# 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 DivisorsInBaseFieldAndFiniteExtensionField 2013-03-22 18:01:35 2013-03-22 18:01:35 pahio (2872) pahio (2872) 6 pahio (2872) Topic msc 13F05 msc 13A18 msc 13A05 DivisorTheory