Any algebraically closed field (ACF) is a model of Modular Arithmetic (MA). MA has the same axioms as first order Peano arithmetic except $\forall x(Sx \neq 0)$ is replaced with $\exists x(Sx=0)$. MA is $\omega$-inconsistent and all infinite models are non-standard. This means any ACF can be a non-standard model of arithmetic.
Are addition and multiplication recursive in an ACF? If so, how would such a model avoid Tennenbaum's theorem? I know there are weak theories with recursive non-standard models, but MA is not a weak theory.
Response to tomasz:
First a quick proof every infinite model of MA is non-standard. Assume $S(n)=0$ and $n$ is a standard natural number. Then the model is finite. $-1$ must be non-standard in any infinite model of MA.
I have been told any algebraically closed field (ACF) is a model of MA. I have yet to prove otherwise. It is clear any ACF satisfies the axioms of MA without induction. ACF's satisfy first order induction because they admit quantifier elimination. I don't pretend to understand the proof or really understand why quantifier elimination implies first order induction is valid. I assume it has something to do with "finiteness". The authors go to a lot of trouble to prove "root diagrams" are definable.
The answer to your first question is yes. Both $\mathbb{A}$ and $\mathbb{C}$ are models of MA. I am not sure what you mean about a field of prime characteristic. A model of MA does not have to be a field. I know practically nothing about ring theory, but I thought all ACF's have characteristic $0$. I recently posted a question about this.
Your second question is tougher and one of the reasons I posted this question. Tennenbaum's theorem does apply to non-standard models of PA. Tennenbaum's theorem is a proof by contradiction. I couldn't find a simple description of it online. I will try to describe the proof without mangling it too much. Any corrections are welcome.
Tennenbaum showed that if either addition or multiplication is recursive in a non-standard model of PA then we can recursively separate two recursively inseparable sets. Let $\mathbb{H}$ be the set of standard natural numbers encoding a Turing machine (TM) that halts on blank input. $\mathbb{H}$ is recursively enumerable and is not recursively separable from the set of numbers representing TM's that don't halt. Let $y=F(n)$ mean $y$ encodes the first $n$ members of $\mathbb{H}$. There are lots of way we can encode members of $\mathbb{H}$. We could define a polynomial $(x-h_1)(x-h_2)...(x-h_n)$. We could use the binary representation of a number such that $2^h$ is true in the representation only if $h \in \mathbb{H}$. Robinson's overspill lemma says if $\exists y (y=F(n))$ is true for all standard natural numbers there must exist a non-standard $n$ for which it is true. Let $y^* = F(n^*)$ be true. We can use the binary expansion of $y^*$ to recursively separate members of $\mathbb{H}$ from representations of non-halting TM's. From this we conclude addition and multiplication are not recursive in non-standard models of PA.
My question is why doesn't this same argument apply to MA? Assume $\mathbb{C}$ is a model of MA. What prevents me from encoding $\mathbb{H}$ using the binary representation of a real number, $r^*$? If I can finitely specify $r^*$ as the $k^{th}$ root of a polynomial of degree $n$, it seems I could recursively determine the elements of $\mathbb{H}$.