What are the parameter-free definable elements of a model of Peano Arithemetic?

Question

Let $M$ be a model of Peano Arithmetic. What are the parameter-free definable elements of $M$? I conjecture that they are precisely the standard natural numbers, meaning, no nonstandard infinite element is parameter-free definable. Is this true? If so, how to prove it?

Answer 1

Nope, this is very false! For example, suppose $M\models\neg Con(\mathsf{PA})$. Then there is an $m\in M$ which $M$ thinks is the (code for the) shortest proof of a contradiction in $\mathsf{PA}$, and this $m$ is nonstandard but parameter-freely-definable.

In fact, by the Tarski-Vaught test and induction in its guise as the least element principle ("every nonempty set has a least element"), if $M\models \mathsf{PA}$ then the substructure of $M$ consisting of parameter-freely-definable elements is an elementary submodel of $M$. In particular, this means that every completion $T$ of $\mathsf{PA}$ has a unique-up-to-isomorphism model in which every element is parameter-freely definable, and unless $T=\mathsf{TA}$ (true arithmetic) this model will be nonstandard.

(More generally, any theory with definable Skolem functions will have this property of "any model's parameter-freely-definable substructure is an elementary submodel." An example of such a theory is $\mathsf{ZFC+V=L}$, with the $L$-ordering serving as a set-theoretic analogue of $<$ in the natural numbers; Hamkins/Linetsky/Reitz, Pointwise definable models of set theory expands on this phenomenon.)