Re: JSH: The "Published" paper he dosen't what you to know about.


Tim Peters wrote:
[W. Dale Hall]
...
Just so you cannot claim not to have seen the reason your
"missing element" argument does not explain the facts, here
it is:

2 and 4 are also coprime in the ring of even integers.

[jstevh@xxxxxxx]
But 2*2 = 4.

So 4 has 2 as a factor.
[...]

The numbers 2 and 4 are *coprime* in the ring of even integers.
The numbers 2 and 8 are *coprime* in the ring of even integers.

The number 2 is *COPRIME TO ITSELF* in the ring of even integers.

The issue that has yer panties in a twist regarding this ring,
namely *that 2 and 6 are coprime in the ring of even integers*,
has nothing to do with the existence or absence of a common factor.

If you knew anything about the definitions, you'd have realized that!

[Tim Peters]
Like that's going to get across ;-)

[Proginoskes]
You can't be subtle with JSH.

Sure you can. It just doesn't do any good <0.5 wink> -- and the reason
that's only half a wink is that being obvious doesn't do any good either :-(

Perhaps I'd better say point-blank what (I hope) is true:

Brave man; wise move.

THE NUMBER 2 HAS NO FACTORS IN THE RING OF EVEN INTEGERS.

No, 2 is not a factor of itself, because there is no _even integer_ N
such that 2*N = 2.

I doubt anyone will disagree with that, but what we should get clear on is
that /in the context of general rings/, "coprime" isn't defined in terms of
divisibility at all (although in some kinds of rings they're intimately
connected).

So 2 cannot share any factors with any other numbers, even with itself,
because it doesn't have any to start with. Thus we have the weird
situation where 2 is coprime with itself.

This is really quite a mess, because the "ring of evens" doesn't have a
multiplicative identity. I'm not sure there's universal agreement about
what "coprime" means in such rings, or even that it makes /sense/ to talk
about coprimeness in such rings. I'll assume ring multiplication is
commutative (so I can talk about "ideals" instead of endlessly qualifying
whether I mean "left ideals" or "right ideals").

In general, in the context of rings, one talks about whether /ideals/ are
coprime, not about whether individual elements are coprime. I'm aware of
two definitions of coprimeness then:

Well, that's two more than I know. 8-) I don't recall having studied
rings without identities in my abstract algebra course. Actually, after
my first aa course, I thought that Galois theory was a "communist
plot"; I took it again in grad school from someone who gave it a more
combinatorial approach, and I don't feel so bad about it now. Of
course, now my field of research is combinatorics/graph theory, and I
use very little algebra in my research. (Technically, in my research,
if I ever got any time to _do_ research.)

Nevertheless, it's interesting to see what's going on in other fields.

--- Christopher Heckman

1. Two ideals I and J of a ring R are said to be coprime iff I+J = R.
When applied to elements x and y, the question is whether the sum of
the ideals generated by x and y is R: does xR + yR = R? If the ring
is Z, turns out this corresponds exactly to saying that x and y are
coprime iff gcd(x, y) = 1: that's true iff xZ + yZ = Z.

2. Two ideals I and J of a ring R are said to be coprime iff I and J
aren't both contained in some prime ideal of R. Again when applied
to elements, the question is whether xR and yR are both contained in
some prime ideal of R.

In a ring with identity, it's possible that those are equivalent, but I only
have experience working with #1 (if I ever studied rings without identity,
it was in a footnote). Under #1, /no/ two elements in R=2Z ("the ring of
evens") are coprime. For example, 2 is not coprime to 2 under #1 because

2R = {i*2: i in R} =
{..., -6, -4, -2, 0, 2, 4, 6, ...} * 2 =
{..., -12, -8, -4, 0, 4, 8, 12, ...} = 4Z

so 2R + 2R = 4Z + 4Z = 4Z =/= 2Z = R. Similarly 2 isn't coprime to 6 either
under #1, since 2R + 6R = 4Z + 12Z = 4Z =/= 2Z = R.

I assume (but don't know) that Dale is using #2, and I expect that's the
"grown up" definition ... luckily for me ;-), Arturo Magidin explained how
that applies to 2Z very clearly 3 years(!) ago:

From: magi...@xxxxxxxxxxxxxxxxx (Arturo Magidin)
Subject: Re: Scientific method, mathematics, ring of algebraic integer
Date: Tue, 26 Aug 2003 15:52:07 +0000 (UTC)
Newsgroups: sci.math
Message-ID: bifvn7$h3p$1@xxxxxxxxxxxxxxxxxx

By that argument, 2 is indeed coprime to everything in R=2Z (the only ideals
of R containing 2R are R itself and 2R itself, neither of which is a prime
ideal of 2Z). More, 4 is coprime to 8 in 2Z, "despite that 8/2 = 4 and 4/2
= 2 are both in 2Z so they share a non-unit common factor", because again
there are no prime ideals of 2Z containing 4R = 8Z or 8R = 16Z.

Hmm ... what about 6 and 12? I think they're not coprime in 2Z: 6R and 12R
are both contained in the ideal 6Z of R (note that this is not generated by
3, since 3 isn't an element of 2Z; and it's not generated by 6, since 6R
consists of multiples of 12; nevertheless, 6Z is a subset of R that
satisfies the definition of an ideal in R). 6Z is a prime ideal of 2Z,
since i*j in 6Z for i and j in R implies (2*i')*(2*j') = 6*k for some
integers i', j' and k, so 3 divides at least one of i' and j', so at least
one of i=2*i' and j=2*j' is a multiple of 6: i*j in 6Z implies i in 6Z or j
in 6Z. That, along with that 6Z is a proper subset of R, means that 6Z is a
prime ideal of R.

Anyway, since all the rings James actually cares about are unital, his 2Z
example is supremely irrlevant to any "larger" issues here. William Hughes
recently said he's using "coprime means no non-unit common divisor", and
that sounds right (seems closest what I guess JSH might intend). But on
that count 2Z is also a terrible example, since as you noted at the start 2
is coprime to everything in 2Z by that reasoning too.

.

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