The Freiman-Vosper theorem

IAI+IBI - 1

Y- A\X-leEA:B(e)-B)

2.8 The Freiman-Vosper theorem

k-1

E cos(2n a j ) j-0

E cos 2na j + E cos 2na

0<a, < 1 /4 1/4<a) < 1 /2

+ cos 2naj + cos 2,raj

1/2<a1 <3/4 3/4<a,<I

E cos2naj - E cos2n(1/2-aj)

0<a,<1/4 1/4<a,<1/2

-

cos27r(aj - 1/2)+ cos2n(1 - c j)

1 /2<a1 <3/4 3/4<a1 <1

f

^1/4

^f

^1/4

sin2ntdt-2n E sin2ntdt

0<a1<1/4 1 1/4<a1<1/2 /2-a1

1/4 1/4

-2n E f sin2ntdt+2n sin2ntdt

1/2<a1<3/4 1-1/2 3/4<a,<l

I

^-a1

rI/4

2,r (n1(t) - n2(t) - n3(t) + n4(t)) sin 2ntdt

< 27r

I Ok sin 2ntdt

I

-

9k,

which contradicts condition (2.9). Therefore, N(AB) > (I +9)k/2 for some,6 E R.

This completes the proof.

68 2. Sums of congruence classes

and 2c, - 3 1 - coci

(2.12)

3 <

c1/2_i

Let p be an odd prime number, and let A be a nonempty set of congruence classes modulo p such that

3<k-IAl:5 cop

^(2.13)

and

12AI <c,k-3.

(2.14)

Define the integer b by 12A I - 2k - 1 + b. Then A is contained in an arithmetic progression in Z/pZ of length k + b.

Theorem 2.11 Let A be a nonempty set of congruence classes modulo p such that

IAI-k<35

and

12A1 <

lsk - 3.

Define the integer r by 12AI - 2k - 1 + r. Then A is contained in an arithmetic progression in Z/pZ of length k + r.

Proof. If co - 1/35 and c, - 12/5, then

2c, - 3 1 - coc,

- 0.6 < 0.601 <

3 c,12

The result follows from Theorem 2.10.

Proof of Theorem 2.10. Inequalities (2.10), (2.11), and (2.12) imply that cl(2c, - 3)2 < 9.

Since the polynomial x(2x - 3)2 is strictly increasing for x > 3/2, it follows that

cl < 2.5.

(2.15)

Let 12A1 - t. By inequalities (2.13) and (2.10), we have

2k-I<2cop<6

and so, by the Cauchy-Davenport theorem,

C-12AI-2k-1+b>mir(p,2k-1)-2k-1

and b > 0. Moreover, by inequality (2.14),

f-12A1 <c,k <coc,p.

(2.16)

By inequality (2.12), we can choose a positive real number 9 such that

I 3

1-cocl

2c1

(2 17 cI/2

. )

Then 3(1 +9)

cI < 2 (2.18)

and

coc1 +9ci/2 < 1. (2.19)

Let A - {ao, al, ... , ak_I } c Z/pZ. Choose rj E 10, 1, ... , p - 1) such that aj - rj + pZ for j -0,1 , ... , k - 1, and let R - {ro, rI, ... , rk_I } c

Z. We consider the exponential sums SA(x) and S2A(x) defined by

k-I

SA(x) - E

^{e2niai/P -}

E

^e2niri.r/p

aEA j-0

and

S2A(x) - ^e2nibxlP

bE2A

We shall prove that there exists an integer z 0- 0 (mod p) such that ISA (z)I > 9k.

If not, thenISA(x)I <9kforallx 00 (mod p). Using Lemma 2.15, the Cauchy- Schwartz inequality, and inequalities (2.17) and (2.16), we obtain

p-1

k2P

-

SA(X)2S2A(X)

.r-0

P-1

SA(0)2S2A(0)+ E SA(X)2S2A(X) .r-1

P-1

k2e + SA(X)2S2A(X) X-1

P-1

< k2e+EISA(X)121S2A(X)I

X-1

P-1

< k2e+tOk E ISA(X)IIS2A(X)1

.r-1 P-I

< k 21 + 0k ISA(X)I IS2A(X)I -0

P-I ^11/2 P-1 ^1/2

k2t_{+ 9k(kp)1}/2(ep)1 /2

70 2. Sums of congruence classes

-

k2f+9k3/2fij2p

< coclk2p+9c1l2k2P

-

(cocl +OcII l2)k2p

< k2p,

which is absurd. Therefore, ISA(Z)I -

P-1

E

^e2aii)zlP

E

^{e2 riazlP} j-0 aEA

> Ok

for some integer z 0 0 (mod p).

Applying Theorem 2.9 to the real numbers aj - rjz/ p for j - 1, ... , k, we

obtain a real number lB and a subset R' C R such that

k'-IR'I> (2 ^>2

and

rjz

E +2¹ (mod 1)

)

for all rj E R'. Since p is odd, the interval [ fi,18+ 1/2) contains (p ± 1)/2 fractions with denominator p, and these fractions are consecutive. This means that there is an integer uo such that the fractions in the interval can be written in the form (uo + s)/p with

S E 0,1,..., P 2

Therefore, for each rj E R' there exist integers mj and sj such that

sjE 0,1,...,P2

and

Then

rjz

UO + sj 1

- mj -

</3+2

rjz ° uo + sj

(mod p).

Since z # 0

(mod p), there exists an integer v1 such that viz = 1 Let u1 - v1uo. Then

r,

u1 + vlsj Reorder the elements of R st; that

(mod p)-

R' -(ro, -1,...,rk,-1) S R - (ro,r1,...,rk,-1, rk... rk-1)

(mod p).

and

05 sp <S1 <... <gk,-1 < P-1

Let and let

for j -O, 1,...,k' - 1. Then

and

The set

(ti,...,tk--i)- I.

T' - (to, ti, .... tk,-i }

is in normal form. If rj E R', then

rj =uI+visj -ui+v1(so+dtj)-u2+v2tj

(mod p), where u2 - u1 + v1so and v2 - vid # 0 (mod p). Let

A' - (rj + pZ : rj E R'} - (u2 + v2tj + pZ : tj E T'} c A.

The following statement, which reduces sums of congruence classes to sums of integers, is the key step in the proof of the theorem. Let j1, j2, .h, j4 E [0, k' - I].

Since each integer tj belongs to the interval [0, (p - 1)/2], it follows that

if and only if if and only if in Z. It follows that

d - (sl - SO,s2 -SO,...,sk-1 -so),

Sj - So

tj d

0-to <ti <... <tk'_i <

^P-

r j, + rj, rj, + rj, (mod p)

tji + t1, = tj, + tj. ^{(mod p)}

tj, + tj, - t j, + tj,

12T'I - 12A'I < 12AI < clk - 3, (2.20) where 2T' is a set of integers and 2A' and 2A are sets of congruence classes modulo P.

If tk,-i > 2k' - 3, then Theorem 1.14 and inequality (2.18) imply that

12T'I>3k'-3>

^{3(l + O)k}²

-3>cik-3,

which contradicts inequality (2.20). Therefore, tk-- i < 2k' - 4 and so

T' c [0, 2k' - 4)

72 2. Sums of congruence classes

and

2T' C [0, 4k' - 81.

Let

T -It E [0, p - 1 ]

: rj = u2 + vet (mod p) for some rJ E R}.

Then T' C T. If there exists an integer t' E T such that

4k'-7<t'<p-2k'+3,

then and

Let r' as u2 + V2t'

T'+{r'}c[4k'-7,p-1]

2T'n(T'+{t'})-0.

(mod p). Since

t'>4k'-7>2k'-2,

it follows that r` E R\R'-{rk',rk'_i,...,rk_,}anda*-r'+pZ E A \ A'. Let

0$JI,J2.J3<k'-I.Since

r;, + rh as 2u2 + v2(tj, + t;,) (mod p)

and

rj,+r' m2u2+v2(tJ,+t)

(mod p),

and since the sets 2T' and T' + {t} are disjoint subsets of [0, p - 1], it follows that no integer in the set

2R'-{rj,+rh, :0 < ji, j2 <k' - 1)

is congruent modulo p to an integer in the set

R'+{r'}-{rj,+r':0<j3<k'-1}.

Since 2R' U (R' + {r'}) is a complete set of representatives of the congruence classes in

2A' U (A'+ {a*)) c 2A C Z/pZ.

it follows from the Cauchy-Davenport theorem and from inequalities (2.18) and (2.20) that

12A1 > 12A'I + IA' U {a'}I

(2k' - l)+k'

3k'-I

> 3(1 +B)k

- l

clk - 1

> cik-3

> 12A1,

which is absurd. Therefore,

TC[0,4k'-8]U[p-(2k'-4),p-1].

The set [0, 4k' - 8] U [p - (2k' - 4), p - I] and the interval

[-(2k' -4),4k'-8] --(2k'-4)+10,6k'- 12]

represent exactly the same integers modulo p, and so for every a E A there exist integers t E T and W E [0, 6k' - 12] such that

a = u2 + vet = u2 + v2(-(2k' - 4) + w) u3 + v2w (mod p), where u3 - U2 - v2(2k' - 4). Let

W-{wE[0,6k'-12]:u3+v2w=a (modp)forsomeaEA).

Since k < co p < p/12 by inequality (2.10), it follows that

6k'-12<6k< 2

Since ci < 2.5 by inequality (2.15), it follows that

12W1=12A1-2k-I +r<cik-3<3k-3,

where 2W is a sumset of integers and 2A is a sumset of congruence classes modulo p. By Theorem 1.16, the set W is contained in an arithmetic progression of length k + b, and so A is contained in an arithmetic progression of length k + b in Z/pZ.