Interpreting peano arithmetic without parameters

Question

I will accept an answer in the form of references to the literature about my question as well as any other information. I am quite ignorant of the area and that will be clear from my question.

I take it that interpreting a system (e.g., PA) in some structure without the use of parameters is harder than doing the same with the use of parameters. I would like to understand the difference of the consequences of the two tasks (interpreting with parameters vs. interpreting without). What does it have to say about the structure in which the interpretation takes place?

Answer 1

Let me give a simple example, which may help to clarify things.

Consider the integer line $\langle\newcommand\Z{\mathbb{Z}}\Z,<\rangle$, and the natural number order $\langle\newcommand\N{\mathbb{N}}\N,<\rangle$. The latter is interpreted in the former, with parameters, since we may fix an arbitrary integer $k\in\Z$ and then the subset of $\Z$ consisting of all things above or equal to $k$ in $\Z$ is a definable set that with the relation $<$ is isomorphic to $\langle \N,<\rangle$. Thus, the structure $\langle\N,<\rangle$ has a definable copy in $\langle\Z,<\rangle$, if we allow a parameter in the definition.

Meanwhile, there is no such definable copy if we do not allow parameters. The reason is that $\Z$ has no definable elements — all elements are automorphic by translation — and so if we had a definable copy of $\N$ in $\Z$ (or even in some finite power) then since $0$ is definable in $\N$ we would have to have a definable element of $\Z$, which we don't.

In general, there is a big difference between a subset of a model being definable with parameters and being definable without parameters, and there are numerous examples covered in almost any introductory model theory book. For example, the only sets definable in $\langle\Z,<\rangle$ without parameters are the whole set and the empty set, but if we allow parameters, we can define any finite collection of intervals.

But you asked about interpretating a theory, rather than definability, so I'm not exactly sure whether that is the kind of answer you seek.

Answer 2

Here is one possible relationship. One common reason to interpret a theory $T_1$ into a second theory $T_2$ is to establish a relative consistency result: if $T_2$ interprets $T_1$ in a syntactic way, then we will probably be able to prove $\text{Con}(T_2) \to \text{Con}(T_1)$.

If we can show that a model $M$ interprets $T_1$ without the use of parameters, then it is also likely, as a heuristic principle, that we can interpret $T_1$ into the theory of $M$, and possible into some weaker theory satisfied by $M$. On the other hand, if we need to use parameters of $M$ in order to interpret $T_1$, then it is not clear that we could interpret $T_1$ into other models of the theory of $M$.

At the same time, at least some nontrivial proofs that a theory is interpretable into a model without using parameters go by first interpreting the theory with parameters, and then showing that those parameters are definable (or, which is not much different, showing that certain parameters are definable and then interpreting the theory using those parameters). For a trivial example, to interpret $(\omega, <)$ into $(\omega_1, <)$, we could first show that $\omega$ is definable using $\omega + 1$ as a parameter, and then show that $\omega + 1$ is definable.

Here is a less trivial example (I wish I knew something more simple). We know from a theorem of Simpson that the theory of $(D, \leq_T)$, where $D$ is the set of Turing degrees and $\leq_T$ is Turing reducibility, is many-one equivalent to the theory $Z^T_2$ of true second-order arithmetic. Part of the proof is an interpretation of $Z^T_2$ into $(D, \leq_T)$. Because $D$ is uncountable, in order to keep the interpretation effective enough to lead to a many-one reduction between the theories, we need to avoid using parameters, or at least only use parameters that are themselves definable.