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 $\displaystyle\frac{3^nu}{v}$ , where $u$ and $v$ are integers not divisible by 3, is defined as $$\nu_0\left(\frac{3^nu}{v}\right) \,:=\, n.$$ 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 $$\pi_2 = \alpha'\beta'$$ where $\alpha' = a_0-a_1\sqrt{-1}$ , $\beta' = b_0-b_1\sqrt{-5}$ . We perceive that $$\pi_1\pi_2 = 6 = \alpha\alpha' \cdot \beta\beta' = (a_0^2+5a_1^2)(b_0^2+5b_1^2),$$ 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 $\alpha\alpha'$ and $\beta\beta'$ , and consequently, e.g. $\alpha\alpha'$ is a unit of $\mathfrak{o}$ and hence of $\mathfrak{O}$ , too. Thus $\alpha$ and $\alpha'$ 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: $$\nu_1(\xi) = m, \quad \nu_2(\xi) = n.$$