Decomposition of Primes: $efg=n$

If $I\subset \O_K$ is any ideal in the ring of integers of a Galois extension $K$ of $\mathbf {Q}$ and $\sigma\in\Gal (K/\mathbf{Q})$, then

$$\sigma(I) = \{\sigma(x) : x \in I\}$$

is also an ideal of $\O_K$.

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 9.2.1 (Residue class degree)   Suppose $P$ is a prime of $\O_K$ lying over $\mathfrak{p}$. Then the residue class degree 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

$$f_{\mathfrak{q}/\mathfrak{p}} = [\O_M/\mathfrak{q}: \O_L/\mathfra... ...t [\O_K/\P: \O_L/\mathfrak{p}] = f_{\mathfrak{q}/\P}\cdot f_{\P/\mathfrak{p}},$$

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 9.2.2   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 $\P_i$, and

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

Morever, if we let $e$ be the common value of the $e_i$, $f$ the common value of the $f_i$, and $n=[K:L]$, then

$$efg=n.$$

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 5.1.4) 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

$$I= \prod_{\sigma\in G} \sigma((a)/\mathfrak{p}) = \prod_{\sigma\i... ...{K/\mathbf{Q}}(a))\O_K}{\displaystyle \prod_{\sigma\in G} \sigma(\mathfrak{p})}$$ (9.1)

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 (9.2.1) 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 (9.2.1) must also be divisibly by $p\O_K$ in order to cancel the $p\O_K$ in the numerator. Thus we have shown that for any $i$,

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

By unique factorization, since every $\mathfrak{p}_j$ appears in the left hand side, we must have that for each $j$ there is a $\sigma$ with $\sigma(\mathfrak{p}_i)=\mathfrak{p}_j$.

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 and using unique factorization, 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 $f_1=f_2=\cdots = f_g$. We have, upon apply CRT and that $\char93 (\O_K/(\mathfrak{p}^m)) = \char93 (\O_K/\mathfrak{p})^m$, that

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

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

which completes the proof. $\qedsymbol$

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