3Multiplicative structure of ideals
II Number Fields
3 Multiplicative structure of ideals
Again, let
L/Q
be a number field. It turns out that in general, the integral
ring
O
L
is not too wellbehaved as a ring. In particular, it fails to be a UFD in
general.
Example. Let L = Q(
√
5). Then O
L
= Z[
√
−5]. Then we find
3 · 7 = (1 + 2
√
−5)(1 − 2
√
−5).
These have norms 9, 49, 21, 21. So none of 3, 7, 1 + 2
√
5 are associates.
Moreover, 3
,
7
,
1
±
2
√
−5
are all irreducibles. The proof is just a straightfor
ward check on the norms.
For example, to show that 3 is irreducible, if 3 =
αβ
, then 9 =
N
(3) =
N
(
α
)
N
(
β
). Since none of the terms on the right are
±
1, we must have
N
(
α
) =
±
3.
But there are no solutions to
x
2
+ 5y
2
= ±3
where x, y are integers. So there is no α = x + y
√
−5 such that N(α) = ±3.
So unique factorization fails.
Note that it is still possible to factor any element into irreducibles, just not
uniquely — we induct on
N
(
α
)

. If
N
(
α
)

= 1, then
α
is a unit. Otherwise,
α
is either irreducible, or
α
=
βγ
. Since
N
(
β
)
N
(
γ
) =
N
(
α
), and none of them are
±1, we must have N (β), N(γ) < N (α). So done by induction.
To fix the lack of unique factorization, we instead look at ideals in
O
L
.
This has a natural multiplicative structure — the product of two ideals
a, b
is
generated by products
ab
, with
a ∈ a, b ∈ b
. The big theorem is that every ideal
can be written uniquely as a product of prime ideals.
Definition
(Ideal multiplication)
.
Let
a, b C O
L
be ideals. Then we define the
product ab as
ab =
X
i,j
α
i
β
j
: α
i
∈ a, β
j
∈ b
.
We write a  b if there is some ideal c such that ac = b, and say a divides b.
The proof of unique factorization is the same as the proof that
Z
is a UFD.
Usually, when we want to prove factorization is unique, we write an object as
a = x
1
x
2
···x
m
= y
1
y
2
···y
n
.
We then use primality to argue that
x
1
must be equal to some of the
y
i
, and
then cancel them from both sides. We can usually do this because we are working
with an integral domain. However, we don’t have this luxury when working with
ideals.
Thus, what we are going to do is to find inverses for our ideals. Of course,
given any ideal
a
, there is no ideal
a
−1
such that
aa
−1
=
O
L
, as for any
b
,
we know
ab
is contained in
a
. Thus we are going to consider more general
objects known as fractional ideals, and then this will allow us to prove unique
factorization.
Even better, we will show that
a  b
is equivalent to
b ⊆ a
. This is a very
useful result, since it is often very easy to show that
b ⊆ a
, but it is usually very
hard to actually find the quotient a
−1
b.
We first look at some examples of multiplication and factorization of ideals
to get a feel of what these things look like.
Example. We have
hx
1
, ··· , x
n
ihy
1
, ··· , y
m
i = hx
i
y
j
: 1 ≤ i ≤ n, 1 ≤ j ≤ mi.
In particular,
hxihyi = hxyi.
It is also an easy exercise to check (ab)c = a(bc).
Example. In Z[
√
−5], we claim that we have
h3i = h3, 1 +
√
−5ih3, 1 −
√
−5i.
So h3i is not irreducible.
Indeed, we can compute
h3, 1 +
√
−5ih3, 1 −
√
−5i = h9, 3(1 + 2
√
−5), 3(1 − 2
√
−5), 21i.
But we know
gcd
(9
,
21) = 3. So
h
9
,
21
i
=
h
3
i
by Euclid’s algorithm. So this is in
fact equal to h3i.
Notice that when we worked with elements, the number 3 was irreducible, as
there is no element of norm 3. Thus, scenarios such as 2
·
3 = (1+
√
−5
)(1
−
√
−5
)
could appear and mess up unique factorization. By passing on to ideals, we can
further factorize
h
3
i
into a product of smaller ideals. Of course, these cannot be
principal ideals, or else we would have obtained a factorization of 3 itself. So
we can think of these ideals as “generalized elements” that allow us to further
break elements down.
Indeed, given any element in
α ∈ O
L
, we obtain an ideal
hαi
corresponding
to α. This map is not injective — if two elements differ by a unit, i.e. they are
associates, then they would give us the same ideal. However, this is fine, as we
usually think of associates as being “the same”.
We recall the following definition:
Definition
(Prime ideal)
.
Let
R
be a ring. An ideal
p ⊆ R
is prime if
R/p
is
an integral domain. Alternatively, for all
x, y ∈ R
,
xy ∈ p
implies
x ∈ p
or
y ∈ p
.
In this course, we take the convention that a prime ideal is nonzero. This is
not standard, but it saves us from saying “nonzero” all the time.
It turns out that the ring of integers
O
L
is a very special kind of rings, known
as Dedekind domains:
Definition (Dedekind domain). A ring R is a Dedekind domain if
(i) R is an integral domain.
(ii) R is a Noetherian ring.
(iii) R
is integrally closed in
Frac R
, i.e. if
x ∈ Frac R
is integral over
R
, then
x ∈ R.
(iv) Every proper prime ideal is maximal.
This is a rather specific list of properties
O
L
happens to satisfy, and it turns
out most interesting properties of
O
L
can be extended to arbitrary Dedekind
domains. However, we will not do the general theory, and just study number
fields in particular.
The important result is, of course:
Proposition.
Let
L/Q
be a number field, and
O
L
be its ring of integers. Then
O
L
is a Dedekind domain.
The first three parts of the definition are just bookkeeping and not too
interesting. The last one is what we really want. This says that
O
L
is “one
dimensional”, if you know enough algebraic geometry.
Proof of (i) to (iii).
(i) Obvious, since O
L
⊆ L.
(ii)
We showed that as an abelian group,
O
L
=
Z
n
. So if
a ≤ O
L
is an ideal,
then
a ≤ Z
n
as a subgroup. So it is finitely generated as an abelian group,
and hence finitely generated as an ideal.
(iii)
Note that
Frac O
L
=
L
. If
x ∈ L
is integral over
O
L
, as
O
L
is integral
over Z, x is also integral over Z. So x ∈ O
L
, by definition of O
L
.
To prove the last part, we need the following lemma, which is also very
important on its own right.
Lemma.
Let
a C O
L
be a nonzero ideal. Then
a ∩ Z 6
=
{
0
}
and
O
L
/a
is finite.
Proof. Let α ∈ a and α 6= 0. Let
p
α
= x
m
+ a
m−1
x
m−1
+ ··· + a
0
be its minimal polynomial. Then
p
α
∈ Z
[
x
]. We know
a
0
6
= 0 as
p
α
is irreducible.
Since p
α
(α) = 0, we know
a
0
= −α(α
m−1
+ a
m−1
α
m−2
+ ··· + a
2
α + a
1
).
We know
α ∈ a
by assumption, and the mess in the brackets is in
O
L
. So the
whole thing is in a. But a
0
∈ Z. So a
0
∈ Z ∩ a.
Thus, we know ha
0
i ⊆ a. Thus we get a surjection
O
L
ha
0
i
→
O
L
a
.
Hence it suffices to show that O
L
/ha
0
i is finite. But for every d ∈ Z, we know
O
L
hdi
=
Z
n
dZ
n
=
Z
dZ
n
,
which is finite.
Finally, recall that a finite integral domain must be a field — let
x ∈ R
with
x 6
= 0. Then
m
x
:
y 7→ xy
is injective, as
R
is an integral domain. So it is a
bijection, as R is finite. So there is some y ∈ R such that xy = 1.
This allows us to prove the last part
Proof of (iv).
Let
p
be a prime ideal. Then
O
L
/p
is an integral domain. Since
the lemma says O
L
/p is finite, we know O
L
/p is a field. So p is maximal.
We now continue on to prove a few more technical results.
Lemma.
Let
p
be a prime ideal in a ring
R
. Then for
a, b C R
ideals, then
ab ⊆ p implies a ⊆ p or b ⊆ p.
Once we’ve shown that inclusion of ideals is equivalent to divisibility, this in
effect says “prime ideals are primes”.
Proof.
If not, then there is some
a ∈ a \ p
and
b ∈ b \ p
. Then
ab ∈ ab ⊆ p
. But
then a ∈ p or b ∈ p. Contradiction.
Eventually, we will prove that every ideal is a product of prime ideals.
However, we cannot prove that just yet. Instead, we will prove the following
“weaker” version of that result:
Lemma.
Let 0
6
=
a C O
L
a nonzero ideal. Then there is a subset of
a
that is a
product of prime ideals.
The proof is some unenlightening abstract nonsense.
Proof.
We are going to use the fact that
O
L
is Noetherian. If this does not hold,
then there must exist a maximal ideal
a
not containing a product of prime ideals
(by which we mean any ideal greater than
a
contains a product of prime ideals,
not that
a
is itself a maximal ideal). In particular,
a
is not prime. So there are
some x, y ∈ O
L
such that x, y 6∈ a but xy ∈ a.
Consider
a
+
hxi
. This is an ideal, strictly bigger than
a
. So there exists
prime ideals p
1
, ··· , p
r
such that p
1
···p
r
⊆ a + hxi, by definition.
Similarly, there exists q
1
, ···q
s
such that q
1
···q
s
⊆ a + hyi.
But then
p
1
···p
r
q
1
···q
s
⊆ (a + hxi)(a + hyi) ⊆ a + hxyi = a
So a contains a product of prime ideals. Contradiction.
Recall that for integers, we can multiply, but not divide. To make life easier,
we would like to formally add inverses to the elements. If we do so, we obtain
things like
1
3
, and obtain the rationals.
Now we have ideals. What can we do? We can formally add some inverse
and impose some nonsense rules to make sure it is consistent, but it is helpful
to actually construct something explicitly that acts as an inverse. We can then
understand what significance these inverses have in terms of the rings.
Proposition.
(i) Let 0 6= a C O
L
be an ideal. If x ∈ L has xa ⊆ a, then x ∈ O
L
.
(ii) Let 0 6= a C O
L
be a proper ideal. Then
{y ∈ L : ya ≤ O
L
}
contains elements that are not in O
L
. In other words,
{y ∈ L : ya ≤ O
L
}
O
L
6= 0.
We will see that the object
{y ∈ L
:
ya ≤ O
L
}
is in some sense an inverse to
a.
Before we prove this, it is helpful to see what this means in a concrete setting.
Example.
Consider
O
L
=
Z
and
a
= 3
Z
. Then the first part says if
a
b
·
3
Z ⊆
3
Z
,
then
a
b
∈ Z. The second says
n
a
b
:
a
b
· 3 ∈ Z
o
contains something not in Z, say
1
3
. These are both “obviously true”.
Proof.
(i)
Let
a ⊆ O
L
. Then since
O
L
is Noetherian, we know
a
is finitely generated,
say by
α
1
, ··· , α
m
. We consider the multiplicationby
x
map
m
x
:
a → a
,
i.e. write
xα
i
=
X
a
ij
α
j
,
where A = (a
ij
) is a matrix in O
L
. So we know
(xI −A)
α
1
.
.
.
α
n
= 0.
By multiplying by the adjugate matrix, this implies
det
(
xI −A
) = 0. So
x
satisfies a monic polynomial with coefficients in
O
L
, i.e.
x
is integral over
O
L
. Since O
L
is integrally closed, x ∈ O
L
.
(ii)
It is clear that if the result is true for
a
, then it is true for all
a
0
⊆ a
. So
it is enough to prove this for
a
=
p
, a maximal, and in particular prime,
ideal.
Let
α ∈ p
be nonzero. By the previous lemma, there exists prime ideals
p
1
, ··· , p
r
such that
p
1
···p
r
⊆ hαi
. We also have that
hαi ⊆ p
by definition.
Assume
r
is minimal with this property. Since
p
is prime, there is some
i
such that
p
i
⊆ p
. wlog, we may as well assume
i
= 1, i.e.
p
1
⊆ p
. But
p
1
is
a prime ideal, and hence maximal. So p
1
= p.
Also, since r is minimal, we know p
2
···p
r
6⊆ hai.
Pick β ∈ p
2
···p
r
\ hai. Then
βp = βp
1
⊆ p
1
p
2
···p
r
⊆ hαi.
Dividing by
α
, we get
β
α
p ⊆ O
L
. But
β 6∈ hαi
. So we know
β
α
6∈ O
L
. So
done.
What is this
{x ∈ L
:
xa ≤ O
L
}
? This is not an ideal, but it almost is. The
only way in which it fails to be an ideal is that it is not contained inside
O
L
. By
this we mean it is closed under addition and multiplication by elements in
O
L
.
So it is an
O
L
module, which is finitely generated (we will see this in a second),
and a subset of L. We call this a “fractional ideal”.
Definition
(Fractional ideal)
.
A fractional ideal of
O
L
is a subset of
L
that is
also an O
L
module and is finitely generated.
Definition
(Integral/honest ideal)
.
If we want to emphasize that
a C O
L
is an
ideal, we say it is an integral or honest ideal. But we never use “ideal” to mean
fractional ideal.
Note that the definition of fractional ideal makes sense only because
O
L
is
Noetherian. Otherwise, the nonfinitelygenerated honest ideals would not qualify
as fractional ideals, which is bad. Rather, in the general case, the following
characterization is more helpful:
Lemma.
An
O
L
module
q ⊆ L
is a fractional ideal if and only if there is some
c ∈ L
×
such that
cq
is an ideal in
O
L
. Moreover, we can pick
c
such that
c ∈ Z
.
In other words, each fractional ideal is of the form
1
c
a
for some honest ideal
a and integer c.
Proof.
(⇐)
We have to prove that
q
is finitely generated. If
q ⊆ L
×
,
c ∈ L
nonzero,
then
cq
∼
=
q
as an
O
L
module. Since
O
L
is Noetherian, every ideal is
finitelygenerated. So cq, and hence q is finitely generated.
(⇒)
Suppose
x
1
, ··· , x
n
generate
q
as an
O
L
module. Write
x
i
=
y
i
n
i
, with
y
i
∈ O
L
and n
i
∈ Z, n
i
6= 0, which we have previously shown is possible.
We let
c
=
lcm
(
n
1
, ··· , n
k
). Then
cq ⊆ O
L
, and is an
O
L
submodule of
O
L
, i.e. an ideal.
Corollary.
Let
q
be a fractional ideal. Then as an abelian group,
q
∼
=
Z
n
, where
n = [L : Q].
Proof.
There is some
c ∈ L
×
such that
cqC O
L
as an ideal, and
cq
∼
=
q
as abelian
groups. So it suffices to show that any nonzero ideal
q ≤ O
L
is isomorphic to
Z
n
. Since
q ≤ O
L
∼
=
Z
n
as abelian groups, we know
q
∼
=
Z
m
for some
m
. But
also there is some
a
0
∈ Z ∩ q
, and
Z
n
∼
=
ha
0
i ≤ q
. So we must have
n
=
m
, and
q
∼
=
Z
n
.
Corollary.
Let
a ≤ O
L
be a proper ideal. Then
{x ∈ L
:
xa ≤ O
L
}
is a
fractional ideal.
Proof.
Pick
a ∈ a
. Then
a · {x ∈ L
:
xa ≤ O
L
} ⊆ O
L
and is an ideal in
O
L
.
Finally, we can state the proposition we want to prove, after all that nonsense
work.
Definition
(Invertible fractional ideal)
.
A fractional ideal
q
is invertible if there
exists a fractional ideal r such that qr = O
L
= h1i.
Notice we can multiply fractional ideals using the same definition as for
integral ideals.
Proposition. Every nonzero fractional ideal is invertible. The inverse of q is
{x ∈ L : xq ⊆ O
L
}.
This is good.
Note that if q =
1
n
a and r =
1
m
b, and a, b C O
L
are integral ideals, then
qr =
1
mn
ab = O
L
if and only if
ab
=
hmni
. So the proposition is equivalent to the statement that
for every a C O
L
, there exists an ideal b C O
L
such that ab is principal.
Proof.
Note that for any
n ∈ O
L
nonzero, we know
q
is invertible if and only if
nq
is invertible. So if the proposition is false, there is an integral ideal
a C O
L
which is not invertible. Moreover, as
O
L
is Noetherian, we can assume
a
is
maximal with this property, i.e. if a < a
0
< O
L
, then a
0
is invertible.
Let
b
=
{x ∈ L
:
xa ⊆ O
L
}
, a fractional ideal. We clearly have
O
L
⊆ b
, and
by our previous proposition, we know this inclusion is strict.
As
O
L
⊆ b
, we know
a ⊆ ab
. Again, this inclusion is strict — if
ab
=
a
, then
for all
x ∈ b
, we have
xa ⊆ a
, and we have shown that this implies
x ∈ O
L
, but
we cannot have b ⊆ O
L
.
So
a ( ab
. By assumption, we also have
ab ⊆ O
L
, and since
a
is not invertible,
this is strict. But then by definition of
a
, we know
ab
is invertible, which implies
a
is invertible (if
c
is an inverse of
ab
, then
bc
is an inverse of
a
). This is a
contradiction. So all fractional ideals must be invertible.
Finally, we have to show that the formula for the inverse holds. We write
c = {x ∈ L : xq ⊆ O
L
}.
Then by definition, we know q
−1
⊆ c. So
O
L
= qq
−1
⊆ qc ⊆ O
L
.
Hence we must have qc = O
L
, i.e. c = q
−1
.
We’re now done with the annoying commutative algebra, and can finally
prove something interesting.
Corollary. Let a, b, c C O
L
be ideals, c 6= 0. Then
(i) b ⊆ a if and only if bc ⊆ ac
(ii) a  b if and only if ac  bc
(iii) a  b if and only if b ⊆ a.
Proof.
(i) (⇒) is clear, and (⇐) is obtained by multiplying with c
−1
.
(ii) (⇒) is clear, and (⇐) is obtained by multiplying with c
−1
.
(iii)
(
⇒
) is clear. For the other direction, we notice that the result is easy if
a
=
hαi
is principal. Indeed, if
b
=
hβ
1
, ··· , β
r
i
, then
b ⊆ hαi
means there
are some β
0
a
, ··· , β
0
r
∈ O
L
such that β
i
= β
0
i
α. But this says
hβ
a
, ··· , β
r
i = hβ
0
1
, ··· , β
0
r
ihαi,
So hαi  b.
In general, suppose we have
b ⊆ a
. By the proposition, there exists an
ideal c C O
L
such that ac = hαi is principal with α ∈ O
L
, α 6= 0. Then
– b ⊆ a if and only if bc ⊆ hαi by (i); and
– a  b if and only if hαi  bc by (ii).
So the result follows.
Finally, we can prove the unique factorization of prime ideals:
Theorem.
Let
a C O
L
be an ideal,
a 6
= 0. Then
a
can be written uniquely as a
product of prime ideals.
Proof.
To show existence, if
a
is prime, then there is nothing to do. Otherwise,
if
a
is not prime, then it is not maximal. So there is some
b ) a
with
b C O
L
.
Hence
b  a
, i.e. there is some
c C O
L
with
a
=
bc
, and
c ⊇ a
. We can continue
factoring this way, and it must stop eventually, or else we have an infinite chain
of strictly ascending ideals.
We prove uniqueness the usual way. We have shown
p  ab
implies
p  a
or
p  b
. So if
p
1
···p
r
=
q
1
···q
s
, with
p
i
, q
j
prime, then we know
p
1
 q
1
···q
s
,
which implies
p
1
 q
i
for some
i
, and wlog
i
= 1. So
q
1
⊆ p
1
. But
q
1
is prime
and hence maximal. So p
1
= q
1
.
Multiply the equation
p
1
···p
r
=
q
1
···q
s
by
p
−1
1
, and we get
p
2
···p
r
=
q
2
···q
s
. Repeat, and we get
r
=
s
and
p
i
=
q
i
for all
i
(after renumbering).
Corollary.
The nonzero fractional ideals form a group under multiplication.
We denote this
I
L
. This is a free abelian group generated by the prime ideals,
i.e. any fractional ideal
q
can be written uniquely as
p
a
1
1
···p
q
r
r
, with
p
i
distinct
prime ideals and a
i
∈ Z.
Moreover, if q is an integral ideal, i.e. q C O
L
, then a
i
, ··· , a
r
≥ 0.
Proof.
We already have unique factorization of honest ideals. Now take any
fractional ideal, and write it as
q
=
ab
−1
, with
a, b ∈ O
L
(e.g. take
b
=
hni
for
some n), and the result follows.
Unimportant side note: we have shown that there are two ways we can
partially order the ideals of
O
L
— by inclusion and by division. We have shown
that these two orders are actually the same. Thus, it follows that the “least
common multiple” of two ideals
a, b
is their intersection
a ∩b
, and the “greatest
common divisor” of two ideals is their sum
a + b = {a + b : a ∈ a, b ∈ b}.
Example. Again let [L : Q] = 2, i.e. L = Q(
√
d) with d 6= 0, 1 and squarefree.
While we proved that every ideal can be factorized into prime ideals, we have
completely no idea what prime ideals look like. We just used their very abstract
properties like being prime and maximal. So we would like to play with some
actual ideals.
Recall we had the example
h3, 1 + 2
√
−5ih3, 1 − 2
√
−5i = h3i.
This is an example where we multiply two ideals together to get a principal ideal,
and the key to this working is that 1 + 2
√
−5
is conjugate to 1
−
2
√
−5
. We will
use this idea to prove the previous result for number fields of this form.
Let
a C O
L
be a nonzero ideal. We want to find some
b C O
L
such that
ab
is principal.
We know
O
L
∼
=
Z
2
, and
a ≤ O
L
as a subgroup. Moreover, we have shown
that we must have
a
∼
=
Z
2
as abelian groups. So it is generated by 2 elements as
subgroups of
Z
2
. Since
Z
is a subring of
O
L
, we know
a
must be generated by
at most 2 elements as
O
L
modules, i.e. as ideals of
O
L
. If it is generated by one
element, then it is already principal. Otherwise, suppose
a
=
hα, βi
for some
α, β ∈ O
L
.
Further, we claim that we can pick
α, β
such that
β ∈ Z
. We write
α
=
a
+
b
√
d
and
β
=
a
0
+
b
0
√
d
. Then let
`
=
gcd
(
b, b
0
) =
mb
+
m
0
b
0
, with
m, m
0
∈ Z
(by
Euclid’s algorithm). We set
β
0
= (mα + m
0
β) ·
−b
0
`
+ β
= (ma + m
0
a
0
+ `
√
d)
−b
0
`
+ a
0
+ b
0
√
d
= (ma + m
0
a
0
)
−b
0
`
+ a
0
∈ Z
using the fact that that −
b
0
`
∈ Z. Then hα, β
0
i = hα, βi.
So suppose a = hb, αi, with b ∈ Z and α ∈ O
L
. We now claim
hb, αihb, ¯αi
is principal (where
α
=
x
+
y
√
d
,
¯α
=
x − y
√
d
). In particular, if
a C O
L
, then
a
¯
a is principal, so the proposition is proved by hand.
To show this, we can manually check
hb, αihb, ¯αi = hb
2
, bα, b¯α, α¯αi
= hb
2
, bα, b tr(α), N (α)i,
using the fact that
tr
(
α
) =
α
+
¯α
and
N
(
α
) =
α¯α
. Now note that
b
2
, b tr
(
α
) and
N
(
α
) are all integers. So we can take the gcd
c
=
gcd
(
b
2
, b tr
(
α
)
, N
(
α
)). Then
this ideal is equal to
= hc, bαi.
Finally, we claim that bα ∈ hci.
Write
bα
=
cx
, with
x ∈ L
. Then
tr x
=
b
c
tr α ∈ Z
since
b
c
∈ Z
by definition,
and
N(x) = N
bα
c
=
b
2
N(α)
c
2
=
b
2
c
N(α)
c
∈ Z.
So x ∈ O
L
. So c  bα in O
L
. So hc, bαi = hci.
Finally, after all these results, we can get to the important definition of the
course.
Definition
(Class group)
.
. The class group or ideal class group of a number
field L is
cl
L
= I
L
/P
L
,
where
I
L
is the group of fractional ideals, and
P
L
is the subgroup of principal
fractional ideals.
If
a ∈ I
L
, we write [
a
] for its equivalence class in
cl
L
. So [
a
] = [
b
] if and only
if there is some γ ∈ L
×
such that γa = b.
The significance is that cl
L
measures the failure of unique factorization:
Theorem. The following are equivalent:
(i) O
L
is a principal ideal domain
(ii) O
L
is a unique factorization domain
(iii) cl
L
is trivial.
Proof.
(i) and (iii) are equivalent by definition, while (i) implies (ii) is wellknown
from IB Groups, Rings and Modules. So the real content is (ii) to (i), which is
specific to Dedekind domains.
If
p C O
L
is prime, and
x ∈ p \ {
0
}
, we factor
x
=
α
1
···α
k
such that
α
i
is
irreducible in
O
L
. As
p
is prime, there is some
α
i
∈ p
. But then
hα
i
i ⊆ p
, and
hα
i
i
is prime as
O
L
is a UFD. So we must have
hα
i
i
=
p
as prime ideals are
maximal. So p is principal.
In the next few chapters, we will come up with methods to explicitly compute
the class group of any number field.