[FOM] New Proof of Fundamental Theorem of Arithmetic

joeshipman@aol.com joeshipman at aol.com
Wed Sep 16 20:01:03 EDT 2009

The way you can tell it is a truly different proof is that it doesn't 
apply to other Euclidean Domains like the GCD proof does. You can 
multiply the factors in a factorization together and mod out by 
principal ideals as before, but the quotient groups are not simple 
because their order is not prime (the norm of the principal ideal in 
question doesn't have to be prime because the additive group of the 
ring has dimension greater than 1 as a Z-module). For example, in 
Z[sqrt(-5)] where 2 * 3  = (1+sqrt(-5)) * (1-sqrt(-5)), you end up 
doing Jordan-Holder on normal series that are not yet composition 
series -- thus 4 x 9 and 6 x 6 both get reduced to a permutation of 2 x 
2 x 3 x 3.

The proof only works for Z because the primes in Z do double duty as 
cardinalities of sets so their irreducibility implies that the 
corresponding groups have no proper subgroups and so are aleady simple.

-- JS

-----Original Message-----
From: Vaughan Pratt <pratt at cs.stanford.edu>
joeshipman at aol.com wrote:
> Since this proof does not use division with remainder or GCD, and 
> the proof of the Jordan-Holder theorem is completely abstract and
> involves only the group operation, and doesn't depend on any 
> of rings which have two operations, should this be counted as
> essentially different from the usual proofs?

More to the point, if you "compile out" the non-abelian part, the core
of this proof is the statement that lcm(p,q) = pq for primes p,q.
... It hadn't
occurred to me to proceed from that end instead of the usual gcd end,
where gcd(p,q) = 1, but it works just as well.  So yes I'd call that a
different proof..

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