Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can we find an integer $n$ such that $\mathbb{Z}[\frac{1}{20},\frac{1}{32}]=\mathbb{Z}[\frac{1}{n}]$?

Thanks in advance!

share|improve this question
add comment

4 Answers

up vote 11 down vote accepted

Rapid answer: since $\frac1{32}=\frac1{2^5}=\frac{2\times5^3}{20^3}\in\mathbb Z[\frac1{20}]$, one can take $n=20$.

More generally the following considerations apply.

Let $R$ be a subring of $\mathbf Q$, and consider the set $D$ of positive integers $n$ such that $\frac1n\in R$. Then for all reduced fractions $\frac nd\in R$ one has $\frac1d\in R$ and therefore $d\in D$: since $n$ is relatively prime to $d$ there exists $s\in \mathbb Z$ with $sn\equiv1\pmod d$, and subtracting an integer from $s\times\frac nd$ will give $\frac1d$. Also the set $D$ is closed under taking arbitrary divisors (since we can multiply $\frac1d$ by any integer), and under multiplication (since we can multiply $\frac1d$ and $\frac1{d'}$). It easily follows that $D$ (and therefore $R$) is determined by the subset $P$ of prime numbers in $D$, as $D$ will be the set of all positive integers all of whose prime factors are in $P$. If $P$ is finite, one can write $R=\mathbb Z[\frac1n]$ for any $n$ such that the set of prime divisors of $n$ is $P$.

In your example $P=\{2,5\}$ and any $n$ with exactly those prime divisors will work; $n=10$ is the smallest positive example of such $n$.

share|improve this answer
add comment

HINT: In order to have $\Bbb Z\left[\frac1{20},\frac1{32}\right]\subseteq\Bbb Z\left[\frac1n\right]$, having integers $a$ and $b$ such that $\frac1{20}=\frac{a}n$ and $\frac1{32}=\frac{b}n$ will work nicely. This implies that $n=20a$ and $n=32b$, so $n$ must be a common multiple of $20$ and $32$. On the other hand, there must also be integers $a$ and $b$ such that $\frac1n=\frac{a}{20}+\frac{b}{32}$; rearrange that to get an equation without fractions involving $a,b$, and $n$, and see what it tells you about $n$. Between the two requirements, you should be able easily to find an $n$ that works.

share|improve this answer
1  
With the ususal sense of square brackets ("ring generated by") the "must" in your first sentence is not warranted. See my answer. In fact your answer works for "additive group generated by". –  Marc van Leeuwen Sep 12 '12 at 7:11
    
@Marc: Fair enough. I was taking the easy route and then carelessly over-generalizing. –  Brian M. Scott Sep 12 '12 at 7:16
add comment

Hint $\ $ Let $\rm\,Z = \Bbb Z$ or any Bezout domain, i.e. a domain with gcds which are linear combinations $\rm\,(a,b) = ja+kb,\ j,k\in Z.\:$ Every ring $\rm\,R\,$ between $\rm\,Z\,$ and its fraction field $\rm\,Q\,$ is equal to the subring generated by $\rm\,Z\,$ and inverses of primes of $\rm\,Z\,$ occuring in denominators of reduced fractions in $\rm\,R.$

First $\rm\,a/b \in R\iff 1/b\in R,\,$ since, wlog $\rm\,(a,b)=1,\,$ thus $\rm\,ja+kb=1\,$ for some $\rm\,j,k,\in Z,\,$ hence $\rm\,j(a/b)+k = (aj+kb)/b = 1/b\in R.\:$ Therefore $\rm\,R\,$ is generated by $\rm\,Z\,$ and the inverses of the denominators of the reduced fractions in $\rm\,R.$

Also $\rm\:1/ab\in R\iff 1/a,\,1/b\,\in R,\:$ by $\rm\:a(1/ab) = 1/b.\,$ Thus if, further, $\rm\,Z\,$ is a UFD, then we can uniquely factor the denominators into primes, whittling the generating set down to $\rm\,Z\,$ and the inverses of said primes. Therefore, for your example

$$\rm \Bbb Z\left[\frac{1}{20},\frac{1}{32}\right] =\, \Bbb Z\left[\frac{1}{5\cdot 2^2},\frac{1}{2^5}\right] =\, \Bbb Z\left[\frac{1}2,\frac{1}5\right] =\, \Bbb Z\left[\frac{1}{10}\right]$$

Remark $\ $ Much is known about rings enjoying this and similar properties, e.g. see


Gilmer, Robert; Ohm, Jack. Integral domains with quotient overrings.
Math. Ann. 153 1964 97--103. MR 28#3051 13.15 (16.00)

An integral domain D with field of quotients K is said to have the QR-property if every ring D' between D and K is a ring of quotients of D. The article studies integral domains with the QR-property. Starting with some simple properties of integral domains with the QR-property, the authors prove the following theorem. Let D be an integral domain with the QR-property, and let P be its arbitrary prime ideal. Then D_P is a valuation ring. If A is a finitely generated ideal of D, then A is invertible and some power of A is contained in a principal ideal.

As for the converse, the authors prove that if every valuation ring of K containing D is a ring of quotients of D and if every prime ideal of D is the radical of a principal ideal, then D has the QR-property. As a particular case, it follows that a Noetherian integral domain D has the QR-property if and only if it is a Dedekind domain in which the ideal class group is a torsion group. Some properties of intermediate rings for such rings are observed.

{Reviewer's note: Though the authors added that the result on Noetherian integral domains with the QR-property was obtained independently also by E. D. Davis, the reviewer is afraid that this special case may be known by some other people. Substantially the same result is also contained in the paper of Goldman reviewed below [#3052].}

Reviewed by M. Nagata


Richman, Fred. Generalized quotient rings.
Proc. Amer. Math. Soc. 16 1965 794--799. MR 31#588013.80 (16.00)

Let A be an integral domain with the quotient field K , and consider an intermediate ring A < B < K . The author calls an intermediate ring B a "generalized quotient ring of A" if B is A-flat. (For example, if B = S^{-1} A for some multiplicative subset S of A, then B is A-flat.) One of the main theorems reads as follows. Let A be an integral domain with the quotient field K. For an intermediate ring A < B < K, the following statements are equivalent: (i) B is a generalized quotient ring of A; (ii) (A:Ab)B = B for all b in B; (iii) B_M = A_{M/\A} for all maximal ideals M of B. The author then studies the correspondence of ideals in A and in its generalized quotient ring, and it is also shown that if B is a generalized quotient ring of A, then the integral closure of B is a generalized quotient ring of the integral closure of A. Finally, the author gives a new characterization of Prüfer rings; namely, an integral domain A is a Prüfer ring if and only if every extension of A in its quotient field is flat.

Reviewed by D. S. Rim

share|improve this answer
add comment

Let $n=\operatorname{lcm}(32,20)=160$ then $R=Z[1/32,1/20]\subset Z[1/n]$ since 1/32=5/160, 1/20=8/160. On the other hand $1/160=4\cdot 1/32\cdot 1/20\in R$. So $n=160$ is a valid answer

share|improve this answer
add comment

Your Answer

 
discard

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.