MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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]:… – K.J. Moi Nov 5 '10 at 16:03
@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
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

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 $S = \{2^a \ | \ a \in \mathbb{Z}^+\}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.