Decomposition of Primes

Fix a prime $\mathfrak{p}\subset \O _K$ and write $\mathfrak{p}\O _K = \P _1^{e_1}\cdots \P _g^{e_g}$, so $S_\mathfrak{p}=\{\P _1,\ldots, \P _g\}$.

Definition 13.2.1 (Residue class degree)   Suppose $P$ is a prime of $\O _K$ lying over $\mathfrak{p}$. Then the of $P$ is

$$f_{\P /\mathfrak{p}} = [\O _K/\P : \O _L/\mathfrak{p}],$$

i.e., the degree of the extension of residue class fields.

If $M/K/L$ is a tower of field extensions and $\mathfrak{q}$ is a prime of $M$ over $P$, then

$$f_{\mathfrak{q}/\mathfrak{p}} = [\O _M/\mathfrak{q}: \O _L/\mathf... ... _K/\P : \O _L/\mathfrak{p}] = f_{\mathfrak{q}/\P }\cdot f_{\P /\mathfrak{p}},$$

so the residue class degree is multiplicative in towers.

Note that if $\sigma\in\Gal (K/L)$ and $\P\in S_p$, then $\sigma$ induces an isomorphism of finite fields $\O _K/\P\to \O _K/\sigma(\P )$ that fixes the common subfield $\O _L/\mathfrak{p}$. Thus the residue class degrees of $P$ and $\sigma(\P )$ are the same. In fact, much more is true.

Theorem 13.2.2   Suppose $K/L$ is a Galois extension of number fields, and let $\mathfrak{p}$ be a prime of $\O _L$. Write $\mathfrak{p}\O _K=\prod_{i=1}^g \P _i^{e_i}$, and let $f_i = f_{\P _i/\mathfrak{p}}$. Then $G=\Gal (K/L)$ acts transitively on the set $S_\mathfrak{p}$ of primes $\P _i$,

$$e_1=\cdots =e_g, \qquad f_1 =\cdots = f_g,$$

and $efg=[K:L]$, where $e$ is the common value of the $e_i$ and $f$ is the common value of the $f_i$.

Proof. For simplicity, we will give the proof only in the case $L=\mathbf{Q}$, but the proof works in general. Suppose $p\in\mathbf{Z}$ and $p\O _K=\mathfrak{p}_1^{e_1}\cdots \mathfrak{p}_g^{e_g}$, and $S=\{\mathfrak{p}_1,\ldots, \mathfrak{p}_g\}$. We will first prove that $G$ acts transitively on $S$. Let $\mathfrak{p}=\mathfrak{p}_i$ for some $i$. Recall that we proved long ago, using the Chinese Remainder Theorem (Theorem 9.1.3) that there exists $a\in\mathfrak{p}$ such that $(a)/\mathfrak{p}$ is an integral ideal that is coprime to $p\O _K$. The product

$$I= \prod_{\sigma\in G} \sigma((a)/\mathfrak{p}) = \prod_{\sigma\i... ...K/\mathbf{Q}}(a))\O _K}{\displaystyle \prod_{\sigma\in G} \sigma(\mathfrak{p})}$$ (13.1)

is a nonzero integral $\O _K$ ideal since it is a product of nonzero integral $\O _K$ ideals. Since $a\in\mathfrak{p}$ we have that $\Norm _{K/\mathbf{Q}}(a) \in \mathfrak{p}\cap\mathbf{Z}=p\mathbf{Z}$. Thus the numerator of the rightmost expression in (13.2.1) is divisible by $p\O _K$. Also, because $(a)/\mathfrak{p}$ is coprime to $p\O _K$, each $\sigma((a)/\mathfrak{p})$ is coprime to $p\O _K$ as well. Thus $I$ is coprime to $p\O _K$. Thus the denominator of the rightmost expression in (13.2.1) must also be divisibly by $p\O _K$ in order to cancel the $p\O _K$ in the numerator. Thus for any $i$ we have

$$\prod_{j=1}^g \mathfrak{p}_j^{e_j} = p\O _K \Big\vert \prod_{\sigma\in G} \sigma(\mathfrak{p}_i),$$

which in particular implies that $G$ acts transitively on the $\mathfrak{p}_i$.

Choose some $j$ and suppose that $k\neq j$ is another index. Because $G$ acts transitively, there exists $\sigma\in G$ such that $\sigma(\mathfrak{p}_k)=\mathfrak{p}_j$. Applying $\sigma$ to the factorization $p\O _K = \prod_{i=1}^g \mathfrak{p}_i^{e_i}$, we see that

$$\prod_{i=1}^g \mathfrak{p}_i^{e_i} = \prod_{i=1}^g \sigma(\mathfrak{p}_i)^{e_i}.$$

Taking $\ord _{\mathfrak{p}_j}$ on both sides we get $e_j = e_k$. Thus $e_1=e_2=\cdots = e_g$.

As was mentioned right before the statement of the theorem, for any $\sigma\in G$ we have $\O _K/\mathfrak{p}_i\cong \O _K/\sigma(\mathfrak{p}_i)$, so by transitivity $f_1=f_2=\cdots = f_g$. Since $\O _K$ is a lattice in $K$, we have

$$[K:\mathbf{Q}] = \dim_{\mathbf{Z}} \O _K = \dim_{\mathbf{F}_p} \O _K/p\O _K$$

$$= \dim_{\mathbf{F}_p} \left(\bigoplus_{i=1}^g \O _K/\mathfrak{p}_i^{e_i}\right) = \sum_{i=1}^g e_i f_i = efg,$$

which completes the proof. $\qedsymbol$

The rest of this section illustrates the theorem for quadratic fields and a cubic field and its Galois closure.