Definition 13.2.1 (Residue class degree)
Suppose
![$ P$](img63.png)
is a prime of
![$ \O _K$](img200.png)
lying over
![$ \mathfrak{p}$](img357.png)
.
Then the
of
![$ P$](img63.png)
is
i.e., the degree of the extension of residue class fields.
Proof.
For simplicity, we will give the proof only in the case
![$ L=\mathbf{Q}$](img1185.png)
, but
the proof works in general. Suppose
![$ p\in\mathbf{Z}$](img470.png)
and
![$ p\O _K=\mathfrak{p}_1^{e_1}\cdots \mathfrak{p}_g^{e_g}$](img1186.png)
, and
![$ S=\{\mathfrak{p}_1,\ldots, \mathfrak{p}_g\}$](img1187.png)
. We
will first prove that
![$ G$](img31.png)
acts transitively on
![$ S$](img121.png)
. Let
![$ \mathfrak{p}=\mathfrak{p}_i$](img1188.png)
for
some
![$ i$](img48.png)
. Recall that we proved long ago, using the Chinese
Remainder Theorem (Theorem
9.1.3) that there exists
![$ a\in\mathfrak{p}$](img1189.png)
such that
![$ (a)/\mathfrak{p}$](img1190.png)
is an integral ideal that is
coprime to
![$ p\O _K$](img472.png)
. The product
![$\displaystyle I= \prod_{\sigma\in G} \sigma((a)/\mathfrak{p}) = \prod_{\sigma\i...
...K/\mathbf{Q}}(a))\O _K}{\displaystyle \prod_{\sigma\in G} \sigma(\mathfrak{p})}$](img1191.png) |
(13.1) |
is a nonzero integral
![$ \O _K$](img200.png)
ideal since it is a product of nonzero
integral
![$ \O _K$](img200.png)
ideals.
Since
![$ a\in\mathfrak{p}$](img1189.png)
we have that
![$ \Norm _{K/\mathbf{Q}}(a) \in \mathfrak{p}\cap\mathbf{Z}=p\mathbf{Z}$](img1192.png)
. Thus the numerator of
the rightmost expression in (
13.2.1) is
divisible by
![$ p\O _K$](img472.png)
. Also, because
![$ (a)/\mathfrak{p}$](img1190.png)
is coprime
to
![$ p\O _K$](img472.png)
, each
![$ \sigma((a)/\mathfrak{p})$](img1193.png)
is coprime to
![$ p\O _K$](img472.png)
as well. Thus
![$ I$](img151.png)
is coprime to
![$ p\O _K$](img472.png)
. Thus the
denominator of the rightmost expression in (
13.2.1)
must also be divisibly by
![$ p\O _K$](img472.png)
in order to cancel the
![$ p\O _K$](img472.png)
in the numerator. Thus for any
![$ i$](img48.png)
we have
which in particular implies that
![$ G$](img31.png)
acts transitively on the
![$ \mathfrak{p}_i$](img408.png)
.
Choose some
and suppose that
is another index. Because
acts transitively, there exists
such that
. Applying
to the factorization
, we see that
Taking
![$ \ord _{\mathfrak{p}_j}$](img1200.png)
on both sides we get
![$ e_j = e_k$](img1201.png)
. Thus
![$ e_1=e_2=\cdots = e_g$](img1202.png)
.
As was mentioned right before the statement of the theorem, for any
we have
, so by transitivity
.
Since
is a lattice in
, we have
which completes the proof.
The rest of this section illustrates the theorem for quadratic fields
and a cubic field and its Galois closure.