Modular arithmetic (MA) has the same axioms as first order Peano arithmetic (PA) except $\forall x(Sx \neq 0)$ is replaced with $\exists x(Sx=0)$. MA has finite models based on modular arithmetic. MA is $\omega$-inconsistent and all infinite models are non-standard. A number of people have stated that any algebraically closed field (ACF) is a model of MA.
Removing first order induction from the other axioms of MA leaves a weak sub-theory of ring theory. Obviosly, every ACF is a model of ring theory. ACF's satisfy first order induction because they admit quantifier elimination. I don't pretend to understand the proof. I do know if ACF's are models of MA they must have some amazing properties.
Wikipedia describes the order type of a countable non-standard model of PA. The standard natural numbers are followed by a countably infinite number of copies of structures with the same order type as the integers. These structures are sometimes call Z-blocks. The Z-blocks are dense in that between any two distinct Z-blocks there is another Z-block.
Models of MA don't really have an order type because they can't be well ordered. But they do have some order. Let $\mathbb{M}$ be a ring $\mathbb{Z} /x \mathbb{Z}$ where $x$ is a non-standard natural number in some countable non-standard model of PA. $\mathbb{M}$ does have a cyclic order. This cyclic order can be "cut" to form a linear order. We can order each element of $\mathbb{M}$: 0 < 1 < ... < x-2 < x-1. It seems reasonable this linear order would have a similar order type as the countable non-standard model of PA. If so, a countably infinite model of MA would be a ring with a countably infinite number of Z-blocks dense in the rationals.
The algebraic numbers, $\mathbb{A}$, are a countably infinite ACF. Does $\mathbb{A}$ have the structure I describe above? Is $\mathbb{A}$ isomorphic to some ring $\mathbb{Z} /x \mathbb{Z}$ where $x$ is a countable non-standard natural number? There are a number of ways I can ask this question. I know little about ring theory, but I think I am asking if an ACF can have a non-standard (and non-zero) characteristic.
$\forall x(Sx=x+S0)$ is a theorem of both MA and PA. Since we are talking about complex numbers, this give us $\forall x (Sx = x +(1.0... + 0.0... \cdot \sqrt{-1}))$. Notice the imaginary part of $S0 = (1.0... + 0.0... \cdot \sqrt{-1})$ is $0$ and no matter how many times we apply successor to $(0.0... + 0.0... \cdot \sqrt{-1})$ the imaginary part will be $0.0...$. Assume we apply successor to $0 \sqrt{-1}$ times. We must get $(\sqrt{-1} + 0.0... \cdot \sqrt{-1})$. $\sqrt{-1}$ has to be like any other hyper-integer. It must be reachable from $0$ with successor.
In both MA and PA every number is the sum of distinct powers of 2. Binary representations are not unique in MA. For example, $2^3 \equiv 2^1+2^0 \equiv 3 \bmod 5$. Does $\sqrt{-1}$ have a binary representation in $\mathbb{A}$? I can already prove every ACF has an odd number of elements which means $\sqrt{-1}$ must be even. If $\sqrt{-1}$ has a binary expansion then it is equivalent to some hyper-integer, $x$. The size of the field would be $x^2+1$.
Response to Lawrence Wong
I am not sure how to answer your question. The short answer is successor is a total function. The way it was explained to me is $n=S_n(0)$ defines $n \in \mathbb{N}^*$ where $S_n(0)$ means applying successor to $0$ $n$ times. Just like "$3 = SSS0$". Outside of the model, we know if $t$ is a non-standard natural number larger than any standard natural number then $S_t(0)$ will require us to apply successor a countably infinite number of times. Inside the model, $t$ is just another "finite" natural number and must be reachable from $0$ with a "finite" number of applications of successor.
Assume $t$ is not reachable from $0$ by successor. Then induction fails. I could define a predicate true for all successors of $0$ and false for $t$.