On Primes in Arithmetic Progressions (II)

Annals of Mathematics

On Primes in Arithmetic Progressions (II)Author(s): Harold N. ShapiroSource: Annals of Mathematics, Second Series, Vol. 52, No. 1 (Jul., 1950), pp. 231-243Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1969521 .

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ANNALa OF MATSEMATCS Vol. 52, No. 1, July, 1950

ON PRIMES IN ARITHMETIC PROGRESSION (II)

BY HAROLD N. SHAPIRO

(Received April 26, 1949)

1. Introduction In this paper we propose to give another proof of Dirichlet's theorem on the

infinitude of primes in arithmetic progressions. This proof will be to a certain extent simpler and more revealing than the classical proofs, and yet will have much in common with such proofs. Though we shall make use of "characters" we shall not use Dirichlet series. Whereas in the classical proofs one is concerned with such functions as L'(s, X)/L(s, x), where

(1.1) L(s, x) = F x(n) (s > 1), n-=1 no'

and studies the behavior of these functions as s approaches 1, the only series which enter our proof are

(1.2) Lo(%) = E x ni-1 n

and

(1.3) Li(x) = E x(n) log n nil n

where x is a non-principal character. At the expense of some simplicity we shall even indicate how to eliminate the use of these infinite series.

Our method will give directly that for any integer B, relatively prime to A)

(1.4) log P 1 log x + O(1). pmB(A) p (p(A)

poz

In [11 it was proved in completely elementary fashion that

2_ (1.5) E log2 p + E log p log q = x log x + O(x). pB (A) pqwA(A)

Taking (1.5) together with (1.4) we shall show that for

OB(x) = (B, A, x)= E log p p-B (A)

we have for each BA (B, A) = 1,

(1.6) lim OB(x) + lim OB(x) 2

231



232 HAROLD N. SHAPIRO

and for any two such B, B1, B2, (B1, A) = (B2, A) 1,

(1.7) urnO IB,(X) _ ._Of,(x) O7 IB(X) oil, 03( (1.7) lim x - lim im = lim X X X X

From this we conclude that the "prime number theorem for progressions modulo A)" would follow from the prime number theorem for any one progression modulo A. Since the prime number theorem for primes congruent to one modulo A is a direct consequence of the "prime-ideal theorem," we then see that the prime number theorem for progressions may be deduced from the prime-ideal theorem.

2. The main lemma

The device which enables us to rid the proof of (1.4) of all reference to Dirichlet series is a rather simple inversion formula which we now prove.

LEMMA 2.1. Let P(n) be any completely multiplicative arithmetic function (i.e. P(mn) = P(m)P(n) for-all integers m and n), P(1) 1; and let f(x) be any complex valued function defined for all real x > 0. Then if we define g(x) by

(2.1) g(x) = E P(n)f( )

we can conclude that

(2.2) f(x) = E2 (n)P(n)g x

PROOF. Using (2.1) we have

X,(n)P(n)g(&-) = IA (n)P(n) P (m)f(-)

- : jA(n)P(m)P(n)f(i)

-=~2 , (n)P(mn)f(- (msf?mn Setting c = mn this becomes

E M(n)P(n)g(z) = I P(c)f() X

;(n) n_2 n C<z cnC

=f(X).

3. The non-vanishing of Lo(x)

The one main element which our proof has in common with the classical proof is the necessity of knowing that for x a non-principal real character modulo A, Lo(x) # 0. However, this can be done directly without using the theory of Dirichlet series. Such a proof appears in [31, page 434, and for the sake of com- pleteness we shall give a sketch of this proof here.



ON PRIMES IN ARITHMETIC PROGRESSION. II. 233

LEMMA 3.1. Let f(x) be such that for x sufficiently large f(x) goes monotonically to zero as x approaches infinity, then for x any non-principal character modulo A, we have (3.1) nI x(n)f(n) = 0(f(z)).

Thus in particularEA. i x(n)f(n) converges. PROOF. Since x is a non-principal character we have S(x) = n X(n) 0(1),

and hence via partial summation

2 x(n)f(n) = I {S(n) - S(n - 1)}f(n) n>s na ?

- 1 S(n) {f(n) - f(n + 1) I - S(z - 1)f(z)

- 0(2 {f(n) - f(n + 1)}) + 0(f(z)) = 0(f(z)). nfl? Taking the special cases f(x) = 1/x, log x/x, 1/V'x we obtain the COROLLARY. For x a non-principal character modulo A

(a) E-n = Lo(x) + (

(b) E

x(n) log n = i(x) + (log x)

(c) X X( (n) _ 0(1).

LEMMA 3.2. For x a non-principal real character modulo A, Lo(X) 5 0.

PROOF. Defining F(n) = Ed/n x(d), it is easily seen that F(n) is multiplicative and hence that (3.2) F (n) >= fofor all n

I for n a square. From (3.2) it follows that

(3.3) G(x) = F(_) _*c as x-co.

On the other hand

_( = 1 Ex(d) x + x(d) d'5Xv1 dsll~s-n2 dl/2;Ssd d1/2d 1 '

dsVi d'I2 { (T#=) + o + xd x(d) + C + O + 2 {

d-1 d' d




where C is some constant. This gives

G(x) = 2Vx> E (d + C d)2 + 0(1) d=1 d d-1 d

=2v Vx xd)1 + 0 (1) = 2 "x- {Lo (x) + ? + OM}

(3.4) G(x) = 2 /x Lo(x) + 0(1).

Thus if Lo(x) = 0, we would have G(x) = 0(1) which contradicts (3.3).

4. Proof of (1.4) In this section we shall deduce (1.4) from Lemmas 2.1 and 3.2, without the use

of Dirichlet series. We begin by applying Lemma 2.1 in the case where f(x) = x, P(n) = x(n)

(x a non-principal character). Then

g(x) = E x(n) * = x x(n) 1n&? n n&S T n

= Lo(%) x + 0(1). From this we get

x f(x) = A, (n)x(n) QLo(x) +0(1)

- Lo(%)x E (n)x(n) + O(x) n,= n

whence

(4.1) L4(x) x: M(n)x(n) = 0(1).

We note that (4.1) gives us no information aboutl^S (A(n)x(n))/n if it should happen that Lo(%) = 0. Thus in this case we must proceed differently. Assuming Lo(x) = 0, we now apply Lemma 2.1 again with

f(x) = x log x, and P(n) =x(n).

Then

g (x)= x() o n E xn n ln n= f fl

=x log x E: x (n) X. x: x(8) log n ntzx n nsx n

- Lo(x) x log x - L(x) x + O(log x)

-Ll(x) x + O(log x).




From this we get

x log x = (n)x(n) -Ll(X) +0 (log n)} = -Li(X)x E s(n)X(n) + O(x)

n~z ln

whence

(4.2) Li(x) (n)x(n) = -log x + 0(1).' nzx n

It is interesting to note at this point that this proves that if Lo(x) = 0, then Li(X) $ 0. Thus from (4.1) and (4.2) we have

LEMMA 4.1. For x a non-principal character modulo A,

ug(n)X(n) O(i) if Lo(x) s 0, it n-z n -log x + 0(1) if Lo(x) =0.

Next we relate the sums which appear above to primes, to wit LEMMA 4.2. For x a non-principal character modulo A,

(4.3) 1: x(p) log p - Ll(x) z A(n)x(n) + 0(1). p x p nzx n

PROOF.

Ex(p) log p X(E') log P + 0(1) p zX p pa X pa

- aX(n)a,(d) log n + 0(1) nx n din d

E x(dd')A(d) log d' + 0(1) dd'z ddM

- ,Iu(d)x(d) X(d') log d 0(1) dex d d'zx/d d

=E '(d) %(d) {Li(x) + log d)} + 0(1)

= Li(X) E g(d)x(d) + 0(1). d~x d

Combining Lemmas 4.1 and 4.2 we obtain LEMMA 4.3. For x a non-principal character modulo A,

E x(P) log p - 0(1) if Lo(x) $ 0 PS; p -logx + 0(1) if Lo(x) = 0.

1 We might note here that (4.1) and (4.2) also result if in applying Lemma 2.1 we use f(x) = 1, log x, respectively; together with P(n) = x(n)/n.




Letting IV be the number of non-principal characters x for which Lo(x) 0, we have

(o(A) E log p x(P) log p p :5 x PzX p

p I1 (mod A)

(4.4) E lgp+ E Ex(P) log p Vp P x p XAX1 P P

(pA)-1

-(1-N) log x + 0(1)

where in addition to Lemma 4.3 we have used the well-known fact that

E logp = logx+ 0(1). P-z p

Since p(A)Fpg (log p/p) > 0 we see that (4.4) implies that p-1(A)

(4.5) 0 ? N ? 1.

From (4.5) it follows immediately that for x a complex character Lo(x) # 0. For if for such a character Lo(x) = 0, then Lo(x) = 0, and we would have N 2 2. Since Lemma 3.2 provides that Lo(x) < 0 for x a real character, we can conclude that for x any non-principal character modulo, A, Lo(x) $) 0. Then from Lemma 4.3 we have for all non-principal characters x,

(4.6) ~ ~ ~ Fx (P) log P = 0 (1). (4.6) t5 ~)

Using (4.6) we have for (B, A) = 1

p(A) E, log P - , %(B) , x (P) log p Pat P x Rsu P

= log P + ?: %(B) E X(P) log P pmi(A) ~ (V,)-

= log x + 0(1)

so that finally we have

(4.7) log P - - log x + 0(1). pnB(A)

5. Some remarks on the above proof As has been noted we have eliminated from the proof of Dirichlet's theorem

the presence of the variable s, and the consequent passage to the limit as s -+ 1. However, in the above proof, there still remains the infinite process represented by the use of the infinite series (1.2) and (1.3). It thus suggests itself to attempt to replace the use of (1.2) and (1.3) in the proof, by the use of finite sums. This




can be done, though at the expense of some simplicity, and we shall sketch the procedure here.

Instead of Lo(x) and L1(X) we consider

(5.1) Lo(x, x, y) = E x(n) z<R?v n

and

(5.2) Li(x, x, y) = E x(n) log n z<nS,, n

Then by an argument similar to that used in proving Lemma 3.1, we obtain LEMMA 5.1. For x a non-principal character modulo A,

(5.3) Lo(x, x, y) = 0(1/x) + 0(1/y) and (5.4) Li(x, x, y) = O(log x/x) + O(log y/y)

Next we must rephrase the statement and proof of Lemma 3.2 to LEMMA 5.2. For x a real non-principal character modulo A, Lo(X, 0, x) =

Lo(X, x), there exists a 6 > 0, and an xo such that for x > xo

(5.5) 1 Lo(x, x) I > 6. PROOF. Following the proof of Lemma 3.2, we note that (3.3) yields

G(x) = F(n) > > c log x, (c>0), XRgS n12 m?~M

Then from the equation just preceding (3.4) we see that z x(d) 20xExd + 0 (1) = a (z) > cl log x

-. d which implies that

(5.6) I~~~~o~x~v'-) I> C2 log - c>0 (5.6) ILO(X, ) I VX 10 / (C2 > ?)

Setting x = z2 in (5.6) yields

(5.7) ILo(xIz)I>c2 g - 2

Thus for any y > z we have

Lo(x, y) | 2 ILo(x, z) |- ILo(x, z, y) I > ci lg z + o + (

(5.8) ~~~~= c2 log z + O z

log z 2




for z sufficiently large. Fixing z as any one of these, say z = zo > 1, for which (5.8) holds, we see that the Lemma follows with xo = , and 5 = c3 log zO/zO.

Using Lemma 2.1, we find as in Section 4 that for a non-principal character x,

(59 A(n)x(n) E -(m) 0(1). neax n m?,In m

Thus we get for y > x

(: E(n)x( 1: x (m)( _ E)(n))x((n) Exx(m) nSx n m:y M nSx n MSx IsM

(n~x n ) 5 (mvm)

so that using (5.3) yields

Lo(x, y) i ,. (n)x(n) + 0(1) = E t(n)x(n) x n~x fl nx fl \fl

O(1) + 0 (og x) 0(1).

Thus we have for all y > x,

(5.10) Lo (, y) E t(n)x(n) 0(1). nSx n

We note that (5.10) is the finite version of (4.1), the only difference being that the Lo(x) appearing in (4.1) is replaced by the partial sum Lo(X, y). Whereas before we were faced with the alternatives arising from whether or not Lo(x) was zero, we now have the two cases

(a) Lo (x, y) F6 o(l)

(A) Lo(X, y) = o(1). Case (a): Lo(x, y) 54 o(l).

This implies that there is an infinite sequence of yi -X and a constant 8 > 0, such that

(5.11) j Lo(xI yi) > 5.

The function y = y) = mini Iy I y 2 x} is then defined for all x, and we have

ILo(x Y(x) > 5.

Since y(x) > x, (5.10) holds, and we can conclude that

(5.12) E (n)x(n) = 0(1). nxz n

Case (A): Lo(x, y) = o(l).




From Lemma 2.1 we obtain

(5.13) E X( log xr = log x. n<f n m<z/n m mn

Then for y 2 x,

( nxn _ x(m) log iu) -log x

(5.14) n?z n m<Y m mn

A(n)x(n) l x(m)log x n5z fn (z/n)<m"v m mn

Since by (5.3) and (5.4)

xm) log- = log-Lo -x,,y -Li x,-) (/n)<my mn n n n

=o (log x/n) + o (log x/n) + o (log y)

we see that

W'(n)x(n) x(m) l = 0(1) + o (log2 x) + o (og x log y) n?z n (x/n)<m?2y m mn y Y

and hence since y x,

IAL(n)x(n) X(m) log x = 0(1). n~z n (z/n)<m5y m mn

Thus from (5.14), for y 2 x,

-I (n)X(n)) ( xm) logm + ( ? M(n)x(n) log x/n) (E X(mi

=logx+ 0(1).

But IA(n)x(n) log x/n Lo(X, y) ? c log2 x j Lo(X Y) I

ng z n

and since Lo(, y) = o(l), there exists a yi = y1(x) such that

ILo(x, Y) I ' clog'

for y 2 yi . Thus for y 2 y* = max tyi, x}

(5.15) L (,,) E IA(n)x(n) - -log x + 0(1). ngxz n

(5.15) thus gives us a finite analogue of (4.2).




Finally we note from the proof of Lemma 4.2 that

x(P) log p Eu (d)x(d) E x(d') log d' + 0(1). pVz P d:!s d d'px/d d'

Hence for y > x,

_ 1:x(P) log P + Li(x, y) E (d)X(d) -= ji(d)x(d) ( Xy) + O(M)

1,5X p d!z d d x d d

- 0(l) + o(log Y log X) + 0(1)

- 0(1),

so that for y 2 y*,

(5.16) E x(p) log p - Lu( , n) E f(n)x(n) + 0(1).

PAzX P n&S n From here the argument proceeds exactly as in Section 4. Letting N' = the number of non-principal characters for which Lo(x, y) = o(1) we find that N' is 0 or 1. Consequently, if Lo(x, y) o(l) then x is real; and this possibility is excluded by Lemma 5.2. Hence N' = 0, and we obtain (4.7).

6. Some consequences of (1.5) In this section we propose to deduce (1.6) and (1.7) from (1.5) by means of

(1.4). THEOREM 6.1. (a) For any two integers B, B', (B, A) = (B', A) = 1,

lim -B(X) = hIm 8 X X

lim 018 W -lim O's I )

- X - X

(b) lim 8(Z) + Fl:a( )- (x) 2

PROOF. Let

aB = lim (), B =limfOB(x) - x x

Then given any e > 0, we can find an xo = xo(e) such that for x > ze,

(6.1) 0B(z) 2 aB(l - e)x

foraUBA 1 < B ? A - 1, (A, B) 1. From [jl]itiseasilyseen that (1.5) may be rewritten as

(6.2) 0a(x) log x + E log poi,-I (-) = ()log X + O(X). (VpA ) 1




Applying (6.1) we obtain

() x log x + O(x) = B(x) log x + log pOBp 1 (p) (P,A)-1

+ I log pO,2, > ( B(x) log x + (1 - O)X log 2Pajp zlXo< p~z P p-52z0o P

(p,A)-1 ( p,A)=1

+ O( E log p) _OB (x) log x + (1 -e)x a, E log p zIZO< pz:g 1?i A .I-1 p 5z P

(iA)1 ulA p-B- (A)

+ 0(x).

Using (1.4) this yields

x2 x log x + 0(x) ? GB(x) log x + lg(A) x + -l $

~~~~~p(A~ ~~~~~~~~F,) 1iA1

From this we get

2(A) _ (A) 1; i5A-l (s,. )-1

and letting e -O 0 gives

(6.3) p <-2(A) (A) ii- .

Similarly we can prove that

(6.4) aB = 2 _ E

Noting that we have demonstrated (6.3) and (6.4) for all B such that (B, A) = 1, we may combine them as follows:

f3B+ (A) =at (A)

2 1 ___

(A- p(A)

so that

#j 2 OS_ -(A) p(A)

or

(6.5) B ? (A)<




Similarly we can show that

(6.6) aB= ? ia,.

Next let m8 = A)x 1' a = (iA) ai 1' so that from (6.5) we get

/3- (A) = pi-A

and hence PB = # for all (B, A) = 1. In the same way, using (6.6) we get that aB = a for all (B, A) = 1. This completes the proof of part (a) of the theorem. Using this (6.3) and (6.4) now become

a+ < (A) and _ 2

whence we get

a + ~ 2

(A) COROLLARY.

lim (x) < 1 < 6 x, B) - x -so(A) x

We can now prove THEOREM 6.2. The prime number theorem for progressions, i.e.

(6.7) lim OB( ) = (for (B1 A) = 1), x-( - ~p (A) ' (fr(,A=1)

may be derived as a consequence of the prime-ideal theorem. PROOF. We note from conclusion (a) of Theorem 6.1 that we need only prove

(6.1) for a single B, (B, A) = 1, in order to establish it for all B relatively prime to A.

Making use of the prime-ideal theorem as applied to the cyclotomic field K of the Ath roots of unity we shall now derive (6.7) in the case B = 1. From the prime-ideal theorem we have

E log No = x +o(x), Npxz

which implies (6.8) S log Np = x + o(x).

} Np:X A, p of degree I

But the prime ideals of K which do not divide (A), and which are of the first degree, are exactly those for which

n/p, p -1 (mod A).




Furthermore, the corresponding rational primes p -1 (mod A) split completely in K into sp(A) such prime ideals of degree one. Thus we obtain from (6.8)

x + o(x)= E log Np =v(A) E log p + 0(1) Np5z p5z

}A, p of degree 1 pml(A)

which may be rewritten as

lim 01( _00 x so(A)

7. Bounds for OB(x)

We note that from the corresponding result for all primes it is clear that OB(x) = O(x). Using (1.4) and the elementary theorem proved in [21 it follows that there exists a cl > 0, such that for all sufficiently large x,

OB(x) >

X

for all B relatively prime to A. Combining this with (6.2) it follows immediately that there exists a c2 , 0 < C2 < 2, such that

OB (X)< 012 +0 X ) p(A) + log x)

NEW YORK UNIVERSITY

REFERENCES

[11 HAROLD N. SHAPIRO, On Primes In Arithmetic Progression (I). Ann. of Math., Vol. 51 (1950) pp 217-230.

[2] HAROLD N. SHAPIRO, On The Number of Primes Less Than Or Equal x (submitted to Bulletin of American Mathematical Society).

[3] E. LANDAU, Vorlesungen uiber Zahlentheorie, Vol. I.



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On Primes in Arithmetic Progressions (II)