# Subrings Between Integers and Rationals

I'm trying to come up with an example of a ring that is bound strictly between the integers and the rational numbers, but I'm finding this construction very difficult.

If anyone has any advice on how I might approach this problem differently, I'd really appreciate it.

Thank you all for your time!

$$\Bbb Z\left[\frac12\right]:=\left\{f\left(\frac12\right)\;;\;\;f(x)\in\Bbb Z[x]\right\}\;?$$

Your attempt was not as unnatural as you think!

The ring of elements of the form $\dfrac{n}{2^m}$ (verify this is indeed a ring) is an example of what you want, and is the localization of $\mathbb{Z}$ at 2. You'll learn more about localization when you study Commutative Algebra, but it's a very natural and essential construction.

• On the other extreme, all fractions (in their simplest form) having odd denominators form a subring of the rationals and contain all integers. – P Vanchinathan Apr 17 '14 at 8:23

If the task is difficult (which depends on how you approach it), it is certainly not so for lack of examples. The first example I usually give is the ring of rational numbers (that can be written) with an odd denominator. Check that this property is retained under addition, subtraction, multiplication. The next example is the set of rationals that have a finite decimal expansion. Again it is easy to see this is a subring.

Now the first example forbids factors $2$ in denominators, while the second example only allows prime factors $2,5$ in denominators. If you think of it, one could allow any subset of the primes one likes, and forbid the rest. That gives uncountably many examples! Now a last step is to show that these are all possible examples; read the answer by Bill Dubuque for that.

Such subrings $\,R\subseteq \Bbb Q$ are characterized by the subset of $S$ of primes that become invertible in $\,R,\,$ i.e. $\,R = \Bbb Z[1/p\ :\ p\in S]$ is obtained by adjoing the inverses of all primes in $S$. One easily checks that this is a ring since fractions having denominators being products of such primes are closed under addition and multiplication.

Conversely, let $\,R\,$ be such a proper subring $\,\Bbb Z\subsetneq R\subset \Bbb Q,\,$ and let $\,r = a/b\in R\,$ be a noninteger. Wlog $\,(a,b) = 1\,$ so by Bezout $\,aj+bk = 1\,$ for some $\,j,k\in\Bbb Z.\,$ Thus $\,j(a/b) + k = 1/b\in R.\,$ Hence $\,a/b\in R\iff 1/b\in R,\,$ so $R$ is generated by adjoining inverses of integers to $\,\Bbb Z.\,$ However $n$ is invertible iff all its prime factors are invertible, so we can restrict to inverses of primes.

This generalizes to any Bezout domain, i.e. rings like $\,\Bbb Z\,$ where gcds enjoy a Bezout identity. See here for further discussion. Said in the language of commutative algebra, every overring (of fractions) of a Bezout domain is a localization, i.e. is generated by adjoining inverses of elements.

• @Downvoter If something is not clear then please feel welcome to ask questions and I will be glad to elaborate. – Bill Dubuque Apr 18 '14 at 12:32