Analytic number theory and quadratic reciprocity

Transcription

1 Analytic number theory and quadratic recirocity Levent Aloge March 31, 013 Abstract What could the myriad tools of analytic number theory for roving bounds on oscillating sums ossibly have to say about algebra? Quite a lot! We take the vaunted quadratic recirocity law of Gauss and rove it by bounding exonential sums sufficiently well. In the rocess we introduce some of the standard techniques of analytic number theory. Secretly the oint of this article is to motivate these surely analytic methods with an algebraic roblem. 1 Introduction Learning analytic number theory is difficult, in no small art because many of the methods seem totally ad hoc to the uninitiated learner. It is tremendously easier to learn the subject with a roblem in mind, and surely the law of quadratic recirocity qualifies as a nice roblem. Of course, it is not at all clear how analysis might be brought to bear on quadratic residues modulo rimes. eally the oint of this article is to resent a bunch of beautiful things to the reader in the guise of an analytic roof of an algebraic fact. Of course there are much faster and clearer roofs of the law! But certainly there can never be enough. We ress on. Theorem 1 (Quadratic ecirocity). Let q be odd rimes. ( ) ( ) q = ( 1) 1 q 1. q ( ) a As a reminder, by we mean the usual Legendre symbol: 1 if a reduces to a square ( ) modulo, 0 if divides a, and 1 otherwise. (So, for instance, 1 + is the number of square roots of a modulo.) By a theorem of Euler, this is congruent to a 1 modulo. Gauss sums Some notation. We write e(x) := e πix, so that the usual Fourier transform of a nice function f is ˆf(ξ) = f(x)e( xξ)dx. 1 a

2 Observe that e( ) : C descends to a ma out of /Z. So, for instance, given an element α Z/nZ, it makes sense to write e(α/n). Finally, we ll write (Z/nZ) for the grou of multilicative units of Z/nZ. Here is a brilliant lea forward, due to Gauss. Definition. Let be an odd rime. The quadratic Gauss sum modulo is: g() := e(k /). k Z/Z We will abuse notation and generalize this definition slightly. Definition 3. Let n be a ositive integer, and χ : (Z/nZ) C a homomorhism (or character ). The Gauss sum associated to χ is: G(χ) := χ(k)e(k/n). k (Z/nZ) The first thing to remember is that ( ) : (Z/Z) C is a homomorhism. Whence generalization? Proosition. Let be an odd rime. G (( )) = g(). Proof. Let s write out the left-hand side. (( )) G = χ(k)e(k/) k (Z/Z) = e(k/), 0 k= e(k/) k where all sums are over k (Z/Z) and our joking notation k = has the (hoefully) obvious meaning (that k is a square). Now, note that e(k/) = 0. k Z/Z (Multily the left-hand side by e(1/) 1 for an easy roof.) So that sum over nonsquares, which remember! did not include zero, is therefore: e(k/). 0 k e(k/) = 1 k= The only thing left to note is that, in (Z/Z), every nonzero square has exactly two square roots, while 0 Z/Z has exactly one.

3 emark 5. The Gauss sums have brothers: Jacobi sums. The two together are analogues of the Gamma and Beta functions of analysis. Using these Jacobi sums, one can, for instance, show that is a cube modulo an odd rime 1 (mod 3) if and only if = A + 7B has a solution over Z. Go read Ireland and osen for more! (I think this observation was first due to Elkies.) These are the usual Gauss sums that come u everywhere in number theory, but we won t see them again it will be enough to deal with the g(). Great, so I ve told you a totally ad hoc definition after waxing oetic about how hard it is to learn from such stuff. But at least we see the Legendre symbol in there so that s a start. Here is another generalization that might have been tugging at you. Definition 6. Let n be a ositive integer. The quadratic Gauss sum modulo n is: g(n) := k Z/nZ e(k /n). Now, writing down the very first definition robably does take a Gauss, because the following miracle relates it to our question. Proosition 7. Let q be odd rimes. ( ) ( ) q q = g()g(q) g(q). (We ll see in a moment that g(q) 0 and so such a division actually makes sense.) Proof. There s really only one thing to do: look at g()g(q) and hoe for the best. So we do that. We get: via g()g(q) = a Z/Z b Z/qZ Now remember the Chinese remainder theorem: is an isomorhism. But under this ma ( a q + b e q Z/qZ Z/Z Z/qZ α (α mod, α mod q) a q + b ( a q, b ). But the squares of Z/qZ are recisely those that reduce to squares in each slot: = (, ). So maybe we should have been more clever to start. Since and q are relatively rime, we can find integers P and Q for which: Q + P q = 1. ). 3

4 Note that, then, and Let s take such P, Q. Note that just because k Z/Z Q 1 P q 1 e(k /) = k Z/Z (mod q), (mod ). ( ) q e(k P/), k Z/Z e(k/) = 0, as noted before (remember the minus sign!). Hence we might have started with: ( ) ( ) q ( a P q + b ) Q g()g(q) = e. q q a Z/Z b Z/qZ But now we are in good shae. The reduction ma takes a P q + b Q ( a, b ) = (a, b), and so as a and b run over Z/Z and Z/qZ, resectively, a P q + b Q runs over the squares of Z/qZ taken with the aroriate multilicities. That is to say, this last exression is exactly as desired. ( q OK, now for a big theorem. Theorem 8. Let n be an odd integer. ) ( ) q e(α /q) = α Z/qZ g(n) = n. ( q ) ( ) q g(q), In articular, g() 0 always, and so g()g(q) 0 for q odd rimes. So indeed the division above was not susect, since we showed that 0 g()g(q) = ±g(q) above. Proof. We just follow our noses and calculate what we can. So we ll instead figure out g(n) = g(n)g(n) = k Z/nZ l Z/nZ = k Z/nZ l Z/nZ ( k l ) e n ( (k l)(k + l) e n ).

5 Via (k, l) (k + l, k l), this becomes k Z/nZ l Z/nZ ( kl e n The fact that n is odd tells us that that factor of can be absorbed into e.g. l. In any case, the inner sum is zero for k 0 mod n (again, multily by e(k/n) 1). Otherwise it is, of course, n. (Interjection: k = 0 now corresonds to k = l in the earlier variables. Thus it stands to reason that this is called the diagonal contribution, which is evidently dominant. This sort of dichotomy occurs all over the lace.) That is to say, our calculation reveals: as desired. g(n) = n, ). Let s slow down for a second and take stock of what we ve just roved. We have some strange sum of q hases on the unit circle and it ends u being of absolute value q. But actually the hases are really comlicated: they secretly deend on some mysterious χ (the Legendre symbol) and also kee twisting around according to this exonential. What a random combination! Well, random is a funny word: let s actually act like we d been choosing (uniformly) random hases off the unit circle and summing u q of them. Well, if q is large, this should (sort of) ring a bell. That is, we re summing a bunch of indeendent and identically distributed random variables (with, say, finite variance), and the central limit theorem (ardon the bell un earlier... ) tells us to exect the thing to be recisely on the order of n. So we could have guessed the answer by giving u on the algebraic structure in advance! So it stands to reason that the absolute value of g() can t ossibly tell us anything dee like quadratic recirocity. And there is only one thing left: the hase. So this big theorem has been reduced to calculating the hase of some exonential sums. 1 3 The sign of the Gauss sum Why sign and not hase? Proosition 9. Let n be a ositive integer and χ : (Z/nZ) = {±1} a homomorhism. G(χ) = χ( 1)G(χ). Proof. G(χ) = k (Z/nZ) χ(n)e( k/n). Now change variables via k k. 1 Now we can see analysis creeing in: n is very far from n! That was essentially my line of thinking when coming u with this... 5

6 So we find that: Corollary 10. Let q be odd rimes. g() = { ± 1 mod, ±i 3 mod, and g(q) = { ± q q 1 mod, ±i q q 3 mod. Let s write ɛ(n) for the sign of the Gauss sum: that is, so that { g() = ɛ() 1 1 mod, i 3 mod. So we re almost there. Gauss also got to this oint retty easily and wanted to know the sign as well (likely for the exact same reason: quadratic recirocity). So he did a bunch of examles and came u with a conjecture. Go ask your comuter for ɛ() for = 3, 5, 7, 11, 13,.... I did, and here s what I got: See a attern? How about ɛ(q)? Well, and So maybe: ɛ(3) = +1, ɛ(5) = +1, ɛ(7) = +1, ɛ(11) = +1 ɛ(13) = +1. ɛ(15) = +1, ɛ(1) = +1, ɛ(33) = +1. Conjecture 11. Let q be odd rimes. ɛ() = ɛ(q) = +1. What does this have to do with quadratic recirocity? Everything! Assume the above conjecture. Proof of Theorem 1 assuming Conjecture 11. By Proosition 7, ({ ) ({ ) 1 1 mod, 1 q 1 mod, ( ) ( ) q i 3 mod. i q 3 mod. = ({ ). q 1 q 1 mod, i q 3 mod. As strange as this may seem, that s it. 6

7 So we just have to rove the conjecture. I should note here that it took Gauss several years to do it. So it is no joke. Poisson summation and smoothing sums Thankfully, we have Fourier analysis nowadays (actually, so did Gauss!). More recisely, we have the Poisson summation formula, which is robably one of the most imortant formulas in mathematics. It states: Proosition 1. Let f : be a function of the Schwartz class. ˆf(n). n Z f(n) = n Z This is, for instance, the key in roving the analytic continuation of the iemann ζ function (and actually all Dirichlet L-functions, Hecke L-functions,... ). Its analogue over F q is the iemann-och formula. One can even comute the secial values ζ(k) with it! (And much more, of course.) So it is no surrise that it is the crux of the calculation. But how can we ossibly aly it? Namely, we have on our hands g(n) = e(k /n), k Z/nZ a finite sum! Well, we re not in such bad shae. The first ste in realizing this as a sum over integers is choosing reresentatives for the elements of Z/nZ inside Z we had might as well take them to lie in an interval. So choose some α Z (you ll see why we write the inequalities this way in a bit), and note that the set of integers k satisfying α < k < α + n is a comlete set of reresentatives for Z/nZ. Then g(n) = e(k /n) α<k<α+n is still a sum over integers, just not all of them. Thinking about it a little, we can of course extend this to a sum over all the integers: just write χ [α,α+n] for the characteristic function of the interval [α, α + n], and observe that: g(n) = k Z χ [α,α+n] (k)e(k /n). But now the summand isn t Schwartz it isn t even differentiable! But that s no roblem. Let Cc ((α, α + n)) be comactly suorted in the given interval and such that 0 1, [ α, α +n] = 1. Then, extended by 0 to have domain all of, is a Schwartz function, and we still have: g(n) = (k)e(k /n). k Z We haven t changed anything excet for the way we ve written our sum: this is called smoothing a sum out. (That such a exists is an absolutely fundamental fact in mathematics, articularly in geometry and analysis.) That is to say, f is infinitely differentiable, and for every a, b N, x a f (b) (x) 0 as x. Actually, much less stringent conditions are required of f, but we ll get away with this statement. 7

8 Now we can aly Poisson summation, and so we do: g(n) = α+n ( ) t (t)e k Z α n kt dt. 5 The stationary hase method The integrals we now have to deal with are of the shae f(x)e(φ(x))dx, with f comactly suorted. Intuitively, if φ is moving quickly i.e., φ is bounded away from zero the factor e(φ(x)) is flying randomly around the unit circle. It is moving so fast that by moving x just a little we can fli its sign. But then the contribution of the first oint, say f(x), is subtracted from f(x + ɛ) for ɛ very small. So now if f varies slowly, we get a bunch of terms like f(x + ɛ) f(x) 0, so this integral should be tiny. This is made recise by integrating by arts: when φ (x) 0 always, f(x)e(φ(x))dx = 1 πi = 1 πi ( ) f(x) ( φ πiφ (x)e πiφ(x)) dx (x) ( ) f(x) φ e(φ(x))dx. (x) eeating this does even better. But it s hard to can methods in analysis, so this is as recise as I can get. Suose now that φ = 0 somewhere, say at X. A Taylor series exansion of φ would then start with φ(x) = φ(x) + φ (X) (x X) +. Oftentimes we are lucky enough to also have φ (X) 0 (let s assume that φ (X) > 0, otherwise comlex conjugate everything), so that φ infinitesimally looks like a quadratic lus higher-order terms about X (its critical oint is called nondegenerate). In our case, we are so lucky. Since the higher-order terms are moving quite a bit slower, and we are assuming f to be slowly-varying, we can kill off the arts of the integral even just a little bit away from X. Very close to X, we may relace f by f(x) and φ by φ(x)+ φ (X) (x X). So our integral is basically equal to f(x)e(φ(x)) x X <O(1) Changing variables, this becomes: f(x)e(φ(x)) φ (X) ( φ ) (X) e (x X) dx. O(1) O(1) e(x )dx. Actually by the same integration by arts argument the bounds of the integral don t really matter, and this is essentially f(x)e(φ(x)) φ e(x )dx. (X) 8

9 This integral, the rotated version of the Gaussian integral, is called the Fresnel integral. One can calculate it by e.g. deforming it slightly to e πi(1+iɛ)x dx and then observing that this is the Fourier transform of a nice function evaluated at zero (whence one may aly analytic continuations of standard Fourier transform identities lus the self-duality of the Gaussian this amounts to a change of variable and a contour shift). Anyway, the end result is that 3 : Proosition 13 (Fresnel). e(x )dx = 1 + i. Hence, in the end, we see that, so long as φ > 0 where φ = 0, f(x)e(φ(x)) f(x)e(φ(x))dx = ζ 8 +, (1) φ (X) φ (X)=0 where ζ 8 := 1+i = e ( 1 8) is a rimitive eighth root of unity. So back to our sum (with which we have to use rigor!). The k-th summand is: α+n ( ) t (t)e n kt dt. Alying what we ve just learned, we study the zeroes of i.e., we wonder if Of course this only haens for α t n k α n k α + n. α n k α n +, so we are in retty good shae: at most three of the integrals will be negligible! Since we are lazy, let s choose α := n (for us n will always be odd) to make the inequalities read 1 k 3, leaving just k = 0 and k = 1. So we exect 3n ( ) t 3n ( ) t g(n) = (t)e dt + (t)e n n n n t dt +. 3 Something else that suggested that this method would work: the reader will notice that there are no minus signs: the real and imaginary arts of the numerator are +1. 9

10 But now, if we believe (1), the first integral becomes and the second becomes That is to say, 1 + i n, 1 + i n e( n/). ( ) 1 + i g(n) = (1 + e( n/)) n +. This is exactly what we hoed for: for n 1 mod, (1 + i)(1 i) =. For n 3 mod, (1 + i)(1 + i) = i. I still can t get over the fact that this works. 6 A tensor ower trick and Hensel s lemma But of course we still have to actually make the thing mathematically rigorous. emember we imagined n was large so we could write. But n isn t large! It could be as small as! Oftentimes in mathematics one can basically rove a bound, u to let s say a constant: f(x) c g(x). If one can deform f and g into a family and rove the bound for all f and g in such a family (with the same constant!), then, if f n and g n lie in this family, we have that f(x) n c g(x) n. Taking n-th owers and taking n removes the constant. Here we have a situation where we can almost rove the result, excet there may be lower-order terms, and so we d like to take a limit as n (and take advantage of the fact that we know that g(n) = n exactly). After some thought the natural thing is to relace n by n k and take k. Why? Let s focus on the case of n = an odd rime for the moment. The claim is that the sum g( a ) = e(k / a ) k Z/ a Z is somewhat related to g() = e(k /). k Z/Z But we already know the absolute values of both: a and 1, resectively. So let s try out what haens when a is odd the best case scenario is that: Proosition 1. g( a+1 ) = a g(). Even though calculations that would qualify as heuristic don t rove anything, they are tremendously imortant in actually doing mathematics. So ractice them! 10

11 But what are the squares modulo e? That is to say, given a k Z/ e Z, how can we tell if the equation t k = 0 has a solution? Certainly by reducing further via Z/ e Z Z/Z a solution modulo e will lead to one modulo. But a lemma of Hensel tells us that that s all: once we have a solution modulo, we can lift it to a solution modulo e for any e. (To rove the thing (and understand the extra condition), remember Newton s method of finding roots of equations!) Lemma 15 (Hensel). Let P (Z/ e Z) [t] be a olynomial such that P (Z/Z) [t], the olynomial with reduced coefficients, has a root ξ Z/Z. Suose there is a unique Ξ Z/ e Z such that P (ξ) 0. Ξ = ξ and P (Ξ) = 0. In our case, we have the olynomial P (t) = t k. Its reduction is P (t) = t k, with (formal) derivative P (t) = t. So once k 0, we re in good shae. Another way of saying this is that the set of squares modulo e certainly mas to the set of squares modulo under the reduction ma. Hensel s lemma tells us that this ma is surjective. That is to say, the squares modulo e are of the form a b, for b 0 and a square modulo. Proof of Proosition 1. OK. So now that we know the squares modulo a+1, we can just calculate. Namely, in general g(n) = k= # {a Z/nZ a = k} e(k/n), and so (writing the nonzero squares of Z/ a+1 Z as e (k + l) with k a square modulo and l free) g( a+1 ) = # {α Z/ a+1 Z α = 0} a + # {α Z/ a+1 Z α = e (k + l)} ( e ) (k + l) e e=0 0<k<, l Z/ (a e) Z k mod = # {α Z/ a+1 Z α = 0} a ( ) k + e (a e)+1 e=0 0<k<, k mod l Z/ (a e) Z a+1 # {α Z/ a+1 Z α = e (k + l)} ( ) l e. (a e) But if α = e (k + l), then α must be of the form e β for some β Z/ a+1 e Z, with β 0 mod. And indeed we are free to choose β so long as it lies over a square root of k + l modulo (a e)+1. But, since is an odd rime, there are at most two square roots to choose from (even modulo a ower of ) of a square: once we ve found one, we write 11

12 down its negative, and that s it. That is to say, there are (at most) two choices of square root modulo (a e)+1, and then e times as many choices of lifts modulo a+1 e. So the count of square roots is e # {α Z/Z α = k} for each! Counting the number of square roots of zero is even easier: ( r s) 0 for s 0 if and only if r a + 1 i.e., the square roots of zero are recisely a+1 s with s Z/ a Z free to roam. That is, there are a many square roots of zero. So we get: g( a+1 ) = a + a e=0 e 0<k<, k mod # {α Z/Z α = k} e ( k (a e)+1 ) l Z/ (a e) Z ( ) l e. (a e) But now that inner sum is zero unless e = a! ( in which case there was no l sum to start with!) That is, g( a+1 ) = a 1 + # {α Z/Z α = k} e(k/) as desired. = a g(), 0 k= mod In fact we have the same deal for g((q) a+1 ): Exercise 16. Show, using the same methods as above (and the Chinese remainder theorem), that: g((q) a+1 ) = (q) a g(q). 7 Proof So now we can finally make everything rigorous. Proof of Conjecture 11 (and hence Theorem 1). As before, by Poisson summation, we have that: g(n) = 3n n 3n ( ) (t)e(t t /n)dt + (t)e n n t + k 0,1 3n n ( ) t (t)e n kt dt. elacing by 1 in the first two integrals changes things by at most a constant. That is, g(n) = 3n n 3n ( ) e(t t /n)dt + e n n t + k 0,1 3n n ( ) t (t)e n kt dt + O(1). Via t nt in the first integral, t nt and then t n t in the second (so that the remaining integrals become the same), we get: g(n) = (1 + e( n/)) 3 n n e(t )dt + n k 0,1 1 3n n ( ) t (t)e n kt dt + O(1).

13 But That is to say, 3 n n e(t )dt = = 1 + i e(t )dt + o(1) + o(1). g(n) = (1 + i)(1 + i n ) n + o( n) + k 0,1 Via t nt in each of the remaining integrals, g(n) = (1 + i)(1 + i n ) n + o( n) + n k 0,1 3n n 3 1 ( ) t (t)e n kt dt. (nt)e ( n(t kt) ) dt. But now integrate by arts twice in each integral that remains (we can do this thanks to the fact that t k 0 always!). The boundary terms vanish thanks to the suort conditions on. We get: g(n) = (1 + i)(1 + i n ) n n + o( n) + (πi) k 0,1 3 1 ( (nt) n(t k) ) n(t k) e ( n(t kt) ) dt. But ( (nt) n(t k) ) n(t k) will have four terms: in the worst case, we differentiate (nt) twice, getting n (nt), and have a denominator of n (t k). But even still, (nt) = 0 inside [ 1 + { n } n, 3 { 3n } ] ( n n, 3n ), leaving an interval of length O ( 1 n). Since the integrand is of absolute value at most a 1 constant (deending on ) times (t k) O ( ) 1 k anyway, the end result is that: ( ) 3 (nt) n(t k) e ( n(t kt) ) ( ) 1 dt O n(t k) nk, 1 and so g(n) = (1 + i)(1 + i n ) n + o( n) + O(1) = (1 + i)(1 + i n ) n + o( n). 13

14 That is to say, g(n) = (1 + i)(1 + i n ) + o(1). n Now take n = a+1 and then n = (q) a+1 and take a. eferences [1] K. Ireland and M. osen, A Classical Introduction to Modern Number Theory, nd Edition, GTM 8, Sringer-Verlag, Berlin-Heidelberg-New York,