# continuation of exponent

Theorem.  Let $K/k$ be a finite field extension and $\nu$ an exponent valuation of the extension field $K$.  Then there exists one and only one positive integer $e$ such that the function

 $(1)\qquad\qquad\qquad\nu_{0}(x)\,:=\,\begin{cases}&\infty\quad\mbox{when }\;x=% 0,\\ &\frac{\nu(x)}{e}\;\;\mbox{when }\;x\neq 0,\end{cases}$

defined in the base field $k$, is an exponent (http://planetmath.org/ExponentValuation) of $k$.

Proof.  The exponent $\nu$ of $K$ attains in the set $k\!\smallsetminus\!\{0\}$ also non-zero values; otherwise  $k$ would be included in $\mathcal{O}_{\nu}$, the ring of the exponent $\nu$.  Since any element $\xi$ of $K$ are integral over $k$, it would then be also integral over $\mathcal{O}_{\nu}$, which is integrally closed in its quotient field $K$ (see theorem 1 in ring of exponent); the situation would mean that  $\xi\in\mathcal{O}_{\nu}$ and thus the whole $K$ would be contained in $\mathcal{O}_{\nu}$.  This is impossible, because an exponent of $K$ attains also negative values.  So we infer that $\nu$ does not vanish in the whole $k\!\smallsetminus\!\{0\}$.  Furthermore, $\nu$ attains in $k\!\smallsetminus\!\{0\}$ both negative and positive values, since  $\nu(a)\!+\!\nu(a^{-1})=\nu(aa^{-1})=\nu(1)=0$.

Let $p$ be such an element of $k$ on which $\nu$ attains as its value the least possible positive integer $e$ in the field $k$ and let $a$ be an arbitrary non-zero element of $k$.  If

 $\nu(a)=m=qe+r\quad(q,\,r\in\mathbb{Z},\;\;0\leqq r

then  $\nu(ap^{-q})=m-qe=r$,  and thus  $r=0$  on grounds of the choice of $p$.  This means that $\nu(a)$ is always divisible by $e$, i.e. that the values of the function $\nu_{0}$ in $k\!\smallsetminus\!\{0\}$ are integers.  Because  $\nu_{0}(p)=1$  and  $\nu_{0}(p^{l})=l$,  the function attains in $k$ every integer value.  Also the conditions

 $\nu_{0}(ab)=\nu_{0}(a)+\nu_{0}(b),\quad\nu_{0}(a+b)\geqq\min\{\nu_{0}(a),\,\nu% _{0}(b)\}$

are in , whence $\nu_{0}$ is an exponent of the field $k$.

Definition.  Let $K/k$ be a finite field extension.  If the exponent $\nu_{0}$ of $k$ is tied with the exponent $\nu$ of $K$ via the condition (1), one says that $\nu$ induces $\nu_{0}$ to $k$ and that $\nu$ is the continuation of $\nu_{0}$ to $K$.  The positive integer $e$, uniquely determined by (1), is the ramification index of $\nu$ with respect to $\nu_{0}$ (or with respect to the subfield $k$).

## References

• 1 S. Borewicz & I. Safarevic: Zahlentheorie.  Birkhäuser Verlag. Basel und Stuttgart (1966).
 Title continuation of exponent Canonical name ContinuationOfExponent Date of creation 2013-03-22 17:59:49 Last modified on 2013-03-22 17:59:49 Owner pahio (2872) Last modified by pahio (2872) Numerical id 6 Author pahio (2872) Entry type Definition Classification msc 13A18 Classification msc 12J20 Classification msc 11R99 Classification msc 13F30 Synonym prolongation of exponent Defines induce Defines continuation Defines continuation of the exponent Defines ramification index of the exponent