5Structure of prime ideals
II Number Fields
5 Structure of prime ideals
We can now move on to find all prime ideals of
O
L
. We know that every ideal
factors as a product of prime ideals, but we don’t know what the prime ideals
are. The only obvious way we’ve had to obtain prime ideals is to take a usual
prime, take its principal ideal and factor it in
O
L
, and get the resultant prime
ideals.
It turns out this gives us all prime ideals.
Lemma.
Let
p C O
L
be a prime ideal. Then there exists a unique
p ∈ Z
,
p
prime, with p  hpi. Moreover, N(p) = p
f
for some 1 ≤ f ≤ n.
This is not really too exciting, as soon as we realize that
p  hpi
is the same
as saying hpi ⊆ p, and we already know p ∩ Z is nonempty.
Proof.
Well
p ∩Z
is an ideal in
Z
, and hence principal. So
p ∩Z
=
pZ
for some
p ∈ Z.
We now claim
p
is a prime integer. If
p
=
ab
with
ab ∈ Z
. Then since
p ∈ p
,
either a ∈ p or b ∈ p. So a ∈ p ∩ Z = pZ or b ∈ p ∩ Z = pZ. So p  a or p  b.
Since
hpi ⊆ p
, we know
hpi
=
pa
for some ideal
a
by factorization. Taking
norms, we get
p
n
= N(hpi) = N(p)N(a).
So the result follows.
This is all good. So all we have to do is to figure out how principal ideals
hpi
factor into prime ideals.
We write
hpi = p
e
1
1
···p
e
m
m
for some distinct prime ideals
p
i
, with
N
(
p
i
) =
p
f
i
for some positive integers
e
i
.
Taking norms, we get
p
n
=
Y
p
f
i
e
i
.
So
n =
X
e
i
f
i
.
We start by giving some names to the possible scenarios.
Definition
(Ramification indices)
.
Let
hpi
=
p
e
1
1
···p
e
m
m
be the factorization
into prime ideals. Then e
1
, ··· , e
m
are the ramification indices.
Definition (Ramified prime). We say p is ramified if some e
i
> 1.
Definition
(Inert prime)
.
We say
p
is inert if
m
= 1 and
e
m
= 1, i.e.
hpi
remains prime.
Definition
(Splitting prime)
.
We say
p
splits completely if
e
1
=
···
=
e
m
= 1 =
f
1
= ··· = f
m
. So m = n.
Note that this does not exhaust all possibilities. The importance of these
terms, especially ramification, will become clear later.
So how do we actually compute
p
i
and
e
i
? In other words, how can we factor
the ideal
hpi C O
L
into prime ideals? The answer is given very concretely by
Dedekind’s criterion.
Theorem
(Dedekind’s criterion)
.
Let
α ∈ O
L
and
g
(
x
)
∈ Z
[
x
] be its minimal
polynomial. Suppose
Z
[
α
]
⊆ O
L
has finite index, coprime to
p
(i.e.
p  O
L
/Z
[
α
]

).
We write
¯g(x) = g(x) (mod p),
so ¯g(x) ∈ F
p
[x]. We factor
¯g(x) = ϕ
e
1
1
···ϕ
e
m
m
into distinct irreducibles in F
p
[x]. We define the ideal
p
i
= hp, ˜ϕ
i
(α)i C O
L
,
generated by
p
and
˜ϕ
i
, where
˜ϕ
i
is any polynomial in
Z
[
x
] such that
˜ϕ
i
mod p
=
ϕ
i
. Notice that if
˜ϕ
0
is another such polynomial, then
p 
(
˜ϕ
i
−ϕ
i
), so
hp, ˜ϕ
0
(
α
)
i
=
hp, ˜ϕ(α)i.
Then the p
i
are prime, and
hpi = p
e
1
1
···p
e
m
m
.
Moreover, f
i
= deg ϕ
i
, so N(a) = p
deg ϕ
i
.
If we are lucky, we might just find an
α
such that
Z
[
α
] =
O
L
. If not, we
can find something close, and as long as
p
is not involved, we are fine. After
finding
α
, we get its minimal polynomial, factor it, and immediately get the
prime factorization of hpi.
Example.
Consider
L
=
Q
(
√
−11
). We want to factor
h
5
i
in
O
L
. We consider
Z
[
√
−11
]
⊆ O
L
. This has index 2, and (hopefully) 5

2. So this is good
enough. The minimal polynomial is
x
2
+ 11. Taking mod 5, this reduces to
x
2
− 4 = (x −2)(x + 2). So Dedekind says
h5i = h5,
√
−11 + 2ih5,
√
−11 − 2i.
In general, consider
L
=
Q
(
√
d
),
d 6
= 0
,
1 and squarefree, and
p
an odd
prime. Then
Z
[
√
d
]
⊆ O
L
has index 1 or 2, both of which are coprime to
p
. So
Dedekind says factor x
2
− d mod p. What are the possibilities?
(i)
There are two distinct roots mod
p
, i.e.
d
is a square mod
p
, i.e.
d
p
= 1.
Then
x
2
− d = (x + r)(x − r) (mod p)
for some r. So Dedekind says
hpi = p
1
p
2
,
where
p
1
= hp,
√
d − ri, p
2
= hp,
√
d + ri,
and N(p
1
) = N(p
2
) = p. So p splits.
(ii) x
2
− d
is irreducible, i.e.
d
is not a square mod
p
, i.e.
d
p
=
−
1. Then
Dedekind says hpi = p is prime in O
L
. So p is inert.
(iii) x
2
−d
has a repeated root mod
p
, i.e.
p  d
, or alternatively
d
p
= 0. Then
by Dedekind, we know
hpi = p
2
,
where
p = hp,
√
di.
So p ramifies.
So in fact, we see that the Legendre symbol encodes the ramification behaviour
of the primes.
What about the case where p = 2, for L = Q(
√
d)? How do we factor h2i?
Lemma. In L = Q(
√
d),
(i) 2 splits in L if and only if d ≡ 1 (mod 8);
(ii) 2 is inert in L if and only if d ≡ 5 (mod 8);
(iii) 2 ramifies in L if d ≡ 2, 3 (mod 4).
Proof.
–
If
d ≡
1 (
mod
4), then then
O
L
=
Z
[
α
], where
α
=
1
2
(1 +
√
d
). This has
minimal polynomial
x
2
− x +
1
4
(1 − d).
We reduce this mod 2.
◦ If d ≡ 1 (mod 8), we get x(x + 1). So 2 splits.
◦
If
d ≡
5 (
mod
8), then we get
x
2
+
x
+ 1, which is irreducible. So
h
2
i
is prime, hence 2 is inert.
–
If
d ≡
2
,
3 (
mod
4), then
O
L
=
Z
[
√
d
], and
x
2
−d
is the minimal polynomial.
Taking mod 2, we get
x
2
or
x
2
+ 1 = (
x
+ 1)
2
. In both cases, 2 ramifies.
Note how important
p  O
L
/Z
[
α
]

is. If we used
Z
[
√
d
] when
d ≡
1 (
mod
4),
we would have gotten the wrong answer.
Recall
D
L
=
(
4d d ≡ 2, 3 (mod 4)
d d ≡ 1 (mod 4)
The above computations show that
p  D
L
if and only if
p
ramifies in
L
. This
happens to be true in general. This starts to hint how important these invariants
like D
L
are.
Now we get to prove Dedekind’s theorem.
Proof of Dedekind’s criterion. The key claim is that
Claim. We have
O
L
p
i
∼
=
F
p
[x]
hϕ
i
i
.
Suppose this is true. Then since
ϕ
i
is irreducible, we know
F
p
[x]
hϕ
i
i
is a field.
So p
i
is maximal, hence prime.
Next notice that
p
e
1
1
= hp, ˜ϕ
i
(α)i
e
i
⊆ hp, ˜ϕ
i
(α)
e
i
i.
So we have
p
e
1
1
···p
e
m
m
⊆ hp, ˜ϕ
1
(α)
e
1
··· ˜ϕ
m
(α)
e
m
i = hp, g(α)i = hpi,
using the fact that g(α) = 0.
So to prove equality, we notice that if we put
f
i
=
deg ϕ
i
, then
N
(
p
i
) =
p
f
i
,
and
N(p
e
1
1
···p
e
m
m
) = N(p
1
)
e
1
···N(p
m
)
e
m
= p
P
e
i
f
i
= p
deg g
.
Since
N
(
hpi
) =
p
n
, it suffices to show that
deg g
=
n
. Since
Z
[
α
]
⊆ O
L
has finite
index, we know Z[α]
∼
=
Z
n
. So 1, α, ··· , α
n−1
are independent over Z, hence Q.
So deg g = [Q(α) : Q] = n = [L : Q], and we are done.
So it remains to prove that
O
L
p
i
∼
=
Z[α]
p
i
∩ Z[α]
∼
=
F
p
[x]
hϕ
i
i
.
The second isomorphism is clear, since
Z[α]
hp, ˜ϕ
i
(α)i
∼
=
Z[x]
hp, ˜ϕ
i
(x), g(x)i
∼
=
F
p
[x]
h˜ϕ
i
(x), g(x)i
=
F
p
[x]
hϕ
i
(x), ¯g(x)i
=
F
p
[x]
hϕ
i
i
.
To prove the first isomorphism, it suffices to show that the following map is an
isomorphism:
Z[α]
pZ[α]
→
O
L
pO
L
(∗)
x + pZ[α] 7→ x + pO
L
If this is true, then quotienting further by ˜ϕ
i
gives the desired isomorphism.
To prove the claim, we consider a slightly different map. We notice
p 
O
L
/Z[α] means the “multiplication by p” map
O
L
Z[α]
O
L
Z[α]
p
(†)
is injective. But
O
L
/Z
[
α
] is a finite abelian group. So the map is an isomorphism.
By injectivity of (
†
), we have
Z
[
α
]
∩ pO
L
=
pZ
[
α
]. By surjectivity, we have
Z
[
α
] +
pO
L
=
O
L
. It thus follows that (
∗
) is injective and surjective respectively.
So it is an isomorphism. We have basically applied the snake lemma to the
diagram
0 Z[α] O
L
O
L
Z[α]
0
0 Z[α] O
L
O
L
Z[α]
0
p
p
p
Corollary.
If
p
is prime and
p < n
= [
L
:
Q
], and
Z
[
α
]
⊆ O
L
has finite index
coprime to p, then p does not split completely in O
L
.
Proof.
By Dedekind’s theorem, if
g
(
x
) is the minimal polynomial of
α
, then the
factorization of
¯g
(
x
) =
g
(
x
)
mod p
determines the factorization of
hpi
into prime
ideals. In particular,
p
splits completely if and only if
¯g
factors into distinct
linear factors, i.e.
¯g(x) = (x − α
1
) ···(x − α
n
),
where
α
i
∈ F
p
and
α
i
are distinct. But if
p < n
, then there aren’t
n
distinct
elements of F
p
!
Example.
Let
L
=
Q
(
α
), where
α
has minimal polynomial
x
3
− x
2
−
2
x −
8.
This is the case where
n
= 3
>
2 =
p
. On example sheet 2, you will see that 2
splits completely, i.e.
O
L
/
2
O
L
=
F
2
× F
2
× F
2
. But then this corollary shows
that for all
β ∈ O
L
,
Z
[
β
]
⊆ O
L
has even index, i.e. there does not exist an
β ∈ O
L
with O
L
/Z[β] odd.
As we previously alluded to, the following is true:
Theorem. p  D
L
if and only if p ramifies in O
L
.
We will not prove this.