II Number Fields - Number fields

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

, 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

[

]. The integers of the form

are exactly the norms of integers in

[

where the norm of x + iy is

N(x + iy) = |x + iy|

= x

+ y

Then the previous result is obvious — if

(

) and

(

), then

N(zw). So ab is of the form x

+ y

Similarly, in the IB Groups, Rings and Modules example sheet, we found

all solutions to the equation

+ 2 =

by working in

[

√

−2

]. This is a very

general technique — working with these rings, and corresponding fields

(

√

−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

is a vector

space over K, and the degree of the field extension is

[L : K] = dim

(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

itself, one interesting thing to talk about is the integers

It turns out the right generalization to number fields is algebraic integers.

Definition

(Algebraic integer)

Let

be a number field. An algebraic integer

is an

α ∈ L

such that there is some monic

f ∈ Z

[

] with

(

) = 0. We write

for the set of algebraic integers in L.

Example.

It is a fact that if

(

), then

[

]. 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

= 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

On the other hand, let

α ∈ Q

. Then there is some coprime

r, s ∈ Z

such that

α =

. If it is an algebraic integer, then there is some

f(x) = x

+ a

n−1

+ ··· + a

with a

∈ Z such that f(α) = 0. Substituting in and multiplying by s

, we get

+ a

n−1

s + ··· + a

| {z }

divisible by s

= 0,

s | r

. But if

s 6

= 1, there is a prime

such that

p | s

, and hence

p | r

. Thus

p | r

. So

is a common factor of

and

. This is a contradiction. So

= 1, and

α is an integer.

How else is this a generalization of

? We know

is a ring. So perhaps

also is.

Theorem. O

is a ring, i.e. if α, β ∈ O

, then so is α ± β and αβ.

Note that in general O

is not a field. For example, Z = O

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

, then so is

αβ

and

α ± β

, as well

if α 6= 0.

To prove this, we notice that

α ∈ K

is algebraic if and only if

[

] is a finite

extension — if α is algebraic, with

f(α) = a

+ ··· + a

= 0, a

6= 0

then

[

] has degree at most

, since

(and similarly

−1

) can be written as

a linear combination of 1

, α, ··· , α

n−1

, and thus these generate

[

]. On the

other hand, if

[

] is finite, say of degree

, then 1

, α, ··· , α

are independent,

hence some linear combination of them vanishes, and this gives a polynomial

for which

is a root. Moreover, by the same proof, if

is any finite extension

over K, then any element in K

is algebraic.

Thus, to prove the result, notice that if

[

] is generated by 1

, α, ··· , α

and

[

] is generated by 1

, β, ··· , β

, then

[

α, β

] is generated by

{α

}

for

≤ i ≤ n,

≤ j ≤ m

. Hence

[

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

as a ring extension of

. 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

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

(Finitely-generated)

We say

is finitely-generated over

if there

exists elements

, ··· , α

∈ S

such that the function

→ S

defined by

(

, ··· , r

)

7→

is surjective, i.e. every element of

can be written as a

linear combination of elements

, ··· , α

. In other words,

is finitely-generated

as an R-module.

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

is integral over Z.

Notation.

, ··· , α

∈ S

, we write

[

, ··· , α

] for the subring of

generated by

R, α

, ··· , α

. In other words, it is the image of the homomorphism

from the polynomial ring R[x

, ··· , x

] → S given by x

7→ α

Proposition.

(i)

Let

R ⊆ S

be rings. If

[

] and

is integral over

, then

finitely-generated over R.

(ii)

[

, ··· , s

] with

integral over

, then

is finitely-generated

over R.

This is the easy direction in identifying integrality with finitely-generated.

Proof.

(i)

We know

is spanned by 1

, s, s

, ···

over

. However, since

is integral,

there exists a

, ··· , a

∈ R such that

= a

+ a

s + ··· + a

n−1

So the

-submodule generated by 1

, s, ··· , s

n−1

is stable under multiplica-

tion by s. So it contains s

, s

n+1

, s

n+2

, ···. So it is S.

(ii)

Let

[

, ··· , s

]. So

i−1

[

]. Since

is integral over

, it is

integral over

i−1

. By the previous part,

is finitely-generated over

i−1

To finish, it suffices to show that being finitely-generated is transitive.

More precisely, if

A ⊆ B ⊆ C

are rings,

is finitely generated over

and

is finitely generated over

, then

is finitely generated over

. This

is not hard to see, since if

, ··· , x

generate

over

, and

, ··· , y

generate

over

, then

is generated by

}

1≤i≤n,1≤j≤m

over

The other direction is harder.

Theorem. If S is finitely-generated 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

, ···

, 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 multiplication-by-

map:

S → S

γ 7→ sγ

. If

were a finite-dimensional vector space over

, then Cayley-Hamilton tells us

and thus

, satisfies its characteristic polynomial, which is monic. Even though

is not a finite-dimensional vector space, the proof of Cayley-Hamilton will

work.

Proof.

Let

, ··· , α

generate

as an

-module. wlog take

= 1

∈ S

. For

any s ∈ S, write

sα

for some

∈ R

. We write

= (

). This is the “matrix of multiplication by

S”. By construction, we have

(sI − B)













= 0. (∗)

Now recall for any matrix

, we have

adj

(

)

= (

det X

)

, where the

i, j

entry of

adj

(

) 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)













= 0

In particular,

det

(

sI −B

)

= 0. Since we picked

= 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

is a ring.

Proof.

α, β ∈ O

, then

[

α, β

] is finitely-generated by the proposition. But

then

[

α, β

] is integral over

, 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

is finitely-generated over

. For example, if

is the set of all algebraic

integers in

, and

, then by definition

is an integral extension of

, 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.

A ⊆ B ⊆ C

be ring extensions such that

over

and

over

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) =

i=0

∈ B[x]

be a monic polynomial such that

(

) = 0. Let

[

, ··· , b

] and let

[

]. Then

is finitely generated as

, ··· , b

are integral over

Also,

is finitely-generated over

, since

is integral over

. Hence

finitely-generated over

. So

is integral over

. Since

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

(x) ∈ K[x] of minimal degree such that p

(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

| f.

Proof. Write f = p

h + r, with r ∈ K[x] and deg(r) < deg(p

). 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

In particular, this lemma implies

is unique. One nice application of this

result is the following:

Proposition.

Let

be a number field. Then

α ∈ O

if and only if the minimal

polynomial p

(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

and

∈ Q

[

] be the minimal polynomial of

, and

(

)

∈ Z

[

] be a monic polynomial which

satisfies. The idea is to use

show that the coefficients of p

are algebraic, thus in fact integers.

Now there exists a bigger field M ⊇ L such that

(x) = (x − α

) ···(x − α

)

factors in

[

]. But by our lemma,

| h

. So

(

) = 0 for all

. So

∈ O

is an algebraic integer. But

is a ring, i.e. sums and products of the

’s are

still algebraic integers. So the coefficients of

are algebraic integers (in

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

and

is that

is the fraction field of

. This

is true for general number fields

Lemma. We have

Frac O



: α, β ∈ O

, β 6= 0



= L.

In fact, for any α ∈ L, there is some n ∈ Z such that nα ∈ O

Proof.

α ∈ L

, let

(

)

∈ Q

[

] be its monic minimal polynomial. Then there

exists

n ∈ Z

non-zero such that

(

)

∈ Z

[

] (pick

to be the least common

multiple of the denominators of the coefficients of

(

)). Now the magic is to

put

h(x) = n

deg(g)





Then this is a monic polynomial with integral coefficients — in effect, we have

just multiplied the coefficient of

deg(g)−i

! Then

(

nα

) = 0. So

nα

integral.