# A minimal structure that is a model of a complete theory must be a prime model

I am trying to see that if $\underline{A}$ is a minimal $L$-structure that is a model of a complete theory $T$, then $\underline{A}$ is a prime model of $T$. I see that if $T$ is assumed to have a prime model this holds, but I can't seem to get it without this assumption.

Here, $\underline{A}$ a $minimal$ $structure$ means $\underline{A}$ has no proper substructure. A $complete$ $theory$ is a satisfiable, deductively closed set of $L$-sentences such that for every $L$-sentence $\phi$, we must have $\phi$ in $T$ or $¬\phi$ in $T$. $\underline{A}$ is a $prime$ $model$ of $T$ if for any other model $\underline{B}$ of $T$, there exists an elementary embedding $\pi$ : $\underline{A} $$\rightarrow$$\underline{B}$.

Any help/ideas would be greatly appreciated. So would any links to examples of questions about equivalence of certain types of models.

• Minimal should mean that it has no proper elementary substructure, right? Feb 23, 2017 at 14:00
• @AlexKruckman for me a "minimal model" A of a theory T is such that it has no proper elementary substructure/ submodel (of T), but a "minimal L-structure" A is simply a structure such that there is no L-structure embedding from A' -->A where the domain/universe of A' is a proper subset of A, and the map is simply inclusion. Feb 23, 2017 at 16:08
• Oh, but this just means that every element of $A$ is named by a term in the constants. Then there's obviously a map from $A$ to any other model of $T$, and it's elementary since $T$ specifies the type of every tuple from $A$. If this is really the question you meant to ask, I'll convert this comment to an answer. Feb 23, 2017 at 16:24
• This is the question I meant, yes. Thank you, that is much more straightforward than I thought. Feb 23, 2017 at 22:09
• @AlexKruckman Why? The theory of the infinite line graph (with two unary functions symbols to go forward and backward, instead of a binary predicate) satisfies the conditions of the question, but has no constants. Feb 24, 2017 at 12:00

## 1 Answer

First, let's observe that any structure $A$ contains the substructure generated by the empty set: $$\langle \emptyset\rangle_A = \{t^A\mid t\text{ is a term}\}.$$ When I talk about terms in this answer, I mean terms with no free variables or parameters, i.e. terms built up from the constant symbols by applications of the function symbols. If there are no constants, there are no terms in this sense, and $\langle \emptyset\rangle_A$ is empty. $t^A$ is the term $t$ evaluated in the structure $A$.

Now if $A$ is a minimal structure, in the sense that it contains no proper substructure, we have $A=\langle\emptyset\rangle_A$, and hence every element of $A$ is named by a term.

Let $T = \text{Th}(A)$, and suppose $B\models T$. Then there is a map $A\to B$ given by $t^A\mapsto t^B$. This map is well-defined and injective, since for terms $s$ and $t$, we have $$s^A = t^A \iff T\models s = t \iff s^B = t^B.$$ And it is an elementary embedding, since for any tuple $t_1^A,\dots,t_n^A$ and $L$-formula $\varphi(x_1,\dots,x_n)$, we have $$A\models \varphi(t_1^A,\dots,t_n^A) \iff T\models \varphi(t_1,\dots,t_n) \iff B\models \varphi(t_1^B,\dots,t_n^B).$$

There are two ways to make this question less trivial.

1. Maybe you don't allow the empty structure, so "$A$ is minimal" means "$A$ has no proper nonempty substructure". If there are no constant symbols in the language, then a minimal structure $A$ in this sense has the unusual feature that for any element $a\in A$, $\langle a\rangle_A = A$. I view this interpretation of the question (and the banning of the empty structure, more generally) to be wrongheaded - in this case, it introduces an artificial distinction between languages with constant symbols and those without constant symbols.

2. You could ask the analogous question for elementary embeddings. That is, is it the case that a structure with no proper elementary substructure is a prime model for its theory?

The answer is no for both of these questions, and I can give you a single counterexample that works for both.

Let $L = \{s,p\}$, where $s$ and $p$ are unary function symbols, and let $L^* = L\cup \{P\}$, where $P$ is a unary relation symbol.

Let $T$ be the $L$-theory which asserts that $s$ is a bijection with no cycles and $p$ is its inverse. So models of $T$ look like disjoint unions of copies of $\mathbb{Z}$, with $s$ and $p$ interpreted as successor and predecessor.

If we expand a model of $T$ to an $L^*$-structure, we think of tagging each element with $0$ or a $1$, depending on whether $P$ holds. Now let $T^*$ be $T$, together with axioms asserting that any finite binary sequence is realized somewhere. For example, given the binary sequence $1101$, we have an axiom that says $\exists x\, P(x)\land P(s(x)) \land \lnot P(s(s(x)) \land P(s(s(s(x)))$. In fancy language, $T^*$ is the model companion of $T$ in the language $L^*$, and it states that $P$ is a generic predicate on models of $T$.

Here are some things to check:

1. $T^*$ is complete. This can be proven using an Ehrenfeucht–Fraïssé game argument, for example.
2. $T^*$ has a model $A$ whose reduct to $L$ is just a single copy of $\mathbb{Z}$. You can build $A$ by hand, by taking the list of all finite binary sequences and tacking them on one by one.
3. $A$ has the property that it is generated as an $L^*$-structure by any single element. Thus, it has no nonempty proper substructure (and, in particular, no proper elementary substructure). So it is minimal in the sense of both 1. and 2. above.
4. $A$ is not a prime model for $T^*$. Why? Given any particular element $a\in A$, $\text{tp}(a)$ contains the whole binary sequence on $a, s(a), s(s(a)), s(s(s(a))),\dots$. This type is not isolated, and it can be omitted in another model of $T^*$ (in fact, you can build this model by hand too, though the construction is a little tricker, since you have to realize all finite binary sequences, while also blocking any particular element from beginning the infinite binary sequence encoded by $\text{tp}(a)$).