Spencer Leonardis5-29-2015 (Late)

Homework 7Math 118

Q1:a. Let p be an odd prime and let a ∈Z such that (a, p)= 1. Show that if a is a quadratic residue mod p,then a is a quadratic residue modulo pn for any positive integer every n.Proof. Hensel’s Lemma: Let f (x) be a polynomial with integer coefficients, p be a prime, and suppose a isa solution of the congruence f (x)≡ 0 mod p j such that f ′(a) 6≡ 0 mod p. Then there exists an integer t(which is unique modulo p) such that a+ tp j is a solution to the congruence f (x)≡ 0 mod p j+1.

Suppose a is quadratic residue modulo p. Then x2 ≡ a mod p =⇒ x2 −a ≡ 0 (mod p). If we apply Hensel’sLemma to the polynomial x2 −a ≡ 0 mod p, we can raise the powers of p inductively to p2, p3,. . . , pn. Inparticular, if we have a unique solution modulo p then we are guaranteed a solution modulo pn. So if ais a quadratic residue modulo p2, it is a quadratic residue pn for every n. In particular, if x2 ≡ a mod pk,one can find y such that (x+ pk y)2 ≡ a mod pk+1. i.e. (x+ pk y)2 ≡ x2 +2pk y mod pk+1. 2

b. To handle the case when (a, p) 6= 1, solve problem 3.3.12. in the book: Consider the congruencex2 ≡ a mod pα with p a prime, αÊ 1,a = pβb, (b, p)= 1. Prove that if βÊα then the congruence issolvable, and that if β<α then the congruence is solvable if and only if β is even and x2 ≡ b mod pα−β issolvable.Proof. Same argument from (a) works here as well, we don’t need to the overly complicated problem totreat this case.

c. Using CRT and the first part of this problem, conclude that if m is odd and (a,m)= 1, then a is aquadratic residue mod m if and only if a is a quadratic residue mod p for every prime p dividing m.Proof. Suppose that a is a quadratic residue modulo m. Let m = pα1

1 pα22 · · · pαk

k denote the primedecomposition of m. By CRT, we have that Z/mZ∼=Z/pα1

1 Z×·· ·×Z/pαkk Z. Since x2 ≡ a mod m has a

solution, it must be the case that x2 ≡ a mod pα11 pα2

2 · · · pαkk has a solution as well. Hence the system of

congruencesx2

1 ≡ a mod pα11

x22 ≡ a mod pα2

2...

x2k ≡ a mod pαk

kis solvable. By applying Hensel’s Lemma (first part of problem), we can reduce the prime power case ofthis system to the prime case, proving that a is a quadratic residue modulo m for every p dividing m. 2

d. Using CRT and the first two parts of this problem prove the following. Let m be odd and let a ∈Z.Prove that the congruence x2 ≡ a mod m has a solution if and only if for each prime p dividing m, one ofthe following conditions holds, where pα || m and pβ || a (this means that pα is the largest power of pdividing m and similarly for pβ and a):- βÊα;- β<α, β is even, and a/pβ is a quadratic residue mod p.Proof. The first direction is clear: A solution modulo m is also a solution modulo pαi

i . For the otherdirection, for each i let xi be such that x2

i ≡ a (mod pαii ). By the Chinese Remainder Theorem, there is

an x such that x ≡ xi (mod pαii ) for all i from 1 to t. This x is a solution of the congruence x2 ≡ a

(mod m). So now it is sufficient to prove the result for m = pα, where p is prime. Let pβ be the largestpower of p that divides a. We need to show that the congruence has a solution if and only if one of thetwo conditions of the post holds. The case βÊα is easy, since x = 0 is a solution. Suppose now that β<α.Assume that x2 ≡ a (mod pα). and let pγ be the largest power of p that divides x. Then unless γ=β/2,

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x2 −a cannot be divisible by pα. So β= 2γ must be even. Let x = pβ/2 y. Then pβy2 ≡ a (mod pα).Dividing through by pβ we get

y2 ≡ a/pβ (mod pα−β).

In particular, since β−αÊ 1, we havey2 ≡ a/pβ (mod p),

and therefore a/pβ is a quadratic residue of p. Hence showing that a/pβ being a quadratic residue of p issufficient in the case β<α. The solution of the congruence y2 ≡ a/pβ (mod p)=⇒ y2 −a/pβ ≡ 0 (mod p)can be lifted to a solution modulo any power of p, in particular pα−β by applying Hensel’s Lemma (asproved in the first two parts of the problem). This completes the proof. 2

Q2:3.3.14. Suppose that p is a prime, p ≡ 1 mod 4, and that a2 +b2 = p with a odd and positive. Show that(

ap

)= 1.

Proof. Using the laws of quadratic reciprocity, we obtain(

ap

)=

( pa

)=

(a2 +b2

a

)=

(b2

a

)= 1. 2

3.3.17. Let p be an odd prime, and put s(a, p)=p∑

n=1

(n(n+a)

p

). Show that s(0, p)= p−1. Show that

p∑n=1

s(a, p)= 0. Show that if (a, p)= 1, then s(a, p)= s(1, p). Conclude that s(a, p)=−1 if (a, p)= 1.

Proof.

Q3:a. Let Q

(pd)= {a+b

pd : a,b ∈Q}. Show that every element of Q

(pd)

has a multiplicative inverse. In

particular, Q(p

d)

is a field.

Proof. Suppose γ ∈Q(p

d)=⇒ γ= a+b

pd. To prove that Q

(pd)

contains multiplicative inverses, observe

that for a non-square d and a−bp

d 6= 0 one has

1γ= 1

a+bp

d· a−b

pd

a−bp

d= a

a2 −b2d− b

a2 −b2d

pd.

If a,b ∈Q=⇒ a2,b2 ∈Q⇒ b2d ∈Q (since d ∈Z) =⇒ a2 −b2d ∈Q⇒ aa2−b2d , b

a2−b2d ∈Q. Thus 1γ∈Q

(pd). 2

b. A number α ∈C is called algebraic if α is algebraic over Q. In particular, a ∈C⊃Q is algebraic if thereexists 0 6= f (x) ∈Q[x] such that f (α)= 0. The minimal polynomial m(x) of an algebraic number α ∈C isthe monic polynomial of smallest degree, with coefficients in Q such that m(α)= 0. What is the minimalpolynomial of α= a+b

pd ∈Q

(pd)?

Proof. Let α= a+bp

d. Thenα−a = b

pd ⇒ (α−a)2 = b2d ⇒α2 −2αa+a2 = b2d ⇒ m(x)= x2 −2xa+a2 −b2d

is the minimal polynomial of Q(p

d). 2

c. Show that if a rational number a ∈Q satisfies a monic polynomial with integer coefficients, then infact a ∈Z.Proof. Express a = p

q as a completely reduced rational in Q and assume to the contrary thata 6∈Z=⇒ q 6= ±1. Observe that the rational number p

q ∈Q, is a solution to qx− p = 0, where qx− p ∈Z[x].Thus every a ∈Q is algebraic. For an integer monic polynomial 0= m(x)= xn +·· ·+a1x+a0 ∈Z[x], onehas m

(pq

)=

(pq

)n +·· ·+a1pq +a0 =⇒ pn +·· ·+a1 p+a0 = 0=⇒ pn ≡ 0 mod q. Since q 6= ±1=⇒ q has a

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prime factor v =⇒ v | an =⇒ v | a. Thus v is a factor of both p and q contradicting the assumption that pq

is completely reduced. ⇒⇐ 2

d. (Quite difficult and therefore optional.) An element α ∈C is called an algebraic integer if thecoefficients of its minimal polynomial f (x) lie in Z. Generalizing part 3c, show that if α ∈C satisfies amonic polynomial with integer coefficients, then in fact α is an algebraic integer. We won’t need this, butit is a very important fact.Proof.

e. Show that the set of algebraic integers in Q(p

d)

is equal to the ring Od ={Z[

pd

2 ] if d is evenZ[1+pd

2 ] if d is odd.

Proof. Let α= x+ yp

d ∈Q(p

d). This is identical to 2x ∈Z and x2 −d y2 ∈Z. The first condition means x

is half of an ordinary integer, so either x ∈Z or x is half an odd number. If x ∈Z then x2 ∈Z, so d y2 ∈Z.Then y ∈Z because d is squarefree. (If y has a prime factor in its denominator, the square of that primewouldn’t be cancelled by d, and then d y2 6∈Z.) Therefore α has the necessary form, with a = x and b = yif d 6≡ 1 mod 4, or a = x− y and b = 2y if d ≡ 1 mod 4, or a = x− y and b = 2y if d ≡ 1 mod 4. If x is halfan odd number, write x = a/2 with a odd. Then N(α)= x2 −dy2 = a2/4−d y2, so multiplying through by 4implies a2 −d(2y)2 ∈ 4Z In particular, d(2y)2 ∈Z. Because d is squarefree, 2y is in Z so either y is in Z oris half an odd number. If y ∈Z then a2−d(2y)2 = a2−4d y2 is odd. ⇒⇐ Therefore it must be the case that

y= b/2 with b an odd number, so α= x+ yp

d = a2+ b

2

pd = a−b

2+b

(1+p

d2

)which has the necessary

form because a−b is divisible by 2⇒ a−b2 ∈Z. Hence the set of algebraic integers is the ring Od.

f. Let I be an ideal of Od. Show that I = {x : x ∈ I} is also an ideal of Od.Proof. Clearly 0 ∈ I. Note that for α,β ∈ I =⇒α+β ∈ I, so for every α,β ∈ I, one has α+β=α+β ∈ I. Sincefor every α ∈ I we have −α ∈ I, it follows that for every α ∈ I, one has −α ∈ I. Hence I is an additivesubgroup of Od. What is left to show is that I is closed under multiplication by ring elements of Od.Observe that for a ∈Od and for x ∈ I one has ax ∈ I. Thus ax = ax ∈ I. Thus I is closed undermultiplication by elements of Od, and combined with the fact that (I,+)ÉOd, we conclude that I is anideal. 2

g. Let N(I) be the greatest common divisor in Z of N(x) for all x ∈ I. Since N(x) ∈ I I for x ∈ I, it is clearthat (N(I))⊂ I I. Hence to prove that (N(I))= I I, we have to prove the reverse inclusion. Thus, letα,β ∈ I; we will prove that αβ ∈ (N(I)) as follows: (i). First, prove that the polynomial

f (x)= x2 − (αβ+αβ)x+ααββ

has coefficients in Z and that αβ is a root.

Proof (i). Let α,β ∈ I. Clearly 1 ∈Z. We need to show that αβ+αβ and ααββ are in Z. Observe that ααand ββ are both in Z because multiplying by the conjugate removes

pd in the middle terms. Thus

ααββ ∈Z. For the other element, we make an explicit calculation (the cancellation is less obvious). If αis an element of the form a1 +b1

pd and β is an element of the form a2 +b2

pd we have that

αβ+αβ= (a1 +b1p

d)(a2 −b2p

d)+ (a1 −b1p

d)(a2 +b2p

d)= a1a2−a1b2

pd+b1a2

pd−b1b2d+a1a2+a1b2

pd−b1a2

pd−b1b2d

= a1a2 −b1b2d+a1a2 −b1b2dSince a1,a2,b1,b2,d ∈Z it follows that a1a2 −b1b2d+a1a2 −b1b2d =αβ+αβ ∈Z also. The sameargument works for the ring Z

[1+pd

2

]. Hence f (x) ∈Z[x]. Now observe that

f (αβ)= (αβ)2 − (αβ+αβ)(αβ)+ααββ= (αβ)2 − (αβ)2 −ααββ+ααββ= 0.Hence αβ is a root. 2

(ii). Second, using the fact that N(I) divides the integers N(α+β), N(α), and N(β), prove that N(I)divides the coefficient of x in f (x), and that N(I)2 divides the constant coefficient.

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Proof (ii). Since N(I) is the greatest common divisor in Z of N(x) for all x ∈ I, it follows that N(I) dividesN(α+β)=αα+αβ+αβ+ββ, N(α)=αα, and N(β)=ββ. Thus N(I) |αβ+αβ. Since N(I) | N(α) andN(I) | N(β), it follows that N(I)2 | N(α)N(β)=ααββ. Thus N(I) divides the coefficient of x in f (x) andN(I)2 divides the constant coefficient. 2

(iii). Conclude that the polynomial g(x)= 1N(I)2 f (x ·N(I)) is a monic polynomial in Z[x], and note that

the elementαβ

N(I)is a root.

Proof (iii). We have that g(x) is integer monic since1

N(I)2 ∈Z and since f (x) ∈Z[x]. Clearly g

(αβ

N(I)

)= 0.

(iv). Now consider two cases: either αβ is rational and thus an integer by part 3c. Otherwise g(x) is the

minimal polynomial ofαβ

N(I), and hence this number is an algebraic integer. Therefore it lies in Od by

part 3e.

In either case,αβ

N(I)∈Od, so αβ ∈ (N(I)) as desired. Thus I I ⊂Od, and combined with the fact that

I I ⊃Od, we conclude I I =Od. 2

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