# 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

$$(1)\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}{\nu}_{0}(x):=\{\begin{array}{cc}\hfill & \mathrm{\infty}\hspace{1em}\text{when}x=0,\hfill \\ \hfill & \frac{\nu (x)}{e}\text{when}x\ne 0,\hfill \end{array}$$ |

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 ${\mathcal{O}}_{\nu}$, 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 $k\setminus \{0\}$. 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

$${\nu}_{0}(ab)={\nu}_{0}(a)+{\nu}_{0}(b),{\nu}_{0}(a+b)\geqq \mathrm{min}\{{\nu}_{0}(a),{\nu}_{0}(b)\}$$ |

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 |