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Homedetermining the continuations of exponent

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# determining the continuations of exponent

Task. Let $\nu_{0}$ be the 3-adic (triadic) exponent valuation of the field $\mathbb{Q}$ of the rational numbers and let $\mathfrak{o}$ be the ring of the exponent. Determine the integral closure $\mathfrak{O}$ of $\mathfrak{o}$ in the extension field $\mathbb{Q}(\sqrt{-5})$ and the continuations of $\nu_{0}$ to this field.

The triadic exponent of $\mathbb{Q}$ at any non-zero rational number $\displaystyle\frac{3^{n}u}{v}$, where $u$ and $v$ are integers not divisible by 3, is defined as

$\nu_{0}\left(\frac{3^{n}u}{v}\right)\,:=\,n.$ |

Any number of the quadratic field $\mathbb{Q}(\sqrt{-5})$ is of the form

$r+s\sqrt{-5}$ |

with $r$ and $s$ rational numbers. When $\alpha=r+s\sqrt{-5}$ belongs to $\mathfrak{O}$, the rational coefficients of the quadratic equation

$x^{2}-2rx+(r^{2}+5s^{2})=0,$ |

satisfied by $\alpha$, belong to the ring $\mathfrak{o}$, whence one has

$\nu_{0}(-2r)\geqq 0,\quad\nu_{0}(r^{2}+5c^{2})\geqq 0.$ |

The first of these inequalities implies that $\nu_{0}(r)\geqq 0$ since $-2$ is a unit of $\mathfrak{o}$. As for $s$, if one had $\nu_{0}(s)<0$, then $\nu_{0}(5s^{2})=2\nu_{0}(s)<0$, and therefore one had

$\nu_{0}(r^{2}+5s^{2})\;=\;\min\{\nu_{0}(r^{2}),\,\nu_{0}(5s^{2})\}<0.$ |

Thus we have to have $\nu_{0}(s)\geqq 0$, too. So we have seen that for $r+s\sqrt{-5}\in\mathfrak{O}$, it’s necessary that $r,\,s\in\mathfrak{o}$. The last condition is, apparently, also sufficient. Accordingly, we have obtained the result

$\mathfrak{O}=\{r\!+\!s\sqrt{-5}\,\vdots\;\;\;r,\,s\in\mathfrak{o}\}.$ |

Since the degree of the field extension $\mathbb{Q}(\sqrt{-5})/\mathbb{Q}$ is 2, the exponent $\nu_{0}$ has, by the theorem in the parent entry, at most two continuations to $\mathbb{Q}(\sqrt{-5})$. Moreover, the same entry implies that the intersection of the rings of those continuations coincides with $\mathfrak{O}$, whose non-associated prime elements determine the continuations in question.

We will show that there are exactly two of those continuations and that one may choose e.g. the conjugate numbers

$\pi_{1}:=1+\sqrt{-5},\quad\pi_{2}:=1-\sqrt{-5}$ |

for such prime elements.

Suppose that $\pi_{1}$ splits in $\mathfrak{O}$ into factors as

$\pi_{1}=\alpha\beta$ |

where $\alpha=a_{0}+a_{1}\sqrt{-1}$, $\beta=b_{0}+b_{1}\sqrt{-5}$ ($a_{i},\,b_{i}\in\mathfrak{o}$). Then also

$\pi_{2}=\alpha^{{\prime}}\beta^{{\prime}}$ |

where $\alpha^{{\prime}}=a_{0}-a_{1}\sqrt{-1}$, $\beta^{{\prime}}=b_{0}-b_{1}\sqrt{-5}$. We perceive that

$\pi_{1}\pi_{2}=6=\alpha\alpha^{{\prime}}\cdot\beta\beta^{{\prime}}=(a_{0}^{2}+% 5a_{1}^{2})(b_{0}^{2}+5b_{1}^{2}),$ |

but according to the entry ring of exponent, the only prime numbers of $\mathfrak{o}$ are the associates of 3. Now we have factorised the prime number $6$ of $\mathfrak{o}$ into a product of two factors $\alpha\alpha^{{\prime}}$ and $\beta\beta^{{\prime}}$, and consequently, e.g. $\alpha\alpha^{{\prime}}$ is a unit of $\mathfrak{o}$ and hence of $\mathfrak{O}$, too. Thus $\alpha$ and $\alpha^{{\prime}}$ are units of $\mathfrak{O}$, which means that $\pi_{1}$ and $\pi_{2}$ have only trivial factors. The numbers $\pi_{1}$ and $\pi_{2}$ themselves are not units, because $\frac{1}{1\pm\sqrt{-5}}=\frac{1}{6}\mp\frac{1}{6}\sqrt{-5}\not\in\mathfrak{O}$; $\pi_{1}$ and $\pi_{2}$ are not associates of each other, since $\frac{\pi_{1}}{\pi_{2}}=1+\frac{1}{3}\sqrt{-5}\not\in\mathfrak{O}$. So $\pi_{1}$ and $\pi_{2}$ are non-associated prime elements of $\mathfrak{O}$. This ring has no other prime elements non-associated with both $\pi_{1}$ and $\pi_{2}$, because otherwise $\nu_{0}$ would have more than two continuations.

According to the entry ring of exponent, any non-zero element of the field $\mathbb{Q}(\sqrt{-5})$ is uniquely expressible in the form

$\xi=\varepsilon\pi_{1}^{m}\pi_{2}^{n},$ |

with $\varepsilon$ a unit of $\mathfrak{O}$ and $m,\,n$ integers. The both continuations $\nu_{1}$ and $\nu_{2}$ of the triadic exponent $\nu_{0}$ are then determined as follows:

$\nu_{1}(\xi)=m,\quad\nu_{2}(\xi)=n.$ |

## Mathematics Subject Classification

11R99*no label found*13A18

*no label found*12J20

*no label found*13F30

*no label found*

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