# ring of exponent

Definition. Let $\nu $ be an exponent valuation of the field $K$. The subring

$${\mathcal{O}}_{\nu}:=\{\alpha \in K\mathrm{\vdots}\nu (\alpha )\geqq 0\}$$ |

of $K$ is called the $\nu $. It is, naturally, an integral domain^{}. Its elements are called $\nu $.

Theorem 1. The ring of the exponent $\nu $ of the field $K$ is integrally closed^{} in $K$.

Theorem 2. The ring ${\mathcal{O}}_{\nu}$ only one prime element^{} $\pi $, when one does not regard associated elements as different. Any non-zero element $\alpha $ can be represented uniquely with a $\pi $ in the form

$$\alpha =\epsilon {\pi}^{m},$$ |

where $\epsilon $ is a unit of ${\mathcal{O}}_{\nu}$ and $m=\nu (\alpha )\geqq 0$. This means that $\mathcal{O}$ is a UFD.

Remark 1. The prime elements $\pi $ of the ring ${\mathcal{O}}_{\nu}$ are characterised by the equation $\nu (\pi )=1$ and the units $\epsilon $ the equation $\nu (\epsilon )=0$.

Remark 2. In an algebraically closed field $\mathrm{\Omega}$, there are no exponents (http://planetmath.org/ExponentValuation). In fact, if there were an exponent $\nu $ of $\mathrm{\Omega}$ and if $\pi $ were a prime element of the ring of the exponent, then, since the equation ${x}^{2}-\pi =0$ would have a root (http://planetmath.org/Equation) $\varrho $ in $\mathrm{\Omega}$, we would obtain $2\nu (\varrho )=\nu ({\varrho}^{2})=\nu (\pi )=1$; this is however impossible, because an exponent attains only integer values.

Theorem 3. Let ${\U0001d512}_{1},\mathrm{\dots},{\U0001d512}_{r}$ be the rings of the different exponent valuations ${\nu}_{1},\mathrm{\dots},{\nu}_{r}$ of the field $K$. Then also the intersection

$$\U0001d512:=\bigcap _{i=1}^{r}{\U0001d512}_{i}$$ |

is a subring of $K$ with unique factorisation (http://planetmath.org/UFD). To be precise, any non-zero element $\alpha $ of $\U0001d512$ may be uniquely represented in the form

$$\alpha =\epsilon {\pi}_{1}^{{n}_{1}}\mathrm{\cdots}{\pi}_{r}^{{n}_{r}},$$ |

in which $\epsilon $ is a unit of $\U0001d512$, the integers ${n}_{1},\mathrm{\dots},{n}_{r}$ are nonnegative and
${\pi}_{1},\mathrm{\dots},{\pi}_{r}$ are coprime^{} prime elements of $\U0001d512$ satisfying

$${\nu}_{i}({\pi}_{j})={\delta}_{ij}=\{\begin{array}{cc}\hfill & 1\text{for}i=j,\hfill \\ \hfill & 0\text{for}i\ne j.\hfill \end{array}$$ |

Title | ring of exponent |

Canonical name | RingOfExponent |

Date of creation | 2013-03-22 17:59:43 |

Last modified on | 2013-03-22 17:59:43 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 13 |

Author | pahio (2872) |

Entry type | Definition |

Classification | msc 13F30 |

Classification | msc 13A18 |

Classification | msc 12J20 |

Classification | msc 11R99 |

Related topic | DiscreteValuationRing |

Related topic | ValuationRingOfAField |

Related topic | LocalRing |

Defines | ring of an exponent |

Defines | ring of the exponent |

Defines | integral with respect to an exponent |