1Number fields
II Number Fields
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.