1 Introduction

A super-Vandermonde set (short: an sV-set) in GF(q), \(q=p^h\), p a prime, is a set T of size \(1<t<q\) such that

$$\begin{aligned} \pi _k(T):=\sum _{y\in T} y^k=0~, \end{aligned}$$

for \(0< k< t\). It follows from the non-singularity of the Vandermonde matrices \((y^k)_{yk}\), \(y\in T\) and \(k\in [0,t)\) resp. \(k\in (0,t]\) that \(0\not \in T\) and that \(\pi _t(T)\ne 0\) (in particular \(p\!\not \!|\,\,t\)). The Newton identities relating the power sums \(\pi _k(T)\) and the elementary symmetric polynomials \(\sigma _k(T)\) imply that in the polynomial

$$\begin{aligned} f(Z):=\prod _{y\in T}(Z-y)=\sum (-1)^k\sigma _k(T)Z^{t-k}~, \end{aligned}$$

the only possible nonzero coefficients are the constant term \((-1)^t\sigma _t\) and the coefficient of \(Z^{t-k}\): \((-1)^k\sigma _k\) with \(k=0\mod p\). The Newton-identities are given by:

$$\begin{aligned} k\sigma _k = \sum _{m=1}^k (-1)^{m-1}\pi _m\sigma _{k-m}~, \end{aligned}$$

and we see that indeed \(\sigma _k=0\) if k is not divisible by p (and less than t).

In terms of the inverses of the elements in T, we get that being sV is equivalent to

$$\begin{aligned} \phi (Y):=\prod _{y\in T} (Y-y^{-1})=Y^t+g(Y)~, \end{aligned}$$

with g a p-th power.

The underlying notion of Vandermonde set was introduced by Gács and Weiner in [1]. They appear at several places in the investigation of special point sets in finite projective planes. More about this, as well as many examples, can be found in Chapter 1 of the thesis of Takáts [2], or in her paper [3] with Péter Sziklai, which also classifies small and large sV-sets. Here small means \(t<p\), and small sV-sets are cosets of multiplicative subgroups of \({GF(q)}^*\): in this case the polynomial g is constant, so

$$\begin{aligned} \phi (Y)=\prod _{y\in T} (Y-y^{-1})=Y^t-c~, \end{aligned}$$

where \(t\,|\,q-1\) and c is a t-th power, so that T is a coset of the group of t-th roots of unity.

By large we mean \(t>q/p\) and again we get cosets of multiplicative subgroups, corresponding to the case that \(g=-c\) is constant. The proof in this case is much more involved, but in the final section we will give a simpler proof.

2 Super-Vandermonde sets of size \(p+1\)

If T is an sV-set of size \(p+1\), then the polynomial \(\prod _{y\in T} (Z-y)\) is of the form \(f(Z)=Z^{p+1}+aZ+b\), so our problem is to classify the polynomials of this form that are fully reducible over GF(q). Notice that two different polynomials of this form have a gcd of degree at most one, so that two elements of GF(q) are contained in at most one sV-set of size \(p+1\). We will see in fact that two elements are contained in an sV-set of this size precisely when they have the same GF(p)-norm. We will prove in the next theorem that they can all be obtained from 2-dimensional GF(p)-vector subspaces of GF(q).

Theorem 2.1

Let T be an sV-set in GF(q), \(q=p^h\), p prime, of size \(p+1\). Then there exists \(\alpha \in {GF(q)}^*\) such that

$$\begin{aligned} T=\{\alpha x_1^{p-1},\dots ,\alpha x_{p+1}^{p-1}\}, \end{aligned}$$
(1)

where \(\{ x_1,\dots ,x_{p+1}\}\) represent the 1-dimensional subspaces of a 2-dimensional GF(p)-vector subspace of GF(q).

Conversely, every 2-dimensional GF(p)-vector subspace of GF(q) defines a family of \(q-1\) sV-sets of type (1). In particular, the elements of an sV-set of size \(p+1\) have the same norm over GF(p).

Proof

We first observe that if \(T=\{y_1,\dots ,y_t\}\) is an sV-set, then for each \(\gamma \in {GF(q)}^*\), the set \(\gamma T=\{\gamma y_1,\dots ,\gamma y_t\}\) is an sV-set as well (and of the same size of course). We first show that 2-dimensional subspaces give rise to sV-sets. Let U be a 2-dimensional GF(p)-vector subspace of GF(q), then U is the set of zeros of a polynomial of the form

$$\begin{aligned} X^{p^2}+aX^p+bX, \end{aligned}$$
(2)

for some \(a,b\in {GF(q)}\). If \(x_1\) and \(x_2\) are two nonzero roots of (2) which are not proportional over GF(p), then \(x_1^{p-1}\) and \(x_2^{p-1}\) are two different roots of the polynomial \(Z^{p+1}+aZ+b\), which turns out to be fully reducible over GF(q). It follows that for each \(\alpha \in {GF(q)}^*\)

$$\begin{aligned} \alpha T=\{\alpha x^{p-1}:\ x \hbox { is a nonzero root of (2)} \} \end{aligned}$$

is an sV-set of size \(p+1\).

On the other hand let \(T=\{y_1,\dots ,y_{p+1}\}\) be an sV-set of size \(p+1\) and let

$$\begin{aligned} f(Z)=Z^{p+1}+aZ+b \end{aligned}$$
(3)

be the associated polynomial. Then, there exist \(y_i,y_j\in T\), with the same GF(p)-norm \(\delta \). Let \(\alpha \) be an element of \({GF(q)}^*\) with norm \(N(\alpha )=\delta \) and set \(z_k:=y_k/\alpha \), for \(k\in \{1,\dots ,p+1\}\). Then

$$\begin{aligned} \frac{1}{\alpha }T:=\{z_1,\dots ,z_{p+1}\}, \end{aligned}$$

is an sV-set of size \(p+1\) with \(N(z_i)=N(z_j)=1\) and its associated polynomial is

$$\begin{aligned} Z^{p+1}+\frac{a}{\alpha ^p}Z+\frac{b}{\alpha ^{p+1}}. \end{aligned}$$

Denoting by \(x_i\) and \(x_j\) the elements of \({GF(q)}^*\) such that \(z_i=x_i^{p-1}\) and \(z_j=x_j^{p-1}\), then \(x_i\) and \(x_j\) are independent over GF(p) and so \(U:=\langle x_i,x_j\rangle \) is a 2-dimensional GF(p)-vector subspace of GF(q), whose elements are the zeros of the polynomial

$$\begin{aligned} X^{p^2}+\frac{a}{\alpha ^p}X^p+\frac{b}{\alpha ^{p+1}}X. \end{aligned}$$

It follows that the elements of \(\frac{1}{\alpha }T\) are of the form \(x^{p-1}\). This completes the proof. \(\square \)

3 Super-Vandermonde sets of size \(q/p-1\)

Consider the polynomial \({\hbox {Tr}_{q\longrightarrow p}}(aZ)=aZ+a^pZ^p+\cdots +a^{p^{h-1}}Z^{p^{h-1}}\), the trace from GF(q) to GF(p). It is clearly fully reducible over GF(q), and we see that the nonzero roots form an sV-set of size \(q/p-1\). The aim of this section is to prove the converse:

Proposition 3.1

Let T be an sV-set in GF(q), of size \(q/p-1\), (\(q=p^h\)) then

$$\begin{aligned} \prod _{y\in T} (Z-y)=(a_{h-1}Z)^{-1}{\hbox {Tr}_{q\longrightarrow p}}(aZ) \end{aligned}$$

for some \(a\in {GF(q)}^*\).

Proof

Consider as before the polynomial

$$\begin{aligned} \phi (Y)=\prod _{y\in T} (Y-y^{-1})=Y^{q/p-1}+g(Y)~, \end{aligned}$$

where g is a p-th power. Let \(T_a\) be the sV-set corresponding to the hyperplane \(\hbox {Tr}(aZ)=0\) with

$$\begin{aligned} \phi _a(Y):=\prod _{y\in T_a} (Y-y^{-1}) = Y^{q/p-1}+g_a(Y). \end{aligned}$$

The greatest common divisor of \(\phi \) and \(\phi _a\) divides \((g(Y)-g_a(Y))^{1/p}\) of degree at most \(q/p^2-1\). So we find that T has at most \(q/p^2-1\) points in every hyperplane, unless it coincides with it. Since the average size of the intersection of T with a hyperplane equals

$$\begin{aligned} \frac{q/p-1}{q-1}\cdot \left( \frac{q}{p}-1\right) >\frac{q}{p^2}-1~, \end{aligned}$$

we see that for some a, T coincides with \(T_a\). \(\square \)

4 Large super-Vandermonde sets

Proposition 4.1

Let T be an sV-set in GF(q), \(q=p^h\) of size \(t>q/p\), then

$$\begin{aligned} \prod _{y\in T} (Y-y^{-1})=Y^t-c \end{aligned}$$

for some t-th power \(c\in {GF(q)}^*\), so T is coset of a multiplicative subgroup.

Proof

As before \(\phi (Y)=\prod _{y\in T} (Y-y^{-1})=Y^t+g(Y)\), where g is a p-th power. Since this polynomial is fully reducible we may write:

$$\begin{aligned} (Y^t+g)(h_0+Yh_1+\cdots +Y^{p-1}h_{p-1})=Y^q-Y,~~~~q=p^h, \end{aligned}$$

where also the polynomials \(h_i\) are p-th powers. We now equate left and right the terms of degree d mod p, \(d=0,p-1,\dots ,1\), writing \(e=t-q/p\) and \(E=q/p\):

We look at the divisibility by Y. From the last equation we see that \(h_1\) is not divisible by Y, in particular \(h_1\ne 0\), then we see from the other equation involving \(h_1\) that \(h_{1+e}\) is divisible by \(Y^{E}\), next \(h_{1+2e}\) by \(Y^{2E}\), (where of course we take indices mod p) and so on until finally \(h_{1+(p-1)e}=h_{p-e+1}\) is divisible by \(Y^{(p-1)E}=Y^{q-q/p}\). If \(h_{p-e+1}\) is nonzero then the total degree of the left hand side will be at least \(t+1+q-q/p>q\), a contradiction, so \(h_{p-e+1}=0\) and now the last equation tells us that g is constant. \(\square \)