Quadratic Number Fields – Lecture 2 – 05/08/17
Tommy Hofmann
Lemma 1.13. — Let K be a quadratic field and α, β ∈ OK . Then the following hold:
(i) If α | β in OK , then N(α) | N(β) in Z.
(ii) The element α is a unit if and only if N(α) = ±1.
(iii) If N(α) is prime in Z, then α is irreducible in OK .
Proof. (i): Let γ ∈ OK such that αγ = β. Then N(α) N(γ) = N(β). As the norms are
integers, we obtain N(α) | N(β).
×
(ii): Assume that α ∈ OK
is a unit. Then there exists β ∈ OK such that αβ = 1. For
the norms we obtain N(α) N(β) = N(1) = 12 = 1. Since the norms are integers we must
have N(α) = ±1. Now assume that N(α) = ±1. Then ±1 = N(α) = αᾱ. Multiplying
by ±1 we get 1 = α · (±ᾱ). Since ±ᾱ is again an algebraic integer, we conclude that α
has an inverse and is therefor a unit.
(iii): Assume that α is not a unit and factors as α = βγ with β, γ ∈ OK . Then
N(α) = N(β) N(γ). As N(α) is prime we obtain N(β) = ±1 or N(γ) = ±1. Using (ii)
this shows that either β or γ is a unit.
√
Corollary 1.14. — For d ∈ Z negative and squarefree let K = Q( d) be the associated imaginary quadratic field. Then


if d = −1,
{±1, ±i},
√
×
2
OK = {±1, ±ζ3 , ±ζ3 },
if d = −3 and where ζ3 = −1+2 3 ,


{±1},
otherwise.
√
×
Proof. Let α = a + b d ∈ OK
. As the norm is positive, by Lemma 1.13 we must have
N(α) = a2 − b2 d = 1.
(a) The case d ≡ 2, 3 mod 4: Then by Theorem 1.12 we know that a, b ∈ Z.
If d < −1, then −d > 1 and the only integer solutions to the equation are (a, b) ∈
{(±1, 0)}, which correspond to the elements {±1}. If d = −1, then the integer
solutions to a2 + b2 = 1 are (a, b) ∈ {(0, ±1), (±1, 0)}, which correspond to the
elements {±1, ±i}.
(b) The case d ≡ 1 mod 4: Multiplying the equation by 4 we obtain
4 = (2a)2 − (2b)2 d.
By Theorem 1.12 we have 2a, 2b ∈ Z. Thus we are again in a situation where we
need to find integer solutions. If d < −3, then d ≤ −7 (since d ≡ 1 mod 4) and
−d ≥ 7. In this case the only solutions are (a, b) ∈ {(±1, 0)}, corresponding to
{±1}. If d = −3, then the equation becomes 4 = (2a)2 + (2b)2 3 and we see that
all solutions are (2a, 2b) ∈ {(±2, 0), (±1, ±1)}, that is, (a, b) ∈ {±1, 0), (± 12 , ± 12 )}.
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Quadratic Number Fields – Lecture 2 – 05/08/17
Tommy Hofmann
√
Example 1.15.
√ — Consider the quadratic field K = Q( −5). As −5 ≡ 3 mod 4, we
have OK = Z[ −5]. We want to show that OK is not a unique factorization domain.
We have
√
√
√
2 · 3 = 6 = 1 + 1 · 5 = N(1 + −5) = (1 + −5)(1 − −5).
√
√
Since there are no elements of norm 2 and 3, the elements 2, 3, 1 + −5 and 1 − −5
are irreducible, and the element 6 ∈ OK has two factorizations into irreducible elements.
At this moment we cannot conclude that OK is not a unique factorization domain, as
×
the irreducible factors
OK
= {±1} by Corollary 1.14 and
√ could√differ by units. Now
×
therefore 2·α 6= 1+ −5, 1− −5 for all α ∈ OK . Hence OK is not a unique factorization
domain.
√
By finding√two distinct factorizations of 6 ∈ Z[ −5] into irreducible elements, we have
seen that Z[ −5] is not a unique factorization domain. Thus (in general) the rings of
integers of quadratic fields will not be as nice as the ring Z of ordinary rational integers.
To overcome this obstacle, we will prove that by passing to ideals, unique factorization
can be recovered. More precisely, our aim is the following result: If K is a quadratic
field, then every nonzero ideal of OK can be uniquely factored into prime ideals. To
get an idea on how to prove this, let us recall the proof of the following classical unique
factorization result in the integers:
Theorem. — Every nonzero integer n ∈ Z>0 can be factored uniquely into prime
numbers.
Proof (Sketch). We do induction on n, the case n = 1 being trivial. If n is prime, we
are done. Thus we may assume that n is not prime. Then we can write n = ab with
1 ≤ a, b < n. Using the induction hypothesis, we know that a and b factor into prime
numbers. Thus the same holds for the product n = ab. The proof of uniqueness rests
on the following observation: If a, b, c are nonzero integers with ab = ac, then b = c. In order to imitate this proof in the new setting of ideals, we need the following things:
• A notion of multiplication and primeness of ideals (to formulate the statement).
• A notion of size of ideals (for the induction).
• Cancellation of ideals (for the uniqueness).
Definition 1.16. — Let R be a ring and α1 , . . . , αn ∈ R. We define
(α1 , . . . , αn ) = { r1 α1 + · · · + rn αn | ri ∈ R }
to be the ideal generated by α1 , . . . , αr . An ideal A of R is called principal, if there exists
α ∈ R such that A = (α) (we also write A = Rα). In case R is an integral domain and
all ideals of R are principal, we call R a principal ideal domain (PID).
Example 1.17. — (i) Consider the ring Z. Since every ideal of Z is of the form Zd
for some d ∈ Z, we conclude that Z is a principal ideal domain.
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Quadratic Number Fields – Lecture 2 – 05/08/17
Tommy Hofmann
√
(ii) We have seen that Z[ −5] is not a unique factorization domain. As every
√ principal
ideal domain is a unique factorization domain, we conclude that Z[ −5] is not a
principal ideal domain.
√
√
(iii) Consider the ideal A = (2, −5) of Z[ −5]. Then
√ √
1 = 2 · 3 − 5 = 2 · 3 + −5 −5 ∈ A
√
shows that 1 ∈ A. Thus A = (1) = Z[ −5]. Here we were lucky. In general it is
hard to decide whether an ideal given by a finite set of generators is principal or
not.
Definition 1.18. — Let R be a ring and A, B two ideals of R. We define the sum
A + B = { α + β | α ∈ A, β ∈ B },
and the product
AB = A · B =
( r
X
i=1
)
αi βi αi ∈ A, βi ∈ B
of A and B.
Lemma 1.19. — Let R be a ring and A, B, C ideals of R. Then the following hold:
(i) The sets A + B, AB and A ∩ B are ideals of R.
(ii) If A = (α1 , . . . , αr ) and B = (β1 , . . . , βs ), then
A + B = (α1 , . . . , αr , β1 , . . . , βs ),
AB = (αi βj | 1 ≤ i ≤ r, 1 ≤ j ≤ s).
(iii) We have A · R = A.
(iv) We have AB = BA and (AB)C = A(BC).
Proof. (i): Let α + β, α0 + β 0 ∈ A + B, with α, α0 ∈ A, β, β 0 ∈ B and r ∈ R. Then
(α + β) + (α0 + β 0 ) = (α + α0 ) + (β + β 0 ) ∈ A + B,
| {z } | {z }
∈A
∈B
Now let γ, γ 0 ∈ AB and r ∈ R. We can write γ =
rαi , γ 0 αi ∈ A for all 1 ≤ i ≤ r we have
0
γγ=
r
X
0
(γ αi )βi ∈ AB
r(α + β) = |{z}
rα + rβ ∈ A + B.
|{z}
∈A
Pr
i=1
∈B
αi βi with αi ∈ A, βi ∈ B. As
r
X
and rγ =
(rαi )βi ∈ AB.
i=1
i=1
Now let γ, γ 0 ∈ A ∩ B and r ∈ R. As γ ∈ A and γ 0 ∈ A we have γ + γ 0 ∈ A. Similar
γ + γ 0 ∈ B and therefore γ + γ 0 ∈ A ∩ B. The same argument shows that rγ ∈ A ∩ B.
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Quadratic Number Fields – Lecture 2 – 05/08/17
Tommy Hofmann
(ii): We first consider the statement for A + B. Since αi = αi + 0, βi = 0 + βi , the
inclusion ⊇“ is clear. Thus P
let γ = α + β P
∈ A + B with α ∈ A, β ∈ B. There exist
”
elements ri , si ∈ R with α = ri=1 ri αi , β = si=1 si βi . Thus
γ =α+β =
r
X
ri αi +
i=1
s
X
si βi ∈ (α1 , . . . , αr , β1 , . . . , βs ).
i=1
Let us now consider
for AB. Since αi βj ∈ AB for all i, j we are left with
Pm the statement
0
∈ AB with γl ∈ A, γl0 ∈ B. There exist rl,i , sl,j ∈ R with
⊆“. PLet α = l=1 γl γl P
”
r
γl = i=1 rl,i αi and γl0 = sj=1 sl,j βj . Hence
!
!
r
s
r X
s
X
X
X
γl γl0 =
rl,i αi
sl,j βj =
(rl,i sl,j )(αi βj ) ∈ (αi βj | 1 ≤ i ≤ r, 1 ≤ j ≤ s).
i=1
j=1
i=1 j=1
As α is the sum of the γl γl0 we also have α ∈ (αi βj | 1 ≤ i ≤ r, 1 ≤ j ≤ s).
P
(iii): Let α ∈ A. Then α = α · 1 ∈ A · R. If γ = li=1 αi ri ∈ A · R with αi ∈ A, ri ∈ R,
then γ ∈ A, since A is an ideal.
The equation AB = BA follows from the commutativity of R. Now let δ =
P(iv):
r
0
γi γi ∈ (AB)C with γi0 ∈ AB, γi ∈ C. There exist αi,j ∈ A, βi,j ∈ B with
i=1P
0
γi = sj=1 αi,j βi,j . Thus


r
s
X
X

δ=
αij (βi,j γi ) ∈ A(BC)
| {z }
i=1
j=1
∈BC
and we have shown that (AB)C ⊆ A(BC). Changing the role of A and C we also have
(CB)A ⊆ C(BA), which is equivalent to A(BC) ⊆ (AB)C using the commutativity.
Thus we are done.
Example 1.20. — (i) Consider the ring Z and ideals A = (m), B = (n) with
m, n ∈ Z. Then AB = (mn), A + B = (m, n) = (gcd(m, n)) and A ∩ B =
(m) ∩ (n) = (lcm(m, n)). Thus Definition 1.18 can (and should) be seen as a
generalization of greatest common divisor and least common multiple to the ideal
setting.
√
(ii) Consider R = Z[ −5] and the ideals
√
A2 = (2, 1 + −5)
√
A02 = (2, 1 − −5)
√
A3 = (3, 1 + −5)
√
A03 = (3, 1 − −5)
√
√
We want to investigate the equation 2 · 3 = 6 = (1 + −5)(1 − −5) using our
new ideal theoretic tools. In order to do this, we compute a few ideal products.
By Lemma 1.19 (ii) we have
√
√
A2 A02 = (4, 2(1 + −5), 2(1 − −5), 6) ⊆ (2).
4
Quadratic Number Fields – Lecture 2 – 05/08/17
Tommy Hofmann
√
(Every element is a multiple of 2.) Moreover we have 2 = 6 − 4 = (1 + −5)(1 −
√
−5) − 2 · 2 ∈ A2 A02 and therefore A2 A02 = (2). A similar computation shows
A3 A03 = (3). We now turn to A2 A3 . Applying Lemma 1.19 (ii) yields
√
√
√
√
A2 A3 = (6, 2(1 + −5), 3(1 + −5), (1 + −5)2 ) ⊆ (1 + −5)
√
√
√
which in combination
with
1
+
−5
=
3(1
+
−5)
−
2(1
+
−5) ∈ A2 A3 shows
√
√
that A2 A3 = (1 + −5). Similarly we have A02 A03 = (1 − −5). Thus
(6) = (2) · (3) = (A2 A02 )(A3 A03 )
√
√
= (1 + −5) · (1 − −5) = (A2 A3 )(A02 A03 ),
and the nonunique factorization into elements turns into unique factorization of
ideals. The counterexample to being a unique factorization domain comes from
grouping the ideals together differently.
Definition 1.21. — Let K be a quadratic field and A an ideal of OK . We define
Ā = {ᾱ | α ∈ A}
to be the conjugate (ideal) of A.
Proposition 1.22. — Let K be a quadratic field and A an ideal of OK . Then the
following hold:
(i) The set A is an ideal of OK .
(ii) If A = (α1 , . . . , αr ), then A = (α1 , . . . , αr ).
(iii) There exists a unique nonnegative integer n ∈ Z≥0 such that AA = (n).
Proof. (i): The set A is the image of the ideal A under the ring homomorphism : OK →
OK and therefore an ideal
P itself.
(ii): Let α ∈ A, α = ri=1 ri αi . Then
α=
r
X
ri αi ∈ (α1 , . . . , αr ).
i=1
Conversely, if α ∈
Pr
i=1 ri αi
∈ (α1 , . . . , αr ), then
α=
r
X
ri αi ∈ A.
i=1
(iii): If A = {0} then A = {0} and AA = {0} = (0) and 0 is indeed the unique
rational integer generating AA. Thus we may assume that A is nonzero. Let α ∈ A,
α 6= 0. Then 0 6= N(α) = αᾱ ∈ AA ∩ Z. Since AA ∩ Z 6= {0} is an ideal of Z, there
exists n ∈ Z≥0 such that AA ∩ Z = Zn. We claim that AA = (n). Since n ∈ AA, it
remains to show that AA ⊆ (n). Thus let α = α1 β1 + · · · + αr βr ∈ AA with αi , βi ∈ A.
We fix i ∈ {1, . . . , r} and consider the element αi βi .
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Quadratic Number Fields – Lecture 2 – 05/08/17
Tommy Hofmann
We have tr(αi βi ) = αi βi + αi βi ∈ AA ∩ Z = Zn and therefore
αi βi
tr(αi βi )
tr
=
∈ Z.
n
n
For the norm we have N(αi ) = αi αi ∈ AA ∩ Z = Zn and similar N(βi ) ∈ Zn. Thus
αi βi
N(αi )
N(βi )
N
=
·
∈ Z.
n
n
n
Since both trace and norm are integral we conclude that (αi βi )/d ∈ OK , which implies
that αi βi ∈ (n). As α is just the sum of the elements
√ αi βi we obtain α ∈ (n).
It remains to consider uniqueness: Let K = Q( d) with d ∈ Z squarefree. Consider
two elements n, n0 ∈ Z≥0 with AA = (n) = (n0 ). Since n0 ∈ (n) there exists α = a+bωd ∈
OK = Z[ωd ] such that n0 = αn = (a + bωd )n = an + bnωd . Thus b = 0 and n0 = an. In
the same vein we obtain n = a0 n0 for some a0 ∈ Z. Since n and n0 are nonnegative we
conclude that n = n0 .
Definition 1.23. — Let K be a quadratic field and A an ideal of OK . We call the
unique integer n ∈ Z≥0 with AA = (n) the norm of A and denote it by N(A).
Proposition 1.24. — Let K be a quadratic field and A, B ideals of OK . Then the
following hold:
(i) We have N(AB) = N(A) N(B).
(ii) If A = (α), α ∈ OK , then N(A) = |N(α)|.
(iii) We have N(A) = 1 if and only if A = OK .
Proof. Exercise.
Definition 1.25. — Let A, B be ideals of a ring R. We say that A divides B (and
we write A | B) if there exists an ideal C of R such that AC = B.
Example 1.26. — If A divides B, then there exists C with AC = B. This shows that
B = AC ⊆ A · R = A. Thus the divisor A is larger than the divisible ideal B. This is
counter-intuitive to the integer setting (where a divisor is smaller than the number that
it divides) and a constant source of confusion. It is best to always have an easy example
in mind: (2) divides (6) and (6) ⊆ (2).
6