Part II — Number Fields
Based on lectures by I. Grojnowski
Notes taken by Dexter Chua
Lent 2016
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Part IB Groups, Rings and Modules is essential and Part II Galois Theory is desirable
Definition of algebraic number fields, their integers and units. Norms, bases and
discriminants. [3]
Ideals, principal and prime ideals, unique factorisation. Norms of ideals. [3]
Minkowski’s theorem on convex bodies. Statement of Dirichlet’s unit theorem. Deter
mination of units in quadratic fields. [2]
Ideal classes, finiteness of the class group. Calculation of class numbers using statement
of the Minkowski bound. [3]
Dedekind’s theorem on the factorisation of primes. Application to quadratic fields. [2]
Discussion of the cyclotomic field and the Fermat equation or some other topic chosen
by the lecturer. [3]
0 Introduction
Technically, IID Galois Theory is not a prerequisite of this course. However,
many results we have are analogous to what we did in Galois Theory, and we
will not refrain from pointing out the correspondence. If you have not learnt
Galois Theory, then you can ignore them.
1 Number fields
The focus of this course is, unsurprisingly, number fields. Before we define what
number fields are, we look at some motivating examples. Suppose we wanted to
find all numbers of the form
x
2
+
y
2
, where
x, y ∈ Z
. For example, if
a, b
can
both be written in this form, does it follow that ab can?
In IB Groups, Rings and Modules, we did the clever thing of working with
Z
[
i
]. The integers of the form
x
2
+
y
2
are exactly the norms of integers in
Z
[
i
],
where the norm of x + iy is
N(x + iy) = x + iy
2
= x
2
+ y
2
.
Then the previous result is obvious — if
a
=
N
(
z
) and
b
=
N
(
w
), then
ab
=
N(zw). So ab is of the form x
2
+ y
2
.
Similarly, in the IB Groups, Rings and Modules example sheet, we found
all solutions to the equation
x
2
+ 2 =
y
3
by working in
Z
[
√
−2
]. This is a very
general technique — working with these rings, and corresponding fields
Q
(
√
−d
)
can tell us a lot about arithmetic we care about.
In this chapter, we will begin by writing down some basic definitions and
proving elementary properties about number fields.
Definition
(Field extension)
.
A field extension is an inclusion of fields
K ⊆ L
.
We sometimes write this as L/K.
Definition
(Degree of field extension)
.
Let
K ⊆ L
be fields. Then
L
is a vector
space over K, and the degree of the field extension is
[L : K] = dim
K
(L).
Definition
(Finite extension)
.
A finite field extension is a field extension with
finite degree.
Definition (Number field). A number field is a finite field extension over Q.
A field is the most boring kind of ring — the only ideals are the trivial one
and the whole field itself. Thus, if we want to do something interesting with
number fields algebraically, we need to come up with something more interesting.
In the case of
Q
itself, one interesting thing to talk about is the integers
Z
.
It turns out the right generalization to number fields is algebraic integers.
Definition
(Algebraic integer)
.
Let
L
be a number field. An algebraic integer
is an
α ∈ L
such that there is some monic
f ∈ Z
[
x
] with
f
(
α
) = 0. We write
O
L
for the set of algebraic integers in L.
Example.
It is a fact that if
L
=
Q
(
i
), then
O
L
=
Z
[
i
]. We will prove this in
the next chapter after we have the necessary tools.
These are in fact the main objects of study in this course. Since we say this
is a generalization of Z ⊆ Q, the following had better be true:
Lemma. O
Q
= Z, i.e. α ∈ Q is an algebraic integer if and only if α ∈ Z.
Proof. If α ∈ Z, then x − α ∈ Z[x] is a monic polynomial. So α ∈ O
Q
.
On the other hand, let
α ∈ Q
. Then there is some coprime
r, s ∈ Z
such that
α =
r
s
. If it is an algebraic integer, then there is some
f(x) = x
n
+ a
n−1
x
n−1
+ ··· + a
0
with a
i
∈ Z such that f(α) = 0. Substituting in and multiplying by s
n
, we get
r
n
+ a
n−1
r
n−1
s + ···+ a
0
s
n
 {z }
divisible by s
= 0,
So
s  r
n
. But if
s 6
= 1, there is a prime
p
such that
p  s
, and hence
p  r
n
. Thus
p  r
. So
p
is a common factor of
s
and
r
. This is a contradiction. So
s
= 1, and
α is an integer.
How else is this a generalization of
Z
? We know
Z
is a ring. So perhaps
O
L
also is.
Theorem. O
L
is a ring, i.e. if α, β ∈ O
L
, then so is α ± β and αβ.
Note that in general O
L
is not a field. For example, Z = O
Q
is not a field.
The proof of this theorem is not as straightforward as the previous one.
Recall we have proved a similar theorem in IID Galois Theory before with
“algebraic integer” replaced with “algebraic number”, namely that if
L/K
is a
field extension with
α, β ∈ L
algebraic over
K
, then so is
αβ
and
α ± β
, as well
as
1
α
if α 6= 0.
To prove this, we notice that
α ∈ K
is algebraic if and only if
K
[
α
] is a finite
extension — if α is algebraic, with
f(α) = a
n
α
n
+ ··· + a
0
= 0, a
n
6= 0
then
K
[
α
] has degree at most
n
, since
α
n
(and similarly
α
−1
) can be written as
a linear combination of 1
, α, ··· , α
n−1
, and thus these generate
K
[
α
]. On the
other hand, if
K
[
α
] is finite, say of degree
k
, then 1
, α, ··· , α
k
are independent,
hence some linear combination of them vanishes, and this gives a polynomial
for which
α
is a root. Moreover, by the same proof, if
K
0
is any finite extension
over K, then any element in K
0
is algebraic.
Thus, to prove the result, notice that if
K
[
α
] is generated by 1
, α, ··· , α
n
and
K
[
β
] is generated by 1
, β, ··· , β
m
, then
K
[
α, β
] is generated by
{α
i
β
j
}
for
1
≤ i ≤ n,
1
≤ j ≤ m
. Hence
K
[
α, β
] is a finite extension, hence
αβ, α ± β ∈
K[α, β] are algebraic.
We would like to prove this theorem in an analogous way. We will consider
O
L
as a ring extension of
Z
. We will formulate the general notion of “being an
algebraic integer” in general ring extensions:
Definition
(Integrality)
.
Let
R ⊆ S
be rings. We say
α ∈ S
is integral over
R
if there is some monic polynomial f ∈ R[x] such that f(α) = 0.
We say S is integral over R if all α ∈ S are integral over R.
Definition
(Finitelygenerated)
.
We say
S
is finitelygenerated over
R
if there
exists elements
α
1
, ··· , α
n
∈ S
such that the function
R
n
→ S
defined by
(
r
1
, ··· , r
n
)
7→
P
r
i
α
i
is surjective, i.e. every element of
S
can be written as a
R

linear combination of elements
α
1
, ··· , α
n
. In other words,
S
is finitelygenerated
as an Rmodule.
This is a refinement of the idea of being algebraic. We allow the use of rings
and restrict to monic polynomials. In Galois theory, we showed that finiteness
and algebraicity “are the same thing”. We will generalize this to integrality of
rings.
Example.
In a number field
Z ⊆ Q ⊆ L
,
α ∈ L
is an algebraic integer if and
only if α is integral over Z by definition, and O
L
is integral over Z.
Notation.
If
α
1
, ··· , α
r
∈ S
, we write
R
[
α
1
, ··· , α
r
] for the subring of
S
generated by
R, α
1
, ··· , α
r
. In other words, it is the image of the homomorphism
from the polynomial ring R[x
1
, ··· , x
n
] → S given by x
i
7→ α
i
.
Proposition.
(i)
Let
R ⊆ S
be rings. If
S
=
R
[
s
] and
s
is integral over
R
, then
S
is
finitelygenerated over R.
(ii)
If
S
=
R
[
s
1
, ··· , s
n
] with
s
i
integral over
R
, then
S
is finitelygenerated
over R.
This is the easy direction in identifying integrality with finitelygenerated.
Proof.
(i)
We know
S
is spanned by 1
, s, s
2
, ···
over
R
. However, since
s
is integral,
there exists a
0
, ··· , a
n
∈ R such that
s
n
= a
0
+ a
1
s + ··· + a
n−1
s
n−1
.
So the
R
submodule generated by 1
, s, ··· , s
n−1
is stable under multiplica
tion by s. So it contains s
n
, s
n+1
, s
n+2
, ···. So it is S.
(ii)
Let
S
i
=
R
[
s
1
, ··· , s
i
]. So
S
i
=
S
i−1
[
s
i
]. Since
s
i
is integral over
R
, it is
integral over
S
i−1
. By the previous part,
S
i
is finitelygenerated over
S
i−1
.
To finish, it suffices to show that being finitelygenerated is transitive.
More precisely, if
A ⊆ B ⊆ C
are rings,
B
is finitely generated over
A
and
C
is finitely generated over
B
, then
C
is finitely generated over
A
. This
is not hard to see, since if
x
1
, ··· , x
n
generate
B
over
A
, and
y
1
, ··· , y
m
generate
C
over
B
, then
C
is generated by
{x
i
y
j
}
1≤i≤n,1≤j≤m
over
A
.
The other direction is harder.
Theorem. If S is finitelygenerated over R, then S is integral over R.
The idea of the proof is as follows: if
s ∈ S
, we need to find a monic
polynomial which it satisfies. In Galois theory, we have fields and vector spaces,
and the proof is easy. We can just consider 1
, s, s
2
, ···
, and linear dependence
kicks in and gives us a relation. But even if this worked in our case, there is no
way we can make this polynomial monic.
Instead, consider the multiplicationby
s
map:
m
s
:
S → S
by
γ 7→ sγ
. If
S
were a finitedimensional vector space over
R
, then CayleyHamilton tells us
m
s
,
and thus
s
, satisfies its characteristic polynomial, which is monic. Even though
S
is not a finitedimensional vector space, the proof of CayleyHamilton will
work.
Proof.
Let
α
1
, ··· , α
n
generate
S
as an
R
module. wlog take
α
1
= 1
∈ S
. For
any s ∈ S, write
sα
i
=
X
b
ij
α
j
for some
b
ij
∈ R
. We write
B
= (
b
ij
). This is the “matrix of multiplication by
S”. By construction, we have
(sI − B)
α
1
.
.
.
a
n
= 0. (∗)
Now recall for any matrix
X
, we have
adj
(
X
)
X
= (
det X
)
I
, where the
i, j
th
entry of
adj
(
X
) is given by the determinant of the matrix obtained by removing
the ith row and jth column of X.
We now multiply (∗) by adj(sI − B). So we get
det(sI − B)
α
1
.
.
.
α
n
= 0
In particular,
det
(
sI −B
)
α
1
= 0. Since we picked
α
1
= 1, we get
det
(
sI −B
) = 0.
Hence if f (x) = det(xI − B), then f(x) ∈ R[x], and f(s) = 0.
Hence we obtain the following:
Corollary. Let L ⊇ Q be a number field. Then O
L
is a ring.
Proof.
If
α, β ∈ O
L
, then
Z
[
α, β
] is finitelygenerated by the proposition. But
then
Z
[
α, β
] is integral over
Z
, by the previous theorem. So
α ± β, αβ ∈
Z[α, β].
Note that it is not necessarily true that if
S ⊇ R
is an integral extension,
then
S
is finitelygenerated over
R
. For example, if
S
is the set of all algebraic
integers in
C
, and
R
=
Z
, then by definition
S
is an integral extension of
Z
, but
S is not finitely generated over Z.
Thus the following corollary isn’t as trivial as the case with “integral” replaced
by “finitely generated”:
Corollary.
If
A ⊆ B ⊆ C
be ring extensions such that
B
over
A
and
C
over
B
are integral extensions. Then C is integral over A.
The idea of the proof is that while the extensions might not be finitely gener
ated, only finitely many things are needed to produce the relevant polynomials
witnessing integrality.
Proof. If c ∈ C, let
f(x) =
N
X
i=0
b
i
x
i
∈ B[x]
be a monic polynomial such that
f
(
c
) = 0. Let
B
0
=
A
[
b
0
, ··· , b
N
] and let
C
0
=
B
0
[
c
]. Then
B
0
/A
is finitely generated as
b
0
, ··· , b
N
are integral over
A
.
Also,
C
0
is finitelygenerated over
B
0
, since
c
is integral over
B
0
. Hence
C
0
is
finitelygenerated over
A
. So
c
is integral over
A
. Since
c
was arbitrary, we know
C is integral over A.
Now how do we recognize algebraic integers? If we want to show something
is an algebraic integer, we just have to exhibit a monic polynomial that vanishes
on the number. However, if we want to show that something is not an algebraic
integer, we have to make sure no monic polynomial kills the number. How can
we do so?
It turns out to check if something is an algebraic integer, we don’t have to
check all monic polynomials. We just have to check one. Recall that if
K ⊆ L
is a field extensions with
α ∈ L
, then the minimal polynomial is the monic
polynomial p
α
(x) ∈ K[x] of minimal degree such that p
α
(α) = 0.
Note that we can always make the polynomial monic. It’s just that the
coefficients need not lie in Z.
Recall that we had the following lemma about minimal polynomials:
Lemma. If f ∈ K[x] with f (α) = 0, then p
α
 f .
Proof. Write f = p
α
h + r, with r ∈ K[x] and deg(r) < deg(p
α
). Then we have
0 = f(α) = p(α)h(α) + r(α) = r(α).
So if r 6= 0, this contradicts the minimality of deg p
α
.
In particular, this lemma implies
p
α
is unique. One nice application of this
result is the following:
Proposition.
Let
L
be a number field. Then
α ∈ O
L
if and only if the minimal
polynomial p
α
(x) ∈ Q[x] for the field extension Q ⊆ L is in fact in Z[x].
This is a nice proposition. This gives us an necessary and sufficient condition
for whether something is algebraic.
Proof. (⇐) is trivial, since this is just the definition of an algebraic integer.
(
⇒
) Let
α ∈ O
L
and
p
α
∈ Q
[
x
] be the minimal polynomial of
α
, and
h
(
x
)
∈ Z
[
x
] be a monic polynomial which
α
satisfies. The idea is to use
h
to
show that the coefficients of p
α
are algebraic, thus in fact integers.
Now there exists a bigger field M ⊇ L such that
p
α
(x) = (x − α
1
) ···(x − α
r
)
factors in
M
[
x
]. But by our lemma,
p
α
 h
. So
h
(
α
i
) = 0 for all
α
i
. So
α
i
∈ O
M
is an algebraic integer. But
O
M
is a ring, i.e. sums and products of the
α
i
’s are
still algebraic integers. So the coefficients of
p
α
are algebraic integers (in
O
M
).
But they are also in Q. Thus the coefficients must be integers.
Alternatively, we can deduce this proposition from the previous lemma plus
Gauss’ lemma.
Another relation between
Z
and
Q
is that
Q
is the fraction field of
Z
. This
is true for general number fields
Lemma. We have
Frac O
L
=
α
β
: α, β ∈ O
L
, β 6= 0
= L.
In fact, for any α ∈ L, there is some n ∈ Z such that nα ∈ O
L
.
Proof.
If
α ∈ L
, let
g
(
x
)
∈ Q
[
x
] be its monic minimal polynomial. Then there
exists
n ∈ Z
nonzero such that
ng
(
x
)
∈ Z
[
x
] (pick
n
to be the least common
multiple of the denominators of the coefficients of
g
(
x
)). Now the magic is to
put
h(x) = n
deg(g)
g
x
n
.
Then this is a monic polynomial with integral coefficients — in effect, we have
just multiplied the coefficient of
x
i
by
n
deg(g)−i
! Then
h
(
nα
) = 0. So
nα
is
integral.
2 Norm, trace, discriminant, numbers
Recall that in our motivating example of
Z
[
i
], one important tool was the norm
of an algebraic integer
x
+
iy
, given by
N
(
x
+
iy
) =
x
2
+
y
2
. This can be
generalized to arbitrary number fields, and will prove itself to be a very useful
notion to consider. Apart from the norm, we will also consider a number known
as the trace, which is also useful. We will also study numbers associated with
the number field itself, rather than particular elements of the field, and it turns
out they tell us a lot about how the field behaves.
Norm and trace
Recall the following definition from IID Galois Theory:
Definition
(Norm and trace)
.
Let
L/K
be a field extension, and
α ∈ L
. We
write
m
α
:
L → L
for the map
7→ α
. Viewing this as a linear map of
L
vector
spaces, we define the norm of α to be
N
L/K
(α) = det m
α
,
and the trace to be
tr
L/K
(α) = tr m
α
.
The following property is immediate:
Proposition.
For a field extension
L/K
and
a, b ∈ L
, we have
N
(
ab
) =
N(a)N(b) and tr(a + b) = tr(a) + tr(b).
We can alternatively define the norm and trace as follows:
Proposition.
Let
p
α
∈ K
[
x
] be the minimal polynomial of
α
. Then the
characteristic polynomial of m
α
is
det(xI − m
α
) = p
[L:K(α)]
α
Hence if p
α
(x) splits in some field L
0
⊇ K(α), say
p
α
(x) = (x − α
1
) ···(x − α
r
),
then
N
K(α)/K
(α) =
Y
α
i
, tr
K(α)/K
(α) =
X
α
i
,
and hence
N
L/K
(α) =
Y
α
i
[L:K(α)]
, tr
L/K
= [L : K(α)]
X
α
i
.
This was proved in the IID Galois Theory course, and we will just use it
without proving.
Corollary. Let L ⊇ Q be a number field. Then the following are equivalent:
(i) α ∈ O
L
.
(ii) The minimal polynomial p
α
is in Z[x]
(iii) The characteristic polynomial of m
α
is in Z[x].
This in particular implies N
L/Q
(α) ∈ Z and tr
L/Q
(α) ∈ Z.
Proof.
The equivalence between the first two was already proven. For the
equivalence between (ii) and (iii), if
m
α
∈ Z
[
x
], then
α ∈ O
L
since it vanishes
on a monic polynomial in
Z
[
x
]. On the other hand, if
p
α
∈ Z
[
x
], then so is the
characteristic polynomial, since it is just p
N
α
.
The final implication comes from the fact that the norm and trace are just
coefficients of the characteristic polynomial.
It would be nice if the last implication is an if and only if. This is in general
not true, but it occurs, obviously, when the characteristic polynomial is quadratic,
since the norm and trace would be the only coefficients.
Example.
Let
L
=
K
(
√
d
) =
K
[
z
]
/
(
z
2
−d
), where
d
is not a square in
K
. As a
vector space over
K
, we can take 1
,
√
d
as our basis. So every
α
can be written
as
α = x + y
√
d.
Hence the matrix of multiplication by α is
m
α
=
x dy
y x
.
So the trace and norm are given by
tr
L/K
(x + y
√
d) = 2x = (x + y
√
d) + (x − y
√
d)
N
L/K
(x + y
√
d) = x
2
− dy
2
= (x + y
√
d)(x − y
√
d)
We can also obtain this by consider the roots of the minimal polynomial of
α = x + y
√
d, namely (α − x)
2
− y
2
d = 0, which has roots x ± y
√
d.
In particular, if
L
=
Q
(
√
d
), with
d <
0, then the norm of an element is just
the norm of it as a complex number.
Now that we have computed the general trace and norm, we can use the
proposition to find out what the algebraic integers are. It turns out the result is
(slightly) unexpected:
Lemma. Let L = Q(
√
d), where d ∈ Z is not 0, 1 and is squarefree. Then
O
L
=
(
Z[
√
d] d ≡ 2 or 3 (mod 4)
Z
h
1
2
(1 +
√
d)
i
d ≡ 1 (mod 4)
Proof.
We know
x
+
y
√
λ ∈ O
L
if and only if 2
x, x
2
− dy
2
∈ Z
by the previous
example. These imply 4
dy
2
∈ Z
. So if
y
=
r
s
with
r, s
coprime,
r, s ∈ Z
, then we
must have s
2
 4d. But d is squarefree. So s = 1 or 2. So
x =
u
2
, y =
v
2
for some
u, v ∈ Z
. Then we know
u
2
−dv
2
∈
4
Z
, i.e.
u
2
≡ dv
2
(
mod
4). But we
know the squares mod 4 are always 0 and 1. So if
d 6≡
1 (
mod
4), then
u
2
≡ dv
2
(
mod
4) imply that
u
2
=
v
2
= 0 (
mod
4), and hence
u, v
are even. So
x, y ∈ Z
,
giving O
L
= Z[
√
d].
On the other hand, if
d ≡
1 (
mod
4), then
u, v
have the same parity mod 2,
i.e. we can write x + y
√
d as a Zcombination of 1 and
1
2
(1 +
√
d).
As a sanity check, we find that the minimal polynomial of
1
2
(1 +
√
d
) is
x
2
− x +
1
4
(1 − d) which is in Z if and only if d ≡ 1 (mod 4).
Field embeddings
Recall the following theorem from IID Galois Theory:
Theorem
(Primitive element theorem)
.
Let
K ⊆ L
be a separable field extension.
Then there exists an α ∈ L such that K(α) = L.
For example, Q(
√
2,
√
3) = Q(
√
2 +
√
3).
Since
Q
has characteristic zero, it follows that all number fields are separable
extensions. So any number field
L/Q
is of the form
L
=
Q
(
α
). This makes it
much easier to study number fields, as the only extra “stuff” we have on top of
Q.
One particular thing we can do is to look at the number of ways we can
embed
L → C
. For example, for
Q
(
√
−1
), there are two such embeddings — one
sends
√
−1 to i and the other sends
√
−1 to −i.
Lemma.
The degree [
L
:
Q
] =
n
of a number field is the number of field
embeddings L → C.
Proof.
Let
α
be a primitive element, and
p
α
(
x
)
∈ Q
[
x
] its minimal polynomial.
Then by we have
deg p
α
= [
L
:
Q
] =
n
, as 1
, α, α
2
, ··· , α
n−1
is a basis. Moreover,
Q[x]
(p
α
)
∼
=
Q(α) = L.
Since L/Q is separable, we know p
α
has n distinct roots in C. Write
p
α
(x) = (x − α
1
) ···(x − α
n
).
Now an embedding
Q
[
x
]
/
(
p
α
)
→ C
is uniquely determined by the image of
x
,
and
x
must be sent to one of the roots of
p
α
. So for each
i
, the map
x 7→ α
i
gives us a field embedding, and these are all. So there are n of them.
Using these field embeddings, we can come up with the following alternative
formula for the norm and trace.
Corollary.
Let
L/Q
be a number field. If
σ
1
, ··· , σ
n
:
L → C
are the different
field embeddings and β ∈ L, then
tr
L/Q
(β) =
X
σ
i
(β), N
L/Q
(β) =
Y
i
σ
i
(β).
We call σ
1
(β), ··· , σ
n
(β) the conjugates of β in C.
Proof is in the Galois theory course.
Using this characterization, we have the following very concrete test for when
something is a unit.
Lemma. Let x ∈ O
L
. Then x is a unit if and only if N
L/Q
(x) = ±1.
Notation. Write O
×
L
= {x ∈ O
L
: x
−1
∈ O
L
}, the units in O
L
.
Proof.
(
⇒
) We know
N
(
ab
) =
N
(
a
)
N
(
b
). So if
x ∈ O
×
L
, then there is some
y ∈ O
L
such that xy = 1. So N(x)N(y) = 1. So N(x) is a unit in Z, i.e. ±1.
(
⇐
) Let
σ
1
, ··· , σ
n
:
L → C
be the
n
embeddings of
L
in
C
. For notational
convenience, We suppose that
L
is already subfield of
C
, and
σ
1
is the inclusion
map. Then for each x ∈ O
L
, we have
N(x) = xσ
2
(x) ···σ
n
(x).
Now if
N
(
x
) =
±
1, then
x
−1
=
±σ
2
(
x
)
···σ
n
(
x
). So we have
x
−1
∈ O
L
, since
this is a product of algebraic integers. So x is a unit in O
L
.
Corollary. If x ∈ O
L
is such that N(x) is prime, then x is irreducible.
Proof.
If
x
=
ab
, then
N
(
a
)
N
(
b
) =
N
(
x
). Since
N
(
x
) is prime, either
N
(
a
) =
±
1
or N (b) = ±1. So a or b is a unit.
We can consider a more refined notion than just the number of field embed
dings.
Definition
(
r
and
s
)
.
We write
r
for the number of field embeddings
L → R
,
and s the number of pairs of nonreal field embeddings L → C. Then
n = r + 2s.
Alternatively,
r
is the number of real roots of
p
α
, and
s
is the number of pairs of
complex conjugate roots.
The distinction between real embeddings and complex embeddings will be
important in the second half of the course.
Discriminant
The final invariant we will look at in this chapter is the discriminant. It is based
on the following observation:
Proposition.
Let
L/K
be a separable extension. Then a
K
bilinear form
L × L → K
defined by (
x, y
)
7→ tr
L/K
(
xy
) is nondegenerate. Equivalent, if
α
1
, ··· , α
n
are a
K
basis for
L
, the Gram matrix (
tr
(
α
i
α
j
))
i,j=1,···,n
has nonzero
determinant.
Recall from Galois theory that if
L/K
is not separable, then
tr
L/K
= 0, and
it is very very degenerate. Also, note that if
K
is of characteristic 0, then there is
a quick and dirty proof of this fact — the trace map is nondegenerate, because
for any
x ∈ K
, we have
tr
L/K
(
x · x
−1
) =
n 6
= 0. This is really the only case
we care about, but in the proof of the general result, we will also find a useful
formula for the discriminant when the basis is 1, θ, θ
2
, . . . , θ
n−1
.
We will use the following important notation:
Notation.
∆(α
1
, ··· , α
n
) = det(tr
L/K
(α
i
α
j
)).
Proof.
Let
σ
1
, ··· , σ
n
:
L →
¯
K
be the
n
distinct
K
linear field embeddings
L →
¯
K. Put
S = (σ
i
(α
j
))
i,j=1,···,n
=
σ
1
(α
1
) ··· σ
1
(α
n
)
.
.
.
.
.
.
.
.
.
σ
n
(α
1
) ··· σ
n
(α
n
).
Then
S
T
S =
n
X
k=1
σ
k
(α
i
)σ
k
(α
j
)
!
i,j=1,···n
.
We know σ
k
is a field homomorphism. So
n
X
k=1
σ
k
(α
i
)σ
k
(α
j
) =
n
X
k=1
σ
k
(α
i
α
j
) = tr
L/K
(α
i
α
j
).
So
S
T
S = (tr(α
i
α
j
))
i,j=1,···,n
.
So we have
∆(α
1
, ··· , α
n
) = det(S
T
S) = det(S)
2
.
Now we use the theorem of primitive elements to write
L
=
K
(
θ
) such that
1, θ, ··· , θ
n−1
is a basis for L over K, with [L : K] = n. Now S is just
S =
1 σ
1
(θ) ··· σ
1
(θ)
n−1
.
.
.
.
.
.
.
.
.
.
.
.
1 σ
n
(θ) ··· σ
n
(θ)
n−1
.
This is a Vandermonde matrix, and so
∆(1, θ, ··· , θ
n−1
) = (det S)
2
=
Y
i<j
(σ
i
(θ) − σ
j
(θ))
2
.
Since the field extension is separable, and hence
σ
i
6
=
σ
j
for all
i, j
, this implies
σ
i
(
θ
)
6
=
σ
j
(
θ
), since
θ
generates the field. So the product above is nonzero.
So we have this nice canonical bilinear map. However, this determinant is
not canonical. Recall that if
α
1
, ··· , α
n
is a basis for
L/K
, and
α
0
1
, ··· , α
0
n
is
another basis, then
α
0
i
=
X
a
ij
α
j
for some A = (a
ij
) ∈ GL
n
(K). So
∆(α
0
1
, ··· , α
0
n
) = (det A)
2
∆(α
1
, ··· , α
n
).
However, for number fields, we shall see that we can pick a “canonical” basis,
and get a canonical value for ∆. We will call this the discriminant.
Definition
(Integral basis)
.
Let
L/Q
be a number field. Then a basis
α
1
, ··· , α
n
of L is an integral basis if
O
L
=
(
n
X
i=1
m
i
α
i
: m
i
∈ Z
)
=
n
M
1
Zα
i
.
In other words, it is simultaneously a basis for L over Q and O
L
over Z.
Note that integral bases are not unique, just as with usual bases. Given one
basis, you can get any other by acting by GL
n
(Z).
Example.
Consider
Q
(
√
d
) with
d
squarefree,
d 6
= 0
,
1. If
d
∼
=
1 (
mod
4),
we’ve seen that 1
,
1
2
(1 +
√
λ
) is an integral basis. Otherwise, if
d
∼
=
2
,
3 (
mod
4),
then 1,
√
d is an integral basis.
The important theorem is that an integral basis always exists.
Theorem.
Let
Q/L
be a number field. Then there exists an integral basis for
O
L
. In particular, O
L
∼
=
Z
n
with n = [L : Q].
Proof.
Let
α
1
, ··· , α
n
be any basis of
L
over
Q
. We have proved that there is
some
n
i
∈ Z
such that
n
i
α
i
∈ O
L
. So wlog
α
1
, ··· , α
n
∈ O
L
, and are an basis of
L
over
Q
. Since
α
i
are integral, so are
α
i
α
j
, and so all these have integer trace,
as we have previously shown. Hence ∆(
α
1
, ··· , α
n
), being the determinant of a
matrix with integer entries, is an integer.
Now choose a
Q
basis
α
1
, ··· , α
n
∈ O
L
such that ∆(
α
1
, ··· , α
n
)
∈ Z \ {
0
}
has minimal absolute value. We will show that these are an integral basis.
Let x ∈ O
L
, and write
x =
X
λ
i
α
i
for some λ
i
∈ Q. These λ
i
are necessarily unique since α
1
, ··· , α
n
is a basis.
Suppose some λ
i
6∈ Z. wlog say λ
1
6∈ Z. We write
λ
1
= n
1
+ ε
1
,
for n
1
∈ Z and 0 < ε
1
< 1. We put
α
0
1
= x − n
1
α
1
= ε
1
α
1
+ λ
2
α
2
+ ··· + λ
n
α
n
∈ O
L
.
So α
0
1
, α
2
, ··· , α
n
is still a basis for L/Q, and are still in O
L
. But then
∆(α
0
1
, ··· , α
n
) = ε
2
1
· ∆(α
1
, ··· , α
n
) < ∆(α
1
, ··· , α
n
).
This contradicts minimality. So we must have
λ
i
∈ Z
for all
Z
. So this is a basis
for O
L
.
Now if
α
0
1
, ··· , α
0
n
is another integral basis of
L
over
Q
, then there is some
g ∈ GL
n
(
Z
) such that
gα
i
=
α
0
i
. Since
det
(
g
) is invertible in
Z
, it must be 1 or
−1, and hence
det ∆(α
0
1
, ··· , α
0
n
) = det(g)
2
∆(α
1
, ··· , α
n
) = ∆(α
1
, ··· , α
n
)
and is independent of the choice of integral basis.
Definition
(Discriminant)
.
The discriminant
D
L
of a number field
L
is defined
as
D
L
= ∆(α
1
, ··· , α
n
)
for any integral basis α
1
, ··· , α
n
.
Example.
Let
L
=
Q
(
√
d
), where
d 6
= 0
,
1 and
d
is squarefree. If
d
∼
=
2
,
3
(mod 4), then it has an integral basis 1,
√
d. So
D
L
= det
1
√
d
1 −
√
d
2
= 4d.
Otherwise, if d
∼
=
1 (mod 4), then
D
L
= det
1
1
2
(1 +
√
d)
1
1
2
(1 −
√
d)
2
= d.
Recall that we have seen the word discriminant before, and let’s make sure
these concepts are moreorless consistent. Recall that the discriminant of a
polynomial f(x) =
Q
(x − α
i
) is defined as
disc(f) =
Y
i<j
(α
i
− α
j
)
2
= (−1)
n(n−1)/2
Y
i6=j
(α
i
− α
j
).
If
p
θ
(
x
)
∈ K
[
x
] is the minimal polynomial of
θ
(where
L
=
K
[
θ
]), then the roots
of p
θ
are σ
i
(θ). Hence we get
disc(p
θ
) =
Y
i<j
(σ
i
(θ) − σ
j
(θ))
2
.
In other words,
disc(p
θ
) = ∆(1, θ, ··· , θ
n−1
).
So this makes sense.
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