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 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

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

are in force, 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 $k$).