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 $𝔬$ be the ring of the exponent.  Determine the integral closure $𝔒$ of $𝔬$ 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 $𝔒$, the rational coefficients of the quadratic equation

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

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

$${\nu }_0(-2r)\geqq 0,{\nu }_0(r^2+5c^2)\geqq 0.$$

The first of these inequalities implies that  ${\nu }_0(r)\geqq 0$  since $-2$ is a unit of $𝔬$.  As for $s$, if one had  ,  then  ,  and therefore one had

Thus we have to have  ${\nu }_0(s)\geqq 0$,  too.  So we have seen that for  $r+s\sqrt{-5}\in 𝔒$,  it’s necessary that  $r,s\in 𝔬$.  The last condition is, apparently, also sufficient.  Accordingly, we have obtained the result

$$𝔒=\{r+s\sqrt{-5}\mathrm{⋮}r,s\in 𝔬\}.$$

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 $𝔒$, whose non-associated (http://planetmath.org/Associate) 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

$${\pi }_1:=1+\sqrt{-5},{\pi }_2:=1-\sqrt{-5}$$

for such prime elements.

Suppose that ${\pi }_1$ splits in $𝔒$ into factors (http://planetmath.org/DivisibilityInRings) as

$${\pi }_1=\alpha \beta$$

where  $\alpha =a_0+a_1\sqrt{-1}$,  $\beta =b_0+b_1\sqrt{-5}$  ($a_i,b_i\in 𝔬$).  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 $𝔬$ are the associates of 3.  Now we have factorised the prime number $6$ of $𝔬$ 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 $𝔬$ and hence of $𝔒$, too.  Thus $\alpha$ and ${\alpha }^{\prime }$ are units of $𝔒$, which means that ${\pi }_1$ and ${\pi }_2$ have only trivial factors.  The numbers ${\pi }_1$ and ${\pi }_2$ themselves are not units, because  $\frac{1}{1\pm \sqrt{-5}}=\frac{1}{6}\mp \frac{1}{6}\sqrt{-5}\notin 𝔒$; ${\pi }_1$ and ${\pi }_2$ are not associates of each other, since  $\frac{{\pi }_1}{{\pi }_2}=1+\frac{1}{3}\sqrt{-5}\notin 𝔒$.  So ${\pi }_1$ and ${\pi }_2$ are non-associated prime elements of $𝔒$.  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

$$\xi =\epsilon {\pi }_1^m{\pi }_2^n,$$

with $\epsilon$ a unit of $𝔒$ and $m,n$ integers.  The both continuations ${\nu }_1$ and ${\nu }_2$ of the triadic exponent ${\nu }_0$ are then determined as follows:

$${\nu }_1(\xi )=m,{\nu }_2(\xi )=n.$$