*Proof*.
We proved before that every ideal class in

is represented by
an ideal

with

. Write

, with each

. Then by multiplicativity of the
norm, each

also satisfies

. If

, then

, since

is the residue
characteristic of

, so

. Thus

is a product of
primes

that satisfies the norm bound of the lemma, whcih proves
the lemma.

*Example 11.1.2*
We compute the class group of

. We have

so

Thus

is generated by the prime divisors
of

. We have

so

is generated by the principal prime
ideal

. Thus

is trivial.

*Example 11.1.3*
We compute the class group of

.
We have

so

Thus

is generated by the primes that divide

.
We have

, where

satisfies

. The polynomial

is irreducible
mod

, so

is prime. Since it is principal, we see
that

is trivial.

*Example 11.1.4*
In this example, we compute the class group of

.
We have

so

Thus

is generated by the prime ideals lying over

and

.
We have

, and

satisfies

.
Factoring

modulo

and

we see that the class group
is generated by the prime ideals

and

Also,

and

, so

and

define elements of order
dividing

in

.

Is either
or
principal? Fortunately,
there is an easier norm trick that allows us to decide.
Suppose
, where
.
Then

Trying the first few values of

, we see that this
equation has no solutions, so

can not
be principal. By a similar argument, we see that

is not principal either. Thus

and

define
elements of order

in

.

Does the class of
equal the class of
?
Since
and
define classes of order ,
we can decide this by finding the class of
.
We have

The ideals on both sides of the inclusion have norm

,
so by multiplicativity of the norm, they must be the
same ideal. Thus

is principal,
so

and

represent the same element of

.
We conclude that