Hilbert's Nullstellensatz and an Algorithm for Proving Combinatorial Infeasibility

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by  J. A. De Loera

Hilbert’s Nullstellensatz and an Algorithm
for Proving Combinatorial Infeasibility
J.A. De Loera1∗ J. Lee2 P. Malkin1∗ S. Margulies3
8 1Department of Mathematics, Univ. of California, Davis, California, USA
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0 2IBM T.J. Watson Research Center, Yorktown Heights, New York, USA
2 3Department of Computer Science, Univ. of California, Davis, California, USA

n
a
J

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2 Abstract

] Systemsofpolynomialequationsoveranalgebraically-closedfieldKcanbeusedtoconcisely
O
model many combinatorial problems. In this way, a combinatorial problem is feasible (e.g., a
C
graph is 3-colorable, hamiltonian, etc.) if and only if a related system of polynomial equations
h. hasasolutionoverK. Inthispaper,weinvestigateanalgorithmaimedatprovingcombinatorial
t infeasibility based on the observed low degree of Hilbert’s Nullstellensatz certificates for poly-
a
nomial systems arising in combinatorics and on large-scalelinear-algebracomputations over K.
m
Wereportonexperimentsbasedontheproblemofprovingthenon-3-colorabilityofgraphs. We
[
successfully solved graph problem instances having thousands of nodes and tens of thousands

1 of edges.
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1 Introduction
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3
.
1 It is well known that systems of polynomial equations over a field can yield small models of difficult
0
combinatorial problems. For example, it was first noted by D. Bayer that the 3-colorability of
8
0 graphs can be modeled via a system of polynomial equations [2]. More generally, one can easily
: prove the following:
v
i
X
Lemma 1.1 The graph G is k-colorable if and only if the zero-dimensional system of n+m equa-
r
a tions in n variables
xk 1 = 0, for every node i V(G),
i − ∈
xk−1+xk−2x + +x xk−2+xk−1 = 0, for every edge i,j E(G),
i i j ··· i j j { } ∈
has a complex solution. Moreover, the number of solutions equals the number of distinct k-colorings
multiplied by k!.
Although such polynomial system encodings have been used to prove combinatorial results (see
[1, 5] and references within), they have not been widely used for practical computation. The key
issue that we investigate here is the use of such polynomial systems to effectively decide whether a
graph, or other combinatorial structure, has a certain property captured by the polynomial system
and its associated ideal. We call this the combinatorial feasibility problem. We are particularly
∗Research supported in part by an IBM Open Collaborative Research Award and by NSFgrant DMS-0608785
1
interested in whether this can be accomplished in practice for large combinatorial structures such
as graphs with many nodes.
Certainly, usingstandardtools incomputational algebra such as Gro¨bnerbases, onecan answer
the combinatorial feasibility problem by simply solving the system of polynomials. Nevertheless, it
has been shown by experiments that current Gro¨bner bases implementations often cannot directly
solvepolynomialsystemswithhundredsofpolynomials. Thispaperproposesanotherapproachthat
relies instead on the nice low degree of the Hilbert’s Nullstellensatz for combinatorial polynomial
systems and on large-scale linear-algebra computation.
For a hard combinatorial problem (e.g., 3-colorability of graphs), we associate a system of poly-
nomial equations J = f (x) = 0,f (x)= 0,...,f (x) = 0 such that the system J has a solution
1 2 s
{ }
if and only if the combinatorial problem has a feasible solution. The Hilbert Nullstellensatz (see
e.g.,[4]) states that the system of polynomial equations has no solution over an algebraically-closed
field K if and only if there exist polynomials β ,...,β K[x ,...,x ] such that 1 = β f . Thus,
1 s 1 n i i
∈ P
if the polynomial system J has no solution, then there exists a certificate that J has no solution,
and thus a certificate that the combinatorial problem is infeasible.
The key idea that we explore in this article is to use the Nullstellensatz to generate a finite
sequence of linear algebra systems, of increasing size, which will eventually become feasible if and
only if the combinatorial problem is infeasible. Given a system of polynomial equations, we fix
a tentative degree k for the coefficient polynomials β in the certificates. We can decide whether
i
there is a Nullstellensatz certificate with coefficients of degree k by solving a system of linear
≤
equations over the field K whose variables are in bijection with the coefficients of the monomials
of the polynomials β ,...,β . If this linear system has a solution, we have found a certificate;
1 s
otherwise, we try a higher degree for the polynomials β . This process is guaranteed to terminate
i
because, for a Nullstellensatz certificate to exist, the degrees of the polynomials β cannot be more
i
than known bounds (see [8] and references therein). We explain the details of the algorithm, which
we call NulLA, in Section 2.
Our method can be seen as a general-field variation of work by Lasserre [9], Laurent [11] and
Parrilo [14]andmany others, whostudiedtheproblem of minimizingageneral polynomialfunction
f(x) over a real algebraic variety with finitely many points. Laurent proved that when the variety
consists of the solutions of a zero-dimensional radical ideal I, one can set up the optimization
problem min f(x) :x variety(I) as a finite sequence of semidefinite programs terminating with
{ ∈ }
the optimal solution (see [11]). There are two key observations that speed up practical calculations
considerably: (1) when dealing with feasibility, instead of optimization, linear algebra replaces
semidefinite programmingand (2) there are ways of controlling thelength of the sequence of linear-
algebra systems including finite field computation instead of calculations over the reals and the
reduction of matrix size by symmetries. See Section 3 for details.
Our algorithm has good practical performance and numerical stability. Although known the-
oretical bounds for degrees of the Nullstellensatz coefficients are doubly-exponential in the size
of the polynomial system (and indeed there exist examples that attain such a large bound and
make NulLA useless in general), our experiments demonstrate that often very low degrees suffice
for systems of polynomials coming from graphs. We have implemented an exact-arithmetic linear
system solver optimized for these Nullstellensatz-based systems. We performed many experiments
using NulLA, focusing on the problem of deciding 3-colorability of graphs (note that the method
is applicable to any combinatorial problem as long as we know a polynomial system that encodes
it). We conclude with a report on these experiments in Section 4.
2
2 The Nullstellensatz Linear Algebra (NulLA) Algorithm
RecallthatHilbert’sNullstellensatzstatesthatasystemofpolynomialequationsf (x) = 0,f (x) =
1 2
0,...,f (x) = 0, where f K[x ,...,x ] and K is an algebraically closed field, has no solution in
s i 1 n
Kn if and only if there ex∈ist polynomials β ,...,β K[x ,...,x ] such that 1 = β f [4]. The
1 s 1 n i i
∈ P
polynomial identity 1 = β f is called a Nullstellensatz certificate. We say a Nullstellensatz
i i
P
certificate has degree d if max deg(β ) = d.
i
{ }
The Nullstellensatz Linear Algebra (NulLA) algorithm takes as input a system of polynomial
equationsandoutputseitherayes answer, ifthesystemofpolynomialequationshasasolution, ora
no answer, alongwithaNullstellensatzinfeasibility certificate, ifthesystemhasnosolution. Before
stating the algorithm in pseudocode, let us completely clarify the connection to linear algebra.
Suppose for a moment that the polynomial system is infeasible over K and thus there must exist
a Nullstellensatz certificate. Assume further that an oracle has told us the certificate has degree
d but that we do not know the actual coefficients of the degree d polynomials β . Thus, we have
i
the polynomial identity 1 = β f . If we expand the identity into monomials, the coefficients of a
i i
P
monomial are linear expressions in the coefficients of the β . Since two polynomials over a field are
i
identical precisely whenthe coefficients of correspondingmonomials are identical, from theidentity
1 = β f , we get a system of linear equations whose variables are the coefficients of the β . Here
i i i
P
is an example:
Example 2.1 Consider the polynomial system x2 1 = 0,x +x = 0,x +x = 0,x +x = 0.
1 − 1 2 1 3 2 3
Clearly this system has no complex solution, and we will see that it has a Nullstellensatz certificate
of degree one.
1= (c x +c x +c x +c )(x2 1)+(c x +c x +c x +c )(x +x )
0 1 1 2 2 3 3 1− 4 1 5 2 6 3 7 1 2
| {βz1 }| {f1z } | {βz2 }| {f2z }
+(c x +c x +c x +c )(x +x )+(c x +c x +c x +c )(x +x ).
8 1 9 2 10 3 11 1 3 12 1 13 2 14 3 15 2 3
| {βz3 }| {f3z } | {βz4 }| {f4z }
Expanding the tentative Nullstellensatz certificate into monomials and grouping like terms, we
arrive at the following polynomial equation:
1 = c x3+c x2x +c x2x +(c +c +c )x2+(c +c )x2+(c +c )x2+
0 1 1 1 2 2 1 3 3 4 8 1 5 13 2 10 14 3
(c +c +c +c )x x +(c +c +c +c )x x +(c +c +c +c )x x +
4 5 9 12 1 2 6 8 10 12 1 3 6 9 13 14 2 3
(c +c c )x +(c +c c )x +(c +c c )x c .
7 11 0 1 7 15 1 2 11 15 2 3 3
− − − −
From this, we extract a system of linear equations. Since a Nullstellensatz certificate is identically
one, all monomials except the constant term must be equal to zero; namely:
c = 0, c = 0, ..., c +c +c = 0, c +c c = 0, c = 1.
0 1 3 4 8 11 15 2 3
− −
By solving the system of linear equations, we reconstruct the Nullstellensatz certificate from the
solution. Indeed
1 1 1
1 = ( 1)(x2 1)+ x (x +x ) x (x +x )+ x (x +x )
− 1 − 2 1 1 2 − 2 1 2 3 2 1 1 3
Now, of course in general, one does not know the degree of the Nullstellensatz certificate in
advance. What one can do is to start with a tentative degree, say start at degree one, produce
3
the corresponding linear system, and solve it. If the system has a solution, then we have found a
Nullstellensatz certificate demonstratingthattheoriginal inputpolynomials donothave acommon
root. Otherwise,weincrementthedegreeuntilwecanbesurethattherewillnotbeaNullstellensatz
certificate atall, andthuswecanconcludethesystemofpolynomialshasasolution. Thenumberof
iterations of the above steps determines the running time of NulLA. For this, there are well-known
upperboundsonthedegreeoftheNullstellensatzcertificate[8]. Theseupperboundsforthedegrees
of the coefficients β in the Hilbert Nullstellensatz certificates for general systems of polynomials
i
are doubly-exponential in the number of input polynomials and their degree. Unfortunately, these
bounds are known to be sharp for some specially-constructed systems. Although this immediately
saysthatNulLAisnotpracticalforarbitrarypolynomialsystems,wehaveobservedinpracticethat
polynomial systems for combinatorial questions are extremely specialized, and the degree growth
is often very slow — enough to deal with large graphs or other combinatorial structures. Now we
describe NulLA in pseudocode:
********************************************
ALGORITHM (Nullstellensatz Linear Algebra (NulLA) Algorithm)
INPUT: A system of polynomial equations F = f1(x)=0,...,fs(x)=0
OUTPUT: yes, if F has solution, else no with a{Nullstellensatz certificat}e of infeasibility.
Set d=1.
Set K equal to the known upper bounds on degree of Nullstellensatz for F (see e.g., [8])
while d K do
cert≤←Psi=1βifi (where βi are polynomials of degree d,
with unknowns for their coefficients).
Extract a system of linear equations from cert with columns corresponding to unknowns,
and rows corresponding to monomials.
Solve the linear system.
if the linear system is consistent then
cert←Psi=1βifi (with unknowns in βi replaced with linear system solution values.)
print “The system of equations F is infeasible.”
return no with cert.
else
Set d:=d+1.
end if
end while
print “The system of equations F is feasible.”
return yes.
********************************************
This opens several theoretical questions. It is natural to ask about lower bounds on the degree
of the Nullstellensatz certificates. Little is known, but recently it was shown in [5], that for the
problem of deciding whether a given graph G has an independent set of a given size, a minimum-
degree Nullstellensatz certificate for the non-existence of an independent set of size greater than
α(G) (the size of the largest independent set in G) has degree equal to α(G), and it is very dense;
specifically, it contains at least one term per independent set in G. For polynomial systems coming
from logic there has also been an effort to show degree growth in related polynomial systems (see
[3, 6] and the references therein). Another question is to provide tighter, more realistic upper
bounds for concrete systems of polynomials. It is a challenge to settle it for any concrete family of
polynomial systems.
4
3 Four mathematical ideas to optimize NulLA
Since we are interested in practical computational problems, it makes sense to explore refinements
and variations that make NulLA robust and much faster for concrete challenges. The main com-
putational component of NulLA is to construct and solve linear systems for finding Nullstellensatz
certificates of increasing degree. These linear systems are typically very large for reasonably-sized
problems,even for certificate degrees as low as four,which can producelinear systems withmillions
of variables (see Section 4). Furthermore, the size of the linear system increases dramatically with
the degree of the certificate. In particular, the number of variables in the linear system to find a
Nullstellensatz certificate of degree d is precisely s n+d where n is the number of variables in the
(cid:0) d (cid:1)
polynomial system and s is the number of polynomials. Note that n+d is the number of possible
(cid:0) d (cid:1)
monomials of degree d or less. Also, the number of non-zero entries in the constraint matrix is
precisely M n+d where M is the sum over the number of monomials in each polynomial of the
(cid:0) d (cid:1)
system.
Forthisreason,inthissection,weexploremathematicalapproachesforsolvingthelinearsystem
more efficiently and robustly, for decreasing the size of the linear system for a given degree, and
for decreasing the degree of the Nullstellensatz certificate for infeasible polynomial systems thus
significantlyreducingthesizeofthelargestlinearsystemthatweneedtosolvetoproveinfeasibility.
Note that these approaches to reduce the degree do not necessarily decrease the available upper
bound on the degree of the Nullstellensatz certificate required for proving feasibility.
It is certainly possible to significantly decrease the size of the linear system by preprocessing
the given polynomial system to remove redundant polynomial equations and also by preprocessing
thelinear system itself to eliminate many variables. For example, in thecase of 3-coloring problems
for connected graphs, since (x3+1) = (x3+1)+(x +x )(x2 +x x +x2), we can remove all but
i j i j i i j j
one of the vertex polynomials by tracing paths through the graph. However, preprocessing alone
is not sufficient to enable us to solve some large polynomial systems.
The mathematical ideas we explain in the rest of this section can be applied to arbitrary
polynomial systems for which we wish to decide feasibility, but to implement them, one has to look
for the right structures in the polynomials.
3.1 NulLA over Finite Fields
The first idea is that, for combinatorial problems, one can often carry out calculations over finite
fields instead of relying on unstable floating-point calculations were we to be working over the reals
orcomplexnumbers. Weillustratethiswiththeproblemofdecidingwhetherthevertices ofagraph
permit a proper 3-coloring. The following encoding (a variation of [2] over the complex numbers)
allows us to compute over F , which is robust and much faster in practice:
2
Lemma 3.1 The graph G is 3-colorable if and only if the zero-dimensional system of equations
x3+1 = 0, for every node i V(G),
i ∈
x2+x x +x2 = 0, for every edge i,j E(G) ,
i i j j { } ∈
has a solution over F , the algebraic closure of F .
2 2
Before we prove Lemma 3.1, we introduce a convenient notation: Let α be an algebraic element
over F such that α2 + α + 1 = 0. Thus, although x3 + 1 has only one root over F , since
2 i 2
5
x3 +1 = (x +1)(x2 +x +1), the polynomial x3 +1 has three roots over F , which are 1,α and
i i i i i 2
α+1.
Proof: If the graph G is 3-colorable, simply map the three colors to 1,α and α+1. Clearly, the
vertex polynomial equations x3+1= 0 are satisfied. Furthermore, given an edge i,j , x +x = 0
i { } i j 6
since variable assignments correspond to a proper 3-coloring and adjacent vertices are assigned
different roots. This implies that x3 + x3 = (x + x )(x2 +x x +x2) = 1+ 1 = 0. Therefore,
i j i j i i j j
x2+x x +x2 = 0 and the edge polynomial equations are satisfied.
i i j j
Conversely, supposethat there exists a solution to the system of polynomial equations. Clearly,
every vertex is assigned either 1,α or α + 1. We will show that adjacent vertices are assigned
different values. Our proof is by contradiction: Assume that two adjacent vertices i,j are assigned
the same value β. Then, 0 = x2 + x x + x2 = β2 + β2 + β2 = 3β2 = 0. Therefore, adjacent
i i j j 6
vertices are assigned different roots, and a solution to the system corresponds directly to a proper
3-coloring. 2
We remark that this result can be extended to k-colorability and F , when q is relatively prime
q
to k. The following computational lemma will allow us to certify graph non-3-colorability very
rapidly over F instead of working over its algebraic closure.
2
Lemma 3.2 Let K be a field and K its algebraic closure. Given f ,f ,...,f K[x ,...,x ], there
1 2 s 1 n
∈
exists a Nullstellensatz certificate 1 = β f where β K[x ,...,x ] if and only if there exists a
i i i 1 n
Nullstellensatz certificate 1 = β′f wPhere β′ K[x ,.∈..,x ].
P i i i ∈ 1 n
Proof: If there exists a Nullstellensatz certificate 1 = β f where β K[x ,...,x ], via
i i i 1 n
NulLA, construct the associated linear system and solve. SPince f K[x ,...∈,x ], the coefficients
i 1 n
∈
in the linear system will consist only of values in K. Thus, solving the linear system relies only
on computations in K, and if the free variables are chosen from K instead of K, the resulting
Nullstellensatz certificate 1 = β′f has β′ K[x ,...,x ]. The reverse implication is trivial. 2
P i i i ∈ 1 n
Therefore, we have the following corollary:
Corollary 3.3 A graph G is non-3-colorable if and only if there exists a Nullstellensatz certificate
1 = β f where β F [x ,...,x ] where the polynomials f F [x ,...,x ] are as defined in
i i i 2 1 n i 2 1 n
P ∈ ∈
Lemma 3.1.
This corollary enables us to compute over F , which is extremely fast in practice (see Section
2
4).
Finally, the degree of Nullstellensatz certificates necessary to prove infeasibility can be lower
over F than over the rationals. For example, one can prove that over the rationals, every odd-
2
wheel has a minimum non-3-colorability certificate of degree four [5]. However, over F , every
2
odd-wheel has a Nullstellensatz certificate of degree one. Therefore, not only are the mathematical
computations more efficient over F as compared to the rationals, but the algebraic properties of
2
the certificates themselves are sometimes more favorable for computation as well.
3.2 NulLA with symmetries
Let us assume that the input polynomial system F = f ,...,f has maximum degree q and that
1 s
{ }
n is the number of variables present. As we observed in Section 2, for a given fixed positive integer
d serving as a tentative degree for the Nullstellensatz certificate, the Nullstellensatz coefficients
6
come from the solution of a system of linear equations. We now take a closer look at the matrix
equation M y = b defining the system of linear equations. First of all, the matrix M has
F,d F,d F,d
one row per monomial xα of degree less than or equal to q+d on the n variables and one column
per polynomial of the form xδf , i.e., the product of a monomial xδ of degree less than or equal to
i
d and a polynomial f F. Thus, M = (M ) where M equals the coefficient of the
i ∈ F,d xα,xδfi xα,xδfi
monomial xα in the polynomialxδf . Thevariable y has one entry for every polynomial of theform
i
xδf denoted y , and the vector b has one entry for every monomial xα of degree less than or
i xδfi F,d
equal to q+d where (bF,d)xα = 0 if α = 0 and (bF,d)1 = 1.
6
Example 3.4 Consider the complete graph K . The shape of a degree-one Hilbert Nullstellensatz
4
certificate over F for non-3-colorability is as follows:
2
1=c (x3+1)
0 1
+(c1 x +c2 x +c3 x +c4 x )(x2+x x +x2)+(c1 x +c2 x +c3 x +c4 x )(x2+x x +x2)
12 1 12 2 12 3 12 4 1 1 2 2 13 1 13 2 13 3 13 4 1 1 3 3
+(c1 x +c2 x +c3 x +c4 x )(x2+x x +x2)+(c1 x +c2 x +c3 x +c4 x )(x2+x x +x2)
14 1 14 2 14 3 14 4 1 1 4 4 23 1 23 2 23 3 23 4 2 2 3 3
+(c1 x +c2 x +c3 x +c4 x )(x2+x x +x2)+(c1 x +c2 x +c3 x +c4 x )(x2+x x +x2)
24 1 24 2 24 3 24 4 2 2 4 4 34 1 34 2 34 3 34 4 3 3 4 4
Note thatwehavepreprocessedthecertificate byremovingtheredundantpolynomialsx3+1where
i
i = 1 and removing some variables that we know a priori can be set to zero, which results in a
6
matrix with less columns. As we explained in Section 2, this certificate gives a linear system of
equations in the variables c and ck (note that k is a superscriptand not an exponent). This linear
0 ij
system can be captured as the matrix equation M c =b where the matrix M is as follows.
F,1 F,1 F,1
c c1 c2 c3 c4 c1 c2 c3 c4 c1 c2 c3 c4 c1 c2 c3 c4 c1 c2 c3 c4 c1 c2 c3 c4
0 12 12 12 12 13 13 13 13 14 14 14 14 23 23 23 23 24 24 24 24 34 34 34 34
1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
x3 1 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1
x2x 0 1 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 2
x2x 0 0 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 3
x2x 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
1 4
x x2 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0
1 2
x x x 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
1 2 3
x x x 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0
1 2 4
x x2 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0
1 3
x x x 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0
1 3 4
x x2 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0
1 4
x3 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0
2
x2x 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0
2 3
x2x 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0
2 4
x x2 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0
2 3
x x x 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0
2 3 4
x x2 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 1 0 0
2 4
x3 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0
3
x2x 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1
3 4
x x2 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 1
3 4
x3 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1
4
Certainly the matrix M is rather large already for small systems of polynomials. The main
F,d
point of this section is to demonstrate how to reduce the size of the matrix by using a group action
on the variables, e.g., using symmetries or automorphisms in a graph. Suppose we have a finite
permutation group G acting on the variables x ,...,x . Clearly G induces an action on the set of
1 n
7
monomials with variables x ,x ,...,x of degree t. We will assume that the set F of polynomials
1 2 n
is invariant under the action of G, i.e., g(f ) F for each f F. Denote by xδ, the monomial
i i
xδ1xδ2...xδn, a monomial of degree δ + δ +∈ + δ . Deno∈te by Orb(xα),Orb(xδf ) the orbit
1 2 n 1 2 ··· n i
under G of monomial xα and, respectively, the orbit of the polynomial obtained as the product of
the monomial xδ and the polynomial f F.
i
∈
We now introduce a new matrix equation M¯ y¯=¯b . The rows of the matrix M¯ are
F,d,G F,d,G F,d,G
indexed by the orbits of monomials Orb(xα) where xα is a monomial of degree less than or equal to
q+d, and the columns of M¯ are indexed by the orbits of polynomials Orb(xδf ) where f F
F,d,G i i
∈
and the degree of the monomial xδ less than or equal to d. Then, let M¯ = (M¯ )
F,d,G Orb(xα),Orb(xδfi)
where
M¯ = M .
Orb(xα),Orb(xδfi) X xα,xγfj
xγfj∈Orb(xδfi)
Note that M = M for all g G meaning that the coefficient of the monomial xα in
xα,xδfi g(xα),g(xδfi) ∈
the polynomial xδf is the same as the coefficient of the monomial g(xα) in the polynomial g(xδf ).
i i
So,
M = M for all xd Orb(xα),
X xα,xγfj X xd,xγfj ∈
xγfj∈Orb(xδfi) xγfj∈Orb(xδfi)
and thus, M¯ is well-defined. We call the matrix M¯ the orbit matrix. The
Orb(xα),Orb(xδfi) F,d,G
variable y¯ has one entry for every polynomial orbit Orb(xδf ) denoted y¯ . The vector ¯b
i Orb(xδfi) F,d
has one entry for every monomial orbit Orb(xα), and let (¯bF,d)Orb(xα) = (bF,d)xα = 0 if α = 0 and
(¯b ) = (b ) = 1. The main result in this section is that, under some assumpti6ons, the
F,d Orb(1) F,d 1
system of linear equations M¯ y¯=¯b has a solution if and only if the larger system of linear
F,d,G F,d,G
equations M y = b has a solution.
F,d F,d
Theorem 3.5 LetKbeanalgebraically-closed field. Considerapolynomial systemF = f ,...,f
1 s
{ }
K[x ,...,x ] and a finite group of permutations G S . Let M ,M¯ denote the matrices
1 n n F,d F,d,G
⊂ ⊂
defined above. Suppose that the polynomial system F is closed under the action of the group G
permuting the indices of variables x ,...,x . Suppose further that the order of the group G and
1 n
| |
the characteristic of the field K are relatively prime. The degree d Nullstellensatz linear system
of equations M y = b has a solution over K if and only if the system of linear equations
F,d F,d
M¯ y¯=¯b has a solution over K.
F,d,G F,d,G
Proof: To simplify notation, let M = M , b = b , M¯ = M¯ and ¯b =¯b . First, we show
F,d F,d F,d,G F,d,G
that if the linear system My = b has a solution, then there exists a symmetric solution y of the
linearsystemMy = bmeaningthaty isthesameforallxδf inthesameorbit, i.e., y = y
xδfi i xγfj xδfi
for all xγf Orb(xδf ). The converse is also trivially true.
j i
∈
Sincethe rows and columns of thematrix M are labeled by monomials xα and polynomialsxδf
i
respectively, we can also think of the group G as acting on the matrix M, permuting the entries of
M, where g(M) = M . Moreover, since M = M for all g G, we
g(xα),g(xδfi) xα,xδfi xα,xδfi g(xα),g(xδfi) ∈
must have g(M) = M, so the matrix M is invariant under the action of the group G. Also, since
the entries of the variable y are labeled by polynomials of the form xαf , we can also think of the
i
group G as acting on the vector y, permuting the entries of the vector y, i.e., applying g G to y
∈
gives the permuted vector g(y) where g(y) = y . Similarly, G acts on the vector b, and in
g(xδfi) xδfi
particular, g(b) = b. Next, we show that if My = b, then Mg(y) = b for all g G. This follows
∈
since
My = b g(My) = g(b) g(M)g(y) =b Mg(y) = b,
⇒ ⇒ ⇒
8
for all g G. Now, let
∈ 1
y′ = g(y).
G X
| | g∈G
Note we need that G is relatively primeto the characteristic of the field K so that G is invertible.
| | | |
Then,
1 1
My′ = Mg(y) = b = b,
G X G X
| | g∈G | | g∈G
so y′ is a solution. Also, y′ = 1 y , so y′ = y′ for all xγf Orb(xδf ).
xδfi |G| Pg∈G g(xδfi) xδfi xγfj j ∈ i
Therefore, y′ is a symmetric solution as required.
Now, assume that there exists a solution of My = b. By the above argument, we can assume
that the solution is symmetric, i.e., y = y where g(xδf ) = xγf for some g G. From this
symmetric solution of My = b, we canxδfifind axsγofljution of M¯y¯=i ¯b by sejtting ∈
y¯ = y .
Orb(xδfi) xδfi
To show this, we check that (M¯y¯) =¯b for every monomial xα.
Orb(xα) Orb(xα)
(M¯y¯) = M¯ y¯
Orb(xα) X xα,Orb(xδfi) Orb(xδfi)
allOrb(xδfi)
 
= M y¯
X X xα,xγfj Orb(xδfi)
allOrb(xδfi)xγfj∈Orb(xδfi) 
 
= M y
X X xα,xγfj xγfj
allOrb(xδfi)xγfj∈Orb(xδfi) 
= M y
X xα,xδfi xδfi
allxδfi
= (My)xα.
Thus, (M¯y¯)Orb(xα) =¯bOrb(xα) since (My)xα = bxα =¯bOrb(xα).
Next, weestablish theconverse more easily. Recall that the columns of M¯ are labeled by orbits.
If there is a solution for M¯y¯=¯b, then to recover a solution of My = b, we set
y = y¯ .
xδfi Orb(xδfi)
Note that y is a symmetric solution. Using the same calculation as above, we have that (My)xα =
(M¯y¯) , and thus, My = b. 2
Orb(xα)
Example 3.6 (Continuation of Example 3.4) Nowconsidertheaction ofthesymmetrygroup
G generated by the cycle (2,3,4) (a cyclic group of order three). The permutation of variables
permutes the monomials and yields a matrix M . We have now grouped together monomials
F,1,G
and terms within orbit blocks in the matrix below. The blocks will be later replaced by a single
entry, shrinking the size of the matrix.
9
c c1 c1 c1 c2 c3 c4 c3 c4 c2 c4 c2 c3 c1 c1 c1 c2 c3 c4 c2 c3 c4 c2 c3 c4
0 12 13 14 12 13 14 12 13 14 12 13 14 23 34 24 23 34 24 24 23 34 34 24 23
1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
x3 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1
x2x 0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 2
x2x 0 0 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
1 3
x2x 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 4
x x2 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0
1 2
x x2 0 0 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0
1 3
x x2 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0
1 4
x x x 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0
1 2 3
x x x 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0
1 2 4
x x x 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0
1 3 4
x3 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0
2
x3 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0
3
x3 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0
4
x2x 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0
2 3
x2x 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1
3 4
x x2 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0
2 4
x2x 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 1
2 4
x x2 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 0 0
2 3
x x2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 0
3 4
x x x 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1
2 3 4
The action of the symmetry group generated by the cycle (2,3,4) yields an orbit matrix M¯
F,q,G
of about a third the size of the original one:
c¯ c¯1 c¯2 c¯3 c¯4 c¯1 c¯2 c¯2 c¯2 c¯ c¯1 c¯2 c¯3 c¯4 c¯1 c¯2 c¯2 c¯2
0 12 12 12 12 23 23 24 34 0 12 12 12 12 23 23 24 34
Orb(1) 1 0 0 0 0 0 0 0 0 Orb(1) 1 0 0 0 0 0 0 0 0
Orb(x3) 1 3 0 0 0 0 0 0 0 Orb(x3) 1 1 0 0 0 0 0 0 0
1 1
Orb(x2x ) 0 1 1 1 1 0 0 0 0 Orb(x2x ) 0 1 1 1 1 0 0 0 0
1 2 1 2
Orb(x x2) 0 1 1 0 0 2 0 0 0 (mod2) Orb(x x2) 0 1 1 0 0 0 0 0 0
1 2 1 2
Orb(x x x ) 0 0 0 1 1 1 0 0 0 ≡ Orb(x x x ) 0 0 0 1 1 1 0 0 0
1 2 3 1 2 3
Orb(x3) 0 0 1 0 0 0 1 1 0 Orb(x3) 0 0 1 0 0 0 1 1 0
2 2
Orb(x2x ) 0 0 0 1 0 0 1 1 1 Orb(x2x ) 0 0 0 1 0 0 1 1 1
2 3 2 3
Orb(x2x ) 0 0 0 0 1 0 1 1 1 Orb(x2x ) 0 0 0 0 1 0 1 1 1
2 4 2 4
Orb(x x x ) 0 0 0 0 0 0 0 0 3 Orb(x x x ) 0 0 0 0 0 0 0 0 1
2 3 4 2 3 4
If G is not relatively prime to the characteristic of the field K, then it is still true that, if
M¯y =|¯b|has a solution, then My = b has a solution. Thus, even if G is not relatively prime to
| |
the characteristic of the field K, we can still prove that the polynomial system F is infeasible by
finding a solution of the linear system M¯y =¯b.
3.3 Reducing the Nullstellensatz degree by appending polynomial equations
We have discovered that by appending certain valid but redundant polynomial equations to the
system of polynomial equations described in Lemma 3.1, we have been able to decrease the degree
of the Nullstellensatz certificate necessary to prove infeasibility. A valid but redundant polynomial
equation is any polynomial equation g(x) = 0 that is true for all the zeros of the polynomial
system f (x) = 0,...,f (x) = 0, i.e., g √I, the radical ideal of I, where I is the ideal generated
1 s
∈
by f ,...,f . Technically, we only require that g(x) = 0 holds for at least one of zeros of the
1 s
polynomial system f (x) = 0,...,f (x) = 0 if a zero exists. We refer to a redundant polynomial
1 s
10