determining the continuations of exponent

Task.  Let ν0 be the 3-adic (triadic) (  exponent valuation of the field of the rational numbers and let 𝔬 be the ring of the exponent.  Determine the integral closureMathworldPlanetmath 𝔒 of 𝔬 in the extension fieldMathworldPlanetmath (-5) and the continuations of ν0 to this field.

The triadic exponent ( of at any non-zero rational number 3nuv, where u and v are integers not divisible by 3, is defined as


Any number of the quadratic fieldMathworldPlanetmath (-5) is of the form


with r and s rational numbers.  When  α=r+s-5  belongs to 𝔒, the rational coefficients of the quadratic equation


satisfied by α, belong to the ring 𝔬, whence one has


The first of these inequalities implies that  ν0(r)0  since -2 is a unit of 𝔬.  As for s, if one had  ν0(s)<0,  then  ν0(5s2)=2ν0(s)<0,  and therefore one had


Thus we have to have  ν0(s)0,  too.  So we have seen that for  r+s-5𝔒,  it’s necessary that  r,s𝔬.  The last condition is, apparently, also sufficient.  Accordingly, we have obtained the result


Since the degree ( of the field extension (-5)/ is 2, the exponent ν0 has, by the theorem in the parent entry (, at most two continuations to (-5).  Moreover, the same entry ( implies that the intersectionDlmfMathworldPlanetmath of the rings of those continuations coincides with 𝔒, whose non-associated ( prime elementsMathworldPlanetmath determine the continuations in question.

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


for such prime elements.

Suppose that π1 splits in 𝔒 into factors ( as


where  α=a0+a1-1,  β=b0+b1-5  (ai,bi𝔬).  Then also


where  α=a0-a1-1,  β=b0-b1-5.  We perceive that


but according to the entry ring of exponent, the only prime numbersMathworldPlanetmath of 𝔬 are the associates of 3.  Now we have factorised the prime number 6 of 𝔬 into a productPlanetmathPlanetmath of two factors ( αα and ββ, and consequently, e.g. αα is a unit of 𝔬 and hence of 𝔒, too.  Thus α and α are units of 𝔒, which means that π1 and π2 have only trivial factors.  The numbers π1 and π2 themselves are not units, because  11±-5=1616-5𝔒; π1 and π2 are not associates of each other, since  π1π2=1+13-5𝔒.  So π1 and π2 are non-associated prime elements of 𝔒.  This ring has no other prime elements non-associated with both π1 and π2, because otherwise ν0 would have more than two continuations.

According to the entry ring of exponent (, any non-zero element of the field (-5) is uniquely in the form


with ε a unit of 𝔒 and m,n integers.  The both continuations ν1 and ν2 of the triadic exponent ν0 are then determined as follows:

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