first integrals

(if any) depend on the coefficientsai . But if every such combination has a first integral,then the previous corollary asserts that we can choose a first integralf that is independentof theai . In particulardi(f ) = 0 for anyi ∈ I .

P. Bonnet / Journal of Pure and Applied Algebra 199 (2005) 11–26 13

Definition 1.5. Let k be an algebraically closed field of characteristic zero. LetE be ak-vector space and� a subset ofE. The set� is residual inE if, for any finite dimensionalk-subspaceF of E, � ∩ F is a countable intersection of Zariski open sets ofF (possiblyempty).

Note that if k = C and � ∩ F �= ∅, then� ∩ F is dense inF for the Zariski andmetric topologies onF. This latter assertion is based on Baire’s theorem about countableintersection of dense open sets in a complete space, and also on the fact that every non-empty Zariski open set is dense inF for the metric topology. So residuality can be seen asa version of density adapted to infinite dimensional spaces. With this definition, Theorem1.3 yields the following result.

Theorem 1.6. Let k be an algebraically closed field of characteristic zero. Let A be anintegral k-algebra of finite type. LetF be a family of k-derivations on A. ThenMF,min isa non-empty residual subset ofMF.

At the end of this paper, we will give a proof of Theorem 1.2 based on Theorem 1.3, andwe will illustrate the notion of residuality with an example. Note that in this paper, we havenot investigated the field of rational first integrals, i.e. the elementsf of the fraction fieldK(A) of A such thatd(f ) = 0 for anyd in F. This field can be extremely large comparedto the kernel ofF, as is the case for the Euler vector field:

d = x1�

�x1+ · · · + xn

��xn

on kn, but it can also be reduced to the fieldk (see for instance[11] or [5]). One questioncould be to try and find an analogue to Theorem 1.3. More precisely, given a collection ofk-derivations{di} on a fieldK of finite transcendence degree, withL = ∩ker di |K, doesthere exist aK-linear combinationd of thedi havingL as its kernel?

2. Reduction to a couple of derivations

Throughout this paper,A will be an integralk-algebra of finite type over a fieldk ofcharacteristic zero. LetF be a family ofk-derivations onA. We will say thatthe triplet(k, A,F) enjoys the propertyP if there exists anA-linear combinationd = ∑

aidi ofelementsdi of F such that

ker � = ker F.

In this section, we are going to see how to restrict the proof of Theorem 1.3 to the case ofa couple of derivations enjoying some remarkable properties. More precisely:

Proposition 2.1. P holds for any triplet(k, A,F) if and only ifP holds for any triplet(k′, A′,F′),whereF′={d ′

1, d′2} is a couple ofk′-derivations satisfying the two conditions:

(1) ker d ′1 ∩ ker d ′

2 = k′ and(2) there exist two elementsx1, x2 of A′ such thatd ′i (xi) = 1

andd ′i (xj ) = 0 if i �= j .


The proof of this proposition is a consequence of the following lemmas.

Lemma 2.2. P holds for any triplet(k, A,F) if and only if it holds for any(k′, A′,F′),whereF′ is a finite family.

Proof. One direction is clear. For the other, consider the triplet(k, A,F)and letDerk(A,A)

be theA-module ofk-derivations onA. SinceA is a finitek-algebra,Derk(A,A) is a noethe-rianA-module. So theA-submoduleMF of Derk(A,A) spanned by the elements ofF isfinitely generated. LetF′ = {d1, . . . , dr} be a finite subset ofF whose elements span theA-moduleMF. Then we have:

ker F = ker F′ = ker d1 ∩ · · · ∩ ker dr .

The inclusionker F ⊆ ker F′ is obvious. Conversely letf be an element ofA such thatd1(f )=· · ·=dr(f )=0. For anyk-derivationdof F, there exist some elementsa1, . . . , arof A such that:

d = a1d1 + · · · + ardr .

Therefored(f ) = a1d1(f ) + · · · + ardr(f ) = 0, f belongs toker F andker F= ker F′.Assume thatP holds for any(k′, A′,F′), whereF′ is finite. Apply this property to thetriplet (k, A,F′). Then there exists anA-linear combination� of elements ofF′ such thatker � = ker F′. Sinceker F′ = ker F andF′ is a subset ofF, the result follows. �

Lemma 2.3. P holds for any triplet(k, A,F),whereF is finite, if and only if it holds forany(k′, A′,F′), whereF′ = {d1, d2}.

Proof. One direction is clear. The other will be proved by induction on the orderr of F.If r = 1, thenF consists of one derivationd1 and we choose� = d1. If r = 2, thenF is acouple of derivations and the result follows by assumption. Assume the property holds tothe orderr�2, and let(k, A,F) be a triplet such thatF={d1, . . . , dr+1}. By assumptionthere exists ak-derivation�′ = a1d1 + a2d2, where everyai belongs toA, such that:

ker �′ = ker d1 ∩ ker d2.

Consider the family{�′, d3, . . . , dr+1}. Since the property holds to the orderr, there existsak-derivation� = b′�′ + · · · + br+1dr+1, whereb′, b3, . . . , br+1 belong toA, such that

ker � = ker �′ ∩ ker d3 ∩ · · · ∩ dr+1 = ker d1 ∩ ker d2 ∩ · · · ∩ ker dr+1.

Since�′ is aA-linear combination ofd1, d2, � belongs to theA-module spanned byd1, . . . ,

dr+1, and the result follows. �

Lemma 2.4. P holds for any triplet(k, A,F),whereF= {d1, d2}, if and only if it holdsfor any(k′, A′,F′),whereF′ = {d ′

1, d′2} is a couple ofk′-derivations for which there exist

two elementsx1, x2 ofA′ such thatd ′i (xi) = 1 andd ′

i (xj ) = 0 if i �= j .

Proof. One direction is clear. For the other, assume thatP holds for any(k′, A′,F′),whereF′ = {d ′

1, d′2} is a couple ofk′-derivations for which there exist two elementsx1, x2


of A′ such thatd ′i (xi) = 1 andd ′

i (xj ) = 0 if i �= j . Let (k, A,F) be a triplet for whichF = {d1, d2}. If ker d1 ⊆ ker d2 or ker d2 ⊆ ker d1, then we have:

ker F = ker d1 or ker F = ker d2

andP holds by choosing either� = d1 or � = d2. So we may assume that:

ker d2�ker d1 and ker d1�ker d2.

By assumption, there exist two elementsx1, x2 of A such that:

d1(x1) �= 0, d2(x1) = 0, d2(x2) �= 0, d1(x2) = 0.

We setp = d1(x1)d2(x2) and consider the triplet

(k′, A′,F′) =(k,A

[1

p

],

{d1

d1(x1),

d2

d2(x2)

}).

By constructionA′ is a k′-algebra of finite type, and it is a domain. Moreoverd ′1, d

′2 act

on A′ ask′-derivations,d ′i (xi) = 1 andd ′

i (xj ) = 0 if i �= j . So there exists anA′-linearcombination�′ of d ′

1, d′2 such that

ker F′ = ker �′.

Up to replacing�′ bypn�′ fornbig enough, we may assume that�′ is anA-linear combinationof d1, d2. Let � be the restriction of�′ toA. Let us prove thatker F = ker �.

The inclusionker F ⊆ ker � is clear. Now letx be an element ofA such that�(x) = 0.Then�′(x)=0 inA′, andxbelongs toker d ′

1∩ker d ′2∩A. Sinced1 (resp.d2) is proportional

to d ′1 (resp.d ′

2), we find

d1(x) = d2(x) = 0.

Sincex belongs toA, x belongs toker d1 ∩ ker d2 = ker F and the result follows. �

Lemma 2.5. The following assertions are equivalent:

• P holds for any triplet(k, A,F), whereF = {d1, d2} is a couple of k-derivations forwhich there exist two elementsx1, x2 of A such thatdi(xi) = 1 anddi(xj ) = 0 if i �= j .

• P holds for any triplet(k′, A′,F′), whereF′ = {d ′1, d

′2} is a couple ofk′-derivations

satisfying the two conditions: (1) ker d ′1 ∩ ker d ′

2 = k′ and (2) there exist two elementsx1, x2 ofA′ such thatd ′

i (xi) = 1 andd ′i (xj ) = 0 if i �= j .

Proof. The first assertion implies clearly the second. Assume now that the second holds,and let(k, A,F) be a triplet satisfying the conditions of the first assertion. Letk′ be thefraction field ofker F, and consider the following triplet:

(k′, A′,F′) = (k′, A⊗kk′, {d ′

1, d′2}),

where everyd ′i acts onA⊗kk

′ according to the following rule:

d ′i (x ⊗ f ) = di(x) ⊗ f.


If S = ker F−{0}, thenA′ =AS andA′ is a domain. Sincek′ containsk, A′ is ak′-algebraof finite type. Moreoverd ′

i (a/b) = di(a)/b for anya/b in AS . In particular, everyd ′i is a

k′-derivation. Letf =a/b be an element ofAS such thatd ′1(f )=d ′

2(f )=0. By constructionwe getd1(a) = d2(a) = 0,a belongs toker F andf lies ink′. Thus we have

ker F′ = ker d ′1 ∩ ker d ′

2 = k′.

So the conditions of the second assertion hold, and there exists anAS-linear combination�′ = a′

1d′1 + a′

2d′2 such that

ker �′ = ker F′ = k′.

Up to a mutiplication by an element ofS, we may assume thata′1 anda′

2 belong toA. Denoteby � the derivationa′

1d1 + a′2d2. Let f be an element ofker �. SinceA is contained inAS

and everyd ′i extendsdi toAS , we have�(f )= �′(f )= 0. Sodi(f )= d ′

i (f )= 0 for i = 1,2andf belongs toker F. Therefore we get

ker � = ker F = ker d1 ∩ ker d2. �

3. Passage to a complete regular local ring

LetA be an integralk-algebra of finite type, andF = {d1, d2} a couple ofk-derivationssatisfying the conditions of Proposition 2.1, i.e. (1)ker d1 ∩ ker d2 = k and (2) there existtwo elementsx1, x2 ofAsuch thatdi(xi)=1 for all i anddi(xj )=0 if i �= j . In this section,we will see how to extend thek-derivationsd1, d2 into a couple ofL-derivations on a formalring L[[t1, . . . , tn]]. This will enable us to rewrite these derivations into a canonical form,that will prove easier to handle.

Proposition 3.1. Let A be an integral k-algebra of finite type, andF={d1, d2} be a coupleof k-derivations satisfying the conditions of Proposition2.1.Let x1, x2 be two elements ofA such thatdi(xi)= 1 for all i and di(xj )= 0 if i �= j .Then there exist two elements�1, �2of k, and an extension L of k such that

• A is a subring ofL[[t1, . . . , tn]],• d1, d2 extend to L-derivations onL[[t1, . . . , tn]],• x1 − �1, x2 − �2 belong to the maximal ideal ofL[[t1, . . . , tn]].

The proof of this proposition will split into several lemmas.

Lemma 3.2. Let x1, x2 be two elements of A satisfying the conditions of Proposition3.1.Thenx1, x2 are algebraically independent in the k-algebra A.

Proof. Assume there exists a non-zero polynomialP in k[u, v] such thatP(x1, x2) = 0.We chooseP of minimal homogeneous degree with respect tou, v. Sincedi(xi)= 1 for all


i anddi(xj ) = 0 if i �= j , we get by derivation

d1(P (x1, x2)) = �P�u

(x1, x2) = 0 and d2(P (x1, x2)) = �P�v

(x1, x2) = 0.

By minimality of the degree, this implies that�P/�u=�P/�v=0. ThereforeP is constantandP(x1, x2) = 0 implies thatP = 0, hence a contradiction.�

Lemma 3.3. There exist two elements�1, �2 of k, and a maximal idealM of A such thatx1 − �1, x2 − �2 belong toM andAM is a regular local ring.

Proof. Up to localizingA with respect to a non-zero elementg of A, we may assume that�A(g)/k � (�A/k)(g) is free. By generic smoothness (see[3]), A′ = A(g) is regular overevery maximal idealM′ not containingg. Sincex1, x2 are algebraically independent, theinclusion induces an injective map

L : k[u, v] −→ A′, P �−→ P(x1, x2).

Therefore the mapL∗ : Spec(A′) → Spec(k[u, v]) is dominant, and there exists an elementf �= 0 of k[u, v] such that every fibreL−1(P) is non-empty for any maximal idealP ofk[u, v] not containingf. Sincef is non-zero, there exists a couple(�1, �2) in k2 such thatf (�1, �2) �= 0. Consider the ideal

P = (u − �1, v − �2).

By construction,P is maximal ink[u, v] and does not containf. So the fibreL−1(P) is notempty. In particular, it contains a maximal idealM′ of A′. If M denotes the intersectionM′ ∩ A, thenM is a maximal ideal not containingg, and we have the isomorphism ofk-algebras

A′M′ � AM.

SinceA′ is regular over every maximal ideal,AM is a regular local ring. By constructionM containsx1 − �1, x2 − �2, and the result follows. �

Since�1, �2 are annihilated by thedi , we may replacexi by xi − �i without changingthe conditions at the beginning of this section. So we may assume thatx1, x2 belong to amaximal idealM of A such that the k-algebraAM is regular. By an easy computation, weget for any positive integerr:

d1(Mr ) ⊆ Mr−1 and d2(M

r ) ⊆ Mr−1.

Therefored1 andd2 are continuous onAM for theM-adic topology, and they uniquelyextend into a couple ofk-derivations on theM-adic completionRof AM. We still denote bydi this extension. SinceAM is regular and contains the fieldk, by Cohen structure theorem(see[3]), there exists an extensionL of k such that

R � L[[t1, . . . , tn]],


where n is the Krull dimension ofA. So thedi can be viewed ask-derivations onL[[t1, . . . , tn]]. In order to get Proposition 3.1, we only need to check that:

Lemma 3.4. d1 andd2 are L-derivations onL[[t1, . . . , tn]].

Proof. It suffices to prove thatdi(L) = {0} for i = 1,2. First note thatL is isomorphic toA/M. SinceA is a finitek-algebra, the fieldL is also a finitek-algebra. SoL is a finiteextension ofk (see[4]). Let � be an element ofL, and letP be a polynomial ink[t] ofminimal degree such thatP(�) = 0. By derivation, we get

di(P (�)) = P ′(�)di(�) = 0.

By minimality of the degree,P ′(�) �= 0 anddi(�) = 0. Since this holds for any� in L, theresult follows. �

4. Canonical form for a couple ofL-derivations

In this section, we consider a couple ofL-derivationsd1, d2 onL[[t1, . . . , tn]], satisfyingthe following condition: there exist two elementsx1, x2 ofL[[t1, . . . , tn]] such thatdi(xi)=1for i = 1,2 anddi(xj ) = 0 if i �= j . We are going to search for a system of parametersfor which these derivations look simpler. Recall that a system of parameters is a family offormal functionss1, . . . , sn generating the maximal ideal(t1, . . . , tn) of L[[t1, . . . , tn]].

Lemma 4.1. Let d be an L-derivation onL[[t1, . . . , tn]],wheren>1.Assume there existsa formal functionx1 such thatd(x1)=1.Then there exists some formal functionsy2, . . . , ynsuch thatx1, y2, . . . , yn is a system of parameters andd(yi) = 0 for any i. In particulard = �/�x1 in this system of parameters.

Proof. LetM be the maximal ideal ofL[[t1, . . . , tn]]. Sinced is anL-derivation, we haved(M2) ⊆ M. Sox1 belongs toM−M2 becaused(x1)= 1. Lety0

2, . . . , y0n be a system of

formal functions such thatx1, y02, . . . , y

0n form a basis ofM/M2. By Nakayama Lemma,

x1, y02, . . . , y

0n is a system of parameters. If�i = d(yi)(0) andy1

i = y0i −�ix1, then we find:

d(y1i )(0) = �i − �id(x1)(0) = 0

andx1, y12, . . . , y

1n is again a system of parameters. For any integerk >0, we are going to

construct a systemyk2, . . . , y

kn of formal functions satisfying the following conditions:

• x1, yk2, . . . , y

kn is a system of parameters,

• for any i, yk+1i − yk

i ≡ 0 [Mk+1],• for any i, d(yk

i ) ≡ 0 [Mk].Assume for the moment that such a construction is possible. Then for anyi, the sequence

(yki )k>0 is Cauchy for theM-adic topology onL[[t1, . . . , tn]]. SinceL[[t1, . . . , tn]] is

complete,(yki )k>0 converges to a formal functionyi . By construction,x1, y2, . . . , yn span


the vector spaceM/M2, hence it is a system of parameters by Nakayama Lemma. Moreoverby passing to the limit, we find

∀i >1, d(yi) = 0

and the result follows. We proceed to this construction by induction onk >0. The casek = 1 has already been treated above. Assume the construction holds up to the orderk.Sinced(yk

i ) ≡ 0 [Mk], there exists a homogeneous polynomialQi,k(u1, . . . , un) of degreek such that

d(yki ) ≡ Qi,k(x1, y

12, . . . , y

1n) [Mk+1].

We setyk+1i = yk

i + Pi,k(x1, y12, . . . , y

1n), where everyPi,k(u1, . . . , un) is defined as

Pi,k(x1, y12, . . . , y

1n) = −

∫ x1

0Qi,k(u, y

12, . . . , y

1n)du.

Since everyQi,k is homogeneous of degreek, everyPi,k is homogeneous of degreek + 1.So yk+1

i − yki ≡ 0 [Mk+1] for any i. In particular,yk+1

i − y1i ≡ 0 [M2] for any i, and

x1, yk+12 , . . . , yk+1

n span the vector spaceM/M2. By Nakayama Lemma,x1, yk+12 , . . . ,

yk+1n is a system of parameters. Moreover we get by derivation

d(yk+1i ) = d(yk

i ) + �Pi,k

�u1(x1, y

12, . . . , y

1n) +

n∑i=2

�Pi,k

�ui

(x1, y12, . . . , y

1n)d(y

1i ).

By construction�Pi,k/�ui(x1, y12, . . . , y

1n) belongs toMk andd(y1

i ) belongs toM. Byreduction moduloMk+1 and construction ofPi,k, we obtain:

d(yk+1i ) ≡ d(yk

i ) + �Pi,k

�u1(x1, y

12, . . . , y

1n) ≡ Qi,k(x1, y

12, . . . , y

1n),

+ �Pi,k

�u1(x1, y

12, . . . , y

1n) ≡ 0 [Mk+1],

thus ending the construction to the orderk + 1, and the result follows. �

Proposition 4.2. Let d1, d2 be a couple of L-derivations onL[[t1, . . . , tn]], wheren>2.Assume there exist two formal functionsx1, x2 such thatdi(xi)= 1 for any i anddi(xj )= 0for i �= j . Then there exist some formal functionsy3, . . . , yn anda3, . . . , an such that:

• x1, x2, y3 . . . , yn is a system of parameters,• d1 = �/�x1 andd2 = �/�x2 + x1

∑i>2 ai�/�yi .

Proof. Sinced1(x1) = 1, Lemma 4.1 asserts there exist some formal functionsy′2, . . . , y

′n

such thatx1, y′2, . . . , y

′n is a system of parameters andd1(y

′i ) = 0 for all i. In particular, in

this system of parameters, we have

d1 = ��x1

.


A fortiori, this implies thatker d1 = L[[y′2, . . . , y

′n]]. Sinced2(x1) = 0, there exist some

formal functionsb′2, . . . , b

′n such that:

d2 =n∑

i=2

b′i

��y′

i

.

Setbi(y′2, . . . , y

′n) = b′

i (0, y′2, . . . , y

′n) for all i, and consider theL-derivation�:

� =n∑

i=2

bi�

�y′i

.

By construction,� acts onL[[y′2, . . . , y

′n]] and�(f )= d2(f )(0, y′

2, . . . , y′n) for any formal

functionf. Sinced1(x2)=0 andd2(x2)=1,x2 belongs toL[[y′2, . . . , y

′n]] and�(x2)=1. By

Lemma 4.1 applied to� andL[[y′2, . . . , y

′n]], there exist some formal functionsy3, . . . , yn

in L[[y′2, . . . , y

′n]] such that

• x2, y3, . . . , yn is a system of parameters ofL[[y′2, . . . , y

′n]],

• for anyi >2, �(yi) = 0.By constructionL[[x2, y3, . . . , yn]]=L[[y′

2, . . . , y′n]] andx1, x2, y3, . . . , yn is a system

of parameters ofL[[t1, . . . , tn]]. Sinceker d1 = L[[y′2, . . . , y

′n]], this implies

d1(x2) = d1(y3) = · · · = d1(yn) = 0.

In this system of parameters, the derivationd1 can be written as

d1 = ��x1

.

Now d2(x1) = 0 andd2(x2) = 1, so thatd2 can be written as

d2 = ��x2

+∑i>2

�i

��yi

,

where all the�i are formal functions. Sinced2(yi) ≡ �(yi) ≡ 0 [x1] for any i, everyfunction�i is divisible byx1. Write �i = x1ai , where everyai belongs toL[[t1, . . . , tn]].By construction, we find

d2 = ��x2

+ x1

∑i>2

ai�

�yi

and the result follows. �

5. Properties of the derivation�m

In this section, we will analyse the properties of some derivations, and we will use themto produce the longly awaited derivation of Theorem 1.3. LetRbe a commutative domain


with identity. For any positive integerm, we define theR-derivation�m onR[x1, x2] as

�m = xm1

��x1

+ xm2

��x2

.

Lemma 5.1. For anym>0,we haveker �m = R.

Proof. Note that�m is homogeneous with respect to the variablesx1, x2. So every elementof its kernel can be uniquely written as a sum of homogeneous elements each belonging toker �m. Considering an element ofker �m, we may therefore assume that it is homogeneouswith respect tox1, x2.

LetP be an homogeneous element ofker �m. If m = 1, then Euler’s formula asserts that�1(P ) = deg(P )P . In this casedeg(P ) = 0 orP = 0. SoP belongs toRandker �1 = R.Assume now thatm>1, and writeP as

P(x1, x2) =n∑

k=0

Pk(x1)xk2,

wherePn(x1) �= 0. Then�m(P ) can be written as

�m(P ) =n∑

k=0

kP k(x1)xk+m−12 +

n∑k=0

xm1 P ′

k(x1)xk2.

If n �= 0, then the leading term of�m(P ) with respect tox2 is equal tonPn(x1)xn+m−12

becausem>1. In particular�m(P ) �= 0 if n �= 0. Son must be equal to 0 andP =P0(x1). But then�m(P )=xm

1 P ′0(x1)=0, which implies thatP0 belongs toR, and the result

follows. �

Lemma 5.2. LetP,Q be two elements ofR[x1, x2], where Q is homogeneous of degree kwith respect tox1, x2. Assume that they satisfy the following relation:

�m(P ) + x1xm2 Q = 0.

If m�k + 4, then P belongs to R andQ = 0.

Proof. Up to adding a constant toP, we may assume thatP(0,0) = 0. We are going toprove thatP = Q = 0. By the previous lemma,�m is injective on the polynomials with noconstant terms. SinceQ and�m are homogeneous,P needs to be homogeneous. We writeP as the sum

P(x1, x2) =∑

i+j=r

ai,j xi1x

j2 .

By derivation, we find

�m(P ) =∑

i+j=r

iai,j xi+m−11 x

j2 +

∑i+j=r

jai,j xi1x

j+m−12 = −x1x

m2 Q(x1, x2).


SinceQ is homogeneous of degreek, we haver − 2 = k andr�m − 2, sincem�k + 4.We havem − 1>0 and the following congruence holds moduloxm−1

2 :

�m(P ) ≡∑

i+j=r

iai,j xi+m−11 x

j2 ≡ 0 [xm−1

2 ].

In this sum, all the indicesj in this sum satisfy�r�m − 2, so thatiai,j = 0 for everycouple(i, j). Thereforeai,j = 0 or i = 0, andP reduces to a polynomial of the formaxr

2,wherea belongs toR. But then we find

�m(P ) = arxm+r−12 = −x1x

m2 Q(x1, x2),

which is impossible unlessP = Q = 0. �

6. Proof of the main theorem

In this section, we work under the assumptions of Section 3. LetAbe an integralk-algebraof finite type over a field of characteristic zero. Letd1, d2 be a couple ofk-derivations onA,andx1, x2 be two elements ofA such thatdi(xi)= 1 for anyi anddi(xj )= 0 for i �= j . Weassume thatk = ker d1 ∩ ker d2. According to Section 3,A is embedded inL[[t1, . . . , tn]],whereL is a finite extension ofk, and thedi extend toL-derivations onL[[t1, . . . , tn]]. ByProposition 4.2, there exist some formal elementsy3, . . . , yn, a3, . . . , an of L[[t1, . . . , tn]]such that:

• x1, x2, y3 . . . , yn is a system of parameters,• d1 = �/�x1 andd2 = �/�x2 + x1

∑i>2 ai�/�yi .

For any positive integerm, we consider the followingk-derivation�m:

�m = xm1 d1 + xm

2 d2.

In order to prove Theorem 1.3, we are going to establish the following result.

Proposition 6.1. Under the previous assumptions, there exists a positive integerm0 suchthat, for anym�m0, we haveker �m = ker d1 ∩ ker d2 = k in A.

Before giving the proof, we make explicit the numberm0. LetRbe the ringL[[y3, . . . , yn]].We provideR[[x1, x2]]=L[[t1, . . . , tn]] with the homogeneous degreedegon the variablesx1, x2. More preciselydeg(x1)= deg(x2)= 1 anddeg(yi)= 0 for anyi�3. Letak

i be thehomogeneous part ofai of degreek. We denote by�k theL-derivation onL[[t1, . . . , tn]]:

�k =∑i>2

aki

��yi

.

Every�k is anL-linear operator of degreek onR[[x1, x2]], and we have by construction

d2 = ��x2

+ x1

∑k�0

�k.


Let D = DerL(L[[t1, . . . , tn]], L[[t1, . . . , tn]]) be the space ofL-derivations onL[[t1, . . . , tn]], and letM be the sub-R[[x1, x2]]-module ofD spanned by the�k. SinceD is noetherian, there exists an integerm′

0 such that

M = R[[x1, x2]]{�0, . . . , �m′0}.

We denote bym0 the integerm0 = m′0 + 4.

Proof of Proposition 6.1. Let f be a non-zero element ofR[[x1, x2]] such that�m(f )=0,and assume thatm�m0. Let us prove by contradiction thatd1(f )=d2(f )=0.We decomposef into its sum of homogeneous parts with respect todeg:

f = f0 + f1 + · · · + fk + · · · .

Assertion1: f0 �= 0. Let i be the smallest index such thatfi �= 0. Since�m(f ) = 0, wehave�m(f ) + x1x

m2

∑k�0 �k(f ) = 0. By considering only the terms of degreei + m − 1

in this equation, we find that�m(fi) = 0. By Lemma 5.1,fi belongs toRandi = 0.Assertion2: f �= f0. Assume on the contrary thatf = f0. Sinced1 = �/�x1, we have

d1(f ) = 0. As �m = xm1 d1 + xm

2 d2 and�m(f ) = 0, we getd1(f ) = d2(f ) = 0, hence acontradiction.Assertion3: Letk be the smallest positive index such thatfk �= 0. Then 2�k�m′

0 + 2.If we consider the terms of degreem + i − 1 in the equation�m(f ) = 0, we obtain

�m(fi) + x1xm2 {�0(fi−2) + �1(fi−3) + · · · + �i−2(f0)} = 0

for any i >1, and�m(f1) = 0. By Lemma 5.1, we havef1 = 0 andk�2. Assume on thecontrary thatk >m′

0+2. Then for any 0< i <k, we havefi =0. This implies that�i (f0)=0for anyi < k − 2, and we have in particular

�0(f0) = �1(f0) = · · · = �m′0(f0) = 0.

By assumption,�0, . . . , �m′0

span the moduleM generated by the�l . For any nonnegativeintegerl, there exist some formal elementsbl,� such that

�l =m′

0∑�=0

bl,��.

This implies that�l (f0)=0 for anyl. In particularf0 is annihilated byd2. Sinced1(f0)=0,f0 belongs toker �m. So(f − f0) lies in ker �m and has no homogeneous part of degree0. By our first assertion,(f − f0) needs to be equal to 0. But that contradicts our secondassertion.Assertion4: The final contradiction. Withk as in Assertion 3, if we consider the terms of

degreem + k − 1 in the equation�m(f ) = 0, we obtain

�m(fk) + x1xm2 �k−2(f0) = 0


becausef1 = · · · = fk−1 = 0. By the previous assertion,k�m′0 + 2 and�k−2(f0) is an

homogeneous polynomial of degree�m′0. In particular, we find

m�m0�m′0 + 4�deg(�k−2(f0)) + 4.

By applying Lemma 5.2 to the relation�m(fk) + x1xm2 �k−2(f0) = 0, we get thatfk = 0,

which is impossible by the third assertion.�Theorem 1.3 is a direct consequence of this proposition. Indeed ifm�m0 and f is an

element ofA such that�m(f ) = 0, thend1(f ) = d2(f ) = 0 as elements ofL[[t1, . . . , tn]].SinceA is a subring ofL[[t1, . . . , tn]], f belongs tok and the result follows.

7. A few consequences and an example

In this section, we are going to derive some consequences of Theorem 1.3. In particular,we will give another proof of Nowicki’s theorem.

7.1. Proof of Theorem 1.6

Let k be an algebraically closed field of characteristic zero. LetA be ak-algebra of finitetype with no zero divisors. LetF be a family ofk-derivations onA, and denote byMF theA-module spanned byF. Recall thatMF,min is the set ofF-minimal derivations, i.e. thesubset ofMF formed by thek-derivationsd such that

ker F = ker d.

Let us prove thatMF,min is a non-empty residual subset ofMF. The non-emptyness isguaranteed by Theorem 1.3. For residuality, letF be any finitek-vector subspace ofMF.SinceA is a finitek-algebra, its dimension is at most countable. So the spaceA/ker Fadmits a filtration{Fn}n∈N, where everyFn is finite dimensional, i.e.

A

ker F=⋃n∈N

Fn and F0 ⊆ F1 ⊆ · · · ⊆ Fn ⊆ · · · .

Consider the set�n defined as

�n = {(d, f ) ∈ F × P(Fn), d(f ) = 0}.Note that� is an algebraic subset ofF × P(Fn) by construction. If : F × P(Fn) → F

denotes the standard projection onF, we find

(�n) = {d ∈ F, ∃d ∈ P(Fn), d(f ) = 0}.In particular, ak-derivationd in F is F-minimal if and only if it does not belong to theunion

⋃n∈N (�n). In other words we have

MF,min ∩ F = F −⋃n∈N

(�n) =⋂n∈N

(F − (�n)).

But every�n is closed inF × P(Fn) for the Zariski topology. SinceP(Fn) is projective, itis a complete variety and the projection is closed. Therefore every set(�n) is closed in


F, andMF,min intersectsF in a countable intersection of Zariski open sets. By definition,MF,min is residual inMF.

7.2. Proof of Nowicki’s result

In this subsection, we will use Theorem 1.3 in order to establish Theorem 1.2. The proofis based on the following lemma.

Lemma 7.1. Let A be a finite k-algebra with no-zero divisors. Let B be a sub-algebra ofA and letdB be a k-derivation on B. Then there exists a non-zero element a of A and ak-derivationdA on A such thatdA = adB on B.

Proof. Letx1, . . . , xn be a set of generators ofAas ak-algebra. IfK=Fr(B)andL=Fr(A),then we haveL = K(x1, . . . , xn). Consider the following chain of fields:

K = K0 ⊆ K1 ⊆ · · · ⊆ Kn, whereKi = K(x1, . . . , xi).

By construction we haveKi+1=Ki(xi+1).Assume there exists ak-derivation�i onKi . Thenthere exists ak-derivation�i+1 onKi+1 that extends the derivation�i (if xi is transcendentaloverKi , see[6, p. 10]and if xi is algebraic overKi , see[7, p. 14]). By a finite induction,we can extend thek-derivationdB = �0 of K = K0 into ak-derivationdL = �n onL = Kn.For any indexi, there exists a non-zero elementai of A such thataidL(xi) belongs toA. Ifa = a1 . . . an, thenadL(x) belongs toA for anyx in A. The derivationdA = adL mapsAintoA and thus defines ak-derivation onA. By construction we havedA = adB onB. �

Proof of Theorem 1.2. Let A be a finitek-algebra with no zero divisors over a fieldk ofcharacteristic zero. Letdbe ak-derivation onAand setB = ker d. By Leibniz rule,Bneedsto be ak-algebra. Letabe any element ofA that is algebraic overB. LetP(t)=b0+· · ·+bnt

n

be a polynomial ofB[t] of minimal degree such thatP(a) = 0. By derivation, we get

d(P (a)) =(

n∑k=1

kak−1

)d(a) = P ′(a)d(A) = 0.

By minimality of the degree, we findP ′(a) �= 0 andd(A) = 0. Soa belongs toB andB isalgebraically closed inA.

Conversely letBbe a sub-algebra ofAand assume thatB is algebraically closed inA. ForanyxoutsideB, denote byBx the algebraB[x] and consider thek-derivationdx = d/dx onBx . Sincex lies outsideB, x is transcendental overBanddx is well-defined. By Lemma 7.1,there exist a non-zero elementax of A and ak-derivation�x onA such that�x = axdx . Inparticular�x(x)=axdx(x)=ax �= 0. Consider the familyF={�x}x∈A−B of k-derivationsonA. By construction, we have

ker F = B.

By Theorem 1.3, there exists anA-linear combinationd of the�x such thatker d = B. InparticularB is the kernel of ak-derivation and the result follows.�


7.3. An example

In this subsection, we will illustrate the notion of residuality in an example. Consider thealgebraA = C[x, y] provided with the following couple ofC-derivations:

F = {d1, d2} ={x

��x

, y��y

}.

Let F be the vector space spanned byd1, d2. Then it is easy to check that an element�1d1 +�2d2 of F admits a first integral if and only if either�2 =0 or�1/�2 is a non-positiverational number. In this latter case, if�1/�2=−p/q, wherep�0, q >0 are coprime integers,then every first integral of�1d1 + �2d2 is a polynomial function of the expression:

fp,q(x, y) = xqyp.

LetD be the union of the coordinate axis ofC2 and of the lines of rational negative slopes.ThenD is a countable union of complex lines inC2. If we identifyd1, d2 with the canonicalbasis ofC2, then we have

F ∩ MF,min = F − D

and this latter set turns out to be residual, since it is a countable intersection of Zariski opensets ofF � C2. Residuality occurs as we cannot find a finite dimensional space inC[x, y]containing all the first integrals of minimal degree of elements ofF. If it were possible,then as in Theorem 1.6, we could prove thatF ∩ MF,min would be a finite intersection ofZariski open sets, hence an open set and this is not the case.

References

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Hermann, 1957.[7] A.R. Magid, Lectures on Differential GaloisTheory, University Lecture Series, vol. 7,American Mathematical

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Families of k-derivations on k-algebras

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