On Wednesday, September 2, 2009 at 8:38:12 PM UTC-5, Bill Dubuque wrote:
> juandiego <sttsc...@tesco.net> wrote:
> >Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> >>juandiego <sttsc...@tesco.net> wrote:
> >>> On 2 Sep, 15:42, Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> >>>> juandiego <sttscitr...@tesco.net> wrote:
> >>>>>On 31 Aug, 15:13, Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> >>>>>>juandiego <sttscitr...@tesco.net> wrote:
> >>>>>>>
> >>>>>>> If you assume that there is a last prime, i.e the primes
> >>>>>>> are finite in number, then w+1 is not divisble by ANY prime
> >>>>>>> (you have exhausted all the possible prime divisors of w+1)
> >>>>>>> Therefore, w+1 has only the trivial divisors 1, w+1
> >>>>>>> This means that w+1 is itself prime - a contradiction.
> >>>>>>> On the other hand, w+1 is >1 and has no prime divisors
> >>>>>>> This implies w+1 is a unit, but there are no units >1
> >>>>>>> - a contradiction. [...]
> >>>>>>> If there were units greater than 1, w+1 could have
> >>>>>>> no prime divisors and the proof would not go through.
> >>>>>>
> >>>>>> Not true.
> >>>>>> Thus, from this more general perspective, excluding the
> >>>>>> possibility that w+1 is a unit is crucial.
> >>>>>
> >>>>> You seem to be contradicting yourself.
> >>>>
> >>>> No. First let's recall what I actually wrote, not the
> >>>> greatly condensed 3-line version that you quoted above.
> >>>>
> >>>> Not true. I had the pleasure of discovering as a teenager
> >>>> that Euclid's proof generalizes to any infinite ring with
> >>>> fewer units than elements, i.e. "fewunit" rings - see [1].
> >>>> The key idea is that Euclid's construction of a new prime
> >>>> generalizes from elements to ideals, i.e. given max ideals
> >>>> P1,...,Pk then a simple pigeonhole argument employing CRT
> >>>> implies that 1 + P1...Pk contains a nonunit, which lies in
> >>>> some maximal ideal P which, by construction, is comaximal
> >>>> (so distinct) from the prior max ideals Pi.
> >>>>
> >>>> Thus, from this more general perspective, excluding the
> >>>> possibility that w+1 is a unit is crucial. Many proofs
> >>>> of Euclid's theorem fail to explicitly exclude this case,
> >>>> so they are (technically) incomplete.
> >>>>
> >>>> The gist of my remark above was to dispute your claim that
> >>>> the proof "would not go through if there were units > 1".
> >>>> As I've pointed out, Euclid's proof can be generalized to
> >>>> infinite fewunits rings (rings with fewer units than elts).
> >>>
> >>> The basis of Euclid's original proof sems to be that given
> >>> any finite set of primes, the product of these primes +1
> >>> gives a number divisible by a prime not in the original set.
> >>>
> >>> If r = sqrt(2), then Z(r) has UFT
> >>>
> >>> r is a prime, NORM =2
> >>> -29 -21r is a prime or associated prime, norm = -41
> >>>
> >>> w+1 = r(-29 -21r )+1 = -41 -29r = unit
> >>>
> >>> So given r, -29-21r, a finite set of primes from Euclid's
> >>> approach I cannot always conculde that there is an "extra" prime.
> >>> So in general, Euclids approach does not work.
> >>>
> >>> How does what you say about fewunit rings apply to
> >>> "infinite" unit rings? If it doesn't, why is my claim wrong ?
> >>
> >> That ring has as many units as elts so its not a fewunit ring.
> >> The fewunits proof works for infinite rings with fewer units
> >> than elts, i.e. smaller cardinality.
> >
> > Then why do you say "... In general 1 + pqr... can fail to
> > produce a new prime (or irreducible) because it may be a unit...".
> Because generally that's the obstruction to the Euclidean proof.
> The fewunits hypothesis is one way of avoiding that obstruction.
> Specializing to Z (or any infinite UFD as in the theorem below)
> we see that the obstruction to this fewunits generalization of
> the Euclidean proof occurs only when there are infinitely many
> units -- not, as you claim, when there are units > 1. Even if
> Z had finitely many units>1 the fewunits proof would still work.
> >>>> Such rings may have units besides +-1, so the w+1 candidates
> >>>> may indeed be units. But since there are fewer units than elts
> >>>> a pigeonhole argument shows that at least one candidate isn't
> >>>> a unit, so the proof does indeed succeed. Thus even for the
> >>>> ring of integers, the proof can succeed without employing
> >>>> the fact that there are no units>1. Instead one may employ
> >>>> the scarcity of units in order to avoid them as need be.
> >>>>
> >>>>> A unit has no prime divisors. If w+1 could be either a prime,
> >>>>> composite or unit, then its non-divisibility by all primes or
> >>>>> any prime leads to no contradiction or no "extra" prime
> >>>>> (w+1 is a unit) and the proof would not go through.
> >>>>
> >>>> But it does go through for infinite fewunit rings like Z.
> >>>> If the proof in [1] is beyond your knowledge of algebra
> >>>> (e.g. it uses simple ideal theory) you can find a much
> >>>> simpler element-wise version in [2]. That 4-line proof
> >>>> should be easily comprehensible to a high-school student.
> >>>> Indeed, here is a simple specialization of said proof:
> >>>>
> >>>> THEOREM If Z is infinite UFD with finitely many units
> >>>> then Z has infinitely many primes.
> >>>>
> >>>> PROOF Via contradiction. Let w = product of all primes of Z
> >>>> Since Z is infinite so is 1 + w Z [ 1+wn = 1+wn' <=> n = n']
> >>>> so it must contain a nonzero nonunit [ since only finite #units ]
> >>>> which has a prime factor p | 1 + w n. But p|w => p|1 =><= QED
> >>>
> >>> Yes, very good. How do you generalize Euclid's approach to rings
> >>> with infinitely many units ?
> >>
> >> See the penultimate paragraph in [2]
> --Bill Dubuque
> [1] http://www.mathlinks.ro/viewtopic.php?p=1209616
> http://google.com/groups?selm=y8zk5f3rn4e.fsf%40nestle.csail.mit.edu
> [2] sci.math, 12 Apr 2005, Euclid in rings: infinitely many primes
> http://google.com/group/sci.math/msg/c502d844bdaeee1e
> http://google.com/groups?selm=y8zbr8kv5vl.fsf_-_%40nestle.csail.mit.edu
> Beware: cut'n'paste error in above: instead of nilradical, please
> read pseudo-radical = intersection of all nonzero prime ideals.
RSA#-(2*first factor)=Whole#1/first factor=Whole#2 IFF TRUE then RSA#/first factor=second factor

Proof:
77-[(7*2)=14]=63/7=9 then 77/7=11
7 and 11 are factors of 77
M u s a t o v