Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let us consider the first-order theory of Peano arithmetic (from now on PA) formulated in the vocabulary with just $+$ (for addition) and $\cdot$ (for multiplication). This vocabulary restriction is not important at all since $0$, succesor $S$, and the order $\leq$ can be defined using first-order formulas.

I am curious what it is known about the dependency, in non standard models, of each one of these basic operations with respect to the other. To be more precise, I am interested on any known answer to the following two questions:

  1. Are there two different (i.e., non-isomorphic) PA models with the same universe and the same interpretation of multiplication?

  2. Are there two different (i.e., non-isomorphic) PA models with the same universe and the same interpretation of addition?

share|improve this question

1 Answer 1

up vote 5 down vote accepted

The following closely related results may interest you.

$1.$ The isomorphism type of the multiplicative semigroup of a model of PA determines, up to isomorphism, the additive semigroup of the model. This is not surprising, since one can define a relation $R(y,x)$ that behaves like $y=2^x$.

$2.$ For countable models of PA, the isomorphism type of the additive semigroup determines the isomorphism type of the multiplicative semigroup.

The above two quite old results are due to Ehrenfeucht.

$3.$ For uncountable models of PA, the isomorphism type of the additive semigrop does not determine the isomorphism type of the multiplicative semigroup.

Information can be found in this paper by Kossak, Nadel, and Schmerl, which unfortunately does not seem to be freely available.

share|improve this answer
    
Great. This paper is exactly what I was looking for. –  boumol May 11 '12 at 18:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.