CH 8 Gaussian integers Flashcards

1
Q

DEF 8.0.1 Gaussian integers

A

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.

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2
Q

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:

A

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].

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3
Q

for reference only: rings

A

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).

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Q

for reference only

examples of rings
Example 8.1.1.

A

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.

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Q

for reference only
Definition 8.1.2 integral domain

A

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

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6
Q

Example 8.1.3. Z, integral domains.

A

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

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7
Q

.
Definition 8.
divides

A

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

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Q

.
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?

A

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

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9
Q

Definition 8.1.6 unit

A

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)

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10
Q

Example 8.1.7. The units of Z are .

A

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!).

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11
Q

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.]

A

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|

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12
Q

Definition 8.1.11
ireducible
prime

A

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)

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13
Q

Example 8.1.12
which primes irreducible in Z

in Z[i[

A

±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

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14
Q

Lemma 8.1.13. irreducible

A

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.

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15
Q

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

Is 2 prime
Is 2 irreducible

A

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]]

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16
Q

lectures α is prime if whenever

A

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

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17
Q

for ref only

A

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

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18
Q

DEF 8.2.1 NORM

defined for Z[i] and Z[√−5]

A

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

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19
Q

N(α) =

A

N(α) =|α|²

In these rings yes in future not always

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20
Q

property of the norm:

A

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

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21
Q

Lemma 8.2.2
property of the norm

A

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). □

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22
Q

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

A

skipped?

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23
Q

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

A

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.

24
Q

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.

A

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.) □

25
Q

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

A

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

26
Q

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

A

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

27
Q

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

A

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!

28
Q

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

α/β

A

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.

29
Q

Gaussian integers

A

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

30
Q

Definition 8.3.3
GCD

A

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)

31
Q

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

A

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

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Q

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

A

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.

33
Q

Theorem 8.3.6 (Bézout’s lemma for Gaussian integers)

A

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

34
Q

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].

A

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.

35
Q

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

A

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).

36
Q

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

A

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

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Q

Corollary 8.3.9
In Z[ı], every irreducible

A

In Z[ı], every irreducible is prime.

38
Q

PROOF
In Z[ı], every irreducible is prime

A

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.

39
Q

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

A

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].

40
Q

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].

A

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)

41
Q

Example 8.3.11. For instance, 5
FTOA for gaussian integers

A

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)

42
Q

PROPn 8.4.1.

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

A

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

43
Q

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

A

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…

44
Q

Proposition 8.4.1.

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

A

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

45
Q

Example 8.4.2. 17
prime? squares?

A

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

46
Q

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

A

(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.

47
Q

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

A

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

48
Q

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].

A

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.

49
Q

Corollary 8.4.4
The Gaussian primes are exactly:

A

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

50
Q

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

A

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

51
Q

in gaussian integers prime means irreducible?

A

true

52
Q

in gaussian integers irreducible means prime?

A

true

53
Q

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

A

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 .

54
Q
A
55
Q

Lemma for alpha beta in R if they divide each other

A

α | β and β |α

IFF

α equals βu for some unit u

56
Q

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.

A

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.