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Homering of exponent

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# ring of exponent

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

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

of $K$ is called the ring of the exponent $\nu$. It is, naturally, an integral domain. Its elements are called integral with respect to $\nu$.

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

Theorem 2. The ring $\mathcal{O}_{\nu}$ contains 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 fixed $\pi$ in the form

$\alpha\;=\;\varepsilon\pi^{m},$ |

where $\varepsilon$ 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 $\varepsilon$ the equation $\nu(\varepsilon)=0$.

Remark 2. In an algebraically closed field $\Omega$, there are no exponents. In fact, if there were an exponent $\nu$ of $\Omega$ 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 $\varrho$ in $\Omega$, 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 $\mathfrak{O}_{1},\,\ldots,\,\mathfrak{O}_{r}$ be the rings of the different exponent valuations $\nu_{1},\,\ldots,\,\nu_{r}$ of the field $K$. Then also the intersection

$\mathfrak{O}\;:=\;\bigcap_{{i=1}}^{r}\mathfrak{O}_{i}$ |

is a subring of $K$ with unique factorisation. To be precise, any non-zero element $\alpha$ of $\mathfrak{O}$ may be uniquely represented in the form

$\alpha\;=\;\varepsilon\pi_{1}^{{n_{1}}}\cdots\pi_{r}^{{n_{r}}},$ |

in which $\varepsilon$ is a unit of $\mathfrak{O}$, the integers $n_{1},\,\ldots,\,n_{r}$ are nonnegative and $\pi_{1},\,\ldots,\,\pi_{r}$ are fixed coprime prime elements of $\mathfrak{O}$ satisfying

$\nu_{i}(\pi_{j})\;=\;\delta_{{ij}}\;=\;\begin{cases}&1\;\;\mbox{for }\,i=j,\\ &0\;\;\mbox{for }\,i\neq j.\end{cases}$ |

## Mathematics Subject Classification

13F30*no label found*13A18

*no label found*12J20

*no label found*11R99

*no label found*

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