Math 306, Spring 2017 Homework 2 — Solutions Due: February 24
· ∗ (1) In class, Narih gave a proof that p : Fp → {±1} is a (well-defined) homomorphism of groups; her proof hinged on the fact that she had already counted the number of quadratic residues and the number of quadratic non-residues. In this problem, you will give an alternative proof to the fact that the Legendre symbol is a homormophism, and then you’ll use that fact to count quadratic residues (and non-residues). (a) Prove that the Legendre symbol is a homomorphism using the content of Presentations 12 and 13 from §1.B.
k 2 Solution. The Legendre symbol p is defined to take the value 1 if x ≡ k (mod p) has a solution, and value −1 if the congruence has no solution. On the other hand, Presentation 12 tells p−1 us that if (k, p) = 1 when k 2 ≡ ±1 (mod p), and presentation 13 tells us that this value is +1 precisely in the case that x2 ≡ k (mod p) is solvable. Hence we have that
k p−1 = k 2 (mod p). p
∗ Furthermore we saw in class that the Legendre symbol is well-defined as a function on Fp; this ∗ means for a given [x] ∈ Fp, we can choose any representatives x1, x2 ∈ [x] we like, and we’ll be x1 x2 · guaranteed that p = p . (As such, we can get away with saying the value of p on [x] is x just p , where x is some representative of [x].) Now let [x], [y] ∈ (Z/pZ)× be given, and let x and y be representatives of these classes. Then we have that
xy p−1 ≡ (xy) 2 (by Euler’s criterion) p p−1 p−1 ≡ x 2 y 2 (since multiplication Z is commutative) x y ≡ (by Euler’s criterion). p p
∗ This is precisely what it means to say the Legendre symbol is a homomorphism on Fp. ∗ ∗ 2 (b) Let f : Fp → Fp be the function f([x]) = [x] . Argue that f is a homomorphism. Prove that ∗ ker(f) 6= {[1]}. Use this to prove that im(f) 6= Fp. ∗ Solution. Let [x], [y] ∈ Fp be given. We have that
f([x][y]) = ([x][y])2 (definition of f) 2 2 = [x] [y] (since Fp∗ is abelian) = f([x])f([y]) (definition of f).
Hence we have that f is a homomorphism. To see that ker(f) 6= {[1]}, note that we have f([−1]) = [−1]2 = [(−1)2] = [1], with the second- ∗ to-last equality holding because this is how we define the product of two classes in Fp. (That this yields a well-defined multiplication operation on Fp comes from how we define quotient rings in 305.) Note that when p > 2 we have [1] 6= [−1], since p - 1 − (−1). Hence [−1] is a nontrivial element in the kernel of f. ∗ Now to see that im(f) 6= Fp, recall that the first isomorphism theorem gives us that
∗ Fp ' im(f). ker(f) http://palmer.wellesley.edu/~aschultz/w17/math306 Page 1 of 5 Math 306, Spring 2017 Homework 2 — Solutions Due: February 24
p−1 In particular, if we take orders on all sides we find that | ker(f)| = |im(f)|. Since we’ve just seen ∗ that | ker(f)| > 1, we have that |im(f)| < p − 1. Hence we have im(f) 6= Fp, and hence f is not surjective. · (c) Prove that p is surjective. Use this to argue that the kernel of the Legendre symbol has order 1 ∗ p−1 p−1 2 |Fp|. Then use this to conclude that there are precisely 2 quadratic residues, as well as 2 quadratic non-residues. (You can use basic facts about homomorphisms that you learned in Math 305.)
1 Solution. Note that we already know p = 1. The only other element in the codomain of the a Legendre symbol is −1, and we know that any non quadratic residue a will satisfy p = −1. So the relevant question is: are there any non quadratic residues modulo p? From above we see that — when p > 2 — the squaring map is not surjective. In particular, this means there exists an ∗ element a ∈ Fp for which the equation f([x]) = [a] is not solvable. Translated according to the definition of f, this means that we cannot solve the equation x2 ≡ a (mod p). In short: a is not a quadratic residue. Hence the Legendre symbol is surjective. Again using the first isomorphism theorem, we get that ∗ Fp ' {±1}. · ker p Taking orders we’re left with p − 1 = 2, ker · p p−1 and rearranging we get that the kernel of the Legendre symbol is order 2 . Since the kernel of the Legendre symbol is — by definition — the set of quadratic residues, we get that there are p−1 2 many quadratic residues. Again, the definition of non-quadratic residues means that we have p−1 p−1 p − 1 − 2 = 2 many non-quadratic residues. [Note: this latter fact can be proved in a slightly different way. In 305, you (probably) learned that if φ : G → H is a homomorphism, then φ is | ker(φ)|-to-1 as a map. Hence the preimage of any element in the image has the same size as the kernel of the map. In our case, the preimage of −1 under the Legendre symbol is the set of all non-quadratic residues; by our 305 fact, this set has to have the same size as the kernel; i.e., the set of non-quadratic residues has to have the same size p−1 as the set of quadratic residues. Since we already argued that there are 2 many elements in the latter, there are the same number of elements in the former.]
(2) In the proofs of reciprocity for n ∈ {−1, −2, −3, 6} we saw in class, there were several times when we had statements of the form ”p ≡ r1 (mod n1) and p ≡ r2 (mod n2).” In these situations, the challenge was to determine whether it’s possible for a number p to satisfy these two congruences simultaneously, and — if so — to find some congruence condition which is equivalent to this simultaneous congruence. (a) Suppose that x and y are integers so that x ≡ ri (mod ni) for i ∈ {1, 2}, and also y ≡ ri (mod ni) for i ∈ {1, 2}. Prove that x ≡ y (mod [n1, n2]), where [n1, n2] is the least common multiple of n1 and n2. [Note: the least common multiple of n1 and n2 is, by definition, the smallest positive integer that is simultaneously a multiple of n1 and n2. It is a fact — that you don’t have to prove — that if x is an integer with n1 | x and n2 | x, then [n1, n2] | x.]
Solution. To make notation simpler, write m for the least a common multiple of n1 and n2. Our goal will be to prove that x ≡ y (mod m). By hypothesis we have x ≡ r1 (mod n1) and y ≡ r1 (mod n1). By transitivity of congruence, if follows that x ≡ y (mod n1). Hence we have n1 | x − y; that is, we have that x − y is a multiple of http://palmer.wellesley.edu/~aschultz/w17/math306 Page 2 of 5 Math 306, Spring 2017 Homework 2 — Solutions Due: February 24
n1. A similar line of argument gives that x − y is a multiple of n2. Therefore, it follows that x − y is a common multiple of n1 and n2. We’re told in the prompt that any common multiple of n1 and n2 is divisible by m. Since we’ve just argued that x − y is a common multiple of n1 and n2, it follows that m | x − y. Hence x ≡ y (mod m), as desired.
(b) Prove that if there exists some x satisfying x ≡ ri (mod ni) for i ∈ {1, 2}, then r1 ≡ r2 (mod (n1, n2))
Solution. Write d = (n1, n2), just to keep notation a bit simpler. Note that since x ≡ r1 (mod n1), we have n1 | x − r1. Since d | n1 and divisibility is transitive, it therefore follows that d | x − r1. Hence x ≡ r1 (mod d). By a similar line of reasoning, we have x ≡ r2 (mod d). Transitivity of congruence gives r1 ≡ r2 (mod d), as desired.
(3)( ?) In this problem you’ll tackle the supplementary laws for quadratic reciprocity for both the Legendre symbol and the Jacobi symbol. p−1 p2−1 −1 2 2 8 (a) For an odd prime p, prove that p = (−1) and that p = (−1) . Solution. For the first result, note that from our solution to 1(a) we have
−1 p−1 = (−1) 2 . p (This is just Euler’s criterion from presentation 13.) For the second, recall that Gina showed in class that 2 1, if p ≡ 1, 7 (mod 8) = . p −1, if p ≡ 3, 5 (mod 8)
p2−1 Our goal will be to show that (−1) 8 is equal to the right-side of this expression; this will then show that 2 p2−1 = (−1) 8 . p Write p = 8k + r, where r ∈ {1, 3, 5, 7}. We then have p2 − 1 (8k + r)2 − 1 64k2 + 16kr + r2 − 1 r2 − 1 = = = 2(4k2 + kr) + , 8 8 8 8 and so 2 2 2 p −1 2(4k2+kr) r −1 r −1 (−1) 8 = (−1) (−1) 8 = (−1) 8 . Now one simply calculates 12 − 1 32 − 1 52 − 1 72 − 1 = 0 = 1 = 3 = 6. 8 8 8 8 Hence the term (r2 − 1)/8 is even when r = 1, 7 and odd when r = 3, 5, and so we have
2 p −1 1, if p ≡ 1, 7 (mod 8) (−1) 8 = . −1, if p ≡ 3, 5 (mod 8) m−1 m2−1 −1 2 2 8 (b) For an odd, positive integer m, prove that m = (−1) and that m = (−1) .
Solution. For the first part, recall that Lucia proved in class that if a1, a2 are odd, then a − 1 a − 1 a a − 1 1 + 2 ≡ 1 2 (mod 2). 2 2 2 We’ll start our proof by using this to prove — by induction — that if a1, ··· , ak are odd integers, then a − 1 a − 1 a − 1 a a ··· a − 1 1 + 2 + ··· + k ≡ 1 2 k (mod 2). 2 2 2 2 http://palmer.wellesley.edu/~aschultz/w17/math306 Page 3 of 5 Math 306, Spring 2017 Homework 2 — Solutions Due: February 24
Lucia’s proof was the base case, so suppose that we know the result for n, and we’ll prove it for n + 1. We have a − 1 a − 1 a − 1 a − 1 a a ··· a − 1 a − 1 1 + 2 + ··· + n + n+1 ≡ 1 2 n + n−1 (by induction) 2 2 2 2 2 2 a ··· a a − 1 ≡ 1 n n+1 (mod 2) (by the base case). 2 m−1 −1 2 Now to return to the identity m = (−1) . Let m = p1 ··· pk be the prime factorization of m; we do not assume these primes are distinct, but note that they must be odd since m is odd. By the definition of the Jacobi symbol and our result from part (a), we get k k −1 Y −1 Y pi−1 Pk pi−1 = = (−1) 2 = (−1) i=1 2 . m p i=1 i i=1 By our extension of Lucia’s lemma from class, the exponent in the last term of this expression satisfies k X pi − 1 p1 ··· pk − 1 m − 1 ≡ ≡ (mod 2). 2 2 2 i=1 So we get −1 Pk pi−1 m−1 = (−1) i=1 2 ≡ (−1) 2 . m 2 For the identity regarding m , we’ll start with a lemma akin to Lucia’s: if r and s are odd integers, then r2 − 1 s2 − 1 r2s2 − 1 + ≡ (mod 2). 8 8 8 Using induction (almost precisely as above), this will then let us conclude the slightly more general result: if r1, ··· , rk are odd integers, then X r2 − 1 r2 ··· r2 − 1 i ≡ 1 k (mod 2). 8 8 r2−1 4a2+4a a(a+1) Now to prove our lemma, let r = 2a + 1 and s = 2b + 1. We then get 8 = 8 = 2 , and s2−1 b(b+1) likewise 8 = 2 . On the other hand, we get r2s2 − 1 (4a2 + 4a + 1)(4b2 + 4b + 1) − 1 = 8 8 16(a2b2 + a2b + ab2 + ab) + 4(a2 + a + b2 + b) = 8 a2 + a + b2 + b ≡ (mod 2) 2 a(a + 1) b(b + 1) = + 2 2 r2 − 1 s2 − 1 = + , 8 8 as desired. 2 Now we’re ready to prove the assertion regarding m . Start by writing m = p1 ··· pk as before, 2 2 2 with the pi all odd primes. Using our result above and the fact that m = p1 ··· pk, we then get k k 2 2 2 2 2 2 Y 2 Y pi −1 Pk pi −1 p1···pk−1 m −1 = = (−1) 8 = (−1) i=1 8 = (−1) 8 = (−1) 8 . m p i=1 i i=1 http://palmer.wellesley.edu/~aschultz/w17/math306 Page 4 of 5 Math 306, Spring 2017 Homework 2 — Solutions Due: February 24
M ∗ (4) Prove that if M ≡ 0 (mod 4), then the Jacobi symbol · is a well-defined function on (Z/MZ) . In ∗ M M other words, argue that if [m] ∈ ( /M ) and n1, n2 ∈ [m] are positive and odd, then = . Z Z n1 n2
Solution. Suppose that n1, n2 ∈ [m] are positive and odd; this gives n2 − n1 = kD for some k ∈ Z. δ e Write M = (−1) 2 p2 ··· pt, where the {pi} are odd primes and — because M ≡ 0 (mod 4) — we have e ≥ 2. Multiplicativity and the 1st and second supplements give, for j ∈ {1, 2}: e δ e δ e δ n2−1 M (−1) 2 p2 ··· pt −1 2 Y pi nj −1 j Y pi = = = (−1) 2 (−1) 8 . n n n n n n j j j j i j i j We will handle two cases: when e is even and when e is odd. Now when e is even we have: δ M n2−1 Y pi = (−1) 2 (e ∈ 2 ) n n Z 2 i 2 δ n2−1 Y n2−1 pi−1 n2 = (−1) 2 (−1) 2 (−1) 2 (QR) p i i δ n1+kD−1 Y n1+kD−1 pi−1 n1 + kD = (−1) 2 (−1) 2 (−1) 2 (since n ≡ n (mod D)) p 1 2 i i δ n1−1 Y n1−1 pi−1 n1 + kD = (−1) 2 (−1) 2 (−1) 2 (D/2 ∈ 2 ) p Z i i δ n1−1 Y n1−1 pi−1 n1 = (−1) 2 (−1) 2 (−1) 2 (congruence properties of Jacobi) p i i δ n1−1 Y pi = (−1) 2 (QR) n i 1 M = (e ∈ 2Z). n1 Note that we’ve phrased this case as the case when e ∈ 2Z, but in reality what we’ve shown is that when n1 ≡ n2 (mod D) and D ≡ 0 (mod 4), then we have δ δ n2−1 Y pi n1−1 Y pi (−1) 2 = (−1) 2 . n n i 2 i 1 (This observation will be useful in our next case.) The other case to handle is when e is odd; note that since e ≥ 2, oddness of e means e ≥ 3. Hence in this case we have D ≡ 0 (mod 8). Hence n2 = n1 + kD = n1 + 8kt. In particular, this means that n2 − 1 n2 + 16kt + 64k2t2 − 1 n2 − 1 n2 − 1 (?) 2 = 1 = 1 + 2 kt + 4k2t2 ≡ 1 mod 2. 8 8 8 8 We compute: δ 2 e M n2−1 n2−1 Y pi = (−1) 2 (−1) 8 (2nd supplement) n n 2 i 2 δ 2 e n2−1 n1−1 Y pi = (−1) 2 (−1) 8 (from (?) above) n i 2 δ 2 e n1−1 n1−1 Y pi = (−1) 2 (−1) 8 (previous case case) n i 1 M = . n1 http://palmer.wellesley.edu/~aschultz/w17/math306 Page 5 of 5