1 Euler s idea: revisiting the infinitude of primes


 Constance MargaretMargaret Simpson
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1 8.785: Aalytic Number Theory, MIT, sprig 27 (K.S. Kedlaya) The prime umber theorem Most of my hadouts will come with exercises attached; see the web site for the due dates. (For example, these are due February 4.) There are likely to be typos i all of my hadouts; it would be helpful if you could report these by (icludig oes I poit out i class). Euler s idea: revisitig the ifiitude of primes To begi our story, we tur to Euler s viewpoit o the fact, origially due to Euclid, that there are ifiitely may prime umbers. Euclid s origial proof was quite simple, ad etirely algebraic: assume there are oly fiitely may primes, multiply them together, add, the factor the result. Euler realized istead that a basic fact from aalysis also leads to the ifiitude of primes. This fact is the divergece of the harmoic series which follows for istace from the fact that , N N 2 log log 2 2 N 2 = = ad the right side teds to as N. (We will usually wat a more precise estimate; see the exercises.) O the other had, if there were oly fiitely may primes, the uique factorizatio of positive itegers ito prime powers would imply that ( ) ( ) = =, p p 2 p = p p which would give the equality betwee a diverget series ad a fiite quatity. Cotradictio. Euler s idea turs out to be quite fruitful: the itroductio of aalysis ito the study of prime umbers allows us to prove distributio statemets about primes i a much more flexible fashio tha is allowed by algebraic techiques. For istace, we will see i a upcomig uit how Dirichlet adapted this idea to prove that every arithmetic progressio whose terms do ot all share a commo factor cotais ifiitely may primes. 2 Riema s zeta fuctio For the momet, however, let us tur to Riema s oe paper i umber theory, i which he fleshes out Euler s idea ad fits it ito the theory of complex fuctios of oe variable.
2 He cosidered the series ζ(s) = s for all s C with Re(s) >. Note that for Re(s) >, the series is absolutely coverget; moreover, it coverges uiformly i ay regio of the form Re(s) + π for π >. Cosequetly, it gives rise to a aalytic fuctio i the halfplae Re(s) >. The boudary Re(s) = is sometimes called the critical lie. I the domai of absolute covergece, we ca also write ( ζ(s) = p ) s, p ad this product coverges absolutely ad uiformly for Re(s) + π for π >. (Remider: a product i (+a i) coverges absolutely if ad oly if i a i coverges absolutely.) It follows that ζ(s) = for Re(s) >. For future referece, we ote that the product represetatio is sometimes more useful i the form log ζ(s) = log( p s ) = p p s =. p = We ow show that ζ exteds somewhat beyod the domai of absolute covergece of the origial series. s Theorem. The fuctio f(s) = ζ(s) s o the domai Re(s) > exteds (uiquely) to a holomorphic fuctio o the domai Re(s) >. Cosequetly, ζ(s) is meromorphic o Re(s) >, with a simple pole at s = of residue ad o other poles. Proof. This is a easy applicatio of oe of the basic tools i this subject, Abel s method of partial summatio (or summatio by parts, as i itegratio by parts). Namely, N N a b = a N+ B N (a + a )B, B = b i. = = i= We apply partial summatio to ζ(s) by takig a = s ad b =, so that B =. Rather, we apply partial summatio to the trucated sum N = s, ad ote that the error term a N+ B N = (N + ) s N teds to for Re(s) >. (Warig: i geeral, I am ot goig to be early so verbose whe applyig partial summatio. So make sure you uderstad this example!) 2
3 With that said, we have ζ(s) = ( s ( + ) s ) = + = s x s dx = = s xx s dx. We ca thus write f(s) = s {x}x s dx, i= for {x} the fractioal part of x; the itegral coverges absolutely for Re(s) >, ad uiformly for Re(s) π for ay π >. This proves the claim. We already kow that ζ(s) caot vaish for Re(s) > ; to prove the prime umber theorem, we eed to also exclude zeroes o the boudary of that halfplae. Theorem 2 (Hadamard, de la ValléePoussi). The fuctio ζ(s) has o zero o the lie Re(s) =. Proof (Mertes). See exercises. We will retur to Riema s memoir, establishig more detailed properties of ζ, i a subsequet uit. 3 Towards the prime umber theorem Usig the aforemetioed properties of the zeta fuctio, Hadamard ad de la ValléePoussi idepedetly established the prime umber theorem i 897. We ll follow here a argumet due to D.J. Newma; our presetatio is liberally plagiarized from D. Zagier, Newma s short proof of the Prime Number Theorem, America Mathematical Mothly 4 (997), For x R, write α(x) = p x λ(x) = log p. p x The prime umber theorem the asserts that x α(x). log x 3
4 This is equivalet to because for ay π >, λ(x) log x = α(x) log x p x λ(x) log x ɛ = ( π)(α(x) + O(x ɛ )) log x. x ɛ p x λ(x) x, What we will prove is that the improper itegral λ(x) x dx () x 2 coverges; remember that this meas that for every π >, there exists N such that for y, z N, z λ(x) x dx < π. x2 y (It is much easier to prove that these itegrals are bouded; see exercises.) To the deduce λ(x) x, suppose that there exists λ > such that λ(x) λx for arbitrarily large x. Sice λ is odecreasig, it the follows that for ay such x, λx λx λ(t) t λx t λ λ t dt dt = dt >, t 2 t 2 t 2 x x cotradictio. Likewise, if there exists λ < such that λ(x) λx for arbitrarily large x, the such x satisfy cotradictio. x x λ(t) t λx t λ t dt dt = dt <, t 2 t 2 t 2 λx λx λ 4 The Tauberia argumet We have thus reduced the prime umber theorem to the covergece of the itegral (); we tur to this ext. Cosider the fuctio Φ(s) = ζ (s)/ζ(s); from the logproduct represetatio for ζ, usig partial summatio as i Theorem, ad substitutig x = e t, we fid Φ(s) = (log p)p s + (log p)p s p p =2 ( ) = s λ(x)x s dx + s λ(x) x s dx =2 2e 2st e 3st = s e st λ(e t ) dt + s λ(e t ) dt ( e st ) 2 4
5 Defie the fuctios by the above, f(t) = λ(e t )e t Φ(z + ) g(z) = ; z + z 2e 2(z+)t e 3(z+)t g(z) = f(t)e zt dt + λ(e t ) dt. ( e (z+)t ) 2 Right ow, we kow that the itegral defiig g(z) makes sese for Re(z) >, but we will deduce () (after substitutig x = e t ) ad hece the prime umber theorem if we ca obtai covergece of g(z) i the case z =. (Note that the secod term coverges absolutely for z =, so we oly have to worry about the first term.) The idea is to do this by leveragig complex fuctiotheoretic iformatio about Φ; this sort of operatio is kow as a Tauberia argumet. To be precise, by what we kow about ζ, Φ(s) is meromorphic o Re(s) >, with a simple pole at s = of residue ad o other poles i Re(s). It follows that f ad g satisfy the coditios of the followig theorem. Theorem 3 (Newma). Let f : [, + ) R be a bouded, locally itegrable fuctio, ad defie g(z) = f(t)e zt dt; ote that this itegral coverges absolutely uiformly for Re(z) π for ay π >. Suppose that g(z) exteds to a holomorphic fuctio o a eighborhood of Re(z). The f(t) dt exists ad equals g(). Proof (Zagier, after Newma). For T >, put g T (z) = T f(t)e zt dt; each fuctio g T is etire, ad we wat lim T g T () = g(). For R large (but fixed util further otice), let C be the boudary of the regio {z C : z R, Re(z) ζ} for some ζ = ζ(r) > chose small eough that C lies iside the domai o which g is holomorphic. By the Cauchy itegral theorem, ( ) zt z 2 dz g() g T () = (g(z) g T (z))e + ; (2) 2αi C R 2 z amely, the oly pole of the itegrad is a simple pole at z =, so we simply pop out the residue there. To boud the right side of (2), we separate the cotour of itegratio C ito C + = C {z C : Re(z) } C = C {z C : Re(z) }. 5
6 Remember that we assumed f is bouded; choose B > so that f(t) B for all t. For Re(z) > with z = R, we have g(z) g T (z) = f(t)e zt dt T B T Be Re(z)T = Re(z) e zt dt ad ( ) z 2 2 Re(z) e zt + = e Re(z)T. R 2 z R 2 Sice the legth of the cotour is at most 2αR, the cotributio over C + to (2) is bouded i absolute value by Be Re(z)T Re(z)T 2 Re(z) 2B (2αR) e =. 2α Re(z) R 2 R Over C, we separate the itegral ito itegrals ivolvig g ad g T. Sice g T is etire, its itegral over C ca istead be calculated over the semicircle C = {z C : z = R, Re(z) }. Sice for Re(z) < we have T g T (z) = f(t)e zt dt B T e zt dt Be Re(z)T =, Re(z) as above we boud this cotributio to (2) by 2B/R. Fially, we cosider the cotributio to (2) from g over C ; we are goig to show that this cotributio teds to as T. By parametrizig the cotour, we ca write ( ) z 2 dz g(z)e zt + = a(u)e b(u)t du, 2αi C R 2 z where a(u) ad b(u) are cotiuous, ad Re(b(u)) < for < u < ; the key poit is that a does ot deped o T, so as T the itegrad teds to poitwise except at the edpoits. Sice the itegrads are all bouded, Lebesgue s domiated covergece theorem implies that the itegral teds to as T. (Agai, I m beig more explicit with the aalysis tha I will be i geeral.) We coclude that 4B lim sup g() g T () ; T R sice R ca be chose arbitrarily large, this yields the desired result. 6
7 You might be thikig at this poit that if oe kew g exteded to a holomorphic fuctio o a regio a bit larger tha Re(s), the maybe oe could prove somethig about the rate of covergece of the itegral f(t) dt. I particular, if oe ca exclude zeroes of ζ i some regio beyod the lie Re(s) =, oe should correspodigly get a prime umber theorem with a improved error term. We will see that this is correct i a subsequet uit, at least if we replace the approximatio α(x) x/(log x) with Gauss s approximatio α(x) li(x) (see exercises). Historical aside: the ErdősSelberg method About 4 years after the origial proof, Erdős ad Selberg gave socalled elemetary proofs of the prime umber theorem, which do ot use ay complex aalysis. The key step i Selberg s proof is to give a elemetary proof of the boud x ( ) log log x R(x) R(x/t) dt + O x, (3) log x log x where R(x) = λ(x) x; I will probably say somethig about this result i the sectio o sievig. Usig (3) ad the fact that x R(t) dt = O() (4) t 2 (much easier tha the covergece of the itegral; see exercises), oe the produces < c < such that if there exists α > such that R(x) < αx for x large, the also R(x) < αcx for x large. I fid this step somewhat uelighteig; if you must kow the details, see A. Selberg, A elemetary proof of the prime umber theorem, Aals of Math. 5 (949), Or see Chapter XXII of HardyWright, or Nathaso s Elemetary Methods i Number Theory. Exercises. Prove that there exists a positive costat ϑ such that log = ϑ + O( ), i i= by comparig the sum to a Riema sum for dx. The umber ϑ is called Euler s x costat, ad it is oe of the most basic costats i aalytic umber theory. However, sice it is defied purely aalytically, we remai astoishigly igorat about it; for istace, ϑ is most likely irratioal (eve trascedetal) but o proof is kow. 7
8 2. Let d() deote the umber of divisors of N. Prove that d(i) = log + (2ϑ ) + O( /2 ), i= by estimatig the umber of lattice poits i the first quadrat uder the curve xy =. 3. (Mertes) Fix t R ozero. Prove that the fuctio Z(s) = ζ(s) 3 ζ(s + it) 4 ζ(s + 2it) exteds to a meromorphic fuctio o Re(s) >. The show that if s R ad s >, the log Z(s) = Re(log Z(s)) ca be writte as a series of oegative terms, so Z(s). 4. Use the previous exercise to prove that ζ(s) has o zeroes o the lie Re(s) =. 5. (Chebyshev) Prove that <p 2 p 2 2 ( ) by cosiderig the cetral biomial coefficiet. The deduce that λ(x) = O(x). 6. Let k be a positive iteger. Prove that for ay c >, if we write C R for the straight cotour from c ir to c + ir, the { x s ds ( ) k x lim = k! x R 2αi C R s(s + ) (s + k) x. (Hit: use a cotourshiftig argumet.) 7. (Gauss) Defie the logarithmic itegral fuctio x dt li(x) =. log t (Warig: there is some disagreemet i the literature about what lower limit of itegratio to use.) Prove that li(x) x/(log x), so that the prime umber theorem is equivalet to α(x) li(x). I fact, Gauss oticed empirically, ad we will prove later, that li(x) gives a somewhat better approximatio to α(x) tha x/(log x). 8. Usig the idetity prove that 2 x x log = the deduce (4) by partial summatio. 2 i= p:p i x p i log p, log p = log x + O(), p p x 8