Any non-standard halting oracles stronger than $\mathbb N$? Let $M$ be some model of PA. Let $H_M$ be the set of codes of standard turing machines $X$ such that $M \models X \text{ halts}$. For example, $H_\mathbb N$ corresponds to the regular halting oracle (where $\mathbb N$ is the standard model).
My question is, is there some model $M$ such that $H_M$ is strictly stronger than $H_\mathbb N$ as a Turing degree?
 A: Yes - in fact, $H_M$ can be arbitrarily complicated. It's a fun exercise to find infinitely many independent existential sentences: that is, conflating sentences and Godel numbers as usual find a computable sequence $(\theta_i)_{i\in\mathbb{N}}$ of $\Sigma_1$-sentences in the language of arithmetic such that for all $S\subseteq\mathbb{N}$ the theory $$T_S:=PA\cup\{\theta_i: i\in S\}\cup\{\neg \theta_j: j\not\in S\}$$ is consistent.
Now since the sequence $(\theta_i)_{i\in\omega}$ is computable, if $M\models T_S$ we can recover $S$ computably from $H_M$; that is, $S\le_TH_M$. And this works for any $S\subseteq\mathbb{N}$, so the set of Turing degrees of sets of the form $H_M$ is cofinal in the set of Turing degrees.

Going a bit further, it's not hard to show that the following are equivalent for a degree d:


*

*d is the degree of $H_M$ for some $M\models PA$.

*d computes a completion of every consistent computable theory $T$.
That is, the degrees of $H_M$s are exactly the PA degrees.  One direction is easy: if d computes a completion of every computable theory $T$, take $T$ to be PA and extract from the computed completion $\hat{T}$ the set $\{e:$  "$\varphi_e(e)\downarrow$" $\in \hat{T}\}$.
The other direction takes a moment of thought. We want to use $H_M$ to build a completion of some consistent computable theory $T$, but we have to be careful: $M$ might think that $T$ is inconsistent! So we have to be careful about how we juggle internal and external considerations here. What we do is the following: we go through the sentences in our language one by one, and we add the sentence currently under consideration to the completion we're building if $H_M$ thinks that its negation would add a strictly shorter contradiction than any contradiction that it adds.
By the Low Basis Theorem, there are models $M$ where $H_M$ is low. Note that this is the hard part: making $H_M$ complicated was just a direct construction, but making $H_M$ simple takes some real effort.

EDIT: You're probably already aware of this, but just for other readers: a somewhat relevant idea here is that of the standard system. (This should really be a comment to the answer, but it's far too long.)
For $M\models$ PA and $a\in M$, we say that $a$ codes a set $X$ of standard natural numbers if for all standard $i$ we have $$p_i\vert a\iff i\in X,$$ where $p_i$ denotes the $i$th prime. Let $SS(M)$ denote the set of sets of standard natural numbers coded by elements of $M$.
If $M$ is standard, $SS(M)$ is just the set of finite sets. If $M$ is nonstandard, however, something cool happens: by an overspill argument, the set $SS(M)$ turns out to be a Scott set! A Scott set is a family of sets of natural numbers which is closed under join and Turing reducibility and such that for every infinite tree $T\subseteq 2^{<\omega}$ in the set, there is a path through $T$ in the set. In particular, every Scott set contians a PA degree, and Scott sets can alternately be characterized as Turing ideals such that for every set in the ideal, there is in the ideal a set of PA degree relative to the original set - so they are related in some sense to the "relativized" version of your question. (Another interesting fact is that if $M_0$ is a nonstandard initial segment of $M$ satisfying PA, then $SS(M_0)=SS(M)$.)
A natural problem at this point is whether every Scott set is the standard system of some model of PA. Scott showed that every countable Scott set is the standard system of some countable model of PA; later, Knight and Nadel extended this to Scott sets of cardinality $\omega_1$, thus solving the problem under the assumption of CH. The general problem, however, is still open (and see this recent article of Gitman for some interesting forcing-related progress).
Incidentally, we can also talk about standard systems relative to nonstandard models of PA: if $M, N$ are arbitrary models of PA with $M\subsetneq_{end}N$, we can still talk about elements of $N$ coding subsets of $M$ and the resulting "$M$-standard system" $SS_M(N)$. The same overspill argument as in the standard case shows that $(M, SS_M(N))\models$ WKL. As far as I know, these generalized standard systems aren't well-studied, but Kossak and Schmerl's book has some information on them (especially chapter 7).
