Proof of the combinatorial nullstellensatz over integral domains in the spirit of Kouba

0.1 MB·English

by  Peter Christian Heinig

PROOF OF THE COMBINATORIAL NULLSTELLENSATZ OVER
INTEGRAL DOMAINS IN THE SPIRIT OF KOUBA
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PETERHEINIG
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2 Abstract. Itisshownthatbyeliminatingdualitytheoryofvectorspacesfrom
arecentproofofKouba(O.Kouba,AdualitybasedproofoftheCombinatorial
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Nullstellensatz. Electron. J. Combin. 16 (2009), #N9) one obtains a direct
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J proof of the nonvanishing-version of Alon’s Combinatorial Nullstellensatz for
polynomials over an arbitrary integral domain. The proof relies on Cramer’s
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rule and Vandermonde’s determinant to explicitly describe a map used by

Koubaintermsofcofactorsofacertainmatrix.
]
C That the Combinatorial Nullstellensatz is true over integral domains is a
well-knownfactwhichisalreadycontainedinAlon’sworkandemphasizedin
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recentarticlesofMicha lekandSchauz;thesolepurposeofthepresentnoteis
. topointoutthatnotonlyisitnotnecessarytoinvokedualityofvectorspaces,
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butbynotdoingsooneeasilyobtainsamoregeneralresult.
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MathematicsSubjectClassification2010: 13G05,15A06
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v 1. Introduction
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The Combinatorial Nullstellensatz is a very useful theorem (see [1]) about mul-
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5 tivariatepolynomialsoveranintegraldomainwhichbearssomeresemblancetothe
0 classical Nullstellensatz of Hilbert.
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1 Theorem1(Alon,CombinatorialNullstellensatz(ideal-containment-version),The-
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orem 1.1 in [1]). Let K be a field, R ⊆K a subring, f ∈R[x ,...,x ], S ,...,S
0 1 n 1 n
1 arbitrary nonempty subsets of K, and gi := s∈Si(xi−s) for every 1 ≤ i ≤ n. If
v: f(s1,...,sn)=0 for every (s1,...,sn)∈S1×Q···×Sn, then there exist polynomi-
als h ∈ R[x ,...,x ] with the property that deg(h ) ≤ deg(f)−deg(g ) for every
i i 1 n i i
X 1≤i≤n, and f = n h g .
i=1 i i
ar Theorem2(Alon,CPombinatorialNullstellensatz(nonvanishing-version),Theorem
1.2 in [1]). Let K be a field, R ⊆ K a subring, and f ∈ R[x ,...,x ]. Let c·
1 n
xd1···xdn be a term in f with c6=0 whose degree d +···+d is maximum among
1 n 1 n
all degrees of terms in f. Then every product S ×···×S , where each S is an
1 n i
arbitrary finite subset of R satisfying |S | = d + 1, contains at least one point
i i
(s ,...,s ) with f(s ,...,s )6=0.
1 n 1 n
Threecommentsareinorder. First,talkingaboutsubringsofafieldisequivalent
to talking about integral domains: every subring of a field clearly is an integral
domain, and, conversely, every integral domain R is (isomorphic to) a subring of
its field of fractions Quot(R). Second, strictly speaking, rings are mentioned in [1]
only in Theorem 1, but Alon’s proof in [1] of Theorem 2 is valid for polynomials
The author was supported by a scholarship from the Max Weber-Programm Bayern and by
theENBgraduateprogramTopMath.
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over integral domains as well. Third, it is intended that the S are allowed to be
i
subsetsofK inTheorem1butrequiredtobesubsetsofRinTheorem2, sincethis
is the slightly stronger formulation: if Theorem 2 is true as it is formulated here,
then by invoking it with R = K and by viewing an f ∈ R[x ,...,x ], R being a
1 n
subring of K, as a polynomial in K[x ,...,x ], it is true as well with the S being
1 n i
allowed to be arbitrary subsets of K.
In[1],Theorem2wasdeducedfromTheorem1. In[3],Koubagaveabeautifully
simple and direct proof of the nonvanishing-version of the Combinatorial Nullstel-
lensatz,bypassingtheuseoftheideal-containment-version. Kouba’sargumentwas
restricted to the case of polynomials over a field and at one step applied a suitably
chosen linear form on the vector space K[x ,...,x ] to the given polynomial f in
1 n
Theorem 2.
However,forKouba’sideatowork,itisnotnecessarytohaverecoursetoduality
theory of vector spaces and in the following section it will be shown how to make
Kouba’s idea work without it, thus obtaining a direct proof of the full Theorem 2.
Finally, two relevant recent articles ought to be mentioned. A very short di-
rect proof of Theorem 2 was given by Micha lek in [5] who explicitly remarks that
the proof works for integral domains as well. Moreover, the differences s−s′ :
{s,s′} ∈ Sk in the proof below play a similar role in Micha lek’s proof(cid:8). In [6],
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Schauz ob(cid:0)tai(cid:1)n(cid:9)ed far-reaching generalizations and sharpenings of Theorem 2, ex-
pressly working with integral domains and generalizations thereof throughout the
paper.
2. Proof of Theorem 2
The proof of the Theorem 2 will be based on the following simple lemma.
Lemma 3. Let R be an integral domain. Let S ={s ,...,s }⊆R be an arbitrary
1 m
finite subset. Then there exist elements λ(S),...,λ(S) of R such that
1 m
λ(S)·(1,s ,s2,...,sm−1)+···+λ(S)·(1,s ,s2 ,...,sm−1)
1 1 1 1 m m m m
=(0,0,0,...,0, (s −s )). (1)
i j
1≤iY<j≤m
Proof. Let [m] := {1,...,m}. Define b to be the right-hand side of the claimed
equation,takenasacolumnvector,andletA=(aij)(i,j)∈[m]2 betheVandermonde
matrix defined by a := si−1. Then the statement of the lemma is equivalent to
ij j
the existence of a solution λ(S) ∈Rm of the system of linear equations Aλ(S) =b.
By the well-known formula for the determinant of a Vandermonde matrix (see [4],
Ch. XIII, §4, example after Prop. 4.10), det(A)= (s −s ).
1≤i<j≤m i j
Since S is a set, all factors of this product are noQnzero, and since R has no zero
divisors, the determinant is therefore nonzero as well. Now let α be the cofactors
ij
of A, i.e. α := (−1)i+jdet(A(ij)), where A(ij) is the (m−1)×(m−1) matrix
ij
obtained from A by deleting the i-th row and the j-th column (see [2], Ch. IX,
§3, before Lemma 1). By Cramer’s rule (see Ch. IX, §3, Corollary 2 of Theorem 6
in [2] or Theorem 4.4 in [4]), for every j ∈[m],
m
det(A)·λ(S) = α b .
j ij i
Xi=1
NULLSTELLENSATZ OVER INTEGRAL DOMAINS 3
Using b = det(A), b = 0 for every 1 ≤ i < m, and the commutativity of an
m i
integral domain, this reduces to
det(A)· λ(S)−α =0.
j mj
(cid:0) (cid:1)
Hence, since det(A) 6= 0 and R has no zero divisors, if follows that the cofactors
λ(S) =α ∈R provide explicit elements with the desired property. (cid:3)
j mj
Usingthislemma, Kouba’sargumentmaynowbecarriedoutwithoutchangein
the setting of integral domains.
Proof of Theorem 2. Let R be an arbitrary integral domain and f ∈R[x ,...,x ]
1 n
be an arbitrary polynomial. Let d ,...,d ∈ N be the exponents of a term
1 n ≥0
c · xd1···xdn with c 6= 0 which has maximum degree in f. For each k ∈ [n],
1 n
choose an arbitrary finite subset S ⊆ R and apply Lemma 3 with S = S and
k k
m=|S|=d +1 to obtain a family of elements (λ(Sk)) of R (where in order
k sk sk∈Sk
to avoid double indices the coefficients λ are now being indexed by the elements of
S directly, not by an enumeration of each S ) with the property that
k k
λ(Sk)·sℓ =0 for every ℓ∈{0,...,d −1}, (2)
sk k k
skX∈Sk
λ(Sk)·sdk = (s−s′)=:r ∈R\{0}. (3)
sk k k
skX∈Sk {s,s′Y}∈(S2k)
Using the coefficient families (λ(Sk)) , define, `a la Kouba, the map
sk sk∈Sk
Φ: R[x ,...,x ]−→R
1 n
g 7−→ λ(S1)···λ(Sn)·g(s ,...,s ). (4)
s1 sn 1 n
(s1,...,sn)X∈S1×···×Sn
Due to the commutativity of an integral domain, Φ is an R-linear form on the
R-module R[x ,...,x ], hence its value Φ(f) on a polynomial f can be evaluated
1 n
termwise as
Φ(f)= c·Φ(t). (5)
c·tatXerminf
If t=c·xd′1···xd′n is an arbitrary term in R[x ,...,x ], then
1 n 1 n
Φ(t)=c·Φ(xd′1···xd′n)=c· λ(S1)···λ(Sn)·sd′1···sd′n
1 n s1 sn 1 n
(s1,...,sn)X∈S1×···×Sn
=c· ··· λ(S1)···λ(Sn)·sd′1···sd′n
s1 sn 1 n
sX1∈S1 snX∈Sn
n
=c· λ(Sk)sd′k, (6)
(cid:18) sk k (cid:19)
kY=1 skX∈Sk
where in the last step again use has been made of the commutativity of an integral
domain. By (6) and (2) it follows that for every term t, if there is at least one
exponent d′ with d′ < d , then Φ(t) = 0. Moreover, by the choice of the term
i i i
c·xd1···xdn, every term c′ ·xd′1···xd′n of f which is different from the term c·
1 n 1 n
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xd1···xdn must, even if it has itself maximum degree in f, contain at least one
1 n
exponent d′ with d′ <d . Therefore
i i i
λ(S1)···λ(Sn)·f(s ,...,s )(=4)Φ(f)(2)=,(6)c·Φ(xd1···xdn)=
s1 sn 1 n 1 n
(s1,...,sn)X∈S1×···×Sn
n n
(3)=,(6)c· (s−s′)=c· r 6=0, (7)
k
kY=1{s,s′Y}∈(S2k) kY=1
since R has no zero divisors. Obviously this implies that there exists at least one
point (s ,...,s )∈S ×···×S where f does not vanish. (cid:3)
1 n 1 n
3. Concluding question
Is there any interesting use for the fact that even in the case of integral domains
the coefficients of Kouba’s map can be explicitly expressed in terms of cofactors of
the matrices (si−1)?
j
Acknowledgement
The author is very grateful to the department M9 of Technische Universit¨at
Mu¨nchen for excellent working conditions.
References
1. N.Alon,Combinatorial Nullstellensatz,Combin.Probab.Comput.8(1999),no.1,7–29.
2. G.D.BirkhoffandS.MacLane,Algebra,3.ed.,AmericanMathematicalSociety,1987.
3. O.Kouba,AdualitybasedproofoftheCombinatorialNullstellensatz,Electron.J.Combin.19
(2009),#N9.
4. S.Lang,Algebra,3.ed.,GraduateTextsinMathematics,vol.211,Springer,2002.
5. M. Micha lek, A short proof of Combinatorial Nullstellensatz, arXiv:0904.4573v1 [math.CO]
(2009).
6. U.Schauz,Algebraically Solvable Problems: Describing Polynomials as Equivalent to Explicit
Solutions,Electron.J.Combin.15(2008),#R10.
Zentrum Mathematik, Lehr- und Forschungseinheit M9 fu¨r Angewandte Geometrie
und Diskrete Mathematik, Technische Universita¨t Mu¨nchen, Boltzmannstraße 3, D-
85747Garching bei Mu¨nchen, Germany
E-mail address: [email protected]