MATH 361: NUMBER THEORY EIGHTH LECTURE

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1 MATH 361: NUMBER THEORY EIGHTH LECTURE 1. Quadratic Recirocity: Introduction Quadratic recirocity is the first result of modern number theory. Lagrange conjectured it in the late 1700 s, but it was first roven by Gauss in From a naive viewoint, there is no aarent reason for it to be true, but now we recognize it as the first of a family of recirocity laws, themselves art of class field theory, itself art of the famous Langlands rogram. Results u to now in the course, such as the Sun Ze Theorem, the cyclic structure of (Z/Z, or Hensel s Lemma may have been leasing, but they have been essentially unsurrising. By contrast, the quadratic recirocity is surrising. Thanks to the results mentioned in the revious aragrah, we have essentially reduced the general congruence in one variable to the case of rime modulus f(x 0 mod, rime. Here f is understood to be a olynomial with integer coefficients. If f has degree 1 then we have a fairly comlete theory. The next case is that f has degree, i.e., f is quadratic. Assuming that is odd (i.e., excluding, the general quadratic congruence modulo reduces as in high school algebra to X a mod. So the question is whether for a given value of a, the congruence has a solution. Clearly this deends only on a mod, and so we may treat a as an equivalence class modulo. Definition 1.1. A nonzero square in Z/Z (i.e., a square in (Z/Z is called a quadratic residue modulo, or just a quadratic residue when is clearly understood. Proosition 1.. Half of the elements of (Z/Z are quadratic residues. Proof. One roof is to observe that there are at most ( 1/ squares because 1 ( 1, (, ( ( Furthermore, these squares are all distinct because for any a,b Z/Z, a b (a b(a+b 0 b ±a. This comletes the argument. A second roof is to recall that (Z/Z is cyclic, (Z/Z {g 0, g 1, g,, g }, 1

2 MATH 361: NUMBER THEORY EIGHTH LECTURE so that (omitting some details the squares are recisely the even owers of the generator g. This second argument shows incidentally that and the roduct of two quadratic residues is again a quadratic residue, the roduct of two quadratic nonresidues is a quadratic residue, and the roduct of a quadratic residue and a quadratic nonresidue is a quadratic nonresidue. The obvious question now is Which values a (Z/Z are the quadratic residues? (It erhas deserves note that in the multilicative grous R + {ositive real numbers} C {nonzero comlex numbers} all elements are squares, while in the multilicative grou (Z/8Z {1,3,5,7} only 1 is a square. The circumstance of half the elements being squares is not general.. Some Results Proved by Multilying Things Together From now on, always denotes an odd rime. The roof of Fermat s Little Theorem roceeds as follows. For any a (Z/Z, observe the equality of sets (with no reference to the sequence in which the elements aear {1,, 3,, 1} {a, a, 3a,, ( 1a}. This is because of the cancellation law in (Z/Z (if xa ya then x y and the fact that (Z/Z is finite. From the set equality it follows that which is to say, 1 3 ( 1 a a 3a ( 1a, 1 3 ( 1 a ( 1. Cancel the nonzero element 1 3 ( 1 to get the desired result, a 1 1. The roof of Wilson s Theorem is very similar. Again working in (Z/Z, the roduct 1 3 ( 1 consists of airwise roducts of elements and their (multilicative inverses excet that each of 1 and 1 is its own inverse. Thus the roduct is 1, i.e., it is 1 since we are working modulo. The roof of Euler s Theorem is virtually identical to the roof of Fermat s Theorem. Let n be any ositive integer, and let a be any element of (Z/nZ. From the set-equality {b (Z/nZ } {ab : b (Z/nZ }

3 MATH 361: NUMBER THEORY EIGHTH LECTURE 3 we have the equality of roducts (recalling that φ(n is by definition the number of elements of (Z/nZ b ab a φ(n b, b (Z/nZ b (Z/nZ b (Z/nZ and we may cancel the roduct b (Z/nZ b to get a φ(n 1. In the next section we will continue to use the idea of the three roofs reviewed here. 3. The Legendre Symbol, Euler s Lemma, and Gauss s Lemma Letabeanyinteger, andletbeanoddrime. DefinetheLegendre symbol(a/ as follows: ( a 0 if a, 1 if a and a is a square modulo, 1 if a and a is not a square modulo. That is, ( a (the number of solutions of X a mod minus one. Determining the number of solutions of the congruence X a mod is equivalent to evaluating the Legendre symbol (a/. The definition of (a/ instantly connotes that As a function of its numerator, (a/ deends only on a mod. Also (see the end of the roof of Proosition 1., the cyclic structure of (Z/Z does most of the work of showing that As a function of its numerator, (a/ is multilicative. That is, ( ( ( ab a b. What is not at all obvious is that also, for ositive a, As a function of its denominator, (a/ deends only on mod 4a. Later we will see that this is one statement of (most of quadratic recirocity. Euler s Lemma rovides a formula for the Legendre symbol. Lemma 3.1 (Euler s Lemma. Let be an odd rime, and let a (Z/Z. Then ( a a ( 1/. Proof. Consider the olynomial factorization (in (Z/Z[X] X 1 1 (X ( 1/ 1(X ( 1/ +1. Since the left side has 1 roots in Z/Z, so does the right side, and so similarly to arguments that we have given in class, each factor of the right side has ( 1/

4 4 MATH 361: NUMBER THEORY EIGHTH LECTURE roots. The ( 1/ squares in (Z/Z satisfy the first factor in the right side because for any square a, by Fermat s Little Theorem, (a ( 1/ a 1 1. Therefore the ( 1/ nonsquares satisfy the second factor in the right side. In sum, for any a (Z/Z, { a ( 1/ 1 if a is a residue, 1 if a is a nonresidue ( a. Euler s Lemma already suffices to comute the Legendre Symbol (a/ in any secific case, esecially since we have a fast raise-to-ower algorithm. For examle, for any odd rime, ( { 1 ( 1 ( 1/ 1 if 1 mod 4, 1 if 3 mod 4. But Euler s Lemma does not rovide structural insight to the behavior of the Legendre symbol. A first ste in that direction is rovided by Gauss s Lemma. Lemma 3. (Gauss s Lemma. Let be an odd rime, and let a (Z/Z. Consider the set T {a, a,, ( 1 a}. Let ν be the number of elements of T that lie in {(+1/,, 1}. Then ( a ( 1 ν. Proof. Notwoelementsoftheset{1,, ( 1/}areequaloradditivelyoosite in Z/Z. Consequently, the observation that for any b,c Z/Z, ba ±ca b ±c, shows that also no two elements of T are equal or additively oosite in Z/Z. Let x 1,, x µ be the elements of T that lie in {1,, ( 1/}, and let x 1,, x ν be the elements of T that lie in {( + 1/,, 1}. Then because no two elements of T are equal or additively oosite in Z/Z, we have the set equality {x 1,, x µ, x 1,, x ν} {1,, 3,, ( 1/}. Now, in the sirit of the roofs of Fermat s Little Theorem, Wilson s Theorem, and Euler s Theorem, multily all the elements of T to comute that a a 3a 1 a x 1 x µ x 1 x ν ( 1 ν x 1 x µ ( x 1 ( x ν ( 1 ν

5 MATH 361: NUMBER THEORY EIGHTH LECTURE 5 That is, a ( 1/( 1 Since the factorial is invertible, it cancels,! ( 1 ν ( 1!. a ( 1/ ( 1 ν, and we are done by Euler s Lemma. As an examle of using Gauss s Lemma, we comute the Legendre symbol ( 5, an odd rime. That is, the a in Gauss s Lemma is now 5, and we need to study the set T { 5, ( 5, 3 ( 5,, 1 ( 5}. The most negative of these is less negative than 5/, and so to count the relevant elements we need to intersect T with the three right halves as Z/Z reeats ever leftward in Z. That is, we need to count the number of T-elements that fall into the union of intervals ( 5/, ( 3/, ( /,0. Using the Division Theorem, write the arbitrary rime as 0q +r, r {1, 3, 7, 9, 11, 13, 17, 19}. (Wait, what th-?!...why twenty? The T-elements that fall into ( /,0 take the form 5k where or equivalently, 0 < 5k < /, 0 < 10k < 0q +r. Similarly, the T-elements that fall into ( 3/, take the form 5k where or equivalently, < 5k < 3/, 40q +r < 10k < 60q +3r. Finally, the T-elements that fall into ( 5/, take the form 5k where or equivalently, < 5k < 5/, 80q +4r < 10k < 100q +5r. We need only to count the total number ν of aroriate k-values modulo, so it suffices to count the k-values such that 10k (0,r (r,3r (4r,5r. (This is why we took 0q + r: only r affects the value of ( 5/, so we are simultaneously calculating ( 5/ for all rather than for one at a time. For

6 6 MATH 361: NUMBER THEORY EIGHTH LECTURE general (a/ we take 4aq +r. Now we can make a table of the ossibilities as mod 0 varies through all ossibilities: r Conditions ν ( 5/ 1 10k (0,1 (,3 (4, k (0,3 (6,9 (1, k (0,7 (14,1 (8, k (0,9 (18,7 (36, k (0,11 (,33 (44, k (0,13 (6,39 (5, k (0,17 (34,51 (68, k (0,19 (38,57 (76, That is, for any odd rime, ( { 5 1 if 1,3,7,9 mod 0, 1 if 11,13,17,19 mod 0. As an exercise, you should show using Gauss s Lemma that for any odd rime, ( { 1 if 1,7 mod 8, 1 if 3,5 mod 8, and then show by an elementary argument that a convenient formula encasulating this result is ( ( 1 ( 1/8. 4. Two Motivating Examles For our first examle, consider the ring R Z[i] of Gaussian integers. Which rimes in the ring Z of rational integers factor further in the larger ring R? One can verify that i(1+i, 3 does not factor, 5 (+i( i, 7 does not factor, 11 does not factor, 13 (3+i(3 i, 17 (4+i(4 i, 19 does not factor, 3 does not factor, 9 (5+i(5 i, 31 does not factor, 37 (6+i(6 i. Ignoring the rime, which is somehow behaving differently from the others, the attern is: The 1 (mod 4 rimes factor and the 3 (mod 4 rimes do not.

7 MATH 361: NUMBER THEORY EIGHTH LECTURE 7 That is, by our calculation immediately after Euler s Lemma: ( 1 The Legendre symbol indicates how behaves in Z[i]. Foroursecondexamle,considertheringR Z[ 5],anonuniquefactorization domain. We have heard that in this ring it is the ideals that factor uniquely. In articular, each rational rime determines an ideal R in the ring. Which such ideals factor in R? One can verify that R (R+(1+ 5R, 3R (3R+(1+ 5R (3R+(1 5R, 5R ( 5R, 7R (7R+(3+ 5R (7R+(3 5R, 11R does not factor, 13R does not factor, 17R does not factor, 19R does not factor, 3R (3R+(15+ 5R (3R+(15 5R, 9R (9R+(13+ 5R (9R+(13 5R, 31R does not factor, 37R does not factor. This time the rimes and 5 are different from the others, but otherwise, it turns out that the attern is: The 1,3,7,9 (mod 0 rimes factor, the 11,13,17,19 (mod 0 rimes do not. That is, by our calculation immediately after Gauss s Lemma: ( 5 The Legendre symbol indicates how behaves in Z[ 5]. Recall that the definition of (a/ instantly connotes that for a fixed odd rime, As a function of its numerator, (a/ deends only on a mod. Also, the cyclic structure of (Z/Z does most of the work of showing that As a function of its numerator, (a/ is multilicative. That is, (ab/ (a/(b/. But what the revious two examles have shown is that for a fixed a, We are interested in (a/ as a function of its denominator. Thus: We want a relation between the Legendre symbol as a function of its numerator (a function that we understand, and the Legendre symbol as a function of its denominator (a function that we care about.

8 8 MATH 361: NUMBER THEORY EIGHTH LECTURE 5. Statements of Quadratic Recirocity From now on, and q denote distinct odd rimes. Euler conjectured the already-mentioned condition that for ositive a, As a function of its denominator, (a/ deends only on mod 4a. More secifically, Euler conjectured that (1 ( q 1 { x mod 4q for some x if 1 mod 4, x mod 4q for some x if 3 mod 4. This is one statement of (most of quadratic recirocity. Note that half the odd values modulo 4q take the form x while the other half take the form x. Another statement, due to Legendre, is that for all distinct odd rimes and q, ( ( q ( ( 1 ( 1/ (q 1/ q or, equivalently, ( ( q q { 1 if at least one of and q is 1 modulo 4, 1 if both and q are 3 modulo 4. This form of quadratic recirocity is leasingly symmetric in and q. Proosition 5.1. Euler s conjecture (1 and Legendre s formulation ( are equivalent. (Note: The roosition does not assert that either formulation of quadratic recirocity is true, only that the truth of either imlies the truth of the other. Proof. Introduce the quantity { ( 1 ( 1/ if 1 mod 4, if 3 mod 4. That is, whichever of ± equals 1 mod 4. With introduced, Legendre s formulation ( of quadratic recirocity, ( ( q ( 1 ( 1/ (q 1/, q becomes ( 1 q ( 1/ ( q or, by the established formula for ( 1/q, which is to say, ( 1 ( 1/ (q 1/ ( ( q ( 1 ( 1/ (q 1/, q ( q ( q ( 1 ( 1/ (q 1/, ( q 1.

9 MATH 361: NUMBER THEORY EIGHTH LECTURE 9 So Legendre s formulation of quadratic recirocity rehrases as an equivalence, ( ( q 1 1. q The equivalence further rehrases as ( q 1 x mod q for some x, and in fact, since x mod q (q x mod q, it rehrases as ( q 1 x mod q for some odd x. Both andx equal1mod4(becausealloddsquaresequal1mod4, solegendre s formulation is ( { } q 1 x mod q x for some x, mod 4 which is, by the Sun-Ze Theorem, ( q 1 x mod 4q for some x. This is Euler s conjecture (1 of quadratic recirocity. 6. Proof that (a/ Deends Only on mod 4a Let be an odd rime, and let a be corime to. Consider the integer multiles ak of a that lie in a right half modulo, (m 1/ < ak < m. We show that the number of such multiles of a deends only on mod 4a. Write Q4a+r with 0 r < 4 a. The revious dislay becomes (m 1Qa+(m 1r/ < ak < 4mQa+mr. Theinterval ( (m 1Qa+(m 1r/,4mQa+(m 1r/ ] containsqmultiles ofa,anevennumber. Soweneedonlytocountthearityofthenumberofintegersk such that 4mQa+(m 1r/ < ak < 4mQa+mr. We may shift both endoints by a multile of a and count instead the arity of the number of integers k such that (m 1r/ < ak < mr. This deends on r but is indeendent of Q, as desired

10 10 MATH 361: NUMBER THEORY EIGHTH LECTURE 7. Lattice Point Counting Proof of Quadratic Recirocity Proosition 7.1. Gauss s Lemma imlies Legendre s formulation ( of quadratic recirocity. Proof. Recall the set T {q, q,, ( 1 q} {qi : 1 i 1 }. Recall that the elements of T {0,, ( 1/} are denoted x 1 through x µ, while the elements of T {(+1/,, 1} are denoted x 1 through x ν, and that Gauss s Lemma asserts that (q ( 1 ν. Write each element of T as qi qi +r i, 0 < r i <. If 1 r i ( 1/ then r i x j for some j, but if (+1/ r i 1 then r i x j for some j. To rove the roosition, we comute the arity of the sum in two different ways. First, ( 1/ i1 Second, the sum is also ( 1/ i1 ( 1/ qi q qi i1 ( 1/ i1 µ x j +ν + j1 ( 1/ i1 ( 1/ i1 ( 1/ i1 qi µ i q x j + j1 ν j1 qi +r i qi + qi + The revious two dislays combine to give and thus x j µ x j + j1 µ x j + j1 ν S(q, mod where S(q, ( q ( 1 S(q,. ν ( x j j1 mod. ν j1 ν j1 x j x j ( 1/ i1 mod. qi,

11 MATH 361: NUMBER THEORY EIGHTH LECTURE 11 Exchanging the roles of and q in the argument now gives ( ( q ( 1 S(,q+S(q,. q But S(,q + S(q, is the number of lattice oints in the box with lower left corner (1,1 and uer right corner ( 1, q 1 (see figure 1. That is, ( ( q ( 1 ( 1/ (q 1/. q This is Legendre s formulation ( of quadratic recirocity. q q Figure 1. Lattice oints 8. First Version of the Algorithm Algorithmically, the useful form of the quadratic recirocity law is that for all distinct odd rimes and q, ( ( q ( 1 ( 1/ (q 1/ q The utility of this formula is that for the left side we may assume that < q, but then on the right side we may relace q by q%. And so evaluating the right side is a strictly smaller roblem, and this rocess can be iterated until and q get small quickly. There are also two auxiliary quadratic recirocity results: For all odd rime, ( 1 ( 1 ( 1/ and ( ( 1 ( 1/8 or, equivalently, ( 1/ is 1 if is 1 modulo 4 but is 1 if is 3 modulo 4, and (/ is 1 if is 1 or 7 modulo 8 but is 1 if is 3 or 5 modulo 8. We have already roved the auxiliary laws.

12 1 MATH 361: NUMBER THEORY EIGHTH LECTURE For an examle, ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( 1 ( 1 1. This calculation tells us that 017 is a square modulo 5003 without finding the square roots. (In fact they are 606 and Seeding U the Algorithm: the Jacobi Symbol For any integer a and any ositive odd integer P the Jacobi symbol (a/p is defined as follows: ( a ( ei a where P ei i P. i i When P is an odd rime (understood to be ositive, the Jacobi symbol is the Legendre symbol. However, for nonrime P, the condition (a/p 1 does not necessarily imly that a is a square modulo P, even though the condition (a/p 1 does imly that a is not a square modulo P. The emty roduct case of the revious dislay says that (a/1 1 for all a. In a moment we will rove that the quadratic recirocity rules extend to the Jacobi symbol. That is, for all odd ositive corime P and Q, ( P Q ( Q P ( 1 (P 1/ (Q 1/ i and for all odd ositive P, ( 1 ( 1 (P 1/ and P ( ( 1 (P 1/8 P The Jacobi symbol rules further seed Legendre symbol calculations since know we can use Jacobi symbols en route. For examle, ( ( ( ( ( ( ( ( ( This examle demonstrates that Jacobi symbol calculations roceed as quickly as the Euclidean algorithm. Determining whether a quadratic congruence has solutions is as fast as solving a linear congruence. The following result hels to rove that the quadratic recirocity rules extend to the Jacobi symbol.

13 MATH 361: NUMBER THEORY EIGHTH LECTURE 13 Lemma 9.1. Let 1,, k be odd rimes, not necessarily distinct, and consider their roduct P i i. Then P 1 i i 1 mod and P 1 8 i i 1 8 mod. Proof. For the first congruence, comute that P i (1+( i 1 1+ i ( i 1+ i<j( i 1( j i ( i 1 mod 4, and so P 1 i i 1 mod. Similarly for the second congruence, P i (1+( i 1 1+ i ( i 1+ i<j( i 1( j i ( i 1 mod 16, and so P 1 8 i i 1 8 mod. It follows from the lemma that for any ositive odd P i i and any ositive odd Q j q j corime to P, ( ( P Q Q P i,j ( i q j i,j ( qj i i,j ( ( i qj ( 1 (i 1/ (qj 1/ ( 1 i,j (i 1/ (qj 1/ i,j ( 1 i (i 1/ j (qj 1/ ( 1 (P 1/ (Q 1/, ( 1 ( 1 ( 1 (i 1/ ( 1 i (i 1/ P i i i ( 1 (P 1/, ( ( ( 1 ( i 1/8 ( 1 i ( i 1/8 P i i i ( 1 (P 1/8. q j i

14 14 MATH 361: NUMBER THEORY EIGHTH LECTURE 10. The Kronecker Symbol The Kronecker symbol further generalizes of the Legendre Jacobi symbol. For denominator, ( a 1 if a 1,7 mod 8, 1 if a 3,5 mod 8, 0 if a 0 mod. Thus (a/ deends on a mod 8, so that the roerty that (a/ deends only on a mod is lost for. Also, (/ (/ for odd rimes. For denominator 1, ( { a 1 if a 0, 1 1 if a < 0. The general Kronecker symbol incororates these two new conventions multilicatively, ( a ( a ( ei a where P sgn(p ei i P sgn(p. i i If P is negative then (a/p is not eriodic in a, and if P is ositive but even then (a/p need not be a function of a mod P. The Kronecker symbol quadratic recirocity law is as follows. Theorem Let P and Q be nonzero corime integers, P e P where P sgn(p and Q f Q odd where Q sgn(q odd e f, where min{e,f } 0 for all. Then ( ( P Q ( 1 (P 1/ (Q 1/+(sgn(P 1/ (sgn(q 1/. Q P The theorem does not decomose P as sgn(p e P, but rather P has the same sign as P (and similarly for Q, of course. The factor ( 1 (sgn(p 1/ (sgn(q 1/ in the recirocity law is 1 if at least one of P and Q is ositive, 1 if both P and Q are negative. We end with the comment that a full set of rimes for Z should be extended to include one so-called Archimedean rime, and some authors (such as Conway Sloane in Shere Packings, Lattices and Grous identify this rime with 1. Under the convention that 1 is rime, P and Q are not corime if both are negative, and so the factor ( 1 (sgn(p 1/ (sgn(q 1/ is unnecessary. That is, it is arguably natural to tidy the recirocity law by excluding the case that P and Q are both negative. i

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