MDS matrices over small fields: A proof of the GM-MDS conjecture
aa r X i v : . [ c s . D M ] M a r MDS matrices over small fields:A proof of the GM-MDS conjecture
Shachar Lovett ∗ March 22, 2018
Abstract
An MDS matrix is a matrix whose minors all have full rank. A question arisingin coding theory is what zero patterns can MDS matrices have. There is a naturalcombinatorial characterization (called the MDS condition) which is necessary over anyfield, as well as sufficient over very large fields by a probabilistic argument.Dau et al. (ISIT 2014) conjectured that the MDS condition is sufficient over smallfields as well, where the construction of the matrix is algebraic instead of probabilistic.This is known as the GM-MDS conjecture. Concretely, if a k × n zero pattern satisfiesthe MDS condition, then they conjecture that there exists an MDS matrix with thiszero pattern over any field of size | F | ≥ n + k −
1. In recent years, this conjecture wasproven in several special cases. In this work, we resolve the conjecture.
An MDS matrix is a matrix whose minors all have full rank. These matrices arise naturallyin coding theory, as they are generating matrices for MDS (Maximally Distance Separable)codes. A question arising in coding theory, motivated by applications in multiple accessnetworks [HHYD14,DSY15] and in secure data exchange [YS13,YSZ14], is what zero patternscan MDS matrices have. Namely, how sparse can MDS matrices be?There is a natural combinatorial characterization on the allowed zero patterns, called theMDS condition. Let A be a k × n MDS matrix with k ≤ n . We can describe its zero/nonzeropattern by a set system S , . . . , S k ⊂ [ n ], where S i = { j ∈ [ n ] : A i,j = 0 } .There are several restrictions on the structure of such set systems. Clearly, any row of A can have at most k − | S i | ≤ k − i . Similarly, any two rows of A can haveat most k − | S i ∩ S j | ≤ k − i = j . In general, this is known as ∗ Department of Computer Science and Engineering, University of California, San Diego. [email protected].
Research supported by NSF CAREER award 1350481 and CCF award 1614023.
MDS condition on the set system: | I | + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)\ i ∈ I S i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ k ∀ I ⊆ [ k ] , I nonempty . ( ⋆ )It is known that the MDS condition is also sufficient for the existence of MDS matriceswith zero pattern given by the set system, if the underlying field is large enough. Concretely,let S , . . . , S k ⊂ [ n ] be a set system which satisfies the MDS condition. Let F be theunderlying field, and assume that | F | > (cid:0) nk (cid:1) . Let A be a randomly chosen k × n matrix over F , where A i,j = 0 if j ∈ S i , and otherwise A i,j ∈ F is chosen uniformly and independently.Such a matrix A is an MDS matrix with positive probability. The reason is that the numberof maximal k × k minors of A is (cid:0) nk (cid:1) , and the MDS condition implies that the determinantsof these minors are not identically zero. So, each minor has a probability of | F | − to besingular, and by the union bound, if | F | > (cid:0) nk (cid:1) , then with positive probability all minors arenonsingular. This bound was improved to | F | > (cid:0) n − k − (cid:1) in [DSDY13].Dau et al. [DSY14] conjectured that the MDS condition is sufficient over small fields aswell. This is known as the GM-MDS conjecture. Concretely, if a k × n zero pattern satisfiesthe MDS condition, then there exists an MDS matrix with this zero pattern over any field ofsize | F | ≥ n + k −
1. Clearly, if this is true then a different argument than the probabilisticargument above would be needed.
Conjecture 1.1 (GM-MDS conjecture [DSY14]) . Let S , . . . , S k ⊂ [ n ] be a set system whichsatisfies the MDS condition. Then for any field F with | F | ≥ n + k − , there exists a k × n MDS matrix A over F with A i,j = 0 whenever j ∈ S i . We prove Conjecture 1.1 in this work.
Theorem 1.2.
Conjecture 1.1 is correct.
First, we describe an algebraic framework introduced by Dau et al. [DSY14] towardsproving Conjecture 1.1.
Dau et al. [DSY14] formulated an algebraic conjecture that implies Conjecture 1.1: if S , . . . , S k is a set system that satisfies the MDS condition, then there exists a General-ized Reed-Muller code with zeros in locations prescribed by the set system. Otherwise put,the matrix A can be factored as the product of a k × k invertible matrix and a k × n Vandermonde matrix. Before explaining these ideas further, we first set up some notations.Let F be a finite field, and let x, a , . . . , a n be formal variables, where we shorthand a = ( a , . . . , a n ). We use the standard notations F [ a , x ] for the ring of polynomials over F in the variables a , x ; F ( a ) for the field of rational functions over F [ a ]; and F ( a )[ x ] for thering of univariate polynomials in x over F ( a ). Given a set S ⊂ [ n ] define a polynomial p = p ( S ) ∈ F [ a , x ] as follows: p ( a , x ) := Y i ∈ S ( x − a i ) . S = { S , . . . , S k } define P ( S ) := { p ( S ) , . . . , p ( S k ) } .Let S = { S , . . . , S k } be a set system which satisfies the MDS condition. It is possible toassume without loss of generality that each S i is maximal, namely that | S i | = k − i ∈ [ k ]. For example, if we are allowed to increase n then we can replace each S i with S i ∪ T i where | T i | = k − − | S i | and T , . . . , T k , [ n ] are pairwise disjoint. An improved reduction isgiven in [DSY14] which does not require increasing n .Either way, under this assumption the polynomials P ( S ) form a set of k polynomials ofdegree k −
1, which we denote by p , . . . , p k . Define the k × n matrix A as A i,j = p i ( a j ).Note that entries of A are polynomials in F [ a ]. The condition that all k × k minors of A arenonsingular is equivalent to the condition that the polynomials P ( S ) are linearly independentover F ( a ) (here, we view the polynomials as elements of F ( a )[ x ] instead of as elements of F [ a , x ]). If this is the case, then one can use the Schwartz-Zippel lemma and show that theformal variables a , . . . , a n can be replaced with distinct field elements from F , while stillmaintaining the property that all k × k minors of A are nonsingular. The bound on the fieldsize | F | ≥ n + k − Conjecture 1.3 (Algebraic GM-MDS conjecture [DSY14]) . Let S , . . . , S k ⊂ [ n ] be a setsystem which satisfies the MDS condition, and where | S i | = k − for all i . Then the set ofpolynomials P ( S ) are linearly independent over F ( a ) . We remark that given any polynomials p , . . . , p k ∈ F [ a , x ] (for example, the polynomialsappearing in P ( S )), an equivalent condition to the polynomials being linearly independentover F ( a ) is the following: for any polynomials w , . . . , w k ∈ F [ a ], not all zero, it holds that k X i =1 w i ( a ) p i ( a , x ) = 0 . Following [DSY14], several works [HHYD14, HS17, YH18a] attempted to resolve the GM-MDS conjecture. They showed that Conjecture 1.3 holds in several special cases, but thegeneral case remained open. In this work we prove Conjecture 1.3, which implies Conjec-ture 1.1.
We start by considering a more general condition. Let v ∈ N n be a vector, where N = { , , , . . . } stands for non-negative integers. The coordinates of v are v = ( v (1) , . . . , v ( n )).We shorthand | v | = P v ( i ). Given vectors v , . . . , v m ∈ N n define V v i ∈ N n to be theircoordinate-wise minimum: ^ i ∈ [ m ] v i := (min( v (1) , . . . , v m (1)) , . . . , min( v ( n ) , . . . , v m ( n ))) . v , . . . , v m ∈ { , } n are indicator vectors of sets S , . . . , S m ⊂ [ n ], then V v i isthe indicator vector of ∩ S i .Given a parameter k > | v | define a set of polynomials in F [ a , x ]: P ( k, v ) := Y j ∈ [ n ] ( x − a j ) v ( j ) x e : e = 0 , . . . , k − − | v | . Note that P ( k, v ) consists of k − | v | polynomials of degree ≤ k −
1, which form a basis for thelinear space of polynomials of degree ≤ k − v ( j ) roots at each a j . Furthermore,note that if v is the indicator vector of a set S ⊂ [ n ] of size | S | = k −
1, then P ( k, v ) = { p ( S ) } .Given a set of vectors V = { v , . . . , v m } ⊂ N n define P ( k, V ) := P ( k, v ) ∪ . . . ∪ P ( k, v m ) . We use in this paper the convention that set union can result in a multiset. So for example, ifthe same polynomial appears in multiple P ( k, v i ) then it appears multiple times in P ( k, V ).Under this assumption we always have the identity: | P ( k, V ) | = | P ( k, v ) | + . . . + | P ( k, v m ) | . The following definition is the natural extension of the MDS condition to vectors.
Definition 1.4 (Property V ( k )) . Let V = { v , . . . , v m } ⊂ N n and k ≥ be an integer. Wesay that V satisfies V ( k ) if it satisfies:(i) | v i | ≤ k − for all i ∈ [ m ] .(ii) For all I ⊆ [ m ] nonempty, P i ∈ I ( k − | v i | ) + (cid:12)(cid:12)V i ∈ I v i (cid:12)(cid:12) ≤ k . Note that when m = k and v , . . . , v k are indicators of sets S , . . . , S k ⊂ [ n ] of size | S i | = k −
1, then property V ( k ) is equivalent to the MDS condition for S , . . . , S k .Observe that in general, if V satisfies V ( k ) then P ( k, V ) contains P mi =1 ( k − | v i | ) ≤ k polynomials of degree ≤ k −
1. The following conjecture is the natural extension ofConjecture 1.3 to vectors.
Conjecture 1.5.
Let
V ⊂ N n and k ≥ . Assume that V satisfies V ( k ) . Then the polyno-mials in P ( k, V ) are linearly independent over F ( a ) . A clarifying remark: as we view the set P ( k, V ) as a multiset, Conjecture 1.5 (andTheorem 1.7 below) imply in particular that the polynomials in P ( k, V ) are all distinct, so P ( k, V ) is in fact a set. 4 .3 An intermediate case We prove Conjecture 1.5 under an additional assumption, which is sufficient to prove Con-jecture 1.1. It is still open to prove Conjecture 1.5 in full generality.
Definition 1.6 (Property V ∗ ( k )) . Let V = { v , . . . , v m } ⊂ N n and k ≥ be an integer. Wesay that V satisfies V ∗ ( k ) if it satisfies V ( k ) , and additionally it satisfies:(iii) v i ∈ { , } n − × N for all i ∈ [ m ] . Namely, all coordinates in v i , except perhaps thelast, are in { , } . Theorem 1.7.
Let
V ⊂ N n and k ≥ . Assume that V satisfies V ∗ ( k ) . Then the polynomials P ( k, V ) are linearly independent over F ( a ) . Conjecture 1.3 follows directly from Theorem 1.7. If S , . . . , S k ⊂ [ n ] are sets whichsatisfy the assumptions of Conjecture 1.3, then their indicator vectors v , . . . , v k ∈ { , } n satisfy the assumptions of Theorem 1.7, and hence P ( { S , . . . , S k } ) = P ( k, { v , . . . , v k } ) arelinearly independent over F ( a ). The rows of a k × n MDS matrix generates a linear code in F n whose minimal distanceis d = n − k + 1. Namely, any vector in the subspace spanned by the rows has at most n − d = k − k ≤ n and d ≤ n − k + 1, what are the necessary and sufficient conditions on the zero pattern of a codewith minimal distance d .As it turns out, this more general question reduces to the one about MDS codes. Corollary 1.8.
Let k ≤ n and d ≤ n − k + 1 . Let S , . . . , S k ⊆ [ n ] . A necessary conditionfor the existence of a k × n matrix A over any field, such that the code spanned by the rowsof the matrix has minimal distance at least d , and such that A i,j = 0 whenever j ∈ S i , is | I | + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)\ i ∈ I S i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n − d + 1 ∀ I ⊆ [ k ] , I nonempty . It is also a sufficient condition over any field F of size | F | ≥ n − d .Proof. We first show that the conditions are necessary. Assume the condition is violated forsome I . Then there are | I | rows with at least n − d + 2 − | I | common zeros. Pick any | I | − I which is zero in thesecoordinates. So this linear combination has ( n − d + 2 − | I | ) + ( | I | −
1) = n − d + 1 manyzeros, a contradiction to the minimal distance being at least d .To show that the conditions are sufficient, consider the set system S , . . . , S k , S k +1 = . . . = S n − d +1 = ∅ . It satisfies that | I | + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)\ i ∈ I S i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n − d + 1 ∀ I ⊆ [ n − d + 1] , I nonempty . The claim follows by applying Theorem 1.2 to this set system.5 .5 Related work
As we already discussed, the GM-MDS conjecture was suggested by [DSY14], and par-tial results were obtained by [HHYD14, HS17, YH18a]. Shortly after posting this result inarXiv [Lov18], we were informed by Yildiz and Hassibi [YH18b] that they too have founda proof of the GM-MDS conjecture. Inspecting their proof, it is similar in spirit to ourproof, in the sense that both proofs generalize the original GM-MDS conjecture, in order tofacilitate an inductive argument. More specifically, our approach is to allow multiple rootsat a distinguished point, while their approach is to allow general multiplicities of sets.
We already discussed Conjecture 1.5. A more general open problem is the following. Let S , . . . , S k ⊂ [ n ] be a set system. Let A be a k × n matrix over some field, such that A i,j = 0whenever j ∈ S i . If we make no assumptions on the set system, then some k × k minors of A are forced to be singular (this happens when the set system, restricted to the minor, violatesthe MDS condition). The question is: what is the minimal field size, for which there existsa matrix where all minors which are not forced to be singular are nonsingular.This question arises naturally in the study of Maximally Recoverable (MR) codes, wherethe minors which are forced to be singular are determined by the underlying topology ofthe code. The GM-MDS conjecture which we prove is the special case where no minoris forced to be singular. In this case, very small fields (of size n + k −
1) are sufficient.However, in general there is no reason for nice algebraic constructions to exist. Two recentworks [KLR17,GGY17] have shown that in specific situations, exponential field size is needed.However, the proof techniques are highly specialized to these specific cases.This raises the following natural conjecture: most set systems require exponential fieldsize.
Conjecture 1.9.
Let S , . . . , S k ⊂ [ n ] be chosen randomly, by including each j ∈ S i inde-pendently with probability / . Assume that there exists a k × n matrix A over a field F thatsatisfies:(i) A i,j = 0 whenever j ∈ S i .(ii) Any k × k minor of A , which is not forced to be singular by (i), is nonsingular.Then with high probability over the choice of the set system, | F | ≥ c (cid:0) nk (cid:1) c , where c > is someabsolute constant. The conjecture basically says that for most set systems, the probabilistic constructionwhich requires field size (cid:0) nk (cid:1) cannot be significantly improved. Acknowledgement.
I thank Hoang Dau and Sankeerth Rao for a careful reading of anearlier version of this paper. 6
Proof of Theorem 1.7
Let n, k ≥
1. Let V = { v , . . . , v m } ⊂ N n which satisfies V ∗ ( k ). We will prove that thepolynomials P ( k, V ) are linearly independent over F ( a ).To that end, we assume that V is a minimal counter-example and derive a contradiction.Concretely, the underlying parameters are n, k, m and d = | P ( k, V ) | = P k − | v i | . We willassume that if V ′ is a set of vectors with corresponding parameters n ′ ≤ n, k ′ ≤ k, m ′ ≤ m, d ′ ≤ d with at least one of the inequalities being sharp, then Theorem 1.7 holds for V ′ .In particular, we assume that m ≥
2, as Theorem 1.7 clearly holds when m = 1.To help the reader, we note that the following lemmas construct such V ′ with the followingparameters: • Lemma 2.2: n, k − , m, d . • Lemma 2.4: n, k, e, d ′ and n, k, m − e + 1 , d ′′ with 2 ≤ e ≤ m − d ′ , d ′′ < d . • Lemma 2.5: n − , k, m, d . • Lemma 2.6: n, k, m, d − v I := ^ i ∈ I v i I ⊆ [ m ] . We introduce sometimes in the proofs an auxiliary set V ′ = { v ′ , . . . , v ′ m ′ } , in which case v ′ I for I ⊆ [ m ′ ] are defined analogously. Below, when we say that V or V ′ satisfy (i), (ii) or (iii),we mean the relevant items in the definition of V ∗ ( k ).Given two vectors u, v ∈ N n we denote u ≤ v if u ( i ) ≤ v ( i ) for all i ∈ [ n ]. Lemma 2.1.
There do not exist distinct i, j ∈ [ m ] such that v i ≤ v j .Proof. Assume the contrary. Applying (i) to j gives | v j | ≤ k −
1. Applying (ii) to I = { i, j } gives ( k − | v i | ) + ( k − | v j | ) + | v i ∧ v j | ≤ k. As v i ≤ v j we have v i ∧ v j = v i , and hence obtain that k − | v j | ≤
0, a contradiction.Lemma 2.1 implies in particular that n ≥
2. This is since if n = 1 then necessarily m = 1,as otherwise there would be i, j for which v i ≤ v j . So we assume n ≥ Lemma 2.2. V i ∈ [ m ] v i = 0 .Proof. Assume not. Then there exists a coordinate j ∈ [ n ] with v i ( j ) ≥ i ∈ [ m ].Define a new set of vectors V ′ = { v ′ , . . . , v ′ m } ⊂ N n as follows: v ′ i := ( v i (1) , . . . , v i ( j − , v i ( j ) − , v i ( j + 1) , . . . , v i ( n )) .
7n words, v ′ i is defined from v i by decreasing coordinate j by 1.We first show that V ′ satisfies V ∗ ( k − | v ′ i | = | v i | −
1. It clearly satisfies(i),(iii). To show that it satisfies (ii) let I ⊆ [ m ]. We have X i ∈ I ( k − − | v ′ i | ) + | v ′ I | = X i ∈ I ( k − | v i | ) + | v I | − ≤ k − . As we showed that V ′ satisfies V ∗ ( k − V implies that the polynomials P ( k − , V ′ ) are linearly independent over F ( a ). The lemma follows as it is simple to verifythat P ( k, V ) = { p ( a , x )( x − a j ) : p ∈ P ( k − , V ′ ) } . In particular, the linear independence of P ( k − , V ′ ) implies the linear independence of P ( k, V ). Definition 2.3 (Tight constraint) . A set I ⊆ [ m ] is tight for V if property (ii) holds withequality for I . Namely if X i ∈ I ( k − | v i | ) + | v I | = k. Note that if | I | = 1 then I is always a tight constraint. The following lemma is anextension of Lemma 2(i) in [YH18a]. It shows that in a minimal counter-example there areno tight sets, except for singletons and perhaps the whole set. Lemma 2.4. If I ⊆ [ m ] is a tight constraint, then | I | = 1 or | I | = m .Proof. Assume towards a contradiction that there exist a tight I with 1 < | I | < m . We willuse the minimality of V to derive a contradiction. Assume for simplicity of notation that I = { e, . . . , m } for 2 ≤ e ≤ m −
1. Define a new set of vectors V ′ = { v ′ , . . . , v ′ e } given by v ′ := v , . . . , v ′ e − := v e − , v ′ e := v I . We first show that V ′ satisfies V ∗ ( k ). It clearly satisfies (i) and (iii). To see that itsatisfies (ii) let I ′ ⊆ [ e ]. If e / ∈ I ′ then V ′ satisfies (ii) for I ′ as it is same condition as for V ,so assume e ∈ I ′ . Let I ′′ = I ′ ∪ { e + 1 , . . . , m } . Then X i ∈ I ′ ( k − | v ′ i | ) + | v ′ I ′ | = X i ∈ I ′′ ( k − | v i | ) + | v I ′′ | ≤ k, where the equality holds since k − | v ′ e | = P i ∈ I ( k − | v i | ) since we assume I is tight, and sinceby definition of I ′′ we have v ′ I = v I ′′ .As we assume that V is a minimal counter-example for Theorem 1.7, the theorem holdsfor V ′ . So, the polynomials P ( k, V ′ ) are linearly independent. Observe that | P ( k, V ′ ) | = | P ( k, V ) | since | P ( k, V ′ ) | = X i ∈ [ e ] ( k − | v ′ i | ) = X i ∈ [ m ] ( k − | v i | ) = | P ( k, V ) | . P ( k, V ) and P ( k, V ′ ) span the same space of polynomialsover F ( a ). To that end, it suffices to prove that F := P ( k, { v e , . . . , v m } ) and F ′ := P ( k, v ′ e )span the same space of polynomials.Let us shorthand v = v ′ e . Define the polynomial p ( a , x ) := Q j ∈ [ n ] ( x − a j ) v ( j ) . Observe that p divides all polynomials in F, F ′ . Moreover, F ′ = { p ( a , x ) x d : d = 0 , . . . , k − − | v |} spansthe linear space of all multiples of p of degree ≤ k −
1. As | F | = | F ′ | it suffices to prove that F are linearly independent over F ( a ), as then they must span the same linear space. However,this follows from the minimality of V , since F = P ( k, V ′′ ) for V ′′ = { v e , . . . , v m } .The following lemma identifies a concrete vector that must exist in a minimal counter-example. It is in its proof that we actually use the assumption that V satisfies (iii), namely V ∗ ( k ) and not merely V ( k ). Lemma 2.5.
There exists i ∈ [ m ] such that v i = (1 , . . . , , .Proof. Lemma 2.2 guarantees that there exists i ∗ ∈ [ m ] for which v i ∗ ( n ) = 0. We will provethat v i ∗ = (1 , . . . , , j ∗ ∈ [ n −
1] be such that v i ∗ ( j ∗ ) = 0.For simplicity of notation assume that i ∗ = m, j ∗ = n −
1. Define a new set of vectors V ′ = { v ′ , . . . , v ′ m } ⊂ N n − as follows: v ′ i := ( v i (1) , . . . , v i ( n − , v i ( n −
1) + v i ( n )) . In words, v ′ i ∈ N n − is obtained by adding the last two coordinates of v i ∈ N n .We first show that V ′ satisfies V ∗ ( k ). Note that | v ′ i | = | v i | . It clearly satisfies (i),(iii).To show that it satisfies (ii) let I ⊆ [ m ]. Note that (ii) always holds if | I | = 1, so we mayassume | I | >
1. We have by definition X i ∈ I ( k − | v ′ i | ) + | v ′ I | − v ′ I ( n −
1) = X i ∈ I ( k − | v i | ) + | v I | − v I ( n − − v I ( n ) . (1)First, consider first the case where | I | < m . Lemma 2.4 gives that I is not tight, andhence X i ∈ I ( k − | v i | ) + | v I | ≤ k − . As V satisfies (iii) we have v i ( n − ∈ { , } for all i . This implies v I ( n − ∈ { , } and v ′ I ( n − ∈ { v I ( n ) , v I ( n ) + 1 } . So Equation (1) gives X i ∈ I ( k − | v ′ i | ) + | v ′ I | ≤ X i ∈ I ( k − | v i | ) + | v I | + 1 ≤ k. Next, consider the case of | I | = m . As v m ( n −
1) = v m ( n ) = 0 we have v ′ m ( n −
1) = 0and hence v I ( n −
1) = v I ( n ) = v ′ I ( n −
1) = 0. Equation (1) then gives X i ∈ I ( k − | v ′ i | ) + | v ′ I | = X i ∈ I ( k − | v i | ) + | v I | ≤ k.
9s we showed that V ′ satisfies V ∗ ( k ), the minimality of V implies that the polynomials P ( k, V ′ ) are linearly independent over F ( a ). We next show that this implies that P ( k, V )are also linearly independent over F ( a ).Let s i := k − | v i | for i ∈ [ m ]. We have P ( k, V ) = { p i,e : i ∈ [ m ] , e ∈ [ s i ] } and P ( k, V ′ ) = { p ′ i,e : i ∈ [ m ] , e ∈ [ s i ] } where p i,e ( a , x ) := x e − Y j ∈ [ n − ( x − a j ) v i ( j ) · ( x − a n − ) v i ( n − ( x − a n ) v ( n ) ,p ′ i,e ( a , x ) := x e − Y j ∈ [ n − ( x − a j ) v i ( j ) · ( x − a n − ) v i ( n − v i ( n ) . Observe that p ′ i,e can be obtained from p i,e by substituting a n − for a n . Namely p ′ i,e ( a , . . . , a n − , x ) = p i,e ( a , . . . , a n − , a n − , x ) . Assume towards a contradiction that { p i,e } are linearly dependent over F ( a ). Equivalently,there exist polynomials w i,e ( a ), not all zero, such that X i ∈ [ m ] X j ∈ [ s i ] w i,e ( a ) p i,e ( a , x ) = 0 . We may assume that the polynomials { w i,e } do not all have a common factor, as otherwisewe can divide them by it. Let w ′ i,e ( a ) be obtained from w i,e ( a ) by substituting a n − for a n .That is, w ′ i,e ( a , . . . , a n − ) = w i,e ( a , . . . , a n − , a n − ). Then we obtain X i ∈ [ m ] X j ∈ [ s i ] w ′ i,e ( a ) p ′ i,e ( a , x ) = 0 . As the polynomials { p ′ i,e } are linearly independent over F ( a ), this implies that w ′ i,e ≡ i, e . That is, the polynomials w i,e satisfy w i,e ( a , . . . , a n − , a n − ) ≡ . This implies that ( a n − − a n ) divides w i,e for all i, e , which is a contradiction to the assumptionthat { w i,e } do not all have a common factor.Lemma 2.5 implies that the vector (1 , . . . , ,
0) belongs to V . Without loss of generality,we may assume that it is v m . This implies that v i ( n ) ≥ i ∈ [ m − v i ≤ v m , violating Lemma 2.1. Lemma 2.6. n = k .Proof. Let v m = (1 , . . . , , n − | v m | ≤ k −
1, so n ≤ k . Assumetowards a contradiction that n < k . Define a new set of vectors V ′ = { v ′ , . . . , v ′ m } ⊂ N n asfollows: v ′ := v , . . . , v ′ m − := v m − , v ′ m := (1 , . . . , , .
10n words, we increase the last coordinate of v m by 1.We claim that V ′ satisfies V ∗ ( k ). It satisfies (i) by our assumption that | v ′ m | = n ≤ k − I ⊆ [ m ]. If m / ∈ I then it clearlysatisfies (ii) for I , as it is the same constraint as for V , so assume m ∈ I . In this case wehave X i ∈ I ( k − | v ′ i | ) + | v ′ I | = X i ∈ I ( k − | v i | ) − ! + ( | v I | + 1) ≤ k. Note that | P ( k, V ′ ) | = | P ( k, V ) | −
1. As V is a minimal counter-example, we have that V ′ satisfies V ∗ ( k ). Let p ( a , x ) := Q j ∈ [ n − ( x − a j ). The construction of V ′ satisfies that P ( k, V )and P ( k, V ′ ) ∪ { p } span the same linear space of polynomials over F ( a ). This is since v ′ i = v i for i = 1 , . . . , m − P ( k, { v m } ) = { px e : e = 0 , . . . , n − k } and P ( k, { v ′ m } ) ∪ { p } = { p ( x − a n ) x e : e = 0 , . . . , n − k − } ∪ { p } both span the linear space of polynomials which are multiples of p and of degree ≤ k − P ( k, V ′ ) by p , . . . , p d − , where d = | P ( k, V ) | . Assume that the polynomials P ( k, V ) are linearly dependent. As P ( k, V ′ ) arelinearly independent, it implies that there exist polynomials w, w , . . . , w d − ∈ F [ a ], where w = 0, such that w ( a ) p ( a , x ) + d − X i =1 w i ( a ) p i ( a , x ) ≡ . Note that by construction, v ′ i ( n ) ≥ i ∈ [ m ]. This implies that p , . . . , p d − are alldivisible by ( x − a n ), while p is not. Substituting x = a n then gives w ≡
0, a contradiction.We can now reach a contradiction to V being a counter-example. We know that v m =(1 , . . . , ,
0) with | v m | = n − k −
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