4Norms of ideals

II Number Fields

4 Norms of ideals

In the previous chapter, we defined the class group, and we know it is generated

by prime ideals of

O

L

. So we now want to figure out what the prime ideals are.

In the case of finding irreducible elements, one very handy tool was the norm

— we know an element of

O

L

is a unit iff it has norm 1. So if

x ∈ O

L

is not

irreducible, then there must be some element whose norm strictly divides

N

(

x

).

Similarly, we would want to come up with the notion of the norm of an ideal,

which turns out to be an incredibly useful notion.

Definition (Norm of ideal). Let a C O

L

be an ideal. We define

|N(a)| = |O

L

/a| ∈ N.

Recall that we’ve already proved that

|O

L

/a|

is finite. So this definition

makes sense. It is also clear that N (a) = 1 iff a = O

L

(i.e. a is a “unit”).

Example. Let d ∈ Z. Then since O

L

∼

=

Z

n

, we have dO

L

∼

=

dZ

n

. So we have

N(hdi) = |Z

n

/(dZ)

n

| = |Z/dZ|

n

= d

n

.

We start with a simple observation:

Proposition. For any ideal a, we have N(a) ∈ a ∩ Z.

Proof.

It suffices to show that

N

(

a

)

∈ a

. Viewing

O

L

/a

as an additive group,

the order of 1 is a factor of

N

(

a

). So

N

(

a

) =

N

(

a

)

·

1 = 0

∈ O

L

/a

. Hence

N(a) ∈ a.

The most important property of the norm is the following:

Proposition. Let a, b C O

L

be ideals. Then N(ab) = N(a)N(b).

We will provide two proofs of the result.

Proof.

By the factorization into prime ideals, it suffices to prove this for

b

=

p

prime, i.e.

N(ap) = N(a)N(p).

In other words, we need to show that

O

L

a

=

O

L

ap

.

O

L

p

.

By the third isomorphism theorem, we already know that

O

L

a

∼

=

O

L

ap

a

ap

.

So it suffices to show that O

L

/p

∼

=

a/ap as abelian groups.

In the case of the integers, we could have, say,

p

= 7

Z

,

a

= 12

Z

. We would

then simply define

Z

7Z

12Z

7 · 12Z

x 12x

However, in general, we do not know that

a

is principal, but it turns out it

doesn’t really matter. We can just pick an arbitrary element to multiply with.

By unique factorization, we know a 6= ap. So we can find some α ∈ a \ ap.

We now claim that the homomorphism of abelian groups

O

L

p

a

ap

x + p αx + ap

is an isomorphism. We first check this is well-defined — if

p ∈ p

, then

αp ∈ ap

since

α ∈ a

. So the image of

x

+

p

and (

x

+

p

)+

p

are equal. So this is well-defined.

To prove our claim, we have to show injectivity and surjectivity. To show

injectivity, since

hαi ⊆ a

, we have

a | hai

, i.e. there is an ideal

c ⊆ O

L

such that

ac = hαi. If x ∈ O

L

is in the kernel of the map, then αx ∈ ap. So

xac ⊆ ap.

So

xc ⊆ p.

As

p

is prime, either

c ⊆ p

or

x ∈ p

. But

c ⊆ p

implies

hαi

=

ac ⊆ ap

,

contradicting the choice of α. So we must have x ∈ p, and the map is injective.

To show this is surjective, we notice that surjectivity means

hαi/ap

=

a/ap

,

or equivalently ap + hαi = a.

Using our knowledge of fractional ideals, this is equivalent to saying (

ap

+

hαi)a

−1

= O

L

. But we know

ap < ap + hαi ⊆ a.

We now multiply by a

−1

to obtain

p < (ap + hαi)a

−1

= p + c ⊆ O

L

.

Since

p

is a prime, and hence maximal ideal, the last inclusion must be an

equality. So ap + hαi = a, and we are done.

Now we provide the sketch of a proof that makes sense. The details are left

as an exercise in the second example sheet.

Proof.

It is enough to show that

N

(

p

a

1

1

···p

a

r

r

) =

N

(

p

1

)

a

1

···N

(

p

r

)

a

r

by unique

factorization.

By the Chinese remainder theorem, we have

O

L

p

a

1

1

···p

a

r

r

∼

=

O

L

p

a

1

1

× ··· ×

O

L

p

a

r

r

where p

1

, ··· , p

r

are distinct prime ideals.

Next, we show by hand that

O

L

p

r

=

O

L

p

×

p

p

2

× ··· ×

p

r−1

p

r

=

O

L

p

r

,

by showing that

p

k

/p

k+1

is a 1-dimensional vector space over the field

O

L

/p

.

Then the result follows.

This is actually the same proof, but written in a much saner form. This is

better because we are combining a general statement (the Chinese remainder

theorem), with a special property of the integral rings. In the first proof, what

we really did was simultaneously proving two parts using algebraic magic.

We’ve taken an obvious invariant of an ideal, the size, and found it is

multiplicative. How does this relate to the other invariants?

Recall that

∆(α

1

, ··· , α

n

) = det(tr

L/Q

(α

i

α

j

)) = det(σ

i

(α

j

))

2

.

Proposition. Let a C O

L

be an ideal, n = [L : Q]. Then

(i) There exists α

1

, ··· , α

n

∈ a such that

a =

n

X

r

i

α

i

: r

i

∈ Z

o

=

n

M

1

α

i

Z,

and

α

1

, ··· , α

n

are a basis of

L

over

Q

. In particular,

a

is a free

Z

-module

of n generators.

(ii) For any such α

1

, ··· , α

n

,

∆(α

1

, ··· , α

n

) = N(a)

2

D

L

.

The prove this, we recall the following lemma from IB Groups, Rings and

Modules:

Lemma.

Let

M

be a

Z

-module (i.e. abelian group), and suppose

M ≤ Z

n

.

Then M

∼

=

Z

r

for some 0 ≤ r ≤ n.

Moreover, if

r

=

n

, then we can choose a basis

v

1

, ··· , v

n

of

M

such that the

change of basis matrix A = (a

ij

) ∈ M

n×n

(Z) is upper triangular, where

v

j

=

X

a

ij

e

i

,

where e

1

, ··· , e

n

is the standard basis of Z

n

.

In particular,

|Z

n

/M| = |a

11

a

12

···a

nn

| = |det A|.

Proof of proposition.

Let

d ∈ a ∩ Z

, say

d

=

N

(

α

). Then

dO

L

⊆ a ⊆ O

L

. As

abelian groups, after picking an integral basis α

0

1

, ··· , α

0

n

of O

L

, we have

Z

n

∼

=

dZ

n

≤ a ≤ Z

n

.

So

a

∼

=

Z

n

. Then the lemma gives us a basis

α

1

, ··· , α

n

of

a

as a

Z

-module. As

a

Q

-module, since the

α

i

are obtained from linear combinations of

α

0

i

, by basic

linear algebra, α

1

, ··· , α

n

is also a basis of L over Q.

Moreover, we know that we have

∆(α

1

, ··· , α

n

) = det(A)

2

∆(α

0

1

, ··· , α

0

n

).

Since

det

(

A

)

2

=

|O

L

/a|

2

=

N

(

a

) and

D

L

= ∆(

α

0

1

, ··· , α

0

n

) by definition, the

second part follows.

This result is very useful for the following reason:

Corollary.

Suppose

a C O

L

has basis

α

1

, ··· , α

n

, and ∆(

α

1

, ··· , α

n

) is square-

free. Then a = O

L

(and D

L

is square-free).

This is a nice trick, since it allows us to determine immediately whether a

particular basis is an integral basis.

Proof. Immediate, since this forces N(a)

2

= 1.

Note that nothing above required

a

to be an actual ideal. It merely had

to be a subgroup of

O

L

that is isomorphic to

Z

n

, since the quotient

O

L

/a

is

well-defined as long as

a

is a subgroup. With this, we can have the following

useful result:

Example.

Let

α

be an algebraic integer and

L

=

Q

(

α

). Let

n

= [

Q

(

α

) :

Q

].

Then a = Z[α] C O

L

. We have

disc(p

α

) = ∆(1, α, α

2

, ··· , α

n−1

) = discriminant of minimal polynomial of α.

Thus if disc(p

α

) is square-free, then Z[α] = O

L

.

Even if

disc

(

p

α

) is not square-free, it still says something: let

d

2

| disc

(

p

α

)

be such that disc(p

α

)/d

2

is square-free. Then |N (Z[α])| divides d.

Let

x ∈ O

L

. Then the order of

x

+

Z

[

α

]

∈ O

L

/Z

[

α

] divides

N

(

Z

[

α

]), hence

d. So d · x ∈ Z[α]. So x ∈

1

d

Z[α]. Hence we have

Z[α] ⊆ O

L

⊆

1

d

Z[α].

For example, if

α

=

√

a

for some square-free

a

, then

disc

(

√

a

) is the discriminant

of x

2

− a, which is 4a. So the d above is 2, and we have

Z[α] ⊆ O

Q(

√

d)

⊆

1

2

Z[α],

as we have previously seen.

We shall prove one more lemma, and start factoring things. Recall that we

had two different notions of norm. Given

α ∈ O

L

, we can take the norm

N

(

hαi

),

or

N

L/Q

(

α

). It would be great if they are related, like if they are equal. However,

that cannot possibly be true, since

N

(

hαi

) is always positive, but

N

L/Q

(

α

) can

be negative. So we take the absolute value.

Lemma. If α ∈ O

L

, then

N(hαi) = |N

L/Q

(α)|.

Proof.

Let

α

1

, ··· , α

n

be an integral basis of

O

L

. Then

αα

1

, .., αα

n

is an integral

basis of hαi. So by the previous lemma,

∆(αα

1

, ··· , αα

n

) = N(hαi)

2

D

L

.

But

∆(αα

1

, ··· , αα

n

) = det(σ

i

(αα

j

)

ij

)

2

= det(σ

i

(α)σ

i

(α

j

))

2

=

n

Y

i=1

σ

i

(α)

!

2

∆(α

1

, ··· , α

n

)

= N

L/Q

(α)

2

D

L

.

So

N

L/Q

(α)

2

= N(hαi)

2

.

But N(hαi) is positive. So the result follows.