Abstract 1 Introduction 2 Incidences between points and algebraic curves 3 A special case 4 Proof of Theorem 5 5 Characterization of rank-1 polynomials 6 Distinct 𝒅-volumes on the moment curve in 𝒅 References

Improved Bound for the 𝒌-Variate Elekes–Rónyai Theorem

Yaara Jahn ORCID Hebrew University of Jerusalem, Israel    Orit E. Raz ORCID Ben-Gurion University, Be’er-Sheva, Israel
Abstract

Let f[x1,,xk], for k2. For any finite sets A1,,Ak, consider the set

f(A1,,Ak):={f(a1,,ak)(a1,,ak)A1××Ak},

that is, the image of A1××Ak under f. Extending a theorem of Elekes and Rónyai, which deals with the case k=2, and the result of Raz, Sharir, and De Zeeuw [9], dealing with the case k=3, it is proved in Raz and Shem Tov [10], that for every choice of finite A1,,Ak, each of size n, one has

|f(A1,,Ak)|=Ω(n3/2), (1)

unless f has some degenerate special form.

In this paper, we introduce the notion of a rank of a k-variate polynomial f, denoted as rank(f). Letting r=rank(f), we prove that

|f(A1,,Ak)|=Ω(n5r42rε), (2)

for every ε>0, where the constant of proportionality depends on ε and on deg(f). This improves the lower bound (1), for polynomials f for which rank(f)3.

We present an application of our main result, to lower bound the number of distinct d-volumes spanned by (d+1)-tuples of points lying on the moment curve in d.

Keywords and phrases:
Polynomial Expansion, Elekes–Rónyai theorem
Copyright and License:
[Uncaptioned image] © Yaara Jahn and Orit E. Raz; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Mathematics of computing Combinatoric problems
Editors:
Hee-Kap Ahn, Michael Hoffmann, and Amir Nayyeri

1 Introduction

In many cases in combinatorial geometry, counting questions involving distances, slopes, collinearity, etc., can be reformulated as analogous counting questions involving grid points lying on certain algebraic varieties. A unified study of such problems began with a question of Elekes [3] about expansion of bivariate real polynomials f(x,y). Specifically, he asked: For a bivariate polynomial f[x,y] and given finite sets A,B, how small can be the image set

f(A,B)={f(a,b)aA,bB}.

Elekes conjectured that the image of f on a n×n Cartesian product must be of cardinality superlinear in n, unless f has a very concrete special form. This was confirmed in 2000 by Elekes and Rónyai [4] who proved the following dichotomy: Either f has one of the forms

f(x,y) =h(p(x)+q(y))or
f(x,y) =h(p(x)q(y)), (3)

for some univariate real polynomials p,q,h, or, otherwise, for every finite A,B, each of size n, we have

|f(A,B)|=ω(n). (4)

In case f is not one of the forms in (3), the lower bound on |f(A,B)| was improved in [8] to be Ω(n4/3), and further improved in [13] to be Ω(n3/2), which is currently the best known lower bound for bivariate polynomials that are not special.

An analogue of the Elekes–Rónyai problem can be formulated for polynomials in more than two variables. The trivariate case was studied by Raz, Sharir, and De Zeeuw [9], and the general k-variate case was established by Raz and Shem Tov [10]. They obtain the following result.

Theorem 1 ([9, 10]).

Let k3 and let f[x1,,xk]. Then one of the following holds:

  1. (i)

    For every A1,,Ak each of size n one has

    |f(A1,,Ak)|=Ω(n3/2)
  2. (ii)

    f is of one of the forms:

    f(x1,,xk) =h(p1(x1)++pk(xk)) (5)
    f(x1,,xk) =h(p1(x1)pk(xk))

Note that the bound in Theorem 1, for non-special polynomials f, is independent of k, and in particular coincides with the bound for k=3. At first glance this may appear to be merely a consequence of the proof. Indeed, the argument in [10] reduces the k-variate case for k4 to the trivariate case by fixing values for k3 of the variables. They then show that if fixing any such subset of k3 variables yields a special trivariate polynomial, then f itself, as a k-variate polynomial, must be special in the sense of (5).

It is natural to expect that increasing the number of variables should force the image of f to grow faster. However, certain polynomials in many variables can in effect behave like polynomials in fewer variables. For example, consider the (k+2)-variate polynomial

f(x,y,z1,zk)=xy+z1+z2++zk.

Let A,B,C1,,Ck, where A,B are arbitrary finite sets of size n and C1==Ck=[n]. Let

C:=C1++Ck={k,k+1,,kn}.

Then |C|=Θ(n), and letting g(x,y,z):=xy+z, we have

f(A,B,C1,,Ck)=g(A,B,C).

In this case, with the current techniques, it is unclear how to obtain a bound on the expansion of f that improves upon the trivariate result for g.

Our results

In this paper, we recognize k-variate polynomials that are, in a precise sense, truly k-variate, and we improve the corresponding expansion bounds for them. More precisely, for a k-variate polynomial f, we introduce the notion of the rank of f. If f has rank r, then, in a rigorous sense, it is essentially (r+1)-variate, and the bound on the size of its image can be improved with an exponent that grows with r.

We now define the rank of a polynomial and then state our main result.

Let f[x1,,xk] and let dx1 stand for the degree of f with respect to the variable x1. Write

f(x1,,xk)=i=0dx1αi(x2,,xk)x1i.

We consider the coefficient map T=Tf,x1:k1dx1+1 given by

(x2,,xk)(α0(x2,,xk),,αdx1(x2,,xk)).

We define the rank of f with respect to the variable x1 to be

rankx1(f):=rank(JT),

where JT stands for the Jacobian matrix of T; here we consider the rank over the field of rational functions in x2,,xk over , which is equivalent to evaluating the rank at a generic point (x2,,xk)k1. Note that

0rankx1(f)k1.

Similarly, define rankxi(f), for every i=2,,k, where xi plays the role of x1.

Finally, define the rank of the polynomial f to be

rank(f):=max1ikrankxi(f).
Example 2.

Let

f(x1,x2,,xk)=x1xk+x2xk2+xk1xkk1.

Then rank(f)=rankxk(f)=k1.

Example 3.

Let

f(x1,x2,,xk)=p1(x1)xk+p2(x1,x2)xk2++pk1(x1,,xk1)xkk1,

where pi is an i-variate polynomial that depends non-trivially on xi. Then rank(f)=rankxk(f)=k1. Indeed, in this case the matrix JTf,xk is upper-triangular.

Example 4.

Let f(x1,,xk)=p1(x1)+p2(x2)++pk(xk). The coefficient map Tf,x1 maps (x2,,xk) to the coefficients of p1(x1), which consist only of a single term depending on the other variables. The Jacobian matrix consists almost entirely of zeros, yielding rank(f)=1. This aligns perfectly with the special degenerate forms described in Theorem 1.

We prove the following main result of the paper.

Theorem 5.

Let k3 and let f[x1,xk]. Assume that rank(f)=r2. Then, for every ε>0, the following holds: Let A1,,Ak be finite, each of size n. Then

|f(A1,,Ak)|=Ω(n5r42rε),

where the constant of proportionality depends on deg(f), on r, and on ε.

Observe that for a generic k-variate polynomial, the rank is maximal, meaning r=k1. In this case, Theorem 5 yields a lower bound of Ω(n5k92k2ϵ). Notably, this relates to Conjecture 1.3 of de Zeeuw [2], which posits that a generic bivariate polynomial has at least Ω(n2ϵ) distinct values. Our theorem pushes towards this super-quadratic growth and currently provides the best known lower bound for the minimal number of distinct values taken by generic k-variate polynomials for k4.

We observe that rank(f)=1 corresponds to the special forms from Theorem 1. Indeed, we have the following theorem.

Theorem 6.

Let k3 and let f[x1,xk]. Assume that f depends non-trivially on each of its variables and that

rank(f)=1.

Then f has one of the forms

f(x1,,xk) =h(p1(x1)++pk(xk)) or
f(x1,,xk) =h(p1(x1)pk(xk)), (6)

for some univariate real polynomials h(x),p1(x),,pk(x).

Finally, we present an application of our results to the following Erdős-type combinatorial geometric problem. Let ν denote the moment curve in d, parameterized by

ν(t)=(t,t2,,td),t.

Let Pν be a finite set of n points. For any distinct p1,,pd+1ν, let σ=σ(p1,,pd+1) denote the d-simplex which is the convex hull of p1,,pd+1 in d, and let vol(σ) denote its d-dimensional volume. Define

Δ(P)={vol(σ(p1,,pd+1))p1,,pd+1P}.

We have the following theorem.

Theorem 7.

Let ν be the moment curve in d and let Pν be any finite set of size n. Then, for every ε>0,

|Δ(P)|=Ω(n5d42dε),

where the implicit constant depends only on ε and on d.

Theorem 7 is obtained by identifying a (d+1)-variate polynomial f whose expansion over a certain n××n grid in d+1 corresponds to the number of distinct volumes of d-simplices spanned by P. We then show that f has rank d and apply our main Theorem 5.

Organization of the paper

The paper is organized as follows. In Section 2 we recall an incidence bound that will serve as a key tool in our arguments. In Section 3, we establish a special case of our main result, Theorem 5, and in Section 4, we complete its proof. The proof of Theorem 6 is provided in Section 5. Finally, Section 6 contains the proof of Theorem 7.

2 Incidences between points and algebraic curves

For a finite set of points 𝒫2 and a finite set of planar curves 𝒞, we let I(𝒫,𝒞) denote the set of point-curve incidences; that is

I(𝒫,𝒞)={(p,γ)𝒫×𝒞pγ}.

The classical Szemerédi–Trotter theorem [14] asserts that, for the special case where 𝒞 is a set of lines, and putting m:=|𝒫| and n:=|𝒞|, one has

|I(𝒫,𝒞)|=O(m2/3n2/3+m+n).

Since the Szemerédi–Trotter result, many alternative proofs and analogue problems have been studied. Today incidence problems play a fundamental role in combinatorial geometry. For our result we will need an extension of the Szemerédi–Trotter theorem to point-curve incidence problems, where the curves are algebraic and come from an s-dimensional family of curves. We now present the definition from Sharir–Zahl [12].

A bivariate polynomial h[x,y] of degree at most D is a linear combination of the form

h(x,y)=0i+jDcijxiyj.

Note that the number of monomials xiyj such that 0i+jD is (D+22). In this sense, every point c(D+22) (other than the all-zero vector) can be associated with a curve in 2, given by the zeroset of the bivariate polynomial whose coefficients are the entries of c. If λ0, then f and λf have the same zero-set. Thus, the set of algebraic curves that can be defined by a polynomial of degree at most D in 2 can be identified with the points in the projective space 𝐏(D+22).

In [12], Sharir and Zahl defined an s-dimensional family of plane curves of degree at most D to be an algebraic variety F𝐏R(D+22) such that dim(F)=s. We will call the degree of the variety F the complexity of the family.

They then proved the following incidence bound:

Theorem 8 (Sharir–Zahl [12]).

Let be an s-dimensional family of plane curves of degree at most D and complexity at most K. Let 𝒫 be a set of m points in the plane and let 𝒞 be a set of n plane curves. Suppose that no two of the curves in 𝒞 share a common irreducible component. Then for every ε>0, we have

I(𝒫,𝒞)=Oε(m2s5s4n5s65s4+ε)+O(m2/3n2/3+m+n),

where the constant of proportionality depends on s,K,D and in the first term also on ε.

3 A special case

In this section we prove the special case of Theorem 5 with r=k1. This will serve as a key ingredient for the proof of the general result.

Proposition 9.

Let k3, f[x1,xk], and assume that rank(f)=k1. Then, for every ε>0, the following holds: Let A1,,Ak be finite, each of size n. Then

|f(A1,,Ak)|=Ω(n5(k1)42(k1)ε),

where the constant of proportionality depends on deg(f), on k, and on ε.

Proof.

Let f,A1,,Ak be as in the statement. Up to renaming of the variables, we may assume without loss of generality that rankx1(f)=k1. Put

B=f(A1,,Ak)

We aim to lower bound |B|. For this, consider:

S={(x1,x2,,xk,y)A1×A2××Ak×By=f(x1,,xk)}.

Notice that, for every (a1,ak)A1××Ak, we have that (a1,ak,f(a1,,ak))S, and so

|S|=nk. (7)

We claim that, for every ε>0, one has

|S|=O(|B|2(k1)5(k1)4n(k1)(1+ε)), (8)

where the constant of proportionality depends on ε, k, and deg(f). Combining (7) and (8), we get

nk=O(|B|2(k1)5(k1)4n(k1)(1+ε))

or

|B|=Ω(n5(k1)42(k1)ε),

where ε=5(k1)42ε, which proves the proposition.

So in order to complete the proof of Proposition 9 we only need to prove (8). Let T=Tf,x1 be the coefficient map defined in the introduction. By assumption, rank(JT)=k1. Thus, there exist indices (i1,,ik1) such that for

T^:(x2,xk)(αi1(x2,xk),,αik1(x2,xk)),

we have

detJT^0.

Define

S0 ={(a1,,ak,b)SdetJT^(a2,,ak)=0},
S =SS0.

Clearly |S|=|S0|+|S|.

Observe that,

|S0|=|A1||(A2××Ak){detJT^=0}|ndeg(detJT^)nk2,

where the inequality is due to the Schwartz–Zippel Lemma (see [11, 16]). Thus, we get

|S0|=O(nk1), (9)

where the constant of proportionality depends on deg(f) and on k.

We now bound |S|. For this, we reduce the problem into a point-curve incidence problem in the plane as follows. With each (a2,,ak)A2××Ak for which detJT^(a2,,ak)0, we associate a curve γa2,,ak in 2 given by the equation

y=f(x,a2,,ak).

Note that γa2,,ak is irreducible for every (a2,,ak)A2××Ak, because it is defined by an equation of the form y=P(x), making it the graph of a polynomial. Let

𝒫 =A1×B,
𝒞 ={γa2,,akdetJT^(a2,,ak)0};

note that curves in 𝒞 are taken without multiplicity. Let I(𝒫,𝒞) denote the set of point-curve incidences between 𝒫 and 𝒞.

Claim 10.

We have

|S|=Θ(|I(𝒫,𝒞)|), (10)

where the constant of proportionality depends only on degf and on k.

Proof.

By definition, if (a1,a2,,ak,b)S then ((a1,b),γa2,,ak)I(𝒫,𝒞). So, to prove the claim, it suffices to show that every (p,γ)I(𝒫,𝒞) corresponds to at most O(1) elements of S.

For γ𝒞, write

m(γ)={(a2,,ak)γa2,,ak=γ and detJT^(a2,,ak)0}.

We need to prove that

|m(γ)|=O(1), (11)

with constant of proportionality that depends only on deg(f) and on k.

By the definition of the set 𝒞, we have that γ is given by an equation of the form

y=i=0dx1cixi,

for some coefficients c0,,cdx1. Let V denote the algebraic variety which is given by the system of equations

αi1(x2,,xk) =ci1
αi2(x2,,xk) =ci2 (12)
αik1(x2,,xk) =cik1.

Then

m(γ)V.

Write V=V0V+, where V0 is the union of all 0-dimensional irreducible components of V, and V+ is the union of all other irreducible components of V. Recall that, by properties of real algebraic varieties, we have that V0 is finite, and

|V0|=O(1),

with a constant that depends only of deg(f) and on k (see e.g. [1, 6, 15] for a bound on the number of connected components of a real algebraic variety).

Thus, to prove (11), it suffices to show that m(γ)V0. Assume, for contradiction, that this is not the case. That is, there exists (a2,,ak)m(γ)V+. Then, by definition, we have

T^ (a2,,ak)=(ci1,,cik1)
det JT^(a2,,ak)0.

By the inverse function theorem, there exists an open neighborhood, N, of (a2,,ak) such that T^ restricted to N is invertible. In particular,

NT^1{(ci1,,cik1)}={(a2,,ak)}.

On the other hand,

NT^1(ci1,,cik1)=NV.

Since N is a neighborhood of (a1,,ak1) and the latter lies on an irreducible component of V of dimension at least 1, the intersection NV must be infinite. This leads to a contradiction, and thus m(γ)V0. This completes the proof of Claim 10.

In view of Claim 10, in order to bound |S| it suffices to bound |I(𝒫,𝒞)|. We have |𝒫|=n|B| and |𝒞|nk1. Since the curves in 𝒞 are irreducible, every two distinct curves γ,γ in 𝒞 intersect in at most dx12 points, by Bezout’s Theorem (see e.g. [5, Corollary 7.8]). We can therefore apply Theorem 8. In our context, the family of curves is parameterized by the k1 variables from A2××Ak, so the dimension is sk1. The degree Dmax(deg(f),1) and the complexity K=O(1) depend strictly on k and deg(f), and crucially, are independent of n. This gives
|I(𝒫,𝒞)| =O((|B|n)2(k1)5(k1)4n(k1)(5(k1)65(k1)4+ε)+(|B|n)2/3(nk1)2/3+|B|n+nk1).

or

|I(𝒫,𝒞)| =O(|B|2(k1)5(k1)4n(k1)(1+ε)+|B|23n2k3+|B|n+nk1).

Note that we may assume without loss of generality that the first summand is dominant. Indeed, the second summand is dominant if

|B|2(k1)5(k1)4n(k1)(1+ε)|B|23n2k3

or

|B|n5(k1)44+O(ε),

which is stronger than the lower bound we wish to prove on |B|, for every k3. Similarly, the third summand is dominant if

|B|2r5r4nr(1+ε)|B|n

or

|B|n(k2)(5(k1)4)3(k1)4+ε(k1)(5(k1)4)3(k1)4,

which is better than the lower we want to prove on |B| for every k, as is easy to verify. Finally, the fourth summand is always subsumed by the first one.

Hence, either the conclusion of Proposition 9 holds, or we obtain

|I(𝒫,𝒞)|=O(|B|2(k1)5(k1)4n(k1)(1+ε)).

In view of Claim 10 and combined with (9), the inequality (8) follows. This completes the proof Proposition 9.

4 Proof of Theorem 5

The following lemma shows that the main Theorem 5 can in fact be reduced to the statement of Proposition 9.

Lemma 11.

Let f[x1,xk] and assume that rankx1(f)=r<k1. Then, up to renaming of the variables x2,,xk, we have that

g(x1,,xr+1):=f(x1,x2,,xr+1,yr+2,,yk)

is a (r+1)-variate polynomial in [x1,,xr+1] with rankx1(g)=r, for all (yr+2,,yk)kr1Z0, where Z0 is some subvariety of kr1 of codimension at least 1 and degree depending only on k and deg(f).

Proof.

Let f be as in the statement. Let T=Tf,x1 be the corresponding coefficient map. By assumption

rank(JT)=r<k1. (13)

Up to renaming the variables, we may assume without the loss of generality, that the first r columns of JT, corresponding to the variables x2,,xr+1, are independent. Observe that the matrix composed of the first r columns of JT, is in fact the Jacobian matrix of the coefficient map Tg,x1, where g is the (r+1)-variate polynomial given by

g(x1,x2,,xr+1):=f(x1,x2,,xr+1,yr+2,,yk);

here x2,,xr+1 are regarded as constant parameters.

Our goal is to show that for JTg,x1, treated as a matrix with entries in (x1,,xr+1), we have that rank(JTg,x1)=r, for every generic choice of (yr+2,,yk)kr1.

Let Δ denote the polynomial corresponding to the sum of squares of the determinants of all the r×r submatrices of JTg,x1. So Δ is a multivariate polynomial, and we can write

Δ(x2,,xr+1,yr+2,,yk)=iβi(yr+2,,yk)gi(x2,,xr+1),

with gi[x2,,xr+1] and βi[yr+2,,yk]. By the definition of Δ we have
rank(JTg,x1(x2,,xr+1,yr+2,,yk))<r if and only if Δ(x2,,xr+1,yr+2,,yk)=0.

Thus, in view of (13), we have Δ0. In particular, the polynomials βi are not all zero. Thus, letting

Z0:={(yr+2,,yk)iβi(yr+2,,yk)=0},

we see that Z0 has codimension at least 1. This completes the proof of the lemma.

We can now complete the proof of our main Theorem 5.

Proof of Theorem 5.

Let f[x1,,xk] and assume without loss of generality that degx1(f)=rk2. Apply Lemma 11 to f. Then, up to renaming of the variables, there exists an algebraic variety Z0kr1, of codimension at least one, and of degree O(1), such that for every (yr+2,,yk)kr1Z0 we have that the polynomial

(x1,,xr+1)f(x1,,xr+1,yr+2,,yk)

is an (r+1)-variate polynomial of rank r.

Observe that, by the Schwartz–Zippel lemma, there exists (ar+2,,ak)(Ar+2××Ak)Z0. Thus

g(x1,,xr+1):=f(x1,,xr+1,ar+2,,ak)

satisfies rankx1(g)=r. Thus, by Proposition 9, we have that

|g(A1,,Ar+1)|=Ω(n5r42rε).

Noting that

g(A1,,Ar+1)=f(A1,,Ar+1,{ar+2},,{ak})f(A1,,Ak),

this completes the proof of the theorem.

5 Characterization of rank-1 polynomials

In this section we prove Theorem 6. For the proof we will use the following lemma from Raz and Shem Tov [10].

Lemma 12 (Raz–Shem Tov [10, Lemma 2.3]).

Let f[x1,,xk]. Assume that

fx1(x1,,xk)r1(x1)==fxk(x1,,xk)rk(xk), (14)

for some univariate real polynomials r1,,rk. Then, f is one of the forms

f(x1,,xk) =h(p1(x1)++pk(xk)) or
f(x1,,xk) =h(p1(x1)pk(xk)),

for some univariate real polynomials h(x),p1(x),,pk(x).

Proof of Theorem 6.

Let f be as in the statement. Since f depends non-trivially on each of its variables, we have in particular that

rankxi(f)=1, for each i=1,,k. (15)

In view of Lemma 12, it suffices to show that f satisfies the differential equation (14), for some univariate polynomials r1,,rk. By symmetry, it suffices to prove that

fx1(x1,,xk)r1(x1)=fx2(x1,,xk)r2(x2) (16)

We write

f(x1,,xk)=i=0dx1αi(1)(x2,,xk)x1i==i=0dxkαi(k)(x1,,xk1)xki, (17)

where dxj stands for the degree of f as a univariate polynomial in the variable xj. Then for every j,{1,,k}, such that j, we have

fxj(x1,,xk)=i=0dxαi()(x1,,x^,,xk)xjxi. (18)

We next show that

fx1fx2=u(x1,x2), (19)

where u is some rational function over .

In other words, we need to show that fx1/fx2 is independent of xi, for every i1,2. By symmetry it suffices to show that this ratio is independent of the variable x3. Consider the first two columns of JTf,x3, corresponding to derivatives with respect to x1 and to x2. Namely,

(α0(3)x1α0(3)x2α1(3)x1α1(3)x2αdx3(3)x1αdx3(3)x2).

Note that neither of the columns is zero, since f depends non-trivially in each of its variables. Moreover, since rank(JTf,x3)=1, by (15), and using the fact that the entries of JTf,x3 are independent of x3, we get that there exists a rational function u(x1,x2,x4,,xk) such that

αi(3)x1=u(x1,x2,x4,,xk)αi(3)x2,for i=0,,dx3.

Using the identity (18), this implies that

fx1fx2=u(x1,x2,x4,,xk);

that is, this ratio is independent of x3. This proves (19).

Repeating the analysis symmetrically for the pair x1,x3 and then to the pair x2,x3 we conclude that there exist real rational functions v(x1,x3) and w(x2,x3) such that

fx1fx3 =v(x1,x3)
fx2fx3 =w(x2,x3).

But then

fx1fx2=v(x1,x3)w(x2,x3)=u(x1,x2),

meaning in particular that the ratio of v,w is independent of x3. Thus, setting an arbitrary value for x3, we get that

fx1fx2=v(x1,0)w(x2,0).

Thus, letting r1(x1):=v(x1,0) and r2(x2):=w(x2,0), this proves (16), and hence completes the proof of the lemma.

6 Distinct 𝒅-volumes on the moment curve in 𝒅

In this section we prove Theorem 7.

Proof of Theorem 7.

The moment curve in d is defined by ν(x)=(x,x2,,xd). Consider d+1 point on ν given by the parameters x1,,xd+1d, and let σ(x1,,xd+1) denote the simplex spanned by these points. Define

f(x1,,xd+1):=volσ(x1,,xd+1).

It suffices to prove that

rankxd+1(f)=d. (20)

Indeed, assume that (20) is true and let Pν be a set of size n. Let A denote the x1-coordinate of the points in P. Then Δ(P)=f(A,,A). By Theorem 5, for every ε>0, we have

|Δ(P)|=|f(A,,A)|=Ω(n5d42dε),

as needed.

So we only need to prove (20). We have

f(x1,,xd+1) =1d!det(1x1x12x1d1x2x22x2d1xd+1xd+12xd+1d),

which is the determinant of the (d+1)×(d+1)-Vandermonde matrix. Thus

f(x1,,xd+1) =1d!1i<jd+1(xjxi)
=1d!1i<jd(xjxi)k=1d(xd+1xk)
=g(x1,,xd)f^(x1,,xd+1),

where g(x1,,xd):=1d!1i<jd(xjxi) is independent of xd+1 and f^(x1,,xd+1):=k=1d(xd+1xk). Note that f^ can be written as

f^(x1,,xd+1)==0ds(x1,,xd)xd+1,

where

s0(x1,,xd) =(1)dx1xd
s1(x1,,xd) =(1)d1(x2xd+x1x3xd++x1xd1)
sd1(x1,,xd) =x1xd
sd(x1,,xd) =1,

are the symmetric polynomials.

Claim 13.

Let M be the d×d matrix M=(Mi,j)0id1, 1jd given by

Mi,j=sixj(0id1, 1jd).

Then rank(M)=d.

The strategy for this proof is adapted from the proof of [ [7], section 4].

Proof.

We prove that detM is not the zero polynomial. Consider the function

P(x1,,xd,t)=k=1d(txk)=k=0dsktk,

which is a monic polynomial in t.

For 1jd we have

Pxj(t)=k=1kjd(txk)=k=0d1skxjtk

Evaluating at t=xi yields

Pxj(xi)=k=0d1skxjxik={0,ij,kj(xjxk),i=j. (21)

The identity (21) implies

(1x1x12x1d11x2x22x2d11xdxd2xdd1)M=diag(Px1(x1),,Pxd(xd)).

Thus

det(M) =(1)dj=1dij(xjxi)1i<jd(xjxi)
=(1)d1ijd(xjxi)1i<jd(xjxi)
=1i<jd(xjxi).

This completes the proof of Claim 13.

We are now ready to prove (20). Write

f(x1,,xd+1)=i=0dαi(x1,,xd)xd+1i.

By the above we have

αi=g(x1,,xd)si(x1,,xd), for i=0,,d

Let T=Tf,xd+1. Denoting partial derivatives with subscripts (e.g., gx1=gx1), we have

JT=(gx1s0+gs0x1gx2s0+gs0x2gxds0+gs0xdgx1sd1+gsd1x1gx2sd1+gsd1x2gxdsd1+gsd1xd1gx1gx2gxd).

Applying row operation to the first d rows, namely, RiRisiRd we get

rankJT =rank(gs0x1gs0x2gs0xdgsd1x1gsd1x2gsd1xd1gx1gx2gxd)
=rank(s0x1s0x2s0xdsd1x1sd1x2sd1xd1gx1/ggx2/ggxd/g).

Note that first d rows of the above matrix are exactly the matrix M from Claim 13. This completes the proof of Theorem 7

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