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

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

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

then  $\nu (a{p}^{-q})=m-qe=r$,  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 $k\setminus \{0\}$ are integers.  Because  ${\nu }_0(p)=1$  and  ${\nu }_0(p^l)=l$,  the function attains in $k$ every integer value.  Also the conditions

$${\nu }_0(ab)={\nu }_0(a)+{\nu }_0(b),{\nu }_0(a+b)\geqq \min \{{\nu }_0(a),{\nu }_0(b)\}$$

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 $k$).