NetNews Usenet Archive 1992 #19/NN_1992_19.iso/spool/sci/math/10649

Text File | 1992-08-27 | 3.9 KB | 96 lines

Newsgroups: sci.math
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From: [e-mail] (Victor Miller)
Subject: Re: not unique factorization
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In-Reply-To: [e-mail]'s message of Thu, 27 Aug 1992 15:11:42 GMT
Date: Thu, 27 Aug 1992 20:24:37 GMT
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>>>>> On Thu, 27 Aug 1992 15:11:42 GMT, [e-mail] (Victor Miller) said:

>>>>> On 26 AUG 92 23:10:07    , [e-mail] (Kermit Rose) said:

Kermit> Hello anyone.

Kermit> I had read that the integers extended by the sqrt(-5) did not
Kermit> have unique factorization into primes.  The example was given

Kermit> 6 = 2 * 3 = (1+sqrt(-5)) * (1 - sqrt(-5))

Kermit> I considered whether or not this ring could be fixed up by additional
Kermit> extension.   Perhaps in a larger ring we could have

Kermit> 6 = a * b * c * d where

Kermit> a * b = 2
Kermit> c * d = 3
Kermit> a * c = 1 + sqrt(-5)       and
Kermit> b * d = 1 - sqrt(-5).

Kermit> This gives

Kermit> b = 2/a

Kermit> c = (1 + sqrt(-5))/a

Kermit> d = (1 - sqrt(-5))/b = (1 - sqrt(-5)) * (a/2)

Kermit> One way to have a,b,c,d to be elements of the ring is to
Kermit> define a = 1/2 to be an element of the ring.  Then all powers
Kermit> of 2 becomes units, and the factorization 6 = 2 * 3 does not
Kermit> count since 2 is not prime.

Kermit> Now for my question.  Does this fix solve all the cases of non-unique
Kermit> factorization in the ring of integers extended by sqrt(-5)?

Victor> Kermit, You are correct about this.  There are basically two ways to
Victor> fix this up: 1) Localize at a finite set of primes, which generate the
Victor> ideal class group (this is a group which measures in a precise way how
Victor> unique factorization fails).  In your case of sqrt(-5), the ideal
Victor> class group is of order two and generated by the prime 2.  So you have

I actually should make a minor correction here: you've localized at
the unique prime above 2: 2 ramifies in Q(sqrt(-5)).  It occured to me
that you might want a proof:

Every ideal in Q(sqrt(-5)) uniquely factors in a product of prime
ideals.  Each of these ideals is either principal, or linearly
equivalent to the ideal above 2.  Change the expression so that every
such ideal is mulitplied and divided by that ideal.  The numerator is
now a product of principal primes ideal (in the localized ring).  The
denominator must be principal, since the LHS is.  The denominator is
in this case a power of the ideal above 2.

Victor> done exactly that.  2) Go to an extension field: the Hilbert class
Victor> field has the property that every ideal downstairs, "capitulates":
Victor> i.e. becomes principal.

Victor> As an intersting aside, fields with class number 2 (one of which is
Victor> given by your example), are characterized by the property that they
Victor> don't have unique factorization, but that every factorization into
Victor> prime ideals has the same NUMBER of factors.

Kermit> [e-mail]          To be sure I see your response, use e-mail.
Kermit> -----------------------------------------------------------------------
Kermit> You may post, repost, or publish ANY communication received from me.
Kermit> To mend an undesirable barrier is to destroy the barrier.
Victor> --
Victor>         Victor S. Miller
Victor>         Bitnet: VICTOR at WATSON
Victor>         Internet: [e-mail]
Victor>         IBM, TJ Watson Research Center
Victor>         "Great artists steal; lesser artists borrow" Igor Stravinsky
--
            Victor S. Miller
            Vnet and Bitnet:  VICTOR at WATSON
            Internet: [e-mail]
            IBM, TJ Watson Research Center