Reducing the number of equations defining a subset of the $n$-space over a finite field

TL;DR

This paper proves that for algebraic sets over finite fields, the number of defining equations can be reduced to the dimension of the space without increasing degrees, simplifying the description of these sets.

Contribution

It introduces a method to reduce the number of defining equations for algebraic sets over finite fields, maintaining degrees and providing a new way to simplify algebraic descriptions.

Findings

Number of equations can be reduced to the space dimension

Reduction preserves the total degree of defining polynomials

Applicable to both affine and projective algebraic sets

Abstract

Let $f_{1}, \ldots, f_{k}$ be polynomials defining an algebraic set in affine $n$-space over a finite field. Suppose $k>n$. We prove that there exists a system of polynomials $g_{1}, \ldots, g_{n}$, each being a linear combination with scalar coefficients of $f_{1}, \ldots, f_{k}$, defining the same algebraic set. In particular, one reduces the number of equations without increasing the total degree. We also have the corresponding result for systems of homogeneous polynomials defining algebraic sets in projective spaces.

Equations

$$\begin{array}[]{c}\mathcal{M}_{n}\to\mathbb{P}^{n}(\mathbb{K})\\ M\mapsto(N(M)\setminus\left\{\theta\right\})_{\sim}\end{array}$$

$$\begin{array}[]{c}\mathcal{N}_{n}\to\mathbb{P}^{n}(\mathbb{K})\\ M\mapsto(N(M)\setminus\left\{\theta\right\})_{\sim}\end{array}$$

$${\mathcal{N}}_n/G{L}_n(\mathbb{K})\to {\mathbb{P}}^n(\mathbb{K})$$

$${\mathcal{M}}_n\to {\mathcal{N}}_n/G{L}_n(\mathbb{K})$$

$$S=\left\{[f_1(x):\dots :f_{n+1}(x)]:x\in X\setminus \mathrm{Z}(f_1,\dots ,f_{n+1})\right\}.$$

$$T={\mathcal{M}}_n\setminus \left\{M_s:s\in S\right\}.$$

$$\left[\begin{array}[]{c}g_{1}\\ \vdots\\ g_{n}\end{array}\right]=M\left[\begin{array}[]{c}f_{1}\\ \vdots\\ f_{n+1}\end{array}\right].$$

$$A=G_1^{-1}{I}_{r,n,n}{G}_3^{-1}{G}_3{I}_{n,n,m}{G}_2^{-1}.$$

$$f_i(x)=\text{the }i\text{th coordinate of }x.$$

$$\left[\begin{array}[]{c}g_{1}\\ \vdots\\ g_{n}\end{array}\right]=A\left[\begin{array}[]{c}f_{1}\\ \vdots\\ f_{n+1}\end{array}\right]$$

$$\begin{array}[]{c}r\colon\mathrm{Map}(\mathbb{P}^{n}({\mathbb{F}_{q}}),{\mathbb{F}_{q}})\to\mathrm{Map}(\mathbb{P}^{n}({\mathbb{F}_{q}})\setminus\left\{\alpha\right\},{\mathbb{F}_{q}})\\ r(f)=f|_{\mathbb{P}^{n}({\mathbb{F}_{q}})\setminus\left\{\alpha\right\}}.\end{array}$$

$$\left|{\mathbb{P}}^n({\mathbb{F}}_q)\setminus \left\{\alpha \right\}\right|=\left|{\mathbb{P}}^n({\mathbb{F}}_q)\right|-1=\frac{q^{n+1}-1}{q-1}-1=\frac{q^{n+1}-q}{q-1}.$$

$$\left[\begin{array}[]{c}\bar{g_{1}}\\ \vdots\\ \bar{g_{n}}\end{array}\right]=A\left[\begin{array}[]{c}\bar{f_{1}}\\ \vdots\\ \bar{f_{k}}\end{array}\right].$$

$$\left[\begin{array}[]{c}{g_{1}}\\ \vdots\\ {g_{n}}\end{array}\right]=A\left[\begin{array}[]{c}{f_{1}}\\ \vdots\\ {f_{k}}\end{array}\right].$$

$$\begin{array}[]{l}\mathrm{Z}(f_{1},\ldots,f_{k})=\left\{\alpha\right\}\cup\mathrm{Z}(\bar{f_{1}},\ldots,\bar{f_{k}}),\,\mathrm{and}\\ \mathrm{Z}(g_{1},\ldots,g_{n})=\left\{\alpha\right\}\cup\mathrm{Z}(\bar{g_{1}},\ldots,\bar{g_{n}}).\end{array}$$

Full text

The theorem that any algebraic set in $n$-dimensional space is the intersection of $n$ hypersurfaces 111The problem dates back to Kronecker. Its rather dramatic story is briefly presented in [2]; for much more detailed vivid account consult N. Schappacher’s available online presentation Political Space Curves. has been proved independently by Storch ([1]), and Eisenbud and Evans ([2]); both short proofs are ring-theoretic, i.e., one reduces the number of generators of radical ideals.

In this note we examine closer the finite fields case of the problem. If just the number of equations needed to describe an algebraic set is in question, then the answer is immediate: it is easy to construct a single polynomial defining it. If, however, the nature of defining polynomials (e.g., their total degree) is to be preserved, this problem becomes more interesting.

It turns out that we can avoid dealing with rings; the vector space structure is sufficient and, as in the theorem cited above, our result again produces $n$ equations; moreover, we show that these new equations can be chosen to be linear combinations with scalar coefficients of the old ones, so, roughly speaking, they remain of the same type (see Corollaries 3 and 4, with accompanying examples), and our proof is surprisingly elementary.

We fix the following notation:

${\mathbb{F}_{q}}$ the finite field with $q$ elements;

$\mathrm{Map}(X,{\mathbb{F}_{q}})$ the vector space of all functions $f:X\to{\mathbb{F}_{q}}$ for a given set $X$;

$\mathrm{Z}(f_{1},\ldots,f_{k})$ the set of common zeros of $f_{1},\ldots,f_{k}\in\mathrm{Map}(X,{\mathbb{F}_{q}})$;

$\mathrm{Span}(f_{1},\ldots,f_{k})$ the subspace of $\mathrm{Map}(X,{\mathbb{F}_{q}})$ generated by $f_{1},\ldots,f_{k}$;

$\mathbb{A}^{n}(\mathbb{K})$ the affine $n$-space over a field $\mathbb{K}$;

$\mathbb{P}^{n}(\mathbb{K})$ the projective $n$-space over a field $\mathbb{K}$;

$\left[\alpha_{1}\colon\ldots\colon\alpha_{n+1}\right]$ a set of homogeneous coordinates for a point in $\mathbb{P}^{n}(\mathbb{K})$.

This theorem is best possible with respect to the cardinality of $X$. Indeed, we have the following.

We have two immediate corollaries of Theorem 1 of interest in algebraic geometry.

The space $\mathcal{F}$ can be, for example, a space of polynomials in $n$ variables of bounded total degree.

The space $\mathcal{F}$ can be a space of homogeneous polynomials in $n+1$ variables of bounded total degree, the space of quadratic (or higher degree) forms in $n+1$ variables, the space of diagonal forms in $n+1$ variables, etc.

Before we present the proofs of Theorem 1 and Proposition 2, we separately state their following ingredient.

Let $\mathbb{K}$ be an arbitrary field, and $n$ be a positive integer. Denote by $\mathcal{M}_{n}$ the set of all matrices in $M_{n,n+1}(\mathbb{K})$ in reduced row echelon form having the rank equal to $n$, by $N(M)$ the null space of a matrix $M$, by $\theta$ the zero vector in $\mathbb{K}^{n+1}$, and by $\sim$ the equivalence relation which identifies points lying on the same line through the origin.

[TABLE]

is bijective.

[TABLE]

Since matrices of the same size have equal null spaces if and only if they are row equivalent, the induced map

[TABLE]

is well-defined and injective. It is also surjective, since every vector subspace of $\mathbb{K}^{n+1}$ having dimension equal to $1$ is the null space of a matrix in $\mathcal{N}_{n}$.

Since the canonical map

[TABLE]

is bijective, the lemma follows. ∎

Denote

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By Lemma 5 every element $s$ of $S$ defines a unique matrix in $\mathcal{M}_{n}$; denote this matrix by $M_{s}$. Examine the set

[TABLE]

By Lemma 5 the number of elements in $\mathcal{M}_{n}$ equals the cardinality of $\mathbb{P}^{n}({\mathbb{F}_{q}})$, i.e., $\frac{q^{n+1}-1}{q-1}$. The number of elements in $S$ is at most the cardinality of $X$, i.e., $\frac{q^{n+1}-q}{q-1}$. Hence the cardinality of $T$ is at least $\frac{q^{n+1}-1}{q-1}-\frac{q^{n+1}-q}{q-1}=1$. So choose a matrix $M\in T$. Our $g_{1},\ldots,g_{n}$ are defined by

[TABLE]

Indeed, the inclusion $\mathrm{Z}(f_{1},\ldots,f_{n+1})\subset\mathrm{Z}(g_{1},\ldots,g_{n})$ is obvious, and by the definition of $T$ the set $\mathrm{Z}(g_{1},\ldots,g_{n})$ is disjoint from $X\setminus\mathrm{Z}(f_{1},\ldots,f_{n+1})$, i.e., $\mathrm{Z}(g_{1},\ldots,g_{n})\subset\mathrm{Z}(f_{1},\ldots,f_{n+1})$. ∎

In order to prove Proposition 2 we need the following.

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Put $B=G_{1}^{-1}I_{r,n,n}G_{3}^{-1}$ and $M=G_{3}I_{n,n,m}G_{2}^{-1}$. ∎

[TABLE]

We have $\mathrm{Z}(f_{1},\ldots,f_{n+1})=\emptyset$.

Let $g_{1},\ldots,g_{n}\in\mathrm{Span}(f_{1},\ldots,f_{n+1})$, i.e.,

[TABLE]

for some matrix $A\in M_{n,n+1}({\mathbb{F}_{q}})$. By Lemma 6 there exist a matrix $M\in M_{n,n+1}({\mathbb{F}_{q}})$ in reduced row echelon form having the rank equal to $n$, and a matrix $B\in M_{n,n}({\mathbb{F}_{q}})$ such that $A=BM$. Hence by Lemma 5 we get that there is $x\in X_{n}$ belonging to $\mathrm{Z}(g_{1},\ldots,g_{n})$. ∎

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For any positive integer $n$ we have

[TABLE]

So we apply Theorem 1 to get $\bar{g_{1}},\ldots,\bar{g_{n}}\in\mathrm{Span}(\bar{f_{1}},\ldots,\bar{f_{k}})$ such that $\mathrm{Z}(\bar{g_{1}},\ldots,\bar{g_{n}})=\mathrm{Z}(\bar{f_{1}},\ldots,\bar{f_{k}})$. Let $A\in M_{k,n}({\mathbb{F}_{q}})$ be such that

[TABLE]

Define $g_{1},\ldots,g_{n}\in\mathrm{Map}(\mathbb{P}^{n}({\mathbb{F}_{q}}),{\mathbb{F}_{q}})$ by

[TABLE]

We are done, since

[TABLE]

∎

Acknowledgements.

We are grateful to Grzegorz Banaszak and Bartosz Naskręcki for discussions and suggestions. We wish to thank an anonymous referee for many improvements; in particular, for suggesting the concise formulation and proof of Lemma 5.