The Kakeya Set conjecture over Z / N Z for general N

: We prove the Kakeya set conjecture for Z / N Z for general N as stated by Hickman and Wright [15]. This entails extending and combining the techniques of Arsovski [1] for N = p k and the author and Dvir [6] for the case of square-free N . We also prove stronger lower bounds for the size of ( m , ε ) -Kakeya sets over Z / p k Z by extending the techniques of [1] using multiplicities as was done in [17, 10]. In addition, we show our bounds are almost sharp by providing a new construction for Kakeya sets over Z / p k Z and Z / N Z .


Introduction
We are interested in proving lower bounds for the sizes of sets in (Z/NZ) n which have large intersections with lines in many directions.We first define the set of possible directions a line can take in (Z/NZ) n .Definition 1.1 (Projective space P(Z/NZ) n−1 ).Let N = p k 1 1 . . .p k r r where p 1 , . . ., p r are distinct primes.The Projective space P(Z/NZ) n−1 consists of vectors u ∈ (Z/NZ) n up to unit multiples of each other such that u (mod p k i i ) has at least one unit co-ordinate for every i = 1, . . ., r.
As mentioned before this also resolves the Minkowski dimension Kakeya Set conjecture over the p-adics.In [2] (which is the arxiv version 2 of the paper [1]) a different approach inspired by the polynomial method proofs in [9,17,10] is used to give the following bound for (m, ε)-Kakeya sets.
As can be seen from the bounds in the Theorems 1.6 and 1.7, the analysis in [2] is looser than the analysis in [1] for the case of (p k , 1)-Kakeya sets and leads to worse bounds in that setting.
In [1,2] tools from p-adic analysis are used to develop the techniques which prove Theorems 1.6 and 1.7.Both the proofs can be thought of as trying to develop a polynomial method on roots of unity in C. Our first result extends the ideas in [1] using multiplicities as in [17,10] to bound the size of (m, ε)-Kakeya sets over Z/p k Z with improved constants.Another advantage of our proof is that it is more elementary and does not require tools from p-adic analysis.
Theorem 1.8 (Stronger (m, ε)-Kakeya Set bounds over Z/p k Z * ).Let k, n, p ∈ N with p prime.Any (m, ε)-Kakeya set S ⊆ (Z/p k Z) n satisfies the following bound, When p > n, we also get the following stronger bound for (m, ε)-Kakeya set S in (Z/p k Z) n , We note that for fields of prime size, the bound above recovers the result of [10] as p tends to infinity.This gives us a new proof for the result with new techniques.
Combining techniques developed for Z/p k Z with techniques from [6], we prove the following lower bound for sizes of Kakeya sets in Z/NZ for general N. Theorem 1.9 (Kakeya set bounds for Z/NZ).Let n ∈ N and R = Z/NZ where N = p k 1 1 . . .p k r r with distinct primes p 1 , . . ., p r and k 1 , . . ., k r ∈ N. Any Kakeya set S in R n satisfies, When p 1 , . . ., p r ≥ n, we also get the following stronger lower bound for the size of a Kakeya set S in * We also note that the techniques presented here can also be used to prove norm bounds for functions f : (Z/NZ) n → C which have rich lines in many directions as was done for N prime (and in general for finite fields) in Theorem 19 of [7].A line L is m-rich with respect to f in this setting if The bound above recovers Theorem 1.5 as the size of the divisors p i grows towards infinity.Note, the techniques here do not naively prove (m, ε)-Kakeya bounds over Z/NZ for general N as the techniques in [6] also do not seem to extend to this setting (even for square-free N).In a follow-up paper [5] we solve this problem by using probabilistic arguments on top of the techniques in this paper and also give Maximal Kakeya bounds over Z/NZ for general N.
Observation 1.10.The number of prime divisors of N satisfies r = O(log N/ log log N).The expression ∏ r i=1 k i is upper bounded by the number of divisors τ(N) of N which satisfies log(τ(N)) = O(log N/ log log N).The proof for the bound on r can be found in [13] and the bound for τ(n) is Theorem 317 in [14].We now see that the expression in Theorem 1.9 is lower bounded by C n N n−O(n/ log log N) and so it indeed proves Conjecture 1.4 for all N.
Note, we could also get Kakeya Set bounds for Z/NZ for general N by combining the techniques from [1] and [6].This would also resolve Conjecture 1.4 by proving a Kakeya set lower bound of C n N n−O(n log(n)/ log log N) with a worse dependence on the dimension n.
We also construct Kakeya sets over Z/p k Z showing that the bound in Theorem 1.8 is close to being sharp.
Theorem 1.11 (Small Kakeya sets over Z/p k Z).Let s, n ∈ N, p ≥ 3 be a prime and k = (p s+1 −1)/(p−1).There exists a Kakeya set S in (Z/p k Z) n such that, The construction here uses ideas from the earlier constructions in [15,12,4] but is quantitatively stronger.
Using the Chinese remainder theorem we also get a construction for N with multiple prime factors showing that the bounds in Theorem 1.9 are also close to being sharp.Corollary 1.12 (Small Kakeya sets over Z/NZ).Let n, s 1 , . . ., s r ∈ N, p 1 , . . ., p r ≥ 3 be primes, k i = (p s i +1 i − 1)/(p i − 1), i = 1, . . ., r and N = p k 1 1 . . .p k r r .There exists a Kakeya set S in (Z/NZ) n such that, Proof.Using the Chinese Remainder Theorem (see Lemma 5.1) we see that the product of Kakeya sets over Z/p k i i Z for i = 1, . . ., r will be a Kakeya set over Z/NZ.We are then done using Theorem 1.11.

Proof Overview
We first start with a brief overview of the approach introduced in [6].Given a Kakeya set S in (Z/NZ) n , we construct a matrix K S whose columns are indexed by points in (Z/NZ) n and its rows are indexed by a direction in P(Z/NZ) n−1 where the dth row would be supported on the line in direction d contained in S.This ensures that the non-zero columns of K S correspond to points in S, which implies that the rank of K S then lower bounds the size of S. The goal is to construct a suitable K S and find a matrix E such that K S E is a matrix independent of S whose rank can be analyzed easily.For prime N = p, E can be a matrix whose columns contain the evaluation of a monomial on F n p .In this case K S ends up being a 'decoder' matrix where each row outputs the evaluation of a monomial on a direction given its evaluations on a line in that direction.This turns out to be a reformulation of Dvir's polynomial method proof [9].When N = pq (or in general is square-free) as Z/NZ ∼ = F p × F q we define E as a tensor of matrices one acting on the F p part and the other on the F q part.This allows for an inductive argument to give Kakeya set lower bounds for square-free N. To get quantitatively stronger bounds we can use the evaluations of (Hasse) derivatives and multiplicities as was done in [17,10].
In [1] for prime power N points in F n p are embedded in C n p (the p-adic complex numbers which is isomorphic to C as a field) using p k th roots of unity.The proof then implements the strategy of [6] using matrices with polynomial entries.U is a matrix whose columns are the evaluations of monomials over the embedding of (Z/p k Z) n in the torus.Let L be a line in direction d.The key statement is that some linear combination (with polynomial coefficients) of the rows in U corresponding to points in L can generate the dth row of a 'Vandermonde' matrix M after applying a mod p operation (using the p-adic structure).The rank of M can be easily lower bounded.The linear combination being taken here corresponds to a row of K S for a Kakeya set S ⊆ (Z/p k Z) n containing L in direction d.The key statement can be reformulated as saying that The argument can be completed by saying that the non-zero columns of S correspond to points in S, alternatively we note that we only use rows in E which correspond to points in S.This shows the rank of M over F p lower bounds the size of S. The argument in [1] is only for (p k , 1)-Kakeya sets.This is because, apriori it is not clear how to define a suitable K S (equivalently how to decode from lines with only some points contained in S).The revised version of the paper [2] has a very different approach to give (m, ε)-Kakeya bounds for the prime power case (the bounds there are quantitatively weaker than the ones here).
We now discuss how we improve the prime power bound quantitatively.Using simple linear algebraic arguments and the Chinese remainder theorem for a suitable polynomial ring we show that given m points on a line L in direction d we can decode the dth row of M up to degree m (see Corollary 3.16).We use multiplicities and (Hasse) derivatives to get stronger quantitative bounds (which Theorem 1.11 shows are almost sharp).We also develop these ideas without using the theory of p-adic numbers.We make this precise in Section 3.
To get bounds for general N we follow an inductive style argument as was done in [6] with some small technical improvements to get better constants.

Organization of the paper
In Section 2 we state definitions and results we need from [6].In Section 3 we state and extend results from [1].In Section 4 we prove Theorem 1.8.In Section 5 we prove Theorem 1.9.Finally, in section 6 we prove Theorem 1.11.

Acknowledgements
The author would like to thank Zeev Dvir for helpful comments and discussion.The author would also like to thank the reviewers for their helpful suggestions.This work was done while the author was supported by the NSF grant DMS-1953807.
2 Rank and crank of matrices with polynomial entries Definition 2.1 (Rank of matrices with entries in F[z]/⟨ f (z)⟩).Given a field F and a matrix M with entries in F[z]/⟨ f (z)⟩ where f (z) is a non-constant polynomial in F[z], we define the F-rank of M denoted rank F M as the maximum number of F-linearly independent columns of M.
In our proof, it will also be convenient to work with the following extension of rank for sets of matrices.
Definition 2.2 (crank of a set of matrices).Let F be a field and f (z) a non-constant polynomial in F[z].Given a finite set T = {A 1 , . . ., A n } of matrices over the ring F[z]/⟨ f (z)⟩ having the same number of columns we let crank F (T ) be the F-rank of the matrix obtained by concatenating all the elements A i in T along their columns.† We will use a simple lemma which follows from the definition of crank.Lemma 2.3 (crank bound for multiplying matrices).Let F be a field and f (z) a non-constant polynomial in F[z].Given matrices A 1 , . . ., A n of size a × b and matrices H 1 , . . ., H n of size c × a with entries in Proof.A dependence on the columns of the matrix obtained by concatenating A i s will be represented by a non-zero vector w ∈ F b such that A i w = 0 for all i = 1, . . ., n.This would imply (H i • A i )w = 0 for all i as well, completing the proof.
Given a matrix A with entries in F[z]/⟨ f (z)⟩ we want to construct a new matrix with entries only in F such that their F-ranks are the same.First, we state a simple fact about with degree strictly less than d and conversely every degree strictly less than d polynomial in we also let it refer to the unique degree strictly less than d polynomial it equals.Definition 2.5 (Coefficient matrix of A).Let F be a field and f (z) a non-constant polynomial in F[z] of degree d > 0. Given any matrix A of size n 1 × n 2 with entries in F[z]/⟨ f (z)⟩ we can construct the coefficient matrix of A denoted by Coeff(A) with entries in F which will be of size dn 1 × n 2 whose rows are labeled by elements in {0, . . ., d − 1} × [n 1 ] such that its (i, j)'th row is formed by the coefficients of z i of the polynomial entries of the j'th row of A.
The key property of the coefficient matrix immediately follows from its definition.Lemma 2.6.Let F be a field and f (z) a non-constant polynomial in F[z] of degree d > 0. Given any matrix A with entries in F[z]/⟨ f (z)⟩ and its coefficient matrix Coeff(A) it is the case that an F-linear combination of a subset of columns of A is 0 if and only if the corresponding F-linear combination of the same subset of columns of Coeff(A) is also 0.
In particular, the F-rank of A equals the F-rank of Coeff(A).
We now need some simple properties related to the crank of tensor products.To that end, we first define the tensor/Kronecker product of matrices.Let [r] = {1, . . ., r}.Definition 2.7 (Kronecker Product of two matrices).Given a commutative ring R and two matrices M A and M B of sizes n 1 × m 1 and n 2 × m 2 corresponding to R-linear maps A : R n 1 → R m 1 and B : R n 2 → R m 2 respectively, we define the Kronecker product M A ⊗ M B as a matrix of size n 1 n 2 × m 1 m 2 with its rows indexed by elements in We will need the following simple property of Kronecker products which follows from the corresponding property of the tensor product of linear maps.

Lemma 2.8 (Multiplication of Kronecker products). Given matrices A
We want to prove a crank bound for a family of matrices with a tensor product structure.This statement is an analog of Lemma 4.8 in [6] for our setting.Indeed we will prove it using Lemma 4.8 in [6] which we now state.Lemma 2.9 (Lemma 4.8 in [6]).Given matrices A 1 , . . ., A n of size a 1 × a 2 over a field F such that and matrices B i, j over the field The above lemma follows easily from simple properties of the tensor product and a proof can be found in [6].We now prove a generalization for our setting.
Lemma 2.10 (crank bound for tensor products).Let F be a field and f (z) ∈ F[z] a non-constant polynomial.Given matrices A 1 , . . ., A n of size a 1 × a 2 over the ring and matrices B i, j over the field Note the asymmetry between the matrices A i having entries in F[z]/⟨ f (z)⟩ and the matrices B i, j only having entries in F. This is important for the proof to work.

Proof. Consider the coefficient matrices Coeff
As B i, j only has entries in F it is easy to see that Coeff(A i ) ⊗ B i, j = Coeff(A i ⊗ B i, j ) (Note this would not be true if B i, j had entries in the ring F[z]/⟨ f (z)⟩).Using Lemma 2.6 now gives us, (2.2) We apply Lemma 2.9 on the family Coeff(A i ) ⊗ B i, j , i ∈ [n], j ∈ [m] and use equations (2.1) and (2.2) to complete the proof.

Polynomial Method on the complex torus
As mentioned in the introduction, the techniques presented here are an extension of the techniques used in [1] and are developed without invoking the tools and the language of p-adic analysis.
We embed (Z/p k Z) n in the complex torus.As mentioned in the proof overview (Section 1.1) we will be working with matrices with polynomial entries.We first define the rings where the entries come from.
where ζ p k is a primitive complex p k 'th root of unity.
As stated in the proof overview (Section 1.1) the goal is decode the dth row of a 'Vandermonde' matrix starting from the evaluations of monomials on a line L in direction d.We set ℓ according to the number of points in L and to what order derivatives we are working with and it controls the size of our 'Vandermonde' matrix as well.If we were not using any multiplicities/derivatives then ℓ could be set to log p (m) to lower bound the size of (m, ε)-Kakeya sets with weaker constants.
We suppress p in the notation as it will be fixed to a single value throughout our proofs.We also let ζ = ζ p k throughout this section for ease of notation.
Note that Z(ζ ) is the ring Z[x]/⟨φ p k (x)⟩ where, is the p k Cyclotomic polynomial.We will also work with the field We need a simple lemma connecting Therefore, the map ψ p k is the map which quotients the ring Z(ζ ) by the ideal ⟨p, ζ − 1⟩.
We note ψ p k can be extended to the rings by mapping z to z.The proof above immediately generalizes to the following corollary.

T k
ℓ /⟨p, ζ − 1⟩ being isomorphic to T ℓ is a special case of the corollary.The ψ p k will correspond to the mod p operation from the overview.
The operation ψ p k corresponds to the mod p operation in (1.1) in the proof overview (Section 1.1).We now prove that the rank of a matrix can only decrease under the quotient map ψ p k .
Lemma 3.4.If A is a matrix with entries in T k ℓ , then we have the following bound, where ψ p k (A) is the matrix with entries in T ℓ obtained by applying ψ p k to each entry of A.
In [1] the above lemma is implicit in the proof of Proposition 4, Arsovski's proof uses tools from p-adic analysis.We provide an alternate proof without using those techniques.
Proof of Lemma 3.4.Let A be a matrix with entries in This also means ψ p k can be applied on Coeff(A).On the other hand ψ p k can be directly applied on A as it has entries in T k ℓ and ψ p k (A) will have entries in Let rank Q(ζ ) A = r.By Lemma 2.6 and simple properties of the determinant, we have that every r + 1 × r + 1 sub-matrix M of Coeff(A) has 0 determinant.As ψ p k is a ring homomorphism it immediately follows that the corresponding r This implies that Coeff(ψ p k (A)) has rank at most r.The statement now follows from applying Lemma 2.6 on ψ p k (A).
We now define the 'Vandermonde' matrix (corresponding to M in Section 1.1) with entries in T ℓ which was defined by Arsovski in [1] and whose F p -rank will help us lower bound the size of Kakeya Sets.
Definition 3.5 (The matrix M p ℓ ,n ).The matrix M p ℓ ,n is a matrix over T ℓ with its rows and columns indexed by points in A lower bound of p ℓn (ℓn) −n is proven within Lemma 5 and Theorem 2 in [1].The same argument can also give the bound above with a slightly more careful analysis.For completeness, we give a proof in Appendix A. For convenience, we now prove that removing the rows in M p ℓ ,n corresponding to elements u ∈ (Z/p ℓ Z) n \ P(Z/p ℓ Z) n−1 does not change the F p -rank of the matrix.
For a given set of elements V ⊆ (Z/p ℓ Z) n let M p ℓ ,n (V ) refer to the sub-matrix obtained by restricting to rows of M p ℓ ,n corresponding to elements in V .In particular, for any given u ∈ R n we let M p k ,n (u) refer to the u'th row of the matrix.Lemma 3.7.
rank F p M p ℓ ,n (P(Z/p ℓ Z) n−1 ) = rank F p M p ℓ ,n .
Proof.Consider the matrix Coeff(M p ℓ ,n ).For any u ∈ (Z/p ℓ Z) n , the u'th row in M p ℓ ,n will correspond to a p ℓ block of rows B u in Coeff(M p ℓ ,n ).Say u doesn't have a unit coordinate.We can find an element u ′ ∈ P(Z/p ℓ Z) n−1 such that for some i, p i u ′ = u.We claim that the block B u can be generated by the block B u ′ via a linear map.This follows because given the coefficient vector of a polynomial f (z) in T ℓ the coefficient vector of the polynomial f (z p ) ∈ T ℓ can be obtained via a F p -linear map.This shows that the span of the rows of Coeff(M p ℓ ,n ) is identical to the span of the rows in Coeff(M p ℓ ,n (V ) where V is the set of vectors in (Z/p k Z) n with at least one unit co-ordinate.Now for any element u ∈ V we can find an element u ′ ∈ P(Z/p ℓ Z) n−1 such that there exists a λ ∈ (Z/p k Z) × for which u = λ u ′ .We now note that the block B u can be generated by the block B u ′ via a linear map as the coefficient vector of a polynomial f (z λ ) ∈ T ℓ can be linearly computed from f (z) ∈ T ℓ .This shows that the span of the rows of Coeff(M p ℓ ,n (V )) is identical to the span of the rows in Coeff(M p ℓ ,n (P(Z/p k Z) n−1 ) completing the proof using Lemma 2.6.
In [1] a row vector which encodes the evaluation of monomials over points in a line in direction u ∈ P(Z/p k Z) n−1 is used to decode the u'th row of M p k ,n .We extend this method by using row vectors which encode the evaluations of Hasse derivative of a monomial over a given point.We first define Hasse derivative.Definition 3.8 (Hasse Derivatives).Given a polynomial f ∈ F[x 1 , . . ., x n ] for any field F and an α α α ∈ Z n ≥0 the α α α'th Hasse derivative of f is the polynomial f (α α α) in the expansion f Definition 3.9 (The evaluation vector U Let F be a field, n, d ∈ Z and α α α ∈ Z n ≥0 .For any given point y ∈ F n we define U (α α α) d (y) to be a row vector of size d n whose columns are indexed by monomials m = x j 1 1 x j 2 2 . . .
We suppress n in the notation as there will be no ambiguity over it in this paper.We define the weight of α α α ∈ Z n ≥0 as wt(α α α) = ∑ n k=1 α k .The following simple fact about uni-variate polynomials illustrates one use of Hasse derivatives and how they correspond to our intuition with regular derivatives in fields of characteristic 0. Lemma 3.10.Let F be a field, y ∈ F and w ∈ N.For any uni-variate polynomial f ∈ F[x] all its Hasse derivatives at the point y of weight strictly less than w are 0 if and only if f ∈ ⟨(x − y) w ⟩ or in other words f is divisible by (x − y) w .
Let L ⊆ (Z/p k Z) n be a line in direction u ∈ P(Z/p k Z) n−1 .A special case of the next lemma proves that, for any polynomial from the evaluation of f on the points f (ζ x ), x ∈ L. This is implicit in the proof of Proposition 4 in [1].We generalize this statement.Let π : L → Z ≥0 be a function on the line such that ∑ x∈L π(x) ≥ p ℓ .We prove that we can decode The lemma below can be thought of as analogous to how over finite fields the evaluation of a polynomial and its Hasse derivatives with high enough weight along a line in direction u can be used to decode the evaluation of that polynomial at the point at infinity along u [10].Lemma 3.11 (Decoding from evaluations on rich lines).Let L = {a + λ u|λ ∈ Z/p k Z} ⊂ (Z/p k Z) n with a ∈ (Z/p k Z) n , u ∈ P(Z/p k Z) n−1 , u ′ ∈ (Z/p ℓ Z) n be such that u ′ (mod p k ) = u and π : L → Z ≥0 be a function which satisfies ∑ x∈L π(x) ≥ p ℓ .
Then, there exist elements c λ ,α α α ∈ Q(ζ )[z] (depending on π, L and u ′ ) for λ ∈ Z/p k Z and α α α ∈ Z n ≥0 with wt(α α α) < π(a + λ u) such that the following holds for all polynomials f ∈ Z[x 1 , . . ., x n ], For the application of ψ p k in the statement of the lemma to make sense we must have its input be an element in T k ℓ .In other words, we need the input of ψ p k to be a polynomial in z with coefficients in Z(ζ ).This is indeed the case and will be part of the proof of the lemma.We need two simple facts for the proof.Lemma 3.12 (Hasse Derivatives of composition of two functions).Let F be a field, n ∈ N. Given a tuple of polynomials C(y) = (C 1 (y),C 2 (y), . . .,C n (y)) ∈ (F[y]) n , w ∈ N and γ ∈ F there exists a set of coefficients b w,α α α ∈ F (which depend on C and γ) where α α α ∈ Z n ≥0 such that for any f ∈ F[x 1 , . . ., x n ] we have, This fact follows easily from the definition of the Hasse derivative.A proof can also be found in Proposition 6 of [10].‡ We also need another fact about the isomorphism between polynomials and the evaluations of their derivatives at a sufficiently large set of points.Lemma 3.13 (Computing polynomial coefficients from polynomial evaluations).Let F be a field and n ∈ N. Given distinct a i ∈ F and m i ∈ Z ≥0 , let h(y , which maps every polynomial f ∈ F[z]/⟨h(z)⟩ to the evaluations ( f ( j i ) (a i )) i, j i where i ∈ {1, . . ., n} and j i ∈ {0, . . ., m i − 1}.This is a simple generalization of Lemma 3.10.It can be proven in several ways, for example it can be proven using Lemma 3.10 and the Chinese remainder theorem for the ring F[y].To prove Lemma 3.11 we will need the following corollary of the fact above.
Corollary 3.14 (Computing a polynomial from its evaluations).Let F be a field and n ∈ N. Given distinct . Then there exists coefficients t i, j ∈ F[z] (depending on h) for i ∈ {1, . . ., n}, j ∈ {0, . . ., m i − 1} such that for any f (y ‡ The idea is that C i (γ + z) can be expanded in terms of its Hasse derivatives.We use that in f (C 1 (γ + z), . . .,C n (γ + z)) and expand it again using the Hasse derivative expansion of f at (C 1 (γ), . . .,C n (γ)).The coefficient of z w is the Hasse derivative h (w) (γ) and it will only get contributions from f (α α α) with wt(α α α) ≤ w.
Proof of Lemma 3.11.As the statement we are trying to prove is linear over Z we see that it suffices to prove the lemma in the case of when f equals a monomial.
where F is an arbitrary field (we will be working with For this proof we use the elements in {0, . . ., p ℓ − 1} to represent the set Z/p ℓ Z (similarly for (Z/p k Z) n ).
We first prove the following claim.
Proof.For every λ ∈ Z/p k Z and w ∈ N, using Lemma 3.12 we can find coefficients b w,α α α (λ ) ∈ Q(ζ ) such that, for every polynomial As u ′ (mod p k ) = u and ζ is a primitive p k 'th root of unity in C we note that We now make the simple observation that for any α α α ∈ Z n ≥0 and v ∈ Z n ≥0 we have The above equation combined with (3.2) for f = m v implies, for all w ∈ N and v ∈ Z n ≥0 .Setting b ′ w,α α α (λ ) = ζ ⟨α α α,a⟩ b w,α α α (λ ) we are done .
Without loss of generality let us assume ∑ x∈L π(x) = p ℓ (if it is greater we can reduce each of the π(x) until we reach equality -this would just mean that our computation was done by ignoring some higher order derivatives at some of the points). Let Using Corollary 3.14 there is a This statement along with Claim 3.15 leads to the following: there exist elements c λ ,α α α ∈ Q(ζ )[z] (depending on π, L and u ′ ) for λ ∈ Z/p k Z and α α α ∈ Z n ≥0 with wt(α α α) < π(a + λ u) such that the following holds for all monomials m v ∈ Z[x 1 , . . ., x n ], v ∈ Z ≥0 we have, We claim that these are coefficients we wanted to construct in the statement of this lemma.All we need to show now is that ψ p k is a ring homomorphism from the ring Z(ζ )/⟨h(z)⟩ to the ring T ℓ = F p (ζ )/⟨z ℓ − 1⟩ and maps ζ ⟨a,v⟩ z ⟨v,u ′ ⟩ to z ⟨v,u ′ ⟩ .This follows from Corollary 3.3 and noting We will use the lemma above in the form of the following corollary.
) n be such that u ′ (mod p k ) = u and π : L → Z ≥0 be a function which satisfies ∑ x∈L π(x) ≥ p ℓ then there exists a Q(ζ )[z]-linear combination (with coefficients depending on π, L and u ′ ) of the vectors U (α α α) p ℓ (ζ x ) for x ∈ L and α α α of weight strictly less than π(x) which under the map ψ p k gives us the vector M p ℓ ,n (u ′ ).
Proof.Lemma 3.11 gives us the required linear combination.
The coefficients defined by the above corollary correspond to the rows of the matrix K S in the proof overview (Section 1.1).

Kakeya Set bounds over Z/p k Z
We first prove two helper lemmas.We recall that GL n (Z/p k Z) is the set of linear isomorphisms over (Z/p k Z) n .They are represented by matrices whose determinants are units in Z/p k Z.
The first lemma implies that given a large subset D ⊆ P(Z/p k Z) n−1 there exists a W in GL n (Z/p k Z) such that M p ℓ ,n (W • D) has high rank where that is the set of directions obtained by rotating the elements in D by W .In general, we also prove that there exists a W such that M p ℓ ,n (D ′ W ) has high rank where that is the set of directions in (Z/p ℓ Z) n which under the mod p k map give an element in W • D. We prove the statement using a random rotation argument.
Lemma 4.1.Let k, n ∈ N, ε ≥ 0 and p be a prime.Given a set D ⊆ P(Z/p k Z) n−1 containing at least an ε fraction of elements in P(Z/p k Z) n−1 then there exists a matrix W ∈ GL n (Z/p k Z) such that, Proof.From Lemma 3.7 we know that the F p rank of M p ℓ ,n ((PZ/p ℓ Z) n−1 ) is at least p ℓ l −1 +n n .This means there exists a set of rows V of size at least p ℓ l −1 +n n in Coeff(M p ℓ ,n ((PZ/p ℓ Z) n−1 )) which are F p -linearly independent.
We pick W ∈ GL n (Z/p k Z) uniformly at random and as in the statement of the lemma consider the set The key claim about D ′ W is the following.
Claim 4.2.Any given row v ∈ V will appear in Coeff(M p ℓ ,n (D ′ W )) with probability at least ε.
Proof.As GL n (Z/p k Z) acts transitively on the set P(Z/p k Z) n−1 and |D| ≥ ε|P(Z/p k Z) n−1 | we see that D ′ W will contain a particular u ∈ P(Z/p ℓ Z) n−1 with probability at least ε.The row v ∈ V will be within Coeff(M p ℓ ,n (u)) for some u ∈ P(Z/p ℓ Z) n−1 .This means that v ∈ V will appear in Coeff(M p ℓ ,n (D ′ W )) with probability at least ε.Now by linearity of expectation we see that the expected number of rows V appearing in Coeff(M p ℓ ,n (D ′ W )) is at least ε|V |.This means there exists a choice of W such that an ε fraction of elements in V do appear in Coeff(M p ℓ ,n (D ′ W )). As all these rows are linearly independent then by Lemma 2.6 we see that the The second lemma lower bounds the size of (m, ε)-Kakeya sets by the rank of a sub-matrix of M p ℓ ,n with rows corresponding to directions with rich lines.
Let π u : L u → Z ≥0 be a weight function which gives weight b to points in S ∩ L u and 0 elsewhere.Using Corollary 3.16 for the line L u and function π u , for all u ′ ∈ (Z/p ℓ Z) n such that u ′ (mod p k ) = u, we can now construct row vectors C u ′ such that ψ p k (C u ′ U L u ,p ℓ ) = M p ℓ ,n (u ′ ).For convenience, for every u ∈ P(Z/p k Z) n−1 we define C u and M u as the matrices whose row vectors are C u ′ and M p ℓ ,n (u ′ ) respectively for all u ′ ∈ (Z/p ℓ Z) n such that u ′ (mod p k ) = u.We note, ψ p k (C u U L u ,p ℓ ) = M p ℓ ,n (u).
We use Lemma 2.3 on {U L u ,p ℓ } u∈D being multiplied by C u , u ∈ D and applying Lemma 3.4 next to get, where we recall D ′ = D ′ I = {u ∈ (Z/p ℓ Z) n |u (mod p k ) ∈ D}.Using the above inequality with (4.1) we have, We are now ready to prove Theorem 1.8.
Theorem 1.8.Let k, n, p ∈ N with p prime.Any (m, ε)-Kakeya set S ⊆ (Z/p k Z) n satisfies the following bound, When p > n, we also get the following stronger bound for (m, ε)-Kakeya set S in (Z/p k Z) n , Proof.Set ℓ = k + ⌈log p (n)⌉.Let D be the set of directions in P(Z/p k Z) n−1 which have m-rich lines.We know D contains at least an ε fraction of directions.By Lemma 4.1 there exists a matrix where D ′ W = {u ∈ (Z/p ℓ Z) n |u (mod p k ) ∈ W • D}.We now note W • S is also an (m, ε)-Kakeya set with W • D as the set of directions with m-rich lines.We now apply Lemma 4.3 with the inequality above to get, For convenience we set a = ⌈log p (n)⌉ which implies p a ≥ n ≥ p a−1 .The above inequality implies, For i ≤ n we have (1 + ip −a ) −1 ≤ 2 which completes the proof.Note, the second half of the theorem follows from observing that for p > n we have (1 + ip −a ) −1 ≤ 1 + n/p and a = ⌈log p (n)⌉ = 1.
5 Kakeya Set bounds over Z/NZ for general N We need some simple facts about Z/p k NZ for p prime and p and N co-prime which follow from the Chinese remainder Theorem.
We briefly discuss the proof strategy first.We use ideas from [6] to prove Theorem 1.9.We give the idea for N = p k 1 q k 2 where p and q are distinct primes.Let S be Kakeya set which contains lines L u in the direction u ∈ P(Z/NZ) n−1 .Using Lemma 5.1 we know that the line L u , u = (u p , u q ) ∈ P(Z/p k 1 Z) n−1 × P(Z/q k 2 Z) n−1 can be decomposed as a product of lines L p (u p , u q ) over Z/p k 1 Z and L q (u p , u q ) over Z/q k 2 Z with directions u p ∈ P(Z/p k 1 Z) n−1 and u q ∈ P(Z/q k 2 Z) n−1 respectively.Note that L p (u p , u q ) and L q (u p , u q ) depend on u = (u p , u q ), and not just on u p or u q respectively.Let ζ be a primitive p k 1 'th root of unity in C and 1 y be the indicator vector of the point y ∈ (Z/q k 2 Z) n .We then examine the span of vectors As there is one vector for each point in S, the dimension of the space spanned by these vectors is at most |S|.We then use the decoding procedure for U p k 1 from Corollary 3.16 to linearly generate vectors M p k 1 ,n (u p ) ⊗ 1 y for y ∈ L q (u p , u q ), (u p , u q ) ∈ P(Z/p k 1 Z) n−1 × P(Z/q k 2 Z) n−1 from the vectors U p k 1 (ζ x ) ⊗ 1 y for x ∈ L p (u p , u q ), y ∈ L q (u p , u q ).This means the dimension of the span of M p k 1 ,n (u p )⊗y for y ∈ L q (u p , u q ), (u p , u q ) ∈ P(Z/p k 1 Z) n−1 × P(Z/q k 2 Z) n−1 lower bounds the size of S. Note that M p k 1 ,n (u p ) only depends on u p and not on u q .
We pick the largest subset of rows V in Coeff(M p k 1 ,n (P(Z/p k 1 Z) n−1 ) which are linearly independent.By Lemma 3.6, V has large size.Any vector v ∈ V will correspond to a row in Coeff(M p k 1 ,n (u p )) for some u p ∈ P(Z/p k Z) n−1 .For that u p , the span of 1 y for y ∈ L q (u p , u q ) as u q varies in P(Z/q k 2 Z) n−1 will have dimension exactly equal to the size of the set S v = u q ∈P(Z/q k 2 Z) n−1 L q (u p , u q ) which is a Kakeya set in (Z/q k 2 Z) n and has large size by Theorem 1.8.Simple properties of the tensor product (Lemma 2.10) now imply that v ⊗ 1 y for v ∈ V and y ∈ S v are linearly independent which gives us a rank bound and hence a lower bound on the size of S Using U (α α α) p ℓ for α α α ∈ Z n ≥0 , wt(α α α) < p ℓ−k as in the proof of Theorem 1.8 gives us better constants.The proof for an arbitrary number of distinct prime factors applies the above argument inductively.Proof.The construction will be inductive.For n = 1 it S is the whole set.For a general n we first construct a set which has lines with directions u = {1, u 2 , . . ., u n } ∈ P(Z/p k Z) n−1 .We will need the following claim.Claim 6.1.There exists a function g : Z/p k Z → Z/p k Z such that for any fixed t ∈ Z/p k Z the function x → tx − g(x), x ∈ Z/p k Z has an image of size at most p k /(k(1 − p −1 )).

Lemma 4 . 3 .
Let k, n ∈ N and p be a prime.Let S ⊆ (Z/p k Z) n be a (m, ε)-Kakeya set and D ⊆ P(Z/p k Z) n−1 , |D| ≥ ε|P(Z/p k Z) n−1 | such that for every u ∈ D we have a m-rich line L u with respect to S in direction u.Then we have the following bound,|S| ⌈p ℓ m −1 ⌉ + n − 1 n ≥ rank F p M p ℓ ,n (D ′ ), for all ℓ ≥ log p (m) where D ′ = D ′ I = {u ∈ (Z/p ℓ Z) n |u (mod p k ) ∈ D}.Proof.We set b = ⌈p ℓ m −1 ⌉.Consider the family of row vectors, U (α α α) p ℓ (ζ x ) where x ∈ S and wt(α α α) < b.For a u ∈ D, let L u be a m-rich line with respect to S in direction u.Let U L u ,p ℓ be the matrix constructed by concatenating the row vectors U (α α α) p ℓ (ζ x ) where x ∈ L u and wt(α α α) < b.By construction as the rows in {U L d ,p ℓ } u∈D correspond to tuples x ∈ S and α α α ∈ Z n ≥0 , wt(α α α) < b we have the following bound,

Lemma 5 . 1 (
Geometry of Z/p k NZ).Let p, N, n, k ∈ N, R = Z/p k NZ, R 1 = Z/NZ, R 0 = Z/p k Zwith p prime and co-prime to N. Using the Chinese remainder theorem we know