# 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 $\U0001d52c$ be the ring of the exponent. Determine the integral closure^{} $\U0001d512$ of $\U0001d52c$ 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 $\frac{{3}^{n}u}{v}$, 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 $\U0001d512$, the rational coefficients of the quadratic equation

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

satisfied by $\alpha $, belong to the ring $\U0001d52c$, whence one has

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

The first of these inequalities implies that ${\nu}_{0}(r)\geqq 0$ since $-2$ is a unit of $\U0001d52c$. 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 $r+s\sqrt{-5}\in \U0001d512$, it’s necessary that $r,s\in \U0001d52c$. The last condition is, apparently, also sufficient. Accordingly, we have obtained the result

$$\U0001d512=\{r+s\sqrt{-5}\mathrm{\vdots}r,s\in \U0001d52c\}.$$ |

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 $\U0001d512$, 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 $\U0001d512$ into factors (http://planetmath.org/DivisibilityInRings) as

$${\pi}_{1}=\alpha \beta $$ |

where $\alpha ={a}_{0}+{a}_{1}\sqrt{-1}$, $\beta ={b}_{0}+{b}_{1}\sqrt{-5}$ (${a}_{i},{b}_{i}\in \U0001d52c$). 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

$${\pi}_{1}{\pi}_{2}=6=\alpha {\alpha}^{\prime}\cdot \beta {\beta}^{\prime}=({a}_{0}^{2}+5{a}_{1}^{2})({b}_{0}^{2}+5{b}_{1}^{2}),$$ |

but according to the entry ring of exponent, the only prime numbers^{} of $\U0001d52c$ are the associates of 3. Now we have factorised the prime number $6$ of $\U0001d52c$ 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 $\U0001d52c$ and hence of $\U0001d512$, too. Thus $\alpha $ and ${\alpha}^{\prime}$ are units of $\U0001d512$, which means that ${\pi}_{1}$ and ${\pi}_{2}$ have only trivial factors. The numbers ${\pi}_{1}$ and ${\pi}_{2}$ themselves are not units, because $\frac{1}{1\pm \sqrt{-5}}=\frac{1}{6}\mp \frac{1}{6}\sqrt{-5}\notin \U0001d512$; ${\pi}_{1}$ and ${\pi}_{2}$ are not associates of each other, since $\frac{{\pi}_{1}}{{\pi}_{2}}=1+\frac{1}{3}\sqrt{-5}\notin \U0001d512$. So ${\pi}_{1}$ and ${\pi}_{2}$ are non-associated prime elements of $\U0001d512$. 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 $\U0001d512$ and $m,n$ integers. The both continuations ${\nu}_{1}$ and ${\nu}_{2}$ of the triadic exponent ${\nu}_{0}$ are then determined as follows:

$${\nu}_{1}(\xi )=m,{\nu}_{2}(\xi )=n.$$ |

Title | determining the continuations of exponent |
---|---|

Canonical name | DeterminingTheContinuationsOfExponent |

Date of creation | 2013-03-22 18:00:16 |

Last modified on | 2013-03-22 18:00:16 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 7 |

Author | pahio (2872) |

Entry type | Example |

Classification | msc 11R99 |

Classification | msc 13A18 |

Classification | msc 12J20 |

Classification | msc 13F30 |

Related topic | ExampleOfRingWhichIsNotAUFD |