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

$\displaystyle \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

$\displaystyle f_{\P/\mathfrak{p}} = [\O_K/\P: \O_L/\mathfrak{p}],$

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

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

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

$\displaystyle 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

$\displaystyle 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

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

$\displaystyle \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

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

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

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

which completes the proof. $ \qedsymbol$

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

William Stein 2012-09-24