(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).