Proving rationals of the form $2^{-n}m$ are a Euclidean ring

I am trying to come up with a proof that numbers of the form $$2^{-n}m$$, where $$~m \in \Bbb Z,~n \in \Bbb Z^+$$ are a Euclidean ring/domain. The division property is easily verifiable (although the quotient and remainder are not unique), but I am having trouble coming up with an appropriate norm for the ring. One should, besides satisfying the axioms, in my opinion have the following features:

1. Since this ring contains $$\Bbb Z$$ as a subring, the norm when restricted to it should become the absolute value,
2. It should produce the same value for any representative of a given equivalence class of rationals (i.e. the same value on $$\frac 1 2$$ and $$\frac 2 4$$),
3. It should produce a nonnegative integral result on the purely fractional elements of the ring (ones of the form $$2^{-n}$$).

I started off with simply trying $$|m|$$ or $$|m+n|$$, but neither clearly satisfies the 2nd property. I then tried something a little more complicated, namely $$|2^{log_2m-n}|$$, which seems to satisfy the 1st and 2nd properties, but not the 3rd, as I now realise. I am now trying to find a way to make it evaluate to $$|2^n|$$, when $$m=1$$, and the previous formula otherwise. Also, having considered this norm, I am not even sure that it satisfies either of the required axioms anyway, namely that $$N(ab)>N(a)$$ and that given a division $$a=qb+r$$, $$N(r).

Am I on the right track or do I err in trying to build a norm like this? Is there a much simpler one out there that satisfies the axioms and has the aforementioned features?

• Hint: you can reformulate your ring as $2^nm$ where $m \in \mathbb Z$ is odd and $n \in \mathbb Z$. Then, |m| or similar should do it. – lisyarus Dec 2 at 23:17
• Hmm, that does make things much easier than before, clearly the odd part doesn't change when the numerator and denominator are scaled by $2$. But what about restriction to $\Bbb Z$, wouldn't, for example, $N(6)=3$ in this case, as opposed to $6$ itself? – V.Ch. Dec 2 at 23:34
• – CopyPasteIt Dec 3 at 0:10

1 Answer

If $$R$$ is an euclidean domain with euclidean function $$N$$ then $$R[a^{-1}]$$ is an euclidean domain with euclidean function $$N(\frac{b}{a^k}) = \min \{ N(\frac{b}{a^m}),\frac{b}{a^m}\in R\}$$

Here $$R[a^{-1}]=\Bbb{Z}[2^{-1}]$$ and $$N(2^k(2m+1)) =N(\frac{2m+1}{2^k}) = |2m+1| = \# \Bbb{Z}[2^{-1}]/(2^k(2m+1))$$

• Please refer to my comment above, doesn't this definition of norm conflict with the absolute value when restricted to $\Bbb Z$, i.g. $N(6)=3$, as opposed to $6$, as when viewed as an element of $\Bbb Z$? – V.Ch. Dec 3 at 18:57