The Dependence Problem for a Class of Polynomial Maps in Dimension Four

(1.College of Science,Civil Aviation University of China,Tianjin,300300)

(2.School of Mathematics,Jilin University,Changchun,130012)

The Dependence Problem for a Class of Polynomial Maps in Dimension Four

JIN YONG1,2AND GUO HONG-BO2

(1.College of Science,Civil Aviation University of China,Tianjin,300300)

(2.School of Mathematics,Jilin University,Changchun,130012)

Communicated by Du Xian-kun

Let h be a polynomial in four variables with the singular Hessian Hh and the gradient∇h and R be a nonzero relation of∇h.Set H=∇R(∇h).We prove that the components of H are linearly dependent when rkHh≤2 and give a necessary and sufficient condition for the components of H to be linearly dependent when rkHh=3.

dependence problem,linear dependence,quasi-translation

1 Introduction

Throughout this paper k denotes a fi eld of characteristic 0,and k[X]:=k[x1,x2,···,xn] denotes the polynomial ring in the variables x1,x2,···,xnover k.

The linear dependence problem asks whether the components of a polynomial map H: kn→ knare linearly dependent over k if the Jacobian matrix JH is nilpotent.Partial positive answers to the problem are obtained in[1–3].By studying quasi-translation De Bondt[4–5]solved the problem negatively for all n≥5 in the homogeneous case and for all n≥4 in the non-homogeneous case.A polynomial map X+H is called a quasi-translation if its inverse is X−H.De Bondt[5]furthermore gave examples of quasi-translations with the components of H linearly independent for n≥6 in the homogeneous case and for n≥4 in the non-homogeneous case,and he also proved that no such examples exist when n≤4 for the homogeneous case and when n≤3 for the non-homogeneous case.

For a polynomial h∈k[X],denote by Hh its Hessian matrix and by∇h its gradient.If R(Y)∈k[Y]is a relation of∇h,that is,R(∇h)=0,we set

De Bondt[5]proved that X+H is a quasi-translation,called quasi-translation corresponding to h,and he asked whether the components of H are linearly dependent.

As mentioned above,for n≤3 the answer to the problem of De Bondt is affirmative and it is also affirmative in the case n=4 and H is homogenous.In this paper,we study the problem for n=4.We prove that the components of H are linearly dependent if the rank rkHh≤2.For the case rkHh=3 and H≠0,we prove that the components of H are linearly dependent if and only if the components of∇g are linearly dependent,where g is a generator of the relation ideal of∇h.Finally,we give an algorithm to decide whether the components of H are linearly dependent.

2 Main Results

For g,h∈k[X],we say that they are linearly equivalent,if there exists a T∈Gln(k)such that g=h(TX).In this case,∇g=Tt∇h(TX),Hg=TtHh(TX)T,and rkHg=rkHh, where Ttdenotes the transpose of T.

Lemma 2.1Suppose thatg,h∈k[X]are linearly equivalent.Then there is a nonzero relationRof∇hsuch that the components of∇R(∇h)are linearly dependent if and only if there is a nonzero relationSof∇gsuch that the components of∇S(∇g)are linearly dependent.

Proof.It suffices to prove the assertion for one direction by the de fi nition of linear equivalence.Let g=h(TX)for some T∈Gln(k)and R∈k[Y]:=k[y1,···,yn]be a nonzero relation of∇h such that the components of H=∇R(∇h)are linearly dependent.Suppose 0≠λ∈knsuch that λH=0.Take S(Y)=R((Tt)−1Y).Then

Let G=∇S(∇g).Note that

Let β=λT.Then β≠0 and

as desired.

For h∈k[X]and a relation R of∇h,let H=∇R(∇h).Taking Jacobian matrix on both sides,we have JH=J(∇R)|X=∇hH(h).Hence rkJH≤rkHh.

Lemma 2.2Forh∈k[x1,x2,x3,x4]and a relationRof∇h,letH= ∇R(∇h).IfrkHh≤2,then the components ofHare linearly dependent.

Proof.If rkHh≤1,then rkJH≤1.By Theorem 3.4.6 and Proposition 3.4.3 in[5]the components of H are linearly dependent.

Owing to Lemma 2.1 we pay our attention to such a g.

that is,the fi rst two components of G are zero.In particular,the components of G are linearly dependent.

where a′(x1)denotes the derivative of a(x1).Thus

Since rkJ(∇g)=2,we know that b′(x1),c′(x1),d′(x1)are not all zero.Without loss of generality,we assume that b′(x1)≠0.Let R∈k[y1,y2,y3,y4]be a nonzero relation of∇g. Write

Then Since b′(x1)≠0,the degree of gi1in x2equals i.Noticing that R(∇g)=0 and x2does not appear in ai(g2,g3,g4)for all i,we deduce that ai(g2,g3,g4)=0 for all i.Hence

In particular,the components of∇R(∇g)are linearly dependent.

Lemma 2.3LetF=(f1,f2,···,fn)∈k[X]nandrkJF=n−1.Then the relation ideal off1,f2,···,fnis a principal and prime ideal ofk[X].

Proof.De fi ne a homomorphism φ:k[x1,x2,···,xn]→k[f1,f2,···,fn]that sends xito fifor all i.It is easy to see that φ is surjective and its kernel P is the relation ideal of f1,f2,···,fn.Since k[x1,x2,···,xn]/P～=k[f1,f2,···,fn]and k[f1,f2,···,fn]is a domain,we see that P is prime.Since rkJF=n−1,we have

Consequently,by Theorem 1.8A in[6],

Hence heightP=1,which implies that P is a principal ideal.

If rkHh=n−1 for h∈k[X],then Lemma 2.3 implies that the relation ideal of∇h is a principal ideal.

Lemma 2.4Leth∈k[X]andrkHh=n−1,Rbe a relation of∇hsuch that∇R(∇h)≠0andgbe a generator of the relation ideal of∇h.Then the components of∇R(∇h)are linearly dependent if and only if the components of∇gare linearly dependent.

Proof.By hypothesis,R=gS for some S∈k[X].It follows that

and hence

Thus

Since∇R(∇h)≠0,we see that S(∇h)≠0.Hence the components of∇R(∇h)are linearly dependent if and only if the components of∇g(∇h)are linearly dependent.Thus in what follows we prove that the components of∇g(∇h)are linearly dependent if and only if so are those of∇g.

The sufficiency is clear.To prove the necessity,suppose that the components of∇g(∇h) are linearly dependent and assume 0≠λ∈knsuch that λT∇g(∇h)=0.Then

By Lemmas 2.2 and 2.4,we have

Theorem 2.1Forh∈k[x1,x2,x3,x4]and a relationRof∇h,letH=∇R(∇h).IfrkHh≤2,then the components ofHare linearly dependent;ifrkHh=3,H≠0,andgis a generator of the relation ideal of∇h,then the components of∇R(∇h)are linearly dependent if and only if the components of∇gare linearly dependent.

Lemma 2.5Leth∈k[X]withrkHh=n−1.Suppose that there existn−1components of∇hthat are algebraically dependent.Then the components of∇R(∇h)are linearly dependent for an arbitrary relationRof∇h.

Proof.We may suppose that∇R(∇h)≠0.Let g be a generator of the relation ideal of∇h. By Lemma 2.4,we only need to prove that the components of∇g are linearly dependent. Since some n−1 components are algebraically dependent,∇h has a nonzero relation that depends on n−1 variables only.Noticing that g divides any relation of∇h,we have that g depends on at most n−1 variables.Consequently,at least one component of∇g is zero, and so the components of∇g are linearly dependent.

By Theorem 2.1 and Lemma 2.5 we get the following corollary.

Corollary 2.1Suppose that there exist3components of∇hforh∈k[x1,x2,x3,x4]that are algebraically dependent.Then the components of∇R(∇h)are linearly dependent for any relationRof∇h.

Lemma 2.6Suppose thath∈k[X]and∇h=(h1,h2,···,hn).Then anyrelements of{h1,h2,···,hn}are algebraically independent if and only if all the principal minors of sizerofHhare none-zero.

＜deg(g),we deduce that

Proof.The sufficiency is clear.Conversely,assume that any r elements of{h1,h2,···,hn} are algebraically independent.It is sufficient to show the corresponding principal minor of Hh is non-zero for any hi1,hi2,···,hir.By Proposition 1.2.9 in[5],

which gives rkJ(hi1,hi2,···,hir)=r.Then the r rows corresponding to hi1,hi2,···,hirin the symmetric matrix Hh are linearly independent(over k(X)).Now by Lemma 5.3.4 in [5]the corresponding principal minor is non-zero.

Now we are able to give the following algorithm which decides whether the components of∇R(∇h)are linearly dependent,where R is a non-zero relation of∇h.

Algorithm

Input h(x1,x2,x3,x4).

Step 1.Compute Hh.

Step 2.Compute rkHh:=r.If r≤2,go to Step 5;if r=3,go to Step 3.

Step 3.Compute the 4 size 3 principal minors of Hh.If one of them equals zero,go to Step 5;otherwise go to Step 4.

Step 4.Compute a generator g of the relation ideal of∇h.If the components of∇g are linearly dependent,go to Step 5;otherwise go to Step 6.

Step 5.Output:linearly dependent.

Step 6.Output:linearly independent.

By the algorithm above and with the help of a Maple procedure we get the following example.

Example 2.1Take

Then any 3 elements of{h1,h2,h3,h4}are algebraically independent and the components of∇R(∇h)are linearly dependent for an arbitrary relation R of∇h.

Remark 2.1This example shows that the inverse of Corollary 2.1 is not true.

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tion:14R99

A

1674-5647(2014)04-0289-06

10.13447/j.1674-5647.2014.04.01

Received date:Nov.10,2011.

Foundation item:The Scienti fi c Research Foundation(2012QD05X)of Civil Aviation University of China and the Fundamental Research Funds(3122014K011)for the Central Universities of China.

E-mail address:kingmeng@126.com(Jin Y).