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Can anyone suggest a good place to read up on the number theoretic properties of and techniques for $\mathbb{Z}[1/p]$, (that is, rational numbers with only powers of a prime $p$ in the denominator)?

I find myself struggling to answer some of the more basic questions about this ring, especially whether or not it is a Euclidean domain, and if so, what the associated Euclidean function/algorithm is.

Thanks in advance!

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Pretty much any arithmetic property of $\mathbf{Z}[1/p]$ is readily deduced from the corresponding property of $\mathbf{Z}$. For instance, a Euclidean function is $\phi(p^r a)=|a|$ for $a\in\mathbb{Z}$ not divisible by $p$. –  Robin Chapman Nov 5 '10 at 15:05
    
@Robin: I don't think this seems to yeild a unique division algorithm: $a=bq+r$, with $q$ and $r$ unique such that $\phi(r)<\phi(b)$. Consider $11/3$ and $7/9 \in \mathbb{Z}[1/3]$. Then $$11/3 = 7/9*3+4/3$$ $$11/3 = 7/9*4+5/9$$ $$11/3 = 7/9*5-2/9$$ $$11/3 = 7/9*6-1$$ and in all cases $\phi(r)<\phi(b)$. –  Aeryk Nov 5 '10 at 15:43
    
See e.g. [About Euclidean rings][1] prop.7, sorry I can't find a pdf anywhere on the web. Now if someone knows what the Galois group of the maximal unramified extension is, then I'm interested in a reference. [1]: ams.org/mathscinet/search/… –  K.J. Moi Nov 5 '10 at 16:03
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@Aeryk: the divison algorithm in $\mathbb{Z}$ is also not unique, for example $1 = 3 \cdot 0 + 1 = 3 \cdot 1 - 2$. It only is unique if you use the convention that you take the remainder $\ge 0$. Besides that, the unit group of $\mathbb{Z}[1/p]$ is much larger than the one of $\mathbb{Z}$, which yields further choice. –  felix Nov 5 '10 at 16:54
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Z[1/p] is just the integers where you made p invertible, so: the unit group changes from {+/-1} to {+/-p^k : k in Z} and the primes drop by 1 since p is no longer prime. Other prime numbers stay prime. Learn what a localization of a ring is and the link between prime ideals in a ring and its localization. –  KConrad Nov 5 '10 at 19:15

1 Answer 1

The general fact here is that any localization of a Euclidean domain is again a Euclidean domain. I will restrict myself to the case where the Euclidean norm on $R$ is multiplicative, i.e., satisfies $|xy| = |x| |y|$ (as does the absolute value on $\mathbb{Z}$, of course), and in this case I will define an explicit Euclidean norm on the localized ring in terms of the given norm and the (let's say saturated, WLOG) multiplicative subset $S$.

For a ring $R$, I write $R^{\bullet}$ for $R \setminus \{0\}$.

Since $R$ is Euclidean, it is a UFD, so to give a function $|\ |: R \setminus \{0\} \rightarrow \mathbb{Z}^{> 0}$ such that $|1| = 1$, $|xy| = |x| |y|$ and $x \in R^{\times} \iff |x| = 1$, it is enough to send every principal prime ideal $(\pi)$ to some integer $n_{\pi} > 1$. (This holds because the multiplicative monoid of principal nonzero $R$-ideals is the free commutative monoid on the principal prime ideals.) Then the norm of an arbitrary nonzero element of $R$ is defined by the uniqueness of factorization into principal prime ideals.

The multiplicative group $R_S^{\bullet}$ of a localization $R_S$ is the free commutative monoid on the principal prime ideals $(\pi)$ such that $(\pi) \cap S = \emptyset$. One can view this naturally as a submonoid of $R^{\bullet}$ and therefore define an induced norm $| \ |_S$. In other words, if $x \in R^{\bullet}$, write $x = s_x x'$ where $s_x \in S$ and $x'$ is prime to $S$. then, for any $s \in S$,

$|\frac{x}{s}|_S = |x|_S = |s_x x'|_S = |x'|_S = |x'|$.

Note that for all $x \in R$, we have $|x|_S \leq |x|$.

Let us now show that if $R$ is Euclidean under $| \ |$, $R_S$ is Euclidean under $|\ |_S$: for $A \in R_S$ and $B \in R_S^{\bullet}$, we must find $Q \in R_S$ such that $|A-QB|_S < |B|_S$. There exist $a,b \in R$ and $s \in S$ such that $A = \frac{a}{s}$, $B = \frac{b}{s}$. Then, since $s \in R_S^{\times}$, $|a-Qb|_S = |\frac{a}{s} - Q \frac{b}{s}|_S = |A - QB|_S$ and $|b|_S = |\frac{b}{s}|_S = |B|_S$, so without loss of generality we may take $s = 1$.

As above, write $b = s_b b'$, and choose $q \in R$ such that $|a-qb'| < |b'|$. Put $Q = \frac{q}{s_b}$. Then

$|a - Q b|_S = |a- \frac{q}{s_b} b|_S = |a-q b'|_S \leq |a-qb'| < |b'| = |b'|_S = |b|_S.$

For your particular question $R = \mathbb{Z}$, the Euclidean norm is the usual absolute value, and $S = \{2^a \ | \ a \in \mathbb{Z}^+\}$.

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