Re: Rings of rationals

On Sun, 17 Jun 2007 00:28:05 -0400, "Stephen J. Herschkorn"
<sjherschko@xxxxxxxxxxxx> wrote:

In a previous thread, I posted the following conjecture:

Every ring of rationals is generated by {1 / p: p in A} for some set
A of primes.


quasi posted a follow-up proof for *unitary* rings. However, my
intended definition of ring does not require a unit element. (It is
difficult for me to quote quasi's post due to lost posts on my server; I
also posted a message about this loss.)

Here's a slightly edited repost of my prior post proving the following
characterization of unitary subrings of Q.

Proposition:

Let R be a unitary subring of Q. Then there exists a set A of primes,
possibly empty, such that R=Z[D] where D={1/p | p in A}.

proof:

Let R be a unitary subring of Q.

Note that x in R implies n*x in R for all integers n. This is true in
any ring (unitary of not).

Since R is unitary, R contains Z.

If R=Z we are done (use A=empty set).

Next, assume R contains Z as a proper subset.

Let A be the set of all prime factors of all denominators of reduced
nonzero elements of R and let D={1/p | p is in A}.

Claim R=Z[D].

Suppose r is in R, r nonzero. Write r=x/y in reduced form and with
y>0. Then y must be a product of primes from A, hence 1/y is in Z[D].
Since 1/y is in Z[D], x*(1/y) is in Z[D], hence r is in Z[D]. It
follows that R is a subset of Z[D].

For the reverse inclusion, it suffices to show D is a subset of R. Let
p be an element of A. Then there must be some nonzero element of R
which, when reduced, has a denominator with a factor of p. Let x/y be
such an element of R. Then y=p*z for some nonzero integer z. Since x/y
is in R, z*(x/y) is in R. Reducing z*(x/y) to lowest terms, it follows
that w/p is in R for some integer w with (w,p)=1. Since (w,p)=1 we can
write wf+pg=1 for some integers f,g. Then (w/p)*f + g = 1/p, hence 1/p
is in R. Thus, since p was an arbitrary element of A, 1/p is in R for
all p in A. It follows that D is a subset of R, hence Z[D] is a subset
of R.

This completes the proof.

quasi
.

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