Finite models are atomic

Question

I want to show that any finite model is atomic. To prove this, it is enough to show that any type realized in a finite model is principal (isolated).

Let $T$ be a theory, $\mathcal{A}$ a finite model of $T$, and $p$ a type realized in $\mathcal{A}$. How can we show that $p$ is principal?

If $T$ is assumed complete, then we can show that $T$ is absolutely categorical, and so $p$ is realized in every model of $T$ and thus is isolated. But how would we show this without assuming $T$ complete? Thanks

Answer 1

A (complete) $n$-type is just a complete theory over the extended language where you adjoin $n$ new constant symbols, and a model together with a realization of the type is just a model of the complete theory in the extended language. So it suffices to show that any complete theory (over a finite language) which has a finite model is generated by a single axiom. This is easy: just write an axiom that completely describes your finite model up to isomorphism.

(As Alex Kruckman's answer shows, you do need to assume the language is finite.)

Answer 2

Let's assume the language is finite. Since $\mathcal{A}$ is finite, there is a sentence $\varphi_{\mathcal{A}}$ which describes $\mathcal{A}$ uniquely up to isomorphism. So $\varphi_\mathcal{A}$ axiomatizes a complete absolutely categorical extension $T_\mathcal{A}$ of $T$. Relative to $T_\mathcal{A}$, any type $p(x)$ realized in $\mathcal{A}$ is isolated by a formula $\psi_p(x)$. Then the formula $\varphi_{\mathcal{A}}\land \psi_p(x)$ isolates $p$ relative to $T$.

If the language isn't finite, it's not true that finite models are atomic. For example, consider the language with countably many unary relation symbols $\{P_i\mid i\in \omega\}$, and let $T$ be the empty theory. A formula $\psi(x)$ can only mention finitely many of the $P_i$, and $T$ doesn't mention any of the $P_i$, so $T\cup \{\psi(x)\}$ can't possibly isolate the complete type of $x$ (which must decide whether $P_i(x)$ holds for all the $P_i$ not mentioned by $\psi$).