Re: Coprimeness - I think I'm confused, but I'm not sure

In article <1112387619.8214cb7fca92c72ca9857edb4554270a@teranews>,
Matt Gutting <matthewdba@xxxxxxxxxxx> wrote:
>In light of some of the comments in the JSH threads involving
>coprimeness and Z[1/2]:
>
>If the definition of coprimeness is something like:
>
>To say that 'p and q are coprime in a ring' is to say
>'there exist a,b in the ring with ap + bq = 1'

This is true in rings with 1.

>then wouldn't 2 and 4 (or any powers of 2) be coprime
>in Z[1/2]? Or, more strongly, are any 2 (non-zero) rationals
>coprime in Q?

Yes, because they are units and a unit is coprime to anything.

>Or am I missing something?

Two elements are coprime if and only if there is no prime ideal
that contains both. By definition, the total ideal is not a prime
ideal.

In rings with 1, maximal ideals are prime; and (assuming the axiom of
choice, at least) every proper ideal is contained in a maximal
ideal. So for rings with 1, "there is no prime ideal containing both
elements" is equivalent to "there is no maximal ideal containing both
elements".

If u is a unit, then any ideal that contains u must contain all of R;
so there is no maximal ideal that contains u; so there is certainly no
maximal ideal that contains u and <anything else>.

Since 2^n is a unit in Z[1/2], 2^n is coprime to everything in
Z[1/2]. Since a nonzero rational is a unit in Q, a nonzero rational is
coprime to everything in Q. In fact, the only pair of rationals which
are not coprime are 0 and itself.

To get from "no maximal ideal contains both" to the expression as a
linear combination, just note that the ideal (p,q) is certainly an
ideal that contains both p and q. Since every proper ideal is
contained in a maximal ideal, and no maximal ideal contains both p and
q, it follows that (p,q) = R. That means that every element of R (in
particular 1) lies in (p,q). The elements of (p,q) are exactly all
elements of the form px + qy for some x and y in R. So there exist
a and b in R such that 1 = ap + bq. So in fact p and q are coprime if
and only if (p,q)=1.

In ->some<- rings, being coprime is equivalent to "having no common
divisors"; e.g., in Z, and in the ring of all algebraic integers. In
some rings, it is not. For example, in Z[x], x and 2 have no common
divisors, but (x,2) is not the total ideal. Likewise, in Z[sqrt(-5)],
2 and 1+sqrt(-5) have no common divisors, but are not coprime.


--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
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Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

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