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

Question

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:

Since this ring contains $\Bbb Z$ as a subring, the norm when restricted to it should become the absolute value,
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 $\frac 2 4$),
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 $a=qb+r$, $N(r)<N(b)$.

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 02 '19 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? — St. Barth, Dec 02 '19 at 23:34

score 1 · Accepted Answer · answered Dec 03 '19 at 05:55

1

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

answered Dec 03 '19 at 05:55

reuns

79,880

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 $\Bbb Z$? – St. Barth Dec 03 '19 at 18:57
You should think to the euclidean function as a map on principal ideals, not on elements. – reuns Jan 16 '20 at 14:01

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

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