divisors in base field and finite extension field

Let $k$ be the quotient field of an integral domain $\mathfrak{o}$ which has the divisor theory $\mathfrak{o}^{*}\to\mathfrak{D}_{0}$ . Let $K/k$ a finite extension, $\mathfrak{O}$ be the integral closure of $\mathfrak{o}$ in $K$ and $\mathfrak{O}^{*}\to\mathfrak{D}$ the uniquely determined divisor theory of $\mathfrak{O}$ (see the parent entry (http://planetmath.org/DivisorTheoryInFiniteExtension)). We will study the of the divisor monoids $\mathfrak{D}_{0}$ and $\mathfrak{D}$ .

Any element $a$ of $\mathfrak{o}^{*}$ , which is a part of $\mathfrak{O}^{*}$ , determines a principal divisor $(a)_{k}\in\mathfrak{D}_{0}$ and another $(a)_{K}\in\mathfrak{D}$ . The (multiplicative) monoid $\mathfrak{o}^{*}$ is isomorphically embedded (via $\iota$ ) in the monoid $\mathfrak{O}^{*}$ . Because the units of the ring $\mathfrak{O}$ , which belong to $\mathfrak{o}$ , are all units of $\mathfrak{o}$ and because associates always determine the same principal divisor, the mentioned embedding defines an isomorphic mapping

\displaystyle(a)_{k}\mapsto(a)_{K}

(1)

from the monoid of the principal divisors of $\mathfrak{o}$ into the monoid of the principal divisors of $\mathfrak{O}$ . One has the

Theorem. There is one and only one isomorphism $\varphi$ from the divisor monoid $\mathfrak{D}_{0}$ into the divisor monoid $\mathfrak{D}$ such that its restriction to the principal divisors of $\mathfrak{o}$ coincides with (1). Then there is the following commutative diagram:

\xymatrix{\mathfrak{o}^{*}\ar[r]^{\iota}\ar[d]&\mathfrak{O}^{*}\ar[d]\\ \mathfrak{D}_{0}\ar[r]_{\varphi}&\mathfrak{D}}

The isomorphism $\varphi\!:\,\mathfrak{D}_{o}\to\mathfrak{D}$ is determined as follows. Let $\mathfrak{p}$ be an arbitrary prime divisor in $\mathfrak{D}_{0}$ and $\nu_{\mathfrak{p}}$ the corresponding exponent valuation of the field $k$ . Let $\nu_{\mathfrak{P}_{1}},\,\ldots,\,\nu_{\mathfrak{P}_{m}}$ be the continuations of the exponent $\nu_{\mathfrak{p}}$ to $K$ , which correspond to the prime divisors $\mathfrak{P}_{1},\,\ldots,\,\mathfrak{P}_{m}$ in $\mathfrak{D}$ . If $e_{1},\,\ldots,\,e_{m}$ are the ramification indices of the exponents $\nu_{\mathfrak{P}_{1}},\,\ldots,\,\nu_{\mathfrak{P}_{m}}$ with respect to $\nu_{\mathfrak{p}}$ , then we have

\nu_{\mathfrak{P}_{i}}(a)=e_{i}\nu_{\mathfrak{p}}(a)\quad\forall a\in\mathfrak% {o}^{*}.

Thus apparently, the factor of the principal divisor $(a)_{K}\in\mathfrak{D}$ , which corresponds to the factor $\mathfrak{p}^{\nu_{\mathfrak{p}}(a)}$ of the principal divisor $(a)_{k}\in\mathfrak{D}_{0}$ , is $(\mathfrak{P}_{1}^{e_{1}}\cdots\mathfrak{P}_{m}^{e_{m}})^{\nu_{\mathfrak{p}}(a)}$ . Then $\varphi$ is settled by

\mathfrak{p}\mapsto\mathfrak{P}_{1}^{e_{1}}\cdots\mathfrak{P}_{m}^{e_{m}}.

When one identifies $\mathfrak{D}_{0}$ with its isomorphic image $\varphi(\mathfrak{D}_{0})$ , we can write

\mathfrak{p}=\mathfrak{P}_{1}^{e_{1}}\cdots\mathfrak{P}_{m}^{e_{m}}\in% \mathfrak{D},

i.e. the prime divisors in $\mathfrak{D}_{0}$ don’t in general remain as prime divisors in $\mathfrak{D}$ . On grounds of the identification one may speak of the divisibility of the divisors of $\mathfrak{o}$ by the divisors of $\mathfrak{O}$ . The coprime divisors of $\mathfrak{o}$ are coprime also as divisors of $\mathfrak{O}$ .

References

1 S. Borewicz & I. Safarevic: Zahlentheorie. Birkhäuser Verlag. Basel und Stuttgart (1966).

Title	divisors in base field and finite extension field
Canonical name	DivisorsInBaseFieldAndFiniteExtensionField
Date of creation	2013-03-22 18:01:35
Last modified on	2013-03-22 18:01:35
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Topic
Classification	msc 13F05
Classification	msc 13A18
Classification	msc 13A05
Related topic	DivisorTheory