Corollary 3.9 Fundamental theorem of finitely generated additive groups)
9.8 Cyclotomic fields, and the uncertainty principle
We now recall some of the elementary theory of cyclotomic fieldsQ(ω), and apply this to obtain an uncertainty principle for the Fourier transform onZp.
Definition 9.47 (Cyclotomic field) Letn≥1 be any positive integer. Annthroot of unityis any complex numberω∈Csuch thatωn =1. Annth root of unityωis said to beprimitiveifωis not anmth root of unity for any 1≤m<n. We define the cyclotomic field of order nto be the fieldQ(ω) obtained by adjoining a primitive nth root of unity to the rationalsQ. We define the nth cyclotomic polynomial n ∈C[z] to be the polynomialn(z) :=
ω(z−ω), whereωranges over the primitiventh roots of unity.
It is easy to see that for eachn, there areφ(n) primitive roots of unity, and they are all powers of each other. Thus there is only one cyclotomic fieldQ(ω) for each ordern. In particular we see thatnis a monic polynomial of degreeφ(n). Some further basic properties ofn as follows.
Lemma 9.48 nhas integer coefficients (thusn∈Z[z]), and is irreducible in Z[z]. Furthermore we haven(1)= p when n is a prime power n= pk,1(1)= 0, andn(1)=1otherwise.
Proof We first observe from the factor theorem that zn−1=
ω:ωn=1
(z−ω) for anyn≥1.
Since everynth root of unity is a primitivedth root of unity for somed, we obtain zn−1=
d|n
d(z). (9.21)
Thus one can obtainn(z) by factoring out
d|n;d<nd(z) fromzn−1. By an easy induction onnthis implies thatnis a monic polynomial with integer coefficients.
Since (zn−1)/1(z)=(zn−1)/(z−1) approaches nas z→1, we obtain the formula
n=
d|n;d>1
d(1).
Taking logarithms and using Exercise 9.4.1 we conclude that
d|n;d>1
(d)=
d|n;d>1
logd(1)
for alln ≥1, where(d) :=logp when d is a prime powerd =pk for some k≥1, and(d)=0 otherwise (cf. Exercise 1.10.6). Another easy induction onn then shows thatn(1)=e(n)for alln>1, which gives the desired formula for n(1).
Now we prove the irreducibility. Whennis prime this can be easily verified from Eisenstein’s criterion (Exercise 9.8.3), but the general case is trickier. We use an argument of Gauss. Suppose for contradiction thatnis reducible inZ[z], then we can partition the primitiventh roots of unity into two disjoint non-empty classes
A and B such that the monic polynomials f(z) :=
ω∈A(z−ω) and g(z) :=
ω∈B(z−ω) lie in f,g∈Z[z]. Of course we have n = f g. Since any two primitiventh roots are powers of each other, we can find anω∈ Asuch thatωm∈ B for some integerm. By decomposingminto primes and arguing by contradiction, we can in fact locate a prime pand anω∈ Asuch thatωp∈ B. This implies that the polynomials f(z) andg(zp) have a common root, and hence by the Euclidean algorithm we can find a non-trivial monic polynomialh(z)∈Z[z] which divides both f(z) andg(zp). This implies thatn(zp)= f(zp)g(zp) contains a factor of h(zp)h(z); by (9.21) we see thatznp−1 also contains a factor ofh(zp)h(z).
Now we work in the finite field Fp. In that setting we haveh(zp)=h(z)pand (zn−1)p=znp−1 (cf. Exercise 9.4.7) and hence (zn−1)pcontains a factor of
h(z)p+1; in particularzn−1 must contain a factor ofh(z)2inFp(cf. Exercise 9.4.1).
Taking formal derivatives, this implies thatzn−1 andnzn−1have a common factor ofh(z); but from the Euclidean algorithm and the fact thatn=0 (mod p) we see that these polynomials have a least common multiple of 1, contradiction.
As a consequence of Lemma 9.48 we obtain a useful criterion for non-vanishing of polynomial expressions of roots of unity, which was already exploited in the proof of Theorem 9.20.
Lemma 9.49 Let p be a prime and q be a power of p. Let P∈Z[t1, . . . ,tk]be a polynomial with integer coefficients such that P(z1, . . . ,zk)=0for some qth roots of unity z1, . . . ,zk. Then the integer P(1, . . . ,1)is divisible by p.
Proof Let ωbe a primitiveqth root of unity, then zi =ωni for some integers ni. If we let Q(t) :=P(tn1, . . . ,tnk), thenQ(ω)=0. ThusQ(t) shares a root in common with the irreducible polynomialq(t), which must then be a factor of Q(t). ThusQ(1)=P(1, . . . ,1) hasq(1)= pas a factor.
We apply this lemma to prove a non-vanishing result on generalized Vander- monde determinants. We first need a coefficient computation.
Proposition 9.50 [355] Let n1, . . . ,nk be non-negative integers, and let P∈ Z[z1, . . . ,zk]be the polynomial
P(z1, . . . ,zk)=
π∈Sk
sgn(π) k i=1
zniπ(i)
(cf. (9.3)). Then we can factor P =kQ, where Q∈Z[z1, . . . ,zk]is such that Q(1, . . . ,1)=k(n1, . . . ,nk)/k(1, . . . ,k).
Proof The expressionP(z1, . . . ,zk) can also be interpreted as the determinant of thek×kmatrix (znij)1≤i,j≤k. This shows in particular thatPvanishes when any two of theziare equal. Dividing out the factors ofzi−zjusing long division and apply- ing Definition 9.6 we conclude the existence of a polynomialQ∈Z[z1, . . . ,zk] such thatP =kQ. It remains to computeQ(1, . . . ,1). To do this we introduce the normalized differentiation operatorsDi :=zi d
dzi, and consider the expression D10D12. . .Dkk−1P(1, . . . ,1). We splitPinto factors
P(z1, . . . ,zk)=
1≤i<j≤k
(zj −zi)×Q(z1, . . . ,zk)
and apply the Leibniz ruleDi(f g)=(Dif)g+ f(Dig) repeatedly. Observe that there are k
2
linear factors in the expression to be differentiated, all of which vanish at (1, . . . ,1). There are alsok
2
derivatives to be applied. Thus the only
terms in the Leibniz rule which do not vanish at (1, . . . ,1) are those in which all the derivatives land on the linear factors. Furthermore each derivative must land on a distinct linear factor to yield a non-zero term. But this means that each of the Dk derivatives must land on one of thezk−zi factors withi <k(and there are (k−1)! ways this can happen); similarly the Dk−1derivatives must then land on one of thezk−1−zi factors withi <k−1 (with (k−2)! ways this can happen), and so forth. We conclude that
D01D21. . .Dkk−1P(1, . . . ,1)=(k−1)!ã ã ã1!0!Q(1, . . . ,1)=k(1, . . . ,k)Q(1, . . . ,1).
On the other hand, since each monomialz1n1ã ã ãznkk is an eigenfunction ofDiwith eigenvalueni, we see from definition ofP that
D10D12ã ã ãDkk−1P(z1, . . . ,zk)=
π∈Sk
sgn(π) k i=1
nπ(i)i−1zinπ(i). Substitutingz1= ã ã ã =zk=1 and applying (9.3) we obtain
D10D21ã ã ãDkk−1P(z1, . . . ,zk)=k(n1, . . . ,nk).
Combining this with the previous identity, the claim follows.
Combining Proposition 9.50 with Lemma 9.49 we obtain
Lemma 9.51 (Chebotarev’s lemma) Let q=pαbe a prime power, let1≤k<
p, and let z1, . . . ,zkbe distinct qthroots of unity. Let n1, . . . ,nkbe integers which are distinct modulo p. Then the k×k matrix(znij)1≤i,j≤khas non-zero determinant.
Indeed, Chebotarev’s lemma follows since k(z1, . . . ,zk) is non-zero and k(n1, . . . ,nk) is not divisible by p. We note that while this result was proved by Chebotarev in 1926 (see [338]), it has been independently rediscovered and reproved a number of times [278], [71], [263], [102], [355], [120], [131]. As a consequence of this lemma, one easily establishes the following uncertainty prin- ciple forZp:
Theorem 9.52 [355] Let p be a prime number. Let f :Zp→Cbe a random variable, and let fˆ:Zp→Cbe its Fourier transform (using the standard bichar- acter e(x, ξ)=exp(2πi xξ/p). Then we have |supp(f)| + |supp( ˆf)| ≥ p+1 Conversely, if A and B are two non-empty subsets ofZ/pZsuch that|A| + |B| ≥ p+1, then there exists a function f such thatsupp(f)= A andsupp( ˆf)=B.
We leave the deduction of Theorem 9.52 from Lemma 9.51 to Exercise 9.8.9.
This result should be compared with (4.21). As an application of this theorem we give yet another proof of the Cauchy–Davenport inequality, this proof being Fourier-analytic (or more precisely Fourier-algebraic) in nature.
Theorem 9.53 (Cauchy–Davenport inequality, yet again) Let F =Fp be a finite field of prime order. If A,B are two additive sets in F, then
|A+B| ≥min(|A| + |B| −1,p).
Proof ([355] and Robin Chapman, private communication) Since AandB are non-empty, we may find two subsetsX andY ofZ/pZsuch that|X| = p+1−
|A|,|Y| = p+1− |B|, and|X∩Y| =max(|X| + |Y| −p,1). By Theorem 9.52 we may find a function f such that supp(f)= Aand supp( ˆf)=X, and a function g such that supp(g)=Band supp( ˆg)=Y. Then f ∗g has support contained in A+B and has Fourier support equal toX∩Y (in particular, f ∗gis non-zero), and hence by Theorem 9.52 again we have|A+B| + |X∩Y| ≥p+1, which gives|A+B| ≥max(|A| + |B| −1,p) as desired.
One can iterate Theorem 9.52 to also apply to the groupZnp for anyn ≥1, which we endow with the standard bilinear form, as in Example 4.2.
Corollary 9.54 [249] Let p be a prime, n≥1be an integer, and f :Znp→Cbe a non-zero random variable. Then we have
pk|supp(f)| + pn−k−1|supp( ˆf)| ≥pn+pn−1 for all0≤k≤n−1.
Remark 9.55 These bounds can be seen to be sharp in a large number of sit- uations, by taking the Cartesian product of the examples in Theorem 9.52 with subgroups of Zp. It has a nice geometric interpretation: if one plots the point (|supp(f)|,|supp( ˆf)|) inZ×Z, then this point lies on or above the convex hull of the points (pj,pn−j) for 0≤ j ≤n, which correspond to the cases where f is the indicator function of a subgroup ofZnp; this convex hull should be contrasted with the hyperbola corresponding to (4.21). In [249], this result was generalized further to arbitrary finite additive groupsZ, see Exercise 9.8.11.
Proof We prove this by induction onn. Forn =1 this is just Theorem 9.52. Now suppose thatn >1, and the Corollary has already been proven for all smaller values ofn. Fix f. We parameterizeZnpasx=(x,xn), wherex∈Znp−1andxn ∈Zp. If g(ξ,xn) is the Fourier transform of f(x,xn) in thexvariable (withxnfixed), then fˆ(ξ, ξn) is the Fourier transform ofg(ξ,xn) in thexnvariable (keepingxfixed).
Let A⊂Zp be the set of allxn such that f(ã,xn) (and henceg(ã,xn)) is not identically zero. Observe that 1≤ |A| ≤pand
|supp(f)| =
xn∈A
|supp(f(ã,xn))|.
Thus by the pigeonhole principle there exists anxnsuch that
|A||supp(f(ã,xn))| ≤ |supp(f)|. (9.22) Fix thisxn. By induction we have
pk|supp(f(ã,xn))| +pn−k−1|supp(g(ã,xn))| ≥pn−1+pn−2 (9.23) for all 0≤k≤n−2. Also, for anyξin the support ofg(ã,xn), we see thatg(ξ,ã) is supported in A, so by Theorem 9.52
|supp( ˆf(ξ,ã))| ≥ p+1− |A|.
Summing this over allξ in the support ofg(ã,xn) we obtain
|supp( ˆf)| ≥(p+1− |A|)|supp(g(ã,xn))|.
Combining this with (9.22) we obtain
pk|supp(f)| + pn−k−1|supp( ˆf)| ≥ pk|A||supp(f(ã,xn))|
+(p+1− |A|)pn−k−1|supp(g(ã,xn))|.
When|A|is equal to 1 or p then the right-hand side here is at least pn+pn−1 thanks to (9.23). Since the right-hand side is linear in|A|, the same is true for the intermediate cases 1<|A|< p. This completes the induction.
Exercises
9.8.1 Letpbe a prime andk≥1. Prove thatp(z)=1+z+z2+ ã ã ã +zp−1 andpk(z)=p(zpk−1).
9.8.2 (Eisenstein’s criterion) Letpbe a prime, and letP(t)=antn+ ã ã ã +a0∈ Z[t] be such thatanis not divisible by p, thatan−1, . . . ,a0are divisible by p, anda0is not divisible byp2. Show that Pis irreducible inZ[t].
9.8.3 Letpbe a prime. Compute the polynomialp(t−1) explicitly, and then use Eisenstein’s criterion to give a proof thatp(t−1), and hencep
itself, is irreducible inZ[t], without using Lemma 9.48.
9.8.4 Letn≥1 be an integer, and suppose thatx∈Fp×is such thatn(x)=0.
Show that ord×(x)=n, and in particularndividesp−1.
9.8.5 Letn,mbe integers. Using Exercise 9.8.4, show that all the prime factors of n(m) are equal to 1 modn and are coprime to m. Using this (and modifying Euclid’s proof of the infinitude of primes) show that there are infinitely many primes equal to 1 mod n; this is a special case of Dirichlet’s theorem.
9.8.6 Let n ≥1, and let ω be a primitive nth root of unity. Show that the cyclotomic fieldQ(ω) is aφ(n)-dimensional vector space overQ, and
that the complex numbers 1, ω, ω2, . . . , ωφ(n)−1 form a linear basis for Q(ω).
9.8.7 Letpbe a prime, and letωbe a primitivepth root of unity. LetZ[ω] be the ring generated byω. Show that the quotient ringZ[ω]/((1−ω)ãZ[ω]) is isomorphic to the fieldFp. (Hint: exploit the fact thatp(1)=p, and hencep(ω)−pcontains a factor of (1−ω).)
9.8.8 [120] Letpbe a prime, letωbe a primitivepth root of unity, letz1, . . . ,zk
be distinct pth roots of unity, and let n1, . . . ,nk be distinct integers in [0,p). Suppose there exists a polynomial P∈Z[ω][z] of degree at most p−1 which vanishes atz1, . . . ,zk and has at mostknon-zero coeffi- cients. Using Exercise 9.8.7 and Lemma 9.26, show thatPis a multiple of (1−ω). Using this and an infinite descent argument, obtain another proof of Lemma 9.51 (at least in the caseq = p, which is all one needs for Theorem 9.52).
9.8.9 [355] Deduce Theorem 9.52 from Lemma 9.51. (Hint: Lemma 9.51 implies that all the minors of the Fourier matrix (e2πi jk/p)1≤j,k≤p are invertible.) Conversely, show that Theorem 9.52 implies theq = pcase of Lemma 9.51.
9.8.10 Letpbe a prime, letG:= {z∈C:zp =1}be thepth roots of unity, and letP∈C[z] be a non-zero polynomial with deg(P)< p. Show that that the number of zeroes ofP inG cannot exceed the number of non-zero coefficients inP.
9.8.11 [249] Given any finite additive groupZand any real numberk, letθ(Z;k) denote the quantity
θ(Z;k) :=inf{|supp( ˆf)|: f ∈L2(Z); f ≡0;|supp(f)| ≤k}.
Show that for every subgroupGof Z and any 1≤k≤ |Z|, we have the inequality
θ(Z;k)≥sup
st=k
θ(G;s)θ(Z/G;t)
by adapting the proof of Corollary 9.54. Conclude via an inductive argument that for any non-zero function f in L2(Z), the lattice point (|supp(f)|,|supp( ˆf)|) lies on or above the convex hull of the points (|G|,|Z|/|G|) asGranges over all subgroups ofZ.
Szemer´edi’s theorem for k = 3
A surprisingly fruitful and deep problem in additive combinatorics is that of deter- mining whether a given setAcontains non-trivial (i.e. proper) arithmetic progres- sions of a given length. We have already seen some special cases of this problem;
in Section 4.7 we saw that sum sets such as A+A, A+A+A, or 2A−2A contained very long arithmetic progressions (and generalized arithmetic progres- sions), while in Section 6.3 we saw that if we colored a large finite group (or a large interval of integers) into a small number of color classes, then one of the color classes must necessarily contain a long arithmetic progression. In this chapter and the next we shall discuss perhaps one of the deepest theorems known to additive combinatorics, namelySzemer´edi’s theorem:
Theorem 10.1 (Szemer´edi’s theorem) [345] Let A be a subset of the positive integers with positive upper density1σ(A)>0. Then A contains arbitrarily long arithmetic progressions.
This theorem was originally proved by Szemer´edi in 1975 by a sophisticated combinatorial argument, introducing for the first time the powerfulSzemer´edi reg- ularity lemma, which we discuss in Section 10.6. There are several other deep and important proofs of this theorem, including the ergodic-theoretic proof of Fursten- berg [125], the additive combinatorial proof of Gowers [138], and the hypergraph regularity proofs of Gowers [140] and Nagle, R¨odl, Schacht, and Skokan [254], [282], [283], [284]. These proofs will be discussed in the next chapter.
One can formulate Szemer´edi’s theorem in a more quantitative manner, using the following definition.
Definition 10.2 (Erd˝os–Tur´an constant) [99] Let Abe an additive set, and let k≥1. We letrk(A) denote the size of the largest subset of A which does not contain any proper arithmetic progressions of lengthk.
1Upper and lower density were defined in Definition 1.21.
369
Examples 10.3 We haver1(A)=0 andr2(A)=1 for any additive setA. Clearly rk(A) is non-decreasing inA, and we have the trivial boundrk(A)≤ |A|for any A. If Alives in a p-torsion group (e.g.A⊆Fpn) thenrk(A)= |A|for allk>n.
Theorem 10.1 is then easily shown to be equivalent to the following version, which was first conjectured by Erd˝os and Tur´an [99].
Theorem 10.4 (Szemer´edi’s theorem, second formulation) Let k≥1 and N ≥1. Then rk([1,N])=oN→∞;k(N)and rk(ZN)=oN→∞;k(N).
One in fact has the following generalization:
Theorem 10.5 (Szemer´edi’s theorem, in an arbitrary group) Let k≥1 and let Z be a finite additive group. Then rk(Z)=o|Z|→∞;k(|Z|).
This generalization either follows from the density Hales–Jewett theorem [124]
or from the hypergraph proofs of Szemer´edi’s theorem [140], [254], [282], [283], [284], and will be discussed in Section 11.6.
A further famous conjecture of Erd˝os and Tur´an remains open:
Conjecture 10.6 (Erd˝os–Tur´an conjecture) [99] Let A⊂Z+ be such that
n∈A 1
n = ∞. Then A contains arbitrarily long proper arithmetic progressions.
Up to very small factors, such as logo(1)N, this conjecture is essentially equiv- alent to asking forrk([1,N])=Ok(N/logN) for allkandN (Exercise 10.0.6).
This conjecture remains unsolved even for progressions of length 3 (though see Theorem 10.30 below). However a special case of this conjecture, restricted to the prime numbersP = {2,3,5, . . .}, has recently been proven by Green and Tao:
Theorem 10.7 (Green–Tao theorem) [158] Let k≥1and N>1. Then rk(P∩ [1,N])=oN→∞;k(|P∩[1,N]|). In particular, the primes contain arbitrarily long arithmetic progressions.
Note from (1.48) that the sum
p 1
p is divergent.
For general k, Szemer´edi’s theorem and the Green–Tao theorem are rather involved and will be treated in Chapter 11. However, thek=3 case is amenable to Fourier-analytic methods, and we have the following famous theorem of Roth:
Theorem 10.8 (Roth’s theorem)[287] We have r3([1,N]),r3(ZN)=oN→∞(N) for all N >1. More generally, for any finite additive group Z of odd order we have r3(Z)=o|Z|→∞(|Z|).
The generalization to arbitrary additive groupsZof odd order is due to Meshu- lam [248]. Note that the restriction thatZbe odd is necessary, since for 2-torsion groups, there are no proper progressions of length three and hencer3(Z)= |Z|in that case.
Both Roth’s theorem and Szemer´edi’s theorem have a surprising diversity of different proofs, using such techniques as harmonic analysis, ergodic theory, graph theory, hypergraph theory, inverse sum set theory, and Ramsey theory. However, they all revolve around a fundamental dichotomy, namely the dichotomy between arithmetically structured sets(e.g. arithmetic progressions, Bohr sets, sets of small doubling, sets of large additive energy, almost periodic sets) and arithmetically unstructured sets(e.g. random sets, pseudo-random sets, “mixing” sets). The point is that one needs very different arguments to deal with either of the two cases, and so any proof of the above theorems must first decompose a general set somehow into a structured component and an unstructured one. To make such a decomposition rigorous, one needs some powerful tools, for instance from harmonic analysis, ergodic theory, or graph theory.
The purpose of this chapter is to give several proofs of Roth’s theorem, both for generalZ and in special cases, and to also discuss some variants of this theorem.
These proofs serve as models for the more difficult Szemer´edi and Green–Tao theorems, to be discussed in the next chapter. It turns out that linear Fourier analysis (as developed in Chapter 4) is a particularly well adapted tool to detect progressions of length 3; as we shall see however in the next chapter, progressions of longer length will require aquadraticor higher-order Fourier analysis.
Exercises
10.0.1 Establish the inequalities
rk([1,N/k))≤rk(ZN)≤rk([1,N])
for anyN >k>1. This shows that the two formsrk(ZN)=oN→∞;k(N) andrk([1,N])=oN→∞;k(N) of Theorem 10.4 are equivalent.
10.0.2 Show that Theorem 10.4 is equivalent to Theorem 10.1. (Hint: to deduce Theorem 10.1 from Theorem 10.4 is rather easy. For the converse direc- tion, argue by contradiction, obtaining dense subsets of [1,N] without any proper arithmetic progressions, and paste those subsets together in some suitable way to contradict Theorem 10.1.)
10.0.3 Show that Theorem 10.1 is equivalent to the statement that every subset of the integers of positive upper density contains infinitely many progres- sions of lengthk, for eachk≥1.
10.0.4 Show that Szemer´edi’s theorem implies van der Waerden’s theorem (Exercise 6.3.7).
10.0.5 Give an example to show that if the positive integersZ+are partitioned into two color classes, then it is not necessarily the case that one of the color class contains aninfinitely longproper arithmetic progression
a+Z+ãr. Thus the properties of containing arbitrarily long proper arith- metic progressions, and infinitely long proper arithmetic progressions, are distinct.
10.0.6 Show that the Erd˝os–Tur´an conjecture is equivalent to the absolute con- vergence of the sum
∞ n=1
rk([1,2n]) 2n .
10.0.7 Show that ifAandBare additive sets which are Freiman isomorphic of order 2, thenrk(A)=rk(B) for allk.
10.0.8 If A and B are additive sets (possibly in different groups), show that rk(A×B)≥rk(A)rk(B).
10.0.9 IfZ,Zare two finite additive groups, show thatrk(Z ×Z)≤rk(Z)|Z|. 10.0.10 Show that to prove Theorem 10.5 for arbitrary groups Z, it suffices to verify it for cyclic groupsZNand for vector spacesZnpover fields of prime order. (Hint: use Corollary 3.8 and the previous exercise.) A similar claim applies of course to Roth’s theorem.
10.0.11 Letn≥1. Define acapsetof ordernto be any subset of the vector space F3n over the finite field F3 which contains no (affine) lines. Show that the largest possible cardinality of a capset of ordern isr3(F3n). Using Exercise 10.0.8, show thatr3(F3n)≥2n.
10.0.12 If Z is a finite additive group whose order is coprime to k!, show that rk(Z)≤(1−1k)|Z|. (Hint: if A⊂Z has cardinality greater than (1−1k)|Z|, choosea ∈Z,r∈ Z\{0}randomly and consider the proba- bility of the eventsa+ jr∈ Afor j =0,1, . . . ,k−1.)