Is number theory $K$ $\aleph_0$-categorical?

Question

Let $K$ be the number theory. The language of $K$ contains the following symbols: $\{0, 1, +, \cdot, <, =\}$. Furthermore, the proper axioms are Peano axioms.

My question is, is $K$ $\aleph_0$-catergorical?

From what I learned, a theory with equality $K$ is $\aleph_0$-categorical if and only if

1. $K$ has at least one countably infinite normal model
2. Any two countably infinite normal models of $K$ are isomorphic.

I believe $K$ has at least one countably infinite normal model. We can just take the standard natural number model, right?

What's left is showing that (2) is true, assuming that $K$ is indeed $\aleph_0$-categorical. But we have to take two arbitrary countably infinite normal models (let's call them $M_1$ and $M_2$) of $K$ and show that they are isomorphic. There is not much information about $M_1$ and $M_2$ I believe, so I was wondering if there is a model $M$ of $K$ that has a denumerable normal model, but $M$ is not isomorphic to a standard natural number model (for example). Any hints/explanation would be appreciated.

Answer 1

This is not even close to $\aleph_0$-categorical.

First of all, per Godel's incompleteness theorem Peano arithmetic isn't even complete. Consequently, per Godel's completeness theorem there are countable models which are not elementarily equivalent to the standard one, so a fortiori not isomorphic to it.

• Strictly speaking, the mere statement of the completeness theorem doesn't say that a countable model exists. However, we can get this by looking at the Henkinization proof of that theorem, which explicitly builds a countable model, or by applying downwards Lowenheim-Skolem (see below) after the fact.

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What if we replace Peano arithmetic with the full theory $\mathsf{TA}$ (true arithmetic) of the structure $(\mathbb{N}; 0,1,+,\cdot,<,=)$? Well, $\mathsf{TA}$ is by definition complete so the above doesn't work. However, via the Lowenheim-Skolem theorem we can show that even $\mathsf{TA}$ is not $\aleph_0$-categorical.

Via the upwards part of the theorem (which is really just a corollary of compactness), let $\mathfrak{A}$ be an uncountable - hence a fortiori nonstandard - model of $\mathsf{TA}$. Now via the downwards part of the theorem (this is the really interesting half of the theorem), fix some infinite element $a\in\mathfrak{A}$ and let $\mathfrak{B}$ be a countable elementary substructure of $\mathfrak{A}$ with $a\in \mathfrak{B}$. This $\mathfrak{B}$ is a countable nonstandard model of $\mathsf{TA}$.