Initial Segments of Modular Arithmetic 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 is $\omega$-inconsistent and all infinite models of MA have non-standard elements.
I have been been trying to figure out if every infinite model of MA is an initial segment of some model is PA. I think the answer is no, but defining initial segments of PA seems to be way above my head. Every infinite model of MA has a subset that is an initial segment of a model of PA. PA would be inconsistent if some initial segment of PA is first order definable in either PA or MA.
This started me thinking about initial segments of MA. It seems obvious that every model of MA has all "smaller" models of MA as definable subsets (definable in the meta-theory). For example, the trivial ring is an initial segment of every model of MA because the trivial ring satisfies all of the axioms of MA. For finite models of MA, the rings $\mathbb{Z} /n \mathbb{Z}$, every initial segment is unique. There is exactly one model for each $n$.
The algebraic numbers, $\mathbb{A}$, is a countably infinite model of MA. Let $\mathbb{M} = \mathbb{Z} /x \mathbb{Z}$ be a model of MA where x is an uncountable non-standard number. Would $\mathbb{A}$ be an initial segment of $\mathbb{M}$?
I know I am not suppose to ask more than one question, but I would be interested in anything that can be said about initial segments of MA. For example, are initial segments of MA first order definable?

Edit in response to Lawrence Wong's questions.
Order can not be defined in MA the same way it is in PA. In PA we can say $x \leq y \rightarrow \exists z(y = x+z)$. This definition doesn't work in MA. MA does have a discrete cyclic order. This order is defined by the successor relation. With a cyclic order we can define [a,b,c] meaning "repeatedly applying successor to a reaches b before c". Since I am working in first order, I am pretty sure [a,b,c] is only definable when a, b, and c are a finite distance from one another.
I don't know much about initial segments which is why I am asking this question. I couldn't find a definition on the net. I think an initial segment is a subset of the model that satisfies the axioms. The standard natural numbers are a subset of every model of PA so they are an initial segment of every model of PA. An uncountable non-standard model would have lots of countable initial segments. Initial segments can be closed under successor and not addition or closed under both successor and addition, etc.
I think we are also allowed to perform transformations on the model. The smallest model of MA is the trivial ring. Given another model of MA, $\mathbb{M}$, I can map every element of $\mathbb{M}$ to $0$. I can then show the model satisfies the axioms under this transformation proving {0} is an initial segment of the model.
The successor relation is defined by the axioms. $\forall x(Sx=x+S0)$ is a theorem of both MA and PA. The successor must be $Sx=x+1$. This defines a cyclic order on the algebraic numbers. Clearly, this is not the same as the "standard" ordering. I will be asking some questions about this soon. I don't think the algebraic numbers are a model of MA, but people smarter than me have said any algebraically closed field is a model of MA.
 A: This probably doesn't qualify as an answer, but it is definitely too long for a comment.
Somewhere hidden in the literature, there is a theory called Peano-with-top, sometimes denotes by PT.  It seems every model of your MA can be tweaked to give a model of PT, and vice versa.  (Do check this properly yourself!)  A quick search on the MathSciNet gives me the following paper.

Patrick Cégielski, Kenneth McAloon, and George Wilmers.
  Modelès recursivement saturés de l'addition et de la multiplication des entier naturels.
  In Logic Colloquium '80, Studies in Logic and the Foundations of Mathematics, vol. 108, pp. 57–68.
  North-Holland, Amsterdam, 1982.

In there, you find a theorem (attributed to Paris) which says that every model of PT is an initial segment of a model of $\mathrm I\Delta_0$.  It seems unlikely that you can require more induction in the extension, because otherwise, our model of PT would be recursively saturated.
I don't know what you may want an initial segment to be in a model of MA, especially when the order is intended to be cyclic.  I suppose you choose the definition that gives you the results you want.  In any case, induction forbids any definable set that contains 0 and is closed under successor to be proper.
