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# 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 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

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

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 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 $k$).

# References

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

## Mathematics Subject Classification

13A18*no label found*12J20

*no label found*11R99

*no label found*13F30

*no label found*

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