Bernoulli numbers. Francesco Chiatti, Matteo Pintonello. December 5, 2016
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1 UNIVERSITÁ DEGLI STUDI DI PADOVA, DIPARTIMENTO DI MATEMATICA TULLIO LEVI-CIVITA Bernoulli numbers Francesco Chiai, Maeo Pinonello December 5, 206 During las lessons we have proved he Las Ferma Theorem for some special cases of regular primes. We could as if he irreguar primes are infinie or no. Indeed, looing a he firs pages of he lis of irregular primes, i is clear hey are ounumbered by he regular ones, bu now we will prove ha here are infiniely many irregular primes. Furher, i is a curious fac ha, even if we now much many regular primes, nobody has proved ha hey are infinie ye! In general we are no even sure ha, in a generical inerval {,..., N } here are more regular han irregular primes. We recall he definiion of regular primes. We define an equivalence relaion on he se of ideals of Z[ω] (ω a primiive p h roo of uniy) aen wo ideals A and B, we will wrie A B if and only if αa βb for some α,β Z[ω]. I is rue ha here are only finiely many equivalence classes of ideals under, so we can define h as he class number of he ring Z[ω], he number of hese classes. Definiion 0.. A prime p is regular if and only if p h. To develop he heory of regular primes we need some heorems abou Bernoulli numbers, which will be defined in he nex secion. These numbers are really imporan for his reason: Theorem 0.2. A prime number p > 3 is regular if and only if all he numeraors of he Bernoulli numbers B 2,B 4,...,B p 3 are no divisible by p. Indeed he second par of he heorem was he original definiion given by Kummer in 950. The proof of he equivalence of he wo definiions is quie long and can be found in [2]. For his reason we develop some heorems abou Bernoulli numbers before giving he proof of he infiniude of irregular primes.
2 BERNOULLI NUMBERS Definiion.. The Bernoulli Numbers, wih m N, are defined by he formula + e m0 m! m (.) Noe ha he series. converges in some neighbourhood of he origin. We shall no worry abou quesions of convergence now, since we are only considering all series formally, bu i will be useful in Theorem.0. Example.2. Le us compue some Bernoulli numbers, by expanding he lef hand side in.: e ( ! + 3 3! +...) + 2! + 2 3! +... ( 2! + 2 ) ( 3! ! + 2 ) 2 ( 3! ! + 2 ) 3 3! One may chec ha B 0, B 2, B 2 6, B 3 0, B 4 30, B 5 0, B 6 42, B 7 0, B 8 30, B 9 0, B , B 0, B , e ceera. Noice ha all he Bernoulli Numbers, wih m > odd, appear o be 0. This is indeed rue, as we will show afer some: Noaion: If f (x) a 0 + a x a n x n is a polynomial, we denoe f (B) : a 0 + a B a n B n Similarly, if f (x, ) + n0 f n (x) n is a power series, wih f n (x) polynomials n N, we denoe f (B, ) : + n0 f n (B) n We can hus resae he definiion of Bernoulli Numbers: Remar.3. I is easily seen ha (one jus muliplies he wo series on he lef hand side). e eb (.2) e a e B e (a+b) a Z (.3) Theorem.4. All he Bernoulli Numbers wih odd index (excep B ) are 0. 2
3 Proof. Consider he ideniy Noice ha he LHS is even. Infac: ( + e + 2 e e + 2 m2 m! m ) (e ) + (e ) (e )(e + ) e + e e e So, since he RHS mus be even, oo, one has ha ( + + m2 ) ( m! m + + m2 ) m! ( )m Hence which implies 0 m 3 odd. 2 m 3 odd m! m 0 Noice ha he Bernoulli Numbers are conneced wih he sums of series of naural numbers. Infac, if we denoe n S (n) : we can sae he following: Theorem.5. The sums S (n) saisfy he formula a a (m + )S m (n) (n + B) m+ + m (.4) Proof. The proof consiss in a simple double-couning. Consider he expansion of e (n+b) e B : ( e (n+b) e B (n + B) + + (n + B)2 2 ) ( B! 2!! + B 2 2 ) ! Where he (m + )-erm is (n + B) m+ m+ m+ + (m + )! (m + )! m+ (n + B)m+ + m+ (.5) (m + )! Bu, since he equaliy.3 holds a Z, one may wrie e (n+b) e B e B ( e n ) ( e n e ) en n e e r r 0 3
4 n + r 0 m0 r m Noice ha he (m + )-erm is m! m + m0 n + n r m m (n r 0 m! 0 + 0! + m (n) ms m ) m! + m Combining.5 and.6, we conclude he proof. (m + )S m (n) (m + )! m+ (m + )S m (n) m+ (.6) (m + )! Anoher imporan resul, in order o prove he infiniy of he irregular primes, is he Von Saud s Theorem. Firs we need a definiion: Definiion.6. A Bernoulli number is said o be p-inegral if p does no appear in he denominaor of. Theorem.7 (Von Saud). Le p be a prime number and m an even ineger. Then If p m, hen is p-inegral If p m, hen p is p-inegral and p (mod p) Example.8. Le us consider m 6. We have seen ha B Le us pic he prime numbers p and p 2 7. Then p 0 6, hen B 6 42 is p-inegral: infac, does no appear in he facorisaion of 42. p 2 6 6, hen p is p-inegral (because 7 6) and p 6 (mod 7) Proof. By.4, for every m we have: (m + )S m (p) ( p + B ) m+ + m+ m m + B p m+ (m + ) p + Dividing by m +, we ge: Hence S m (p) p + p S m (p) m m m + m + m + B p m+ + m m + B p m+ m + B p m+ m + pm pb We wan o show, by inducion on m, ha p is p-inegral. If m, i is rivially rue. For m >, we claim ha all he erms under he summaion sign are p-inegers which are 4
5 divisible by p [in he ring of p-inegral numbers]. By inducive assumpion, he erm pb is a p-ineger < m. Le us consider he erms m + p m : α m + If p 2, hen, since m + is odd, α is a 2-ineger and is divisible by 2. If p 2, we wrie α in he form m(m )...( + ) α (m + )! Noice ha p occurs in (m + )! : r! wih exponen r p + r p < r p + r p p m r p r 2 r m Hence, p m (m +)! is a p-ineger, and is divisible by p. So, since all he summands are p- inegral (and divisible by p) and S m (p) is an ineger, we deduce ha p is p-inegral and p S m (p) (mod p) [in he ring of p-inegral numbers]. Now If p m, hen p S m (p) m (p ) m p (mod p) (and we already proved ha p is p-inegral) If p m, hen, aing a primiive roo g modulo p, one has ha p S m (p) x So is p-inegral x m p r ( g r ) p 2 m r 0 g r m g (p )m g m m g m 0 (mod p) Theorem.9 (Kummer s Congruence). If p is a prime and m N even such ha (p ) m, hen /m is p-ineger and +p m + p (mod p) m We could say ha /m has period p modulo p on m. Proof. We have o consider he funcion F () g e g e g m m m! m! m (g m ) m (.7) m! m0 wih g a primiive roo modulo p,(a generaor of he muliplicaive group of Z p, a (p )h roo of uniy modulo p) so g p. Now we se (e ) u, so ha F () m0 g ( + u) g u G(u), m 5
6 wih g G(u) ( + u) g u g g ( g ) u u u + as U is divisible by u. As all he coefficiens c are p-inegral oo. Now we can wrie ( g ) g ( g g ) 2 g u } {{ } U G(u) G(e ) c (e ) d r e r u ( U+U 2... ) c u are in he ring of p-inegral elemens, all he r 0 m0 A m m! m. (.8) An crucial remar is ha, if we have a collecion of series wih all coefficiens p-inegral and such ha hey have period p modulo p, hen also a linear combinaion of hese series wih p-inegral coefficiens saisfies he propery. We already now ha he c are p-inegral. Then we have used he Newon binomial formula, so he d r are linear combinaions of he c wih inegral coefficiens (he various binomial coefficiens). Now we can consider he funcions e r wih r 0. We can use heir asympoic expansions so we ge e r r n n! n, n0 hence, by he small Ferma heorem (in he case r 0 mod p, if r 0 mod p by direc compuaion), we have r n+p r n (mod p) We have proved ha he funcions r n are p periodic modulo p (on n). Then, as every A m is a sum wih p-inegral coefficiens of some r n (o be precise, wih n m), we ge ha he A m are p-inegral and p periodic modulo p (his ime on m). We highligh ha he only coefficiens which are no p-inegral are he denominaors n! o he expansion of e r, bu hey are no included in he A m bu are colleced in he denominaors m!. Now comparing he coefficiens of.7 and.8 we see ha (g m ) m! A m (m )!, (he difference m and m is because F () G(u), and ha he series in m in.7 sars from, in.8 sars from 0), so we ge m (g m ) A m. 6
7 As g is a primiive roo modulo p and m (p ), for sure g m mod p. By he small Ferma heorem, he sequence g m has period p modulo p on m oo. Collecing ogeher our resuls, we ge ha he numbers /m are equal o A m /(g m ) and as his sequence is p-inegral and p periodic modulo p on m, he Kummer s Congruence is proved. Theorem.0. The Bernoulli numbers saisfy he formula where ζ(s) n /n s. m 2(2m)! B 2m ( ) ζ(2m), (2π) 2m Proof. To prove his resul we need an expansion of classical analysis: co x x 2 x n 2 π 2 x 2. n Now muliplying by x and using he formula of geomeric series (we can use i in a neighborhood of 0), we ge x co x 2 x 2 n n 2 π 2 x 2 2 n x 2 n 2 π 2 x2 n 2 π 2 2 n m ( x 2 n 2 π 2 ) m x ) 2m x ) 2m 2 /n m( 2m 2 ζ(2m) (.9) π n m( π Now we can expand x co x in anoher way, recalling ha cos x (e i x + e i x )/2 and sin x (e i x e i x )/2i, so ( e i x + e i x ) ( e i x ) x co x i x e i x e i x i x + 2 e i x e i x i x + 2i x e 2i x (2i x) n (2i x) n i x + + B n + B n. (.0) n! n! Now we can compare he coefficiens of x 2m of.9 and.0, so we ge n n2 So he heorem is proved. 2 ζ(2m) ( )m 22m π2m (2m)! B 2m, By his heorem we can esimae he growh of B 2m. Indeed, as ζ(2m) > and by Sirling formula (2m)! > (2m/e) 2m, hen ( m ) 2m B 2m > 2. πe So, dividing by 2m, we ge lim B 2m 2m. (.) m 7
8 2 INFINITUDE OF THE NUMBER OF IRREGULAR PRIMES Theorem 2.. There are infiniely many irregular prime numbers. Proof. By conradicion, we assume here are only finiely many irregular primes p,..., p s. We will consruc anoher prime p, differen from every p i, and hen we will show i is irregular. Le n r (p ) (p s ), r N As we have proved in., he Bernoulli numbers B 2 have he propery ha B 2 2 as, we can obviously choose r large enough so ha B n /n >. Le p be a divisor of he numeraor (in lowes erms) of B n /n. If (p ) n, hen by Von Saud s Theorem.7 p would divide he denominaor of B n exacly one ime, bu by definiion p is a divisor of he numeraor in lowes erms so his is impossible. Therefore (p ) n and so p is differen from p i, i s (and from 2). We can wrie n m + a(p ), 2 m p 3 (Noice ha p 2 is no possible because n is even, so m mus be even oo and p 2 is odd). For Kummer s Congruence.9, we ge m B n n (mod p) again in he ring of p-inegral numbers. Bu by definiion of p, B n /n 0 mod p so /m 0 mod p oo. Bu m was one of he even numbers 2, 4,..., p 3, so by he Theorem 0.2 p is irregular. REFERENCES [] Marcus D:A., Number Fields, Springer, 977 [2] Borevich Z.I:, Shafarevich I.R., Number Theory, Academic Press, 966 [3] Ireland K., Rosen M., A Classical INroducion o MOdern Number Theory, Springer, 990 8
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