CH 8 Gaussian integers

Question 1

DEF 8.0.1 Gaussian integers

Answer 1

The Gaussian integers are the complex numbers in the set
Z[ı] := {a + bı : a, b ∈ Z},
where ı =√−1.

We also talk about Z[√−5] = {a + b√−5 : a, b ∈ Z} quite a few times.

Question 2

This section is here for reference only. We are not interested in the general theory of
rings, but need some basic concepts in order to discuss Z[ı], Z[√−5] and similar rings.
Therefore, this section is not directly examinable:

Answer 2

he exam will not ask ‘define what
is a ring’ or ‘give an example of a ring which is not an integral domain’. Rather, you
may be asked about primes, units, factorisations in specific rings of number theoretic
interest such as Z[ı], Z[√−5].

Question 3

for reference only: rings

Answer 3

A ring is a set R equipped with addition ‘α + β’ and multiplication ‘αβ’ satisfying the
usual axioms:
* addition and multiplication are associative; addition is commutative;
* multiplication is distributive over the sum (α(β + γ) = αβ + αγ);
* there is an additive identity (α + 0 = α, α · 1 = α);
* for every α ∈ R, there is an additive inverse −α ∈ R (that is, α + (−α) = 0).

Question 4

for reference only

examples of rings
Example 8.1.1.

Answer 4

Examples you have seen are Z, Q, R, C. The Gaussian integers form a ring. Indeed, commutativity, distributivity, associativity hold in C, so they also hold
in Z[ı] automatically; we only need to verify that Z[ı] is closed under sums, products,
and taking additive inverses. Let a + bı, c + dı ∈ Z[ı], thus a, b, c, d ∈ Z. Then
(a + bı) + (c + dı) = (a + c) + (b + d)ı ∈ Z[ı],
(a + bı)(c + dı) = (ac − bd) + (ad − bc)ı ∈ Z[ı],
(a + bı) + (−a + (−b))ı = 0 + 0ı = 0.
Two-by-two matrices also form a ring.
N is not a ring: no natural number apart from 0 has an additive inverses in N.

Question 5

for reference only
Definition 8.1.2 integral domain

Answer 5

An integral domain is a ring where multiplication is commutative and αβ = 0
happens only if α = 0 or β = 0.

Question 6

Example 8.1.3. Z, integral domains.

Answer 6

Example 8.1.3. Z, Q, R, C, Z[ı] are all integral domains.
Two-by-two matrices are not: multiplication is not commutative. Diagonal two-bytwo matrices do not form an integral domain either: multiplication is commutative,
but for instance
[1 0
0 0 ]
[ 0 0
0 1 ]
=

0 0
0 0
.
From now on, let R be an integral dom

Question 7

.
Definition 8.
divides

Answer 7

Let α, β ∈ R. We say that β divides α (written β | α) if there is γ ∈ R with α = ,βγ

Question 8

.
Example 8.1.5. Within Z, this is the usual definition of divisibil does 2 divide 6 in Z[i]?
Does 1+i divide 2?

Answer 8

for instance, 2 | 6
but 2 ∤ 3.
In Z[ı], we have (1 + ı)(1 − ı) = (1² − ı²) + (1 − 1)ı = 1 + 1 = 2, thus (1 + ı) | 2.

Note that when R ⊆ C, to check if β | α, it suffices to compute γ = α/β
2/(1+i)=2(1-i)/(1+i)(1-i) = 2-2i/2=1+i in Z[i] so true

(a complex number) and check if γ ∈ R. This is analogous to computing a fraction a/b in Q and
checking if a/b ∈ Z

Question 9

Definition 8.1.6 unit

Answer 9

An element α ∈ R is a unit if α | 1, that is to say, if there is β ∈ R such that αβ = 1
(such β is the multiplicative inverse of 1).

(reminder: a unit is s.t there exists a multiplicative inverse that is also a gaussian integer)

Question 10

Example 8.1.7. The units of Z are .

Answer 10

Example 8.1.7. The units of Z are the integers ±1.

The units of Z[ı] are
are +1-1,+i,-i
indeed
i(-i)=-(-1)(1) multiplicative inverses?

all the numbers α such that
1/α=α_overline/{αα_overline)=
α_overline/|α|^2
∈ Z[ı].
For instance, (−ı) is a unit: its inverse is ı itself (note that | − ı|^2 = 1!).

Question 11

Exercise 8.1.8. Show that the units of Z[ı] are exactly ±1 and ±ı. [Hint: show that the
units of Z[ı] must have modulus 1, then find all such Gaussian integers.]

Answer 11

if αβ = 1 for α,β in Z[i]
then | α|^2|β|^2 = 1
property of modulus that it is multiplicative
so |ab|=|a||b|

Question 12

Definition 8.1.11
ireducible
prime

Answer 12

Let R be an integral domain. An element α ∈ R that is not a unit is:
* irreducible if whenever α = βγ, one of β, γ is a unit;
* prime if whenever α | βγ, we have α | β or α | γ

(in Z analagous to divisors being 1 or itself, instead we say divisors relate to units)

(irreducible- whenevr we break into products of two other integers then 1 of the integers is a unit)

(prime divide one of two factors)

Question 13

Example 8.1.12
which primes irreducible in Z

in Z[i[

Answer 13

±2, ±3, ±5, . . . are all primes and irreducible, while for instance
6 = 2 · 3 is not irreducible, nor prime.

in Z[i]
2 is not irreducible
2=(i+1)(1-i) and these arent units
3 is irreducible as we will see later

Question 14

Lemma 8.1.13. irreducible

Answer 14

Lemma 8.1.13. Every prime is irreducible

Suppose that α is prime and α = βγ. Then α | β or α | γ, while of course β | α and γ | α.If α | β, then γ is a unit by Lemma 8.1.10; if α | γ, then β is a unit. Thus one of
β, γ is a unit, showing that α is irreducible.

Question 15

example:
Example 8.1.14. Take R = Z[√−5].

Is 2 prime
Is 2 irreducible

Answer 15

6 = 2 · 3 = (1 +√−5)(1 −√−5).
Clearly, 2 ∤ (1 +√−5) and 2 ∤ (1 −√−5) (if you divide by 2, you’ll get 1/2 which is not in Z).

However, 2 is irreducible in Z[√−5]. This will become clear in the next section

so 2 is not prime in Z[√−5]]

Question 16

lectures α is prime if whenever

Answer 16

α is prime if whenever α| βγ then α | β or α | γ

prime in this case is the same as saying it’s not a unit

Question 17

for ref only

Answer 17

missed some examples of rings in notes but its for reference mostly end of section here

Question 18

DEF 8.2.1 NORM

defined for Z[i] and Z[√−5]

Answer 18

The norm of α = a + bı ∈ Z[ı] is
N(α) := a² + b²
The norm of β = c + d√−5 ∈ Z[√−5] is
N(β) := a² + 5b²

Note that the norm is always in N.
always posive also
Note here N(α) =|α|²

Question 19

N(α) =

Answer 19

N(α) =|α|²

In these rings yes in future not always

Question 20

property of the norm:

Answer 20

multiplicative
N(x)N(y)=N(xy)

Question 21

Lemma 8.2.2
property of the norm

Answer 21

For all α, β ∈ Z[ı] (or Z[√−5]), we have N(αβ) = N(α) N(β).

not given as lemma in lectures, not proven
proof: Proof. Just recall that |αβ| = |α||β|. For completeness:
N((ac − bd) + (ad + bc)ı) = (ac − bd)^2 + (ad + bc)^2
= (ac)^2 + (bd)^2 −2abcd + (ad)^2 + (bc)^2 +2abcd
= (a^2 + b^2)(c^2 + d^2) = N(a + bı) N(c + dı). □

Question 22

Exercise 8.2.3. Show that 2 is irreducible in Z[√−5].

Answer 22

skipped?

Question 23

Lemma 8.2.4. Let α ∈ Z[ı]. Then α = 0 if and only if

Answer 23

Lemma 8.2.4. Let α ∈ Z[ı]. Then α = 0 if and only if N(α) = 0, and α is a unit if and
only if N(α) = 1.

Question 24

proof

Lemma 8.2.4. Let α ∈ Z[ı]. Then α = 0 if and only if N(α) = 0, and α is a unit if and
only if N(α) = 1.

Answer 24

Proof. Write α = a + bı. Of course, N(α) = 0 if and only if a² + b² = 0 thus if and only
if a = b = 0.

If α is a unit, then αβ = 1 for some β, thus N(α) N(β) = 1 which implies N(α) = 1.
(as Norms are positive integers so -1 or +1)

Conversely, if N(α) = 1, then
1/α= α_conjugate / αα_conjugate
=α_conjugate /¦α¦^2 = α_conjugate /N(α) = α_conujgate

which is still an integer as just flipping sign of b in either Z[i]or Z[√−5])

1/(a + bı) =(a − bı)/(a² + b²) =
(a − bı)/N(α)= a − bı ∈ Z[ı].
(This is just the answer to Exercise 8.1.8 rewritten using the norm.) □

Question 25

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

Answer 25

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

Question 26

Lemma 8.3.1 (Division algorithm for Gaussiann integers (∈ Z[ı],

Answer 26

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

Question 27

PROOF
Lemma 8.3.1 (Division algorithm for Gaussian integers (∈ Z[ı],

Answer 27

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!

Question 28

Example 8.3.2. Take α = 10 + 10ı, β = 3 − 2ı

α/β

Answer 28

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.

Question 29

Gaussian integers

Answer 29

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

Question 30

Definition 8.3.3
GCD

Answer 30

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)

Question 31

Example 8.3.4. With this new definition, the greatest common divisors of 4 and 6 in Z

Answer 31

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

Question 32

Lemma 8.3.5.

Let γ be a GCD of α and β in R. Then all GCD δ of α and β are of the form

Answer 32

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.

Question 33

Theorem 8.3.6 (Bézout’s lemma for Gaussian integers)

Answer 33

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

Question 34

Proof

Theorem 8.3.6 (Bézout’s lemma for Gaussian integers)
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 34

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.

Question 35

Example 8.3.7 Use Euclid’s algorithm to find GCD of two gaussian integers
α = 9 − 2ı and β = −5 + 5ı.

Answer 35

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)<N(β )
N(r_3)<N(r_2)
eventually r_k=0
then r_k-1 is a GCD of α, β

α/β=
(9-2i)(-5-5i)/(-5+5i)(-5-5i)=-55/50 - (35/50)i
= -(11/10)-(7/10)i
not gone through just pointed out
but show you can continue to
The last non-zero remainder is a greatest common divisisor, and we set γ = r_2 =−1 − 2ı. Moreover, working backwards,
γ = −1 − 2ı = α − β(−1 − ı).

Question 36

The division algorithm in Z[√−5]! Repeat our previous
argument for Z[ı] and find what breaks

Answer 36

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

Question 37

Corollary 8.3.9
In Z[ı], every irreducible

Answer 37

In Z[ı], every irreducible is prime.

Question 38

PROOF
In Z[ı], every irreducible is prime

Answer 38

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.

Question 39

Theorem 8.3.10 (Fundamental theorem of arithmetic for Gaussian integers) (cont)

Answer 39

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].

Question 40

PROOF
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 40

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)

Question 41

Example 8.3.11. For instance, 5
FTOA for gaussian integers

Answer 41

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)

Question 42

PROPn 8.4.1.

p∈ Z (ring of integers)
then
p is not primein Z[i]
when

Answer 42

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

Question 43

true ot false p∈ Z (ring of integers)
p ≡ 1 (mod 4),
then
p is not prime in Gaussian integers

Answer 43

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…

Question 44

Proposition 8.4.1.

when is a prime in Z not a prime in Z[i]

Answer 44

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

Question 45

Example 8.4.2. 17
prime? squares?

Answer 45

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

Question 46

proof:
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

Answer 46

(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 <N(α), N(β) < N(p) = p²
Since N(α) N(β) = N(p) = p² , we must have N(α) =N(β) = p.
(since not units and bigger than 1, the only way factorisation can happen is if factor p^2 into pxp)
Write α = a + bı: we must have a² + b² = p.

Question 47

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

Answer 47

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

Question 48

PROOF
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 48

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.

Question 49

Corollary 8.4.4
The Gaussian primes are exactly:

Answer 49

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

Question 50

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
PROOF

Answer 50

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

Question 51

in gaussian integers prime means irreducible?

Answer 51

true

Question 52

in gaussian integers irreducible means prime?

Answer 52

true

Question 53

Exercise 8.4.5. Write a shorter proof of Fermat’s Christmas theorem using Gaussian
integers.

Answer 53

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 .

Question 55

Lemma for alpha beta in R if they divide each other

Answer 55

α | β and β |α

IFF

α equals βu for some unit u

Question 56

proof of :
Lemma 8.1.10. Let α, β ∈ R. Then β | α and α | β if and only if α = uβ for some unit u.

Answer 56

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.