# determining the continuations of exponent

Task.  Let $\nu_{0}$ be the 3-adic (triadic) (http://planetmath.org/PAdicValuation)  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 (http://planetmath.org/ExponentValuation) 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 (http://planetmath.org/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 (http://planetmath.org/TheoremsOnContinuation), at most two continuations to $\mathbb{Q}(\sqrt{-5})$.  Moreover, the same entry (http://planetmath.org/TheoremsOnContinuation) 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 (http://planetmath.org/DivisibilityInRings) 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 (http://planetmath.org/Product) $\alpha\alpha^{\prime}$ 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 (http://planetmath.org/RingOfExponent), any non-zero element of the field $\mathbb{Q}(\sqrt{-5})$ is uniquely 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:

 $\nu_{1}(\xi)=m,\quad\nu_{2}(\xi)=n.$
Title determining the continuations of exponent DeterminingTheContinuationsOfExponent 2013-03-22 18:00:16 2013-03-22 18:00:16 pahio (2872) pahio (2872) 7 pahio (2872) Example msc 11R99 msc 13A18 msc 12J20 msc 13F30 ExampleOfRingWhichIsNotAUFD