splitting and ramification in number fields and Galois extensions

Let $F/K$ be an extension of number fields and let $\\mathcal{O}_{F}$ and $\\mathcal{O}_{K}$ be their respective rings of integers. The ring of integers of a number field is a Dedekind domain, and these enjoy the property that every ideal ${\\mathfrak{A}}$ factors uniquely as a finite product of prime ideals (see the entry fractional ideal (http://planetmath.org/FractionalIdeal)). Let ${\\mathfrak{p}}$ be a prime ideal of $\\mathcal{O}_{K}$ . Then ${\\mathfrak{p}}\\mathcal{O}_{F}$ is an ideal of $\\mathcal{O}_{F}$ . Let us assume that the prime ideal factorization of ${\\mathfrak{p}}\\mathcal{O}_{F}$ into primes of $\\mathcal{O}_{F}$ is as follows:

\\displaystyle{\\mathfrak{p}}\\mathcal{O}_{F}=\\prod_{i=1}^{r}{{\\mathfrak{P}}_{i}}%^{e_{i}}

(1)

We say that the primes ${\\mathfrak{P}}_{i}$ lie above ${\\mathfrak{p}}$ and ${\\mathfrak{P}}_{i}|{\\mathfrak{p}}$ (divides). The exponent $e_{i}$ (commonly denoted as $e({\\mathfrak{P}}_{i}|{\\mathfrak{p}})$ ) is the ramification index of ${\\mathfrak{P}}_{i}$ over ${\\mathfrak{p}}$ . Notice that for each prime ideal ${\\mathfrak{P}}_{i}$ , the quotient ring $\\mathcal{O}_{F}/{\\mathfrak{P}}_{i}$ is a finite field extension of the finite field $\\mathcal{O}_{K}/{\\mathfrak{p}}$ (also called the residue field). The degree of this extension is called the inertial degree of ${\\mathfrak{P}}_{i}$ over ${\\mathfrak{p}}$ and it is usually denoted by:

f({\\mathfrak{P}}_{i}|{\\mathfrak{p}})=[\\mathcal{O}_{F}/{\\mathfrak{P}}_{i}:%\\mathcal{O}_{K}/{\\mathfrak{p}}].

Notice that as it is pointed out in the entry “inertial degree (http://planetmath.org/InertialDegree)”, the ramification index and the inertial degree are related by the formula:

\\displaystyle\\sum_{i=1}^{r}e({\\mathfrak{P}}_{i}|{\\mathfrak{p}})f({\\mathfrak{P}%}_{i}|{\\mathfrak{p}})=[F:K]

(2)

where $r$ is the number of prime ideals lying above ${\\mathfrak{p}}$ (as in Eq. (1)). See the theorem below for an improvement of Eq. (2) in the case when $F/K$ is Galois.

Definition 1.

Let $F, K$ and ${\\mathfrak{P}}_{i},{\\mathfrak{p}}$ be as above.

1.
If $e_{i}>1$ for some $i$ , then we say that ${\\mathfrak{P}}_{i}$ is ramified over ${\\mathfrak{p}}$ and ${\\mathfrak{p}}$ ramifies in $F/K$ . If $e_{i}=1$ for all $i$ then we say that ${\\mathfrak{p}}$ is unramified in $F/K$ .
2.
If there is a unique prime ideal ${\\mathfrak{P}}$ lying above ${\\mathfrak{p}}$ (so $r=1$ ) and $f({\\mathfrak{P}}|{\\mathfrak{p}})=1$ then we say that ${\\mathfrak{p}}$ is totally ramified in $F/K$ . In this case $e({\\mathfrak{P}}|{\\mathfrak{p}})=[F:K]$ .
3.
On the other hand, if $e({\\mathfrak{P}}_{i}|{\\mathfrak{p}})=f({\\mathfrak{P}}_{i}|{\\mathfrak{p}})=1$ for all $i$ , we say that ${\\mathfrak{p}}$ is totally split (or splits completely) in $F/K$ . Notice that there are exactly $r=[F:K]$ prime ideals of $\\mathcal{O}_{F}$ lying above ${\\mathfrak{p}}$ .
4.
Let $p$ be the characteristic of the residue field $\\mathcal{O}_{K}/{\\mathfrak{p}}$ . If $e_{i}=e({\\mathfrak{P}}_{i}|{\\mathfrak{p}})>1$ and $e_{i}$ and $p$ are relatively prime, then we say that ${\\mathfrak{P}}_{i}$ is tamely ramified. If $p|e_{i}$ then we say that ${\\mathfrak{P}}_{i}$ is strongly ramified (or wildly ramified).

When the extension $F/K$ is a Galois extension then Eq. (2) is quite more simple:

Theorem 1.

Assume that $F/K$ is a Galois extension of number fields. Then all the ramification indices $e_{i}=e({\\mathfrak{P}}_{i}|{\\mathfrak{p}})$ are equal to the same number $e$ , all the inertial degrees $f_{i}=f({\\mathfrak{P}}_{i}|{\\mathfrak{p}})$ are equal to the same number $f$ and the ideal ${\\mathfrak{p}}\\mathcal{O}_{F}$ factors as:

{\\mathfrak{p}}\\mathcal{O}_{F}=\\prod_{i=1}^{r}{\\mathfrak{P}}_{i}^{e}=({%\\mathfrak{P}}_{1}\\cdot{\\mathfrak{P}}_{2}\\cdot\\ldots\\cdot{\\mathfrak{P}}_{r})^{e}

Moreover:

e\\cdot f\\cdot r=[F:K].