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Homesplitting and ramification in number fields and Galois extensions

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# 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). 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”, 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:

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

## Mathematics Subject Classification

12F99*no label found*13B02

*no label found*11S15

*no label found*

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