Quadratic Number Fields – Lecture 3 – 05/15/17
Tommy Hofmann
Proposition 1.27. — Let K be a quadratic field and A, B ideals of OK with A 6= {0}.
Then A | B if and only if B ⊆ A.
Proof. We already showed one of the directions in the discussion following Definition 1.25.
Now let B ⊆ A. By Proposition 1.22 we know that AA = (n) for some n ∈ Z.
We have n 6= 0 since A 6= {0}. Multiplying the inclusion B ⊆ A with A we obtain
BA ⊆ AA = (n). Thus the set
C :=
nα
o
BA
:=
| α ∈ BA
n
n
is a subset of OK . In fact it is an ideal and we have
AC = CA =
BAA
B(n)
BA
A=
=
= B.
n
n
n
See the second problem set for a justification of these steps.
Proposition 1.28. — Let K be a quadratic field and A, B, C ideals of OK with A 6=
{0}. If AB = AC, then B = C.
Proof. By Proposition 1.22 we know that AA = (n) for some n ∈ Z, n 6= 0. Multiplying
the equality AB = AC with A we obtain (n)B = AAB = AAC = (n)C. Now let β ∈ B.
Then nβ ∈ (n)B = (n)C. Thus there exists γ ∈ C such that nβ = nγ and since OK
is an integral domain this implies β = γ ∈ C. Changing the role of B and C we also
obtain C ⊆ B and therefore B = C.
Definition 1.29. — Let R be a ring. An ideal P 6= R is called prime, if for all ideals
A, B of R the inclusion AB ⊆ P implies A ⊆ P or B ⊆ P .
Example 1.30. — Let R = Z be the ring of integers. Then pZ, p 6= 0, is a prime
ideal if and only if p is a prime number.
The next proposition will give various useful characterizations for the case of rings of
integers of quadratic fields. In particular we generalize the following statement to ideals:
A number n ∈ Z≥2 is prime if and only if the only divisors of n are 1 and n.
Proposition 1.31. — Let K be a quadratic field. For an ideal P 6= OK the following
are equivalent:
(i) The ideal P is a prime ideal.
(ii) If A, B are ideals of OK and P divides AB, then P divides A or P divides B.
(iii) The only divisors of P are OK and P .
(iv) If α, β are elements of OK with αβ ∈ P , then α ∈ P or β ∈ P .
(v) The ring OK /P is an integral domain.
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Quadratic Number Fields – Lecture 3 – 05/15/17
Tommy Hofmann
Proof. (i) ⇔ (ii): This is Proposition 1.27.
(i) ⇔ (iv) ⇔ (v): Exercise.
(ii) ⇒ (iii): Assume that P is prime and A is an ideal dividing P . Then P ⊆ A
and there exists an ideal B with AB = P = P · OK . Thus P divides AB, which by
assumption yields P | A or P | B. In the first case we have A ⊆ P and therefore
A = P . In the second case we obtain B ⊆ P and therefore B = P (since B divides P
we also have P ⊆ B). Hence P = AB = AP , from which we conclude A = OK using
Proposition 1.28. Thus (iii) holds.
(iii) ⇒ (ii): Assume that P | AB. Consider the ideal A + P . Since P ⊆ A + P (that
is, A + P | P ), we have A + P = P or A + P = OK . If A + P = P , then A ⊆ P and
P | A by Proposition 1.27. Thus we may assume that A + P = OK . Since 1 ∈ OK there
exist α ∈ A, β ∈ P such that 1 = α + β. Now let γ ∈ B. Then
γ = 1 · γ = (α + β)γ = αγ + βγ ∈ AB + BP ⊆ P + P = P.
|{z} |{z}
∈AB
∈BP
Thus B ⊆ P , that is, P | B by Proposition 1.27.
Lemma 1.32. — Let K be a quadratic field an P an ideal of OK . If N(P ) ∈ Z is a
prime number, then P is a prime ideal.
Proof. Assume that P is not prime. Then by Proposition 1.31 there exists an ideal
Q 6= OK , Q 6= P with Q | P . Thus N(Q) 6= 1, N(Q) 6= N(P ) and N(Q) | N(Q). Since
both N(Q) and N(P ) are non-negative, this implies that N(P ) is not prime.
Theorem 1.33. — Let K be a quadratic field and A 6= OK a nonzero ideal of OK .
Then there exist prime ideals P1 , . . . , Pg of OK such that
A = P 1 · · · Pg .
Moreover, the factorization is unique up to permutation of the factors.
Proof. Existence: We use induction on n = N(A). We have n > 1 since A 6= OK .
If n = 2, then A is a prime ideal: If B is a divisor of A, then N(B) | n = 2. Thus
N(B) ∈ {1, 2}, which implies B = A or B = OK (since an ideal contains its norm, an
ideal of norm 1 is the whole ring). Thus in this case A has a factorization into prime
ideals. Now let n > 2. If A is prime we are done. Therefore we assume that A is not
prime. Then by Proposition 1.31 the ideal A has a nontrivial divisor and there exist
B, C with A = BC and 1 < N(B), N(C) < N(A) = n. By induction we know that B
and C factor into prime ideals. Thus the same holds for the product A = BC.
Uniqueness: We use induction on n = N(A). For n = 2 the ideal A is prime and
if A = P1 · · · Pg = Q1 · · · Qs are factorizations into prime ideals, applying the norm
shows that g = 1 = s and A = P1 = Q1 . In particular the factorization is unique.
Thus let n > 2 and consider prime factorizations A = P1 · · · Pg = Q1 · · · Qg of A. Then
P1 | (Q1 · · · Qs ) and as P1 is prime, there exists j with P1 | Qj . After reordering the Qi ,
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Quadratic Number Fields – Lecture 3 – 05/15/17
Tommy Hofmann
we may and will assume that j = 1. Thus P1 | Q1 . Since Q1 is prime and P1 6= OK ,
Proposition 1.31 implies that P1 = Q1 . Thus we can cancel this factor on both sites
and obtain P2 · · · Pg = Q2 · · · Qs . Since N(P1 ) > 1 these are factorizations of an ideal
with norm strictly less than N(A). Hence by induction P2 , . . . , Pg are a permutation of
Q2 , . . . , Qs . Since P1 = Q1 this holds also for P1 , . . . , Pg and Q1 , . . . , Qs .
Remark 1.34. —
(i) An integral domain in which every nontrivial ideal factors uniquely into a product
of prime ideals is a Dedekind domain. They form an interesting class of rings and
play a crucial role in number theory and algebraic geometry.
(ii) Let K be a quadratic field and assume that R is a subring of K with Z ( R ( OK .
Then one can show that unique factorization into prime ideals fails for non-zero
ideals√of R. For example, if d is a squarefree integer with d ≡ 1 mod 4, then ideals
of Z[ d] do not factor uniquely into prime ideals.
While we know that everything factors into prime ideals, so far we don’t know much
about prime ideals in rings of integers of quadratic fields.
Lemma 1.35. — Let K be a quadratic field and P a nonzero prime ideal.
(i) If P is a prime ideal of OK , then P ∩ Z is a prime ideal of Z.
(ii) If P is a non-zero prime ideal of OK , then there exists a prime number p ∈ Z such
that P ∩ pOK .
Proof. (i): Easy. (ii): Since N(P ) ∈ P we have P | (N(P )). Now consider the factorization N(P ) = p1 · · · pr of N(P ) ∈ Z into prime numbers pi ∈ Z. Then P | (N(P )) =
(p1 · · · pr ) = (p1 ) · · · (pr ) and as P is prime there exists i with pi ∈ P , that is, P | (pi ).
Thus every prime ideal of OK occurs in the factorization of the ideals (p), p ∈ Z
prime. Using the Legendre symbol, we can make this much more explicit. Recall that
for a prime p ∈ Z and d ∈ Z, the Legendre symbol ( dp ) is defined as follows:


1,
d
= −1,

p

0,
if p - d and there exists a ∈ Z such that a2 ≡ d mod p,
if there does not exist a ∈ Z such that a2 ≡ d mod p,
if p | d.
√
Theorem 1.36. — Let K = Q( d) be a quadratic field and p ∈ Z an odd prime.
Then the following hold:
(i) If ( dp ) = −1, then (p) is a prime ideal with norm p2 .
√
(ii) If ( dp ) = 1, let a ∈ Z be an integer such that a2 ≡ d mod p. Then P = (p, a + d)
is a prime ideal of OK √and (p) = P P . Moreover N(P ) = N(P ) = p.
(iii) If ( dp ) = 0, let P = (p, d). Then P is prime, (p) = P 2 and N(P ) = p.
Proof. Exercise.
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Quadratic Number Fields – Lecture 3 – 05/15/17
Tommy Hofmann
√
Example
1.37. — We want to
√
√ factor the principal ideal A = (−55 + 187 −5) of
Z[ −5] = OK , where K = Q( −5). To narrow down the possible prime factors we
consider the norm of A: Assume that A = Q1 · · · Qs is the prime ideal factorization
of A and N(A) = p1 · · · pr is the prime factorization of N(A) ∈ Z into prime numbers
pi ∈ Z. Then N(Q1 ) · · · N(Qs ) = p1 · · · pr and because the pi are prime we know that
every N(Qj ) is divided by only one of the pi . Thus the only prime ideals which can show
up in the factorization of A are the prime ideals lying over the prime factors of the norm
of A. In our case we have N(A) = 552 + 5 · 1872 = 177870 = 2 · 3 · 5 · 72 · 112 . Hence we
need to determine the prime ideals lying over 2, 3, 5, 7 and 11.
(i) p = 2: We cannot directly apply Theorem 1.36 (2 is not odd). But there exists
a similar theorem
(we do not prove it here) which asserts that (2) = P22 , where
√
P2 = (2, 1 + −5) is a prime ideal with norm N(P2 ) = 2.
(ii) p = 3: We apply Theorem 1.36. Since ( −5
) = ( 31 ) = 1 and 12 ≡ −5 mod 3 we
√3
know that (3) = P3 P3 , where P3 = (3, 1 + −5) is prime and N(P3 ) = N(P3 ) = 3.
(iii) p = 5: We apply Theorem 1.36. Since ( −5
) = 0, we have (5) = P52 , where
5
√
√
P5 = (5, −5) = ( −5) is prime and N(P5 ) = 5.
) = 1 and 32 ≡ −5 mod 7, we know that
(iv) p = 7: We apply Theorem 1.36. Since ( −5
7
√
(7) = P7 P7 , where P7 = (7, 3 + −5) is prime and N(P7 ) = N(P7 ) = 7.
) = −1, we know that P11 = (11) is
(v) p = 11: We apply Theorem 1.36. Since ( −5
11
prime and N(P11 ) = 11.
Combining this with the factorization of N(A) we conclude that
A = P2 · Q3 · P5 · Q7 · Q07 · P11 ,
with Q3 ∈ {P3 , P3 } and Q7 , Q07 ∈ {P7 , P7 }. Here we used the fact that P2 , P5 and P11
are the unique prime ideals with norm 2, 5 and 112 respectively. It remains to determine
Q3 , Q7 as well as Q07 .
Note that Q7 must be equal to Q07 . For if this is not the case, then {Q7 , Q07 } = {P7 , P7 }
0
0
and Q7 · Q√
7 = P7 · P7 = (7). Since both Q7 and Q7 divide A, this implies that 7 divides
−55 + 187 −5, a contradiction. Thus we have
A = P2 · Q3 · P5 · Q27 · P11 .
Recall that in Example 1.20 we had shown that
√
√
√
P2 · P3 = (2, 1 + −5)(3, 1 − −5) = (1 − −5).
√
√
√
√
Thus P3 divides
(1
−
−5).
As
(1
−
−5)(−165
+
22
−5)
=
−55
+
187
−5 we also
√
have that (1 − −5) divides A. Hence by transitivity of divisibility we conclude that P3
divides A and therefore obtain Q3 = P3 . A similar argument shows that Q7 = P7 . We
conclude that
√
√
√
√
√
(−55 + 187 −5) = (2, 1 + −5)(3, 1 − −5)( −5)(7, 3 + −5)2 (11)
is the unique prime ideal factorization.
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