# [FOM] Query on nonstandard models of the integers

joeshipman@aol.com joeshipman at aol.com
Sun Nov 4 18:34:37 EST 2007

```>-----Original Message-----
>From: joeshipman at aol.com
>
>Is there any nonstandard model of the integers which has the unique
>factorization property?

Since some people have expressed confusion, let me clarify my query.

Part of unique factorization is a first-order property -- you can
define "irreducible" to be a number that is not a nontrivial product,
and define "prime" to be a non-unit that divides a or b whenever it
divides ab, and "all irreducible elements are prime" is the key lemma
for proving uniqueness of factorizations.

However it does not prove EXISTENCE of factorizations -- you could have
a number which was not expressible as a finite product of prime numbers
(the fact that you can define exponentiation and prove every number has
a largest prime factor and a largest exponent doesn't help because in a
nonstandard model there could still be infinitely many prime factors
smaller than the largest one and the exponents could still be infinite
numbers).

The unique factorization property for rings involves both existence and
uniqueness. In non-noetherian rings, an element can be divisible by
infinitely many primes (indeed by infinitely many standard primes 2, 3,
5, 7, 11...). You can add a constant c to the language of arithmetic
and the axioms that c is divisible by 2, divisible by 3, divisible by
5, etc., and by compactness there will be a nonstandard model of the
theory of integers where c is divisible by all "finite" primes. If you
then remove c from the language, you've still got a model where unique
factorization fails.

My question was whether ALL nonstandard models of the theory of
integers must have unique factorization fail in this way.

It looks to me now like they do -- it is a true theorem of arithmetic
(after you code to represent sequences etc.) that for any integer n,
there is a number that is divisible by the first n primes greater than
or equal to n (and only by those); a nonstandard model must have
elements greater than all the finite integers, so it must have elements
which have as their "prime factorizations" an infinite set of infinite
primes. It's still true that any prime dividing these elements divides
one of the members of any complementary pair of factors, but this is
not what I count as satisfying "unique factorization".

-- JS

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