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Suppose we have a complete theory $T$ and a model $\mathfrak{M}\models T$. Recall that, if $\varphi(v)$ is a strongly minimal formula with parameters in some set $A_0\subseteq\mathfrak{M}$, then we define a pregeometry $(\varphi(\mathfrak{M}),\text{cl})$ by $\text{cl}(X)=\text{acl}^\mathfrak{M}(X\cup A_0)\cap\varphi(\mathfrak{M})$ for any $X\subseteq\varphi(\mathfrak{M})$. We then denote the dimension of this pregeometry by $\dim_\varphi(\mathfrak{M})$. I am having a small hangup in the proof of the following result:

Let $\mathfrak{M}$ and $\mathfrak{N}$ models of $T$ containing some set $A_0$, and $\varphi(v)$ a strongly minimal formula with parameters in $A_0$. Then, if $\dim_\varphi(\mathfrak{M})=\dim_\varphi(\mathfrak{N})$, there exists an elementary bijection from $\varphi(\mathfrak{M})$ to $\varphi(\mathfrak{N})$ fixing $A_0$ pointwise.

I will run through my understanding of the proof of this. It follows the treatment given by Tent and Ziegler in section 5.7 of their textbook, but I have added in my understanding of a number of details they glazed over. The result in particular is Lemma 5.7.6, on page 84.


Lemma 1: If $\mathfrak{M}\models T$ and $A\subseteq\mathfrak{M}$, and $\varphi(v)$ is a strongly minimal formula with parameters in $A$, then for any elementary extension $\mathfrak{N}\succcurlyeq\mathfrak{M}$ and any $A\subseteq B\subseteq \mathfrak{N}$ there is a unique non-algebraic type in $S_1^\mathfrak{N}(B)$ containing $\varphi(v)$.

Proof: Consider the set $p(v)=\{\psi\in\text{Form}(\mathcal{L}_B):\varphi\wedge\neg\psi\text{ is algebraic}\}$. Since $\varphi$ is non-algebraic, $p(v)$ does not contain any algebraic formulas, and since $\varphi$ is strongly minimal $p(v)$ indeed contains one of $\psi$ and $\neg\psi$ for any $\mathcal{L}_B$ formula $\psi$. If $p(v)$ were not consistent there would be $\psi_1,\ldots,\psi_n$ such that each $\varphi\wedge\neg\psi_i$ is algebraic and $\mathfrak{N}\models\forall v\neg(\varphi(v)\wedge\bigwedge_{i=1}^n\psi_i(v))$, ie $\mathfrak{N}\models\forall v(\varphi(v)\rightarrow\bigvee_{i=1}^n\neg\psi_i(v))$, contradicting that $\varphi$ is non-algebraic. So $p(v)$ is a non-algebraic type in $S_1^\mathfrak{N}(B)$ containing $\varphi$, and it is clearly the unique such type.


Lemma 2: Suppose $A\subseteq\mathfrak{M}$ and that $p(v)\in S_1^\mathfrak{M}(A)$ is a non-algebraic type containing a strongly minimal $\mathcal{L}_A$ formula $\varphi(v)$. Then we have the following for any $\mathfrak{N}\succcurlyeq\mathfrak{M}$: if $x_1,\ldots,x_n\in\mathfrak{N}$ are realizations of $p$ such that each $x_i\notin\text{acl}(\{x_1,\ldots,x_{i-1}\}\cup A)$, and $y_1,\ldots,y_n\in\mathfrak{N}$ are realizations of $p$ satisfying the same, we have $\text{tp}^\mathfrak{N}(x_1,\ldots,x_n/A)=\text{tp}^\mathfrak{N}(y_1,\ldots,y_n/A)$.

Proof: By induction on $n$. The case $n=1$ is clear, for $\text{tp}^\mathfrak{N}(x_1/A)=p(v)=\text{tp}^\mathfrak{N}(y_1/A)$. Now suppose the results holds for $n-1$. To show $\text{tp}^\mathfrak{N}(x_1,\ldots,x_n/A)=\text{tp}^\mathfrak{N}(y_1,\ldots,y_n/A)$ it suffices to show (i) that $\text{tp}^\mathfrak{N}(x_1,\ldots,x_{n-1}/A)=\text{tp}^\mathfrak{N}(y_1,\ldots,y_{n-1}/A)$ and (ii) letting $p(v)=\text{tp}^\mathfrak{N}(x_n/\{x_1,\ldots,x_{n-1}\}\cup A)$ and $q(v)=\text{tp}^\mathfrak{N}(y_n/\{y_1,\ldots,y_{n-1}\}\cup A)$, that the set $r(v):=\{\varphi(y_1,\ldots,y_{n-1},v):\varphi(x_1,\ldots,x_{n-1},v)\in p(v)\}$ equals $q(v)$, where the $\varphi(\bar{w},v)$ are $\mathcal{L}_A$ formulas.

(i) is just the induction hypothesis, and for (ii) we claim that $r(v)$ is a non-algebraic type in $S_1^\mathfrak{N}(A\cup\{y_1,\ldots,y_{n-1}\})$. To see that it is a type, suppose for contradiction that $r(v)$ were not consistent. Then there would be $\varphi_1(\bar{w},v),\ldots,\varphi_k(\bar{w},v)\in\mathcal{L}_A$ such that $\mathfrak{N}\models\forall v(\neg\bigwedge_{i=1}^k\varphi_i(y_1,\ldots,y_{n-1},v))$. But then $\forall v(\neg\bigwedge_{i=1}^k\varphi_i(w_1,\ldots,w_{n-1},v))\in\text{tp}(y_1,\ldots,y_{n-1}/A)$, so by the induction hypothesis $\forall v(\neg\bigwedge_{i=1}^k\varphi_i(w_1,\ldots,w_{n-1},v))\in\text{tp}(x_1,\ldots,x_{n-1}/A)$, a contradiction.

A similar argument shows that $r$ is non-algebraic; if we had $\varphi(y_1,\ldots,y_{n-1},v)\in r(v)$ algebraic – say with size $k\in\omega$ – then we would have $\neg\exists^{>k}v(\varphi(w_1,\ldots,w_{n-1},v))\in\text{tp}(x_1,\ldots,x_{n-1}/A)$, contradicting that $x_n\notin\text{acl}(\{x_1,\ldots,x_{n-1}\}\cup A)$. Thus $r(v)$ is a non-algebraic type, and so $r(v)$ and $q(v)$ are two non-algebraic types in $S_1^\mathfrak{N}(A\cup\{y_1,\ldots,y_{n-1}\})$ containing $\varphi(v)$, whence by lemma 1 we have $q(v)=r(v)$, as desired.


Now we may prove the main result. Let $U\subseteq\varphi(\mathfrak{M})$ and $V\subseteq\varphi(\mathfrak{N})$ be bases for $\varphi(\mathfrak{M})$ and $\varphi(\mathfrak{N})$, respectively. Since the dimensions are the same, we may find $f:U\rightarrow V$ a bijection. Now, this is the point where I get confused: Tent and Ziegler now claim that $f$ is an "$A_0$-elementary" map, which they have not defined but which I assume means that the induced map $f:A_0\sqcup U\rightarrow A_0\sqcup V$ (acting on $A_0$ as the identity) is partial elementary. If this map $f$ is partial elementary, then we have a elementary bijection between $A_0\sqcup U$ and $A_0\sqcup V$, which lifts to an elementary bijection between $\varphi(\mathfrak{M})=\text{acl}(A_0\sqcup U)$ and $\text{acl}(A_0\sqcup V)=\varphi(\mathfrak{N})$ fixing $A_0$ pointwise, where these last equalities hold since $U$ (resp. $V$) is a spanning set for $\varphi(\mathfrak{M})$ (resp. $\varphi(\mathfrak{N})$). So the result would be proved.

However, I'm not quite sure why $f$ is partial elementary. For this to be the case, it suffices to show that $\text{tp}^\mathfrak{M}(u_1,\ldots,u_n/A_0)=\text{tp}^\mathfrak{N}(v_1,\ldots,v_n/A_0)$ for any (pairwise distinct) $u_i\in U$, $v_i\in V$. This is almost but not quite clear to me. Indeed, since $U$ is a basis and thus independent, we have $u\notin\text{cl}(U\setminus\{u\})=\text{acl}(A_0\cup U\setminus\{u\})$ for any $u\in U$. In particular, $u_i\notin\text{acl}(\{u_1,\ldots,u_{i-1}\}\cup A_0)$ for any $i$. The same holds for the $v_i$, and so we are nearly in a position to apply lemma 2.

Lemma 2, however, does not deal with arbitrary substructures, and for it to apply I believe we would in particular need $\text{tp}^\mathfrak{M}(A_0)=\text{tp}^\mathfrak{N}(A_0)$. Then we would have $S_1^\mathfrak{M}(A_0)=S_1^\mathfrak{N}(A_0)$, and so could argue as follows: for any $u\in U$, $u$ is algebraic over $A_0$, and so the type $p(v)=\text{tp}^\mathfrak{M}(u)\in S_1^\mathfrak{M}(A_0)$ is non-algebraic. However, this type also contains the strongly minimal formula $\varphi(v)$, and so by lemma 1 is the unique non-algebraic type in $S_1^\mathfrak{M}(A_0)$ containing $\varphi(v)$. Likewise, for any $v\in V$, $q(v)=\text{tp}^\mathfrak{M}(v)\in S_1^\mathfrak{N}(A_0)$ is the unique non-algebraic type in $S_1^\mathfrak{N}(A_0)$ containing $\varphi(v)$. Since $S_1^\mathfrak{M}(A_0)=S_1^\mathfrak{N}(A_0)$, this means that $p(v)=q(v)$, so all the $u_i$ and $v_i$ are realizations of the same non-algebraic type $p$, and hence we could argue as in lemma 2.

So all we need to show is that $\text{tp}^\mathfrak{M}(A_0)=\text{tp}^\mathfrak{N}(A_0)$. However, I'm at a complete loss as to why this will be the case in general. Can anyone help with this last step of the proof? Sorry for the excessively long post, I wanted to give the context properly to make sure I'm not missing anything.

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When Tent and Ziegler write "Let $\mathfrak{M}$ and $\mathfrak{N}$ be models containing $A_0$", they implicitly mean that $\text{tp}^{\mathfrak{M}}(A_0) = \text{tp}^{\mathfrak{N}}(A_0)$. I agree that this should have been made explicit.

In Section 6.1, you will be introduced to the notion of a monster model $\mathfrak{C}$ and the common convention "all models are elementary submodels of the monster model". Under this convention, the condition on $A_0$ is automatic, since $\mathfrak{M}\prec \mathfrak{C}$ and $\mathfrak{N}\prec\mathfrak{C}$, so the truth value of a formula on a tuple from $A_0$ transfers up from $\mathfrak{M}$ to $\mathfrak{C}$ and then back down to $\mathfrak{N}$. This convention is so commonly used in model theory that I suspect Tent and Ziegler forgot that it was not in place yet when they wrote this section. (But note that we can't read the whole book as if this convention were in place: If it were, the characterization of quantifier elimination given in Theorem 3.2.5 would be trivially satisfied in every theory. For reasons like this, I prefer to avoid talking about two models $M$ and $N$ having a common substructure $A$ whenever possible, replacing the equality $A = A$ by an isomorphism or a partial elementary map between substructures $A\subseteq M$ and $A'\subseteq N$.)

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    $\begingroup$ I'm reading Ziegler's (very beautiful!) paper Model Theory of Modules, and it's really making me think of this answer, since his exposition mostly uses the idea of two modules having a common (pure) submodule rather than framing things in terms of the appropriate (pure) embeddings. I definitely agree that the latter approach seems cleaner $\endgroup$ Jan 17, 2021 at 17:24

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