Proof.
We proved before that every ideal class in
![$ \Cl (K)$](img893.png)
is represented by
an ideal
![$ I$](img151.png)
with
![$ \Norm (I)\leq B_K$](img896.png)
. Write
![$ I=\prod_{i=1}^m
\mathfrak{p}_i^{e_i}$](img897.png)
, with each
![$ e_i\geq 1$](img898.png)
. Then by multiplicativity of the
norm, each
![$ \mathfrak{p}_i$](img408.png)
also satisfies
![$ \Norm (\mathfrak{p}_i)\leq B_K$](img899.png)
. If
![$ \mathfrak{p}_i \cap
\mathbf{Z}=p\mathbf{Z}$](img900.png)
, then
![$ p\mid \Norm (\mathfrak{p}_i)$](img901.png)
, since
![$ p$](img4.png)
is the residue
characteristic of
![$ \O _K/\mathfrak{p}$](img363.png)
, so
![$ p\leq B_K$](img902.png)
. Thus
![$ I$](img151.png)
is a product of
primes
![$ \mathfrak{p}$](img357.png)
that satisfies the norm bound of the lemma, whcih proves
the lemma.
Example 11.1.2
We compute the class group of
![$ K=\mathbf{Q}(i)$](img247.png)
. We have
so
Thus
![$ \Cl (K)$](img893.png)
is generated by the prime divisors
of
![$ 2$](img25.png)
. We have
so
![$ \Cl (K)$](img893.png)
is generated by the principal prime
ideal
![$ \mathfrak{p}=(1+i)$](img909.png)
. Thus
![$ \Cl (K)=0$](img910.png)
is trivial.
Example 11.1.3
We compute the class group of
![$ K=\mathbf{Q}(\sqrt{5})$](img249.png)
.
We have
so
Thus
![$ \Cl (K)$](img893.png)
is generated by the primes that divide
![$ 2$](img25.png)
.
We have
![$ \O _K=\mathbf{Z}[\gamma]$](img913.png)
, where
![$ \gamma=\frac{1+\sqrt{5}}{2}$](img914.png)
satisfies
![$ x^2-x-1$](img252.png)
. The polynomial
![$ x^2-x-1$](img252.png)
is irreducible
mod
![$ 2$](img25.png)
, so
![$ 2\O _K$](img490.png)
is prime. Since it is principal, we see
that
![$ \Cl (K)=1$](img915.png)
is trivial.
Example 11.1.4
In this example, we compute the class group of
![$ K=\mathbf{Q}(\sqrt{-6})$](img450.png)
.
We have
so
Thus
![$ \Cl (K)$](img893.png)
is generated by the prime ideals lying over
![$ 2$](img25.png)
and
![$ 3$](img462.png)
.
We have
![$ \O _K=\mathbf{Z}[\sqrt{-6}]$](img451.png)
, and
![$ \sqrt{-6}$](img461.png)
satisfies
![$ x^2+6=0$](img918.png)
.
Factoring
![$ x^2+6$](img919.png)
modulo
![$ 2$](img25.png)
and
![$ 3$](img462.png)
we see that the class group
is generated by the prime ideals
![$\displaystyle \mathfrak{p}_2 = (2,\sqrt{-6})$](img920.png)
and
Also,
![$ \mathfrak{p}_2^2 = 2\O _K$](img922.png)
and
![$ \mathfrak{p}_3^2=3\O _K$](img923.png)
, so
![$ \mathfrak{p}_2$](img924.png)
and
![$ \mathfrak{p}_3$](img925.png)
define elements of order
dividing
![$ 2$](img25.png)
in
![$ \Cl (K)$](img893.png)
.
Is either
or
principal? Fortunately,
there is an easier norm trick that allows us to decide.
Suppose
, where
.
Then
Trying the first few values of
![$ a, b\in \mathbf{Z}$](img929.png)
, we see that this
equation has no solutions, so
![$ \mathfrak{p}_2$](img924.png)
can not
be principal. By a similar argument, we see that
![$ \mathfrak{p}_3$](img925.png)
is not principal either. Thus
![$ \mathfrak{p}_2$](img924.png)
and
![$ \mathfrak{p}_3$](img925.png)
define
elements of order
![$ 2$](img25.png)
in
![$ \Cl (K)$](img893.png)
.
Does the class of
equal the class of
?
Since
and
define classes of order
,
we can decide this by finding the class of
.
We have
The ideals on both sides of the inclusion have norm
![$ 6$](img463.png)
,
so by multiplicativity of the norm, they must be the
same ideal. Thus
![$ \mathfrak{p}_2\cdot \mathfrak{p}_3=(\sqrt{-6})$](img932.png)
is principal,
so
![$ \mathfrak{p}_2$](img924.png)
and
![$ \mathfrak{p}_3$](img925.png)
represent the same element of
![$ \Cl (K)$](img893.png)
.
We conclude that