CH 8 Gaussian integers Flashcards

Question

I could ask you: what are the units in Z[i sqrt -5]

Answer 1

we can use norms which numbers are + or - 1 here will be units To find units we determine which elements have multiplicative inverse also in this set ie u is unit if there exists v in the set st uv equals 1 Ie expand and set real part to one and imaginary part to zero So solving these a^2 +5b^2 equals 1 The norm must be 1 It’s inverse will also have a norm of one So possible solutions are only a equals +1 or -1 for b equals zero If b was not zero then 5b^2 would be a positive multiple of 5 greater than zero making the norm greater than one

Answer 2

Let α, β ∈ Z[ı], where β ̸= 0. Then there exists q,r ∈ Z[ı] (quotient and remainder) such that α = qβ + r and **0 ≤ N(r) < N(β).** -norm here replaces size of integer in EUCLIDS -norm is an integer so we can use induction

Answer 3

1) compute α/β, a complex number, (1) compute α/β, a complex number, α/β = αβ_conjugate/ββ_conjugate = αβ_conjugate/ N(β) α/β = A+Bi , A,B in Reals not necessarily integers 2)Choose q = a + bı ∈ Z[ı] as close as possible to A + Bı: pick a, b ∈ Z s.t. |a − A| ≤ 1/2, |b − B| ≤ 1/2 Let r = α − qβ ∈ Z[ı]. then |(α/β)− q|= |(a − A)^2 + (b − B)^2| ≤1/4+1/4=1/2 Therefore, N(r) = N(α − qβ) = (N(β) )((A-e)^2 +(B-f)^2) |(α/β)− q|^2N(β) ≤ (1/2)N(β) < N(β). q and r are not necessarily unique!

Answer 4

Lemma 8.3.1 (Division algorithm for Gaussian integers (∈ Z[ı], α/β = multiply by conjugate (10 + 10ı)(3 + 2ı)/(3 − 2ı)(3 + 2ı) = (30 − 20) + (30 + 20)ı/(3^2 + 2^2) = (10 + 50ı)/13 . The nearest Gaussian integer is **q = 1 + 4ı** (in this case, it is unique). Here r = (10 + 10ı) − **(1 + 4ı)**(3 − 2ı) = (10 − 3 − 8) + (10 − 12 + 2)ı = −1. Indeed, N(r) = 1 < N(β) = 13.

Answer 5

in this chapter we dont work with integers anymore, complex numbers with integer parts

Answer 6

Let α, β ∈ R integral domain. A greatest common divisor is some element γ ∈ R such that: (1) γ is a common divisor, namely γ | α and γ | β; (2) if δ is a common divisor (δ | α, δ | β), then δ | γ may not exist, and may not be unique (everytime we have another common divisor it divides gamma itself)

Answer 7

are ±2. In Z we have the luxury of picking the positive greatest common divisor and calling it gcd(a, b)

Answer 8

Let γ be a greatest common divisor of α and β in R. Then all greatest common divisors δ of α and β are of the form δ = uγ for u unit of R.

Answer 9

Any two Gaussian integers α, β ∈ Z[ı] have a greatest common divisor γ. Moreover, γ = sα + tβ for some s, t ∈ Z[ı].

Answer 10

Essentially the same proof as for Bézout’s Lemma for Z. If α = β = 0, just take γ = 0, so suppose that one of α, β is not zero. Among all numbers γ = sα + tβ with s, t ∈ Z[ı], pick one with N(γ) minimum possible. First, we claim that γ | α. By Lemma 8.3.1, α = qγ + r where N(r) < N(γ). Then r = α − qγ = α − q(sα + tβ) = (1 − qs)α − qtβ, thus N(r) = 0 because of the minimality of N(γ). So r=0 Hence γ | α. Similarly, γ | β, so γ is a common divisor. then γ=sα + tβ Now if δ is another common divisor, then δ | (sα + tβ) = γ.

Answer 11

For example, take α = 9 − 2ı and β = −5 + 5ı. Divide r_0 = α by r_1 = β giving quotient q_2 and remainder r_2. N(r_2)

Answer 12

The division algorithm does not work in Z[√−5]! Repeat our previous argument for Z[ı] and find what breaks

Answer 13

In Z[ı], every irreducible is prime.

Answer 14

Let α be irreduciblet with α | βγ, and that α ∤ β. Let δ be a GCD r of α and β. α = δϵ: As α is irreducible, one of δ, ϵ is a unit. Since δ | β and α ∤ β, δ must be the unit. (BECAUSE …otherwise it would contradict the fact that alpha doesn’t divide beta. If it weren’t a unit it would be a non trivial divisor of alpha and of beta, thus …)) We have determined that a GCD of α and β is 1. Write it as 1 = sα + tβ. Then sαγ + tβγ = γ. Since α divides the left hand side, we find that α | γ. This shows that α is prime.

Answer 15

If α ∈ Z[ı] is not zero and not a unit, we can write α = π₁ · · · πᵣ where each πᵢ is prime in Z[ı]. Moreover, if α = σ₁ · · · σₛ is also a product of primes, then r = s and, after rearranging the factors, we have σᵢ = uiπᵢ for some units uᵢ ∈ Z[ı].

Answer 16

i wont go through proof as same as the proof of Fundamental theorem of arithmetic for Z by induction on the norm USES INDUCTION ON N(alpha) product of prime assumed Now suppose that N(α) = k + 1. If α is irreducible, it is also prime, and we are done. Otherwise, suppose that α = βγ where neither β nor γ are units. Then N(β), N(γ) > 1, and since N(α) = N(β) N(γ), we have N(β), N(γ) ≤ k. By induction hypothesis, both can be written as products of primes, and so does N(α). For the uniqueness, suppose that α = π1 · · · πr = σ1 · · · σs . Each irreducible is prime, thus π1 | σi for some i. Rearrange the terms so that i = 1. Then σ1 = π1u1 for some unit u1. Now divide both sides by π1, σ1 and repeat the argument by induction until there are no primes left (in particular, we will find that r = s)

Answer 17

For instance, 5 = (2 + ı)(2 − ı) = (1 + 2ı)(1 − 2ı). The two factorisations are equivalent because 2 + ı = ı(2 − ı) and 2 − ı = (−ı)(1 − 2ı). 5 is prime in Z but 5 not prime in Z[i] as a gaussian integer is not irreducible (2-i) (2+i) are primes or can write as a product of primes (1+2i)(1-2i) here 1+2i= i(2-i) this prime is essentially the same prime just differ by unit i 1-2i=-i(2+i)

Answer 18

IF p∈ Z (ring of integers) then If p is a positive prime of Z with p ≡ 1 (mod 4), then p is not prime in Z[ı] and there is α ∈ Z[ı] such that N(α) = p (equivalently, there are a, b ∈ Z such that p = a^2 + b^2). when p ≡ 1 (mod 4), and there is an element s α ∈ Z[ı] such that N(α) = p ie. prime integers dont remain prime, especially if p ≡ 1 (mod 4), ONLY ONE DIRECTION

Answer 19

false! If p is a positive prime of Z with p ≡ 1 (mod 4), then p is not prime We need the condition that there is no element in the set with the norm equal to this prime??? For example 13 is congruent to 1 mod 4 but 2 +3i has norm equal to 13 But 13 is not prime in Gaussian … BECAUSE IT ONLY WORKS IN ONE DIREXTION IE OUR LEMMA ONLY WORKS TO SHOW IT S NOT PRIME Remember if irreducible then prime WE CAN SHOW ITS NOT ORIME…

Answer 20

If p is a positive prime of Z with p ≡ 1 (mod 4), then p is not prime in Z[ı] and there is α ∈ Z[ı] such that N(α) = p (equivalently, there are a, b ∈ Z such that p = a² + b² ONLY IN ONE DIRECTION

Answer 21

17 = N(4 + ı) = (4 + ı)(4 − ı) = 4² + 1²

Answer 22

(we already know that if p ≡ 1 (mod 4) By Corollary 2.3.3, there is x ∈ Z such that x² ≡ −1 (mod p) -1 is a quadratic residue mod p) So p |(x²+ 1) in Z =(x+i)(x-i) however p does not divide (x+or -i) (if it did then p(a+bi)=(x +/- i) then pb =+1 or -1 which is impossible Clearly p does not divide 1 + xı nor 1 − xı, thus it is not prime in Z[ı], so not irreducible either (cant write as elements using units). Because: N(a+bi) =a²+² p not prime implies p not irreducible Therefore, we can write p = αβ where neither α nor β is a unit. In particular, 1

Answer 23

Proposition 8.4.3. If p is a positive prime of Z and p ≡ 3 (mod 4), then p is prime in Z[ı].

Answer 24

being irreducible and prime are the same thing in the ring of gaussian integers: taking a prime number p≡ 3 mod 4 if not prime in gaussian integers there would be a factorisation of non units p = αβ but then if there is a factorisation then p will be the norm of someone which means it will be the sum of 2 squares but chapter 6 tells us no! because the only primes that are sums of 2 squares are congruent to 1 mod 4! Proof. Suppose by contradiction that p = αβ for some non-units α, β. Just as in the previous proof, we have 1 < N(α), N(β) < N(p) = p² , thus N(α) = N(β) = p. This implies that p = a²+ b² which is impossible because −1 is not a square root (or go back to Corollary 2.3.3) p≡ 1 mod 4 contradiction.

Answer 25

The Gaussian primes are exactly: (1) primes of Z of the form 4k + 3, multiplied by ±1 or ±i; 2) elements of the form a + bı where a² + b² is a prime equal to 2 or of the form 4k + 1. RE-WRITTEN: the primes of Z[i] are a)(±p) or (±ip) where p in Z is a prime b) Norm is a prime N(α) =a² + b² = 2 or 1 mod 4

Answer 26

Proof. It only remains to check that α = a + bı with N(α) = p, where p = 2 or p ≡ 1 (mod 4) is prime in Z, is a prime. But it is easy to see that α is irreducible: if α = βγ, then N(β) N(γ) = p, thus N(β) = 1 or N(γ) = 1, so one of β, γ is a unit. which is exactly the definition of being irreducible remember prime iff irreducible in gaussian integers

Answer 27

SKIPPED Suppose that N = a² + b² for some a, b ∈ Z. Then N = N(α) where α = a + bı ∈ Z[ı]. Let α = π₁ · · · πᵣ be a prime factorisation of α in Z[ı]. Then N = N(α) N(π₁ ) · · · N(πᵣ ). It follows at once that if a prime p of the form 4k + 3 divides N, then p | N(πi ) for some i, thus in fact N(πi ) = p2. Therefore, the maximum power of p dividing N has even exponent. Conversely, write a prime factorisation in Z N = p α1 1 · · · prαr q21β₁ · · · q2s βs where the pi ’s are either 2 or of the form 4k + 1. For each of them, take πi ∈ Z[ı] such that N(πi ) = pi . Then N = N(π1 )α₁ · · · N(πr )αr N(q1 )β1 · · · N(qs )βs = N(α) = a2 + b2 where α = a + bı = π1α1 · · · πrαr q1β1 · · · qsβs .

Answer 28

α | β and β |α IFF α equals βu for some unit u

Answer 29

Proof. Suppose that there are γ, δ such that α = βγ and β = αδ. Then α = αγδ, that is to say, α(1 − γδ) = 0. Since R is an integral domain, we must have γδ = 1, thus γ is a unit. Conversely, suppose that α = uβ for some unit u. Then β | α. Let v be the multiplicative inverse of u. Then vα = β, so α | β.