I'm having a bit of trouble with Lemma 6.3.2. of Tent and Ziegler's model theory textbook. Here is the lemma and their proof, transcribed verbatum.

Lemma. Let $T$ be $\omega$-stable, $M\prec N$ models of $T$, $\varphi(v)$ be strongly minimal and $b_i\in\varphi(N)$. If the $b_i$ are independent over $\varphi(M)$, they are independent over $M$.

Proof: Assume that $b_1,\ldots,b_n$ are algebraically independent over $\varphi(M)$ but dependent over $\bar{a}\in M$. Put $\bar{b}=(b_1,\dots,b_n)$. An argument as in the proof of Theorem 5.5.2 shows that we may assume that $M$ is $\omega$-saturated. Let $p$ be the type of $\bar{b}$ over $M$. We choose a sequence $\bar{b}^0,\bar{b}^1,\dots$ in $\varphi(M)$ such that $\bar{b}^{2i}$ is an $n$-tuple of elements algebraically independent over $\bar{b}^0,\dots,\bar{b}^{2i-1}$ and $\bar{b}^{2i+1}$ realizes $p\upharpoonright_{\bar{a},\bar{b}^0,\dots,\bar{b}^{2i}}$. Let $q$ be the type of $\bar{a}$ over the set $B$ of elements of $(\bar{b}^i)$. Since the sequence $(\bar{b}^i)$ is indiscernible, every permutation $\pi$ of $\omega$ defines a type $\pi(q)$ over $B$. If $\{i:\pi(2i)\text{ even}\}=\{i:\pi'(2i)\text{ even}\}$, we have $\pi(q)\neq\pi'(q)$. So there are uncountably many types over $B$ and $T$ is not $\omega$-stable.

Unfortunately, a few points of this proof are unclear to me. First, I think I have a sketch of a proof that we may assume $M$ is $\aleph_0$-saturated; we essentially enrich our language by including a single unary predicate symbol, call it $U$, such that $U(N)=M$. There are sentences expressing that $U$ is a proper elementary substructure, and we may further form a set of formulas over free variables $v_1,\dots,v_n$ expressing that $(v_1,\dots,v_n)$ is algebraically independent over $\varphi(U)$, by taking formulas of the form $$\forall\bar{w}(\bigwedge_j(\varphi(w_j)\wedge{U(w_j)})\rightarrow((\exists^{\leqslant k}\bar{u}\theta(\bar{u},\bar{w}))\rightarrow\neg\theta(\bar{v},\bar{w})))$$ for all $k\in\omega$ and $\theta\in\operatorname{Form}(\mathcal{L})$. Then $(b_1,\ldots,b_n)$ satisfies these formulas in $N$, and so if we take the union of an elementary chain $(N_i)_{i\in\omega}$ such that $N_0=N$ and all types over finite subsets of $U(N_i)$ are realized in $U(N_{i+1})$, we obtain a new structure $N'$ such that $M':=U(N')$ is $\aleph_0$-saturated and $(b_1,\dots,b_n)$ is not algebraic over $\varphi(M')$. Clearly $(b_1,\dots,b_n)$ will still be dependent over $\bar{a}\in M'$, witnessed by the same formula as in $N$, and so since $M'\prec N'$ we have that all the hypotheses of the theorem are satisfied. So, first question, is this sketch the right idea?

It's at this point that I start to get confused. Constructing the sequence $(\bar{b}^i)$ seems clear to me; the case where $i$ is odd is immediate, and for the even case we can do a similar trick as above, taking the partial type $$\Sigma(\bar{v})=\{\bigwedge_{i=0}^n\varphi(v_i)\}\cup\{(\exists^{\leqslant k}\bar{u}\theta(\bar{u}))\rightarrow\neg\theta(\bar{v}):\theta\in\operatorname{Form}(\mathcal{L}\cup\{\bar{b}^{0},\dots,\bar{b}^{2i-1}\})\}_{k\in\omega}.$$ If this were not finitely satisfiable, there would be algebraic formulas $\theta_1,\ldots,\theta_m\in\operatorname{Form}(\mathcal{L}\cup\{\bar{b}^{0},\dots,\bar{b}^{2i-1}\})$ such that $$M\models\forall\bar{v}(\bigwedge_{i=0}^n\varphi(v_i)\rightarrow\bigvee_{j=1}^m\theta(\bar{v})),$$ which would mean $\varphi(M)^n$ is finite, an obvious contradiction. Thus $\Sigma(\bar{v})$ is consistent and so realized in $M$. Is this right?

However, I have absolutely no idea why the sequence $(\bar{b}^i)$ is indiscernible. We wish to show that, for any $\mathcal{L}$-formula $\psi(\bar{v}_1,\dots,\bar{v}_m)$, and any $i_1<\dots i_m\in\omega$ and $j_1<\dots j_m\in\omega$, we have $$M\models\psi(\bar{b}^{i_1},\dots,\bar{b}^{i_m})\leftrightarrow \psi(\bar{b}^{j_1},\dots,\bar{b}^{j_m}).$$ Certainly it suffices to show the $\rightarrow$ implication, and we clearly need to divide into cases depending on whether $i_m$ and $j_m$ are odd or even. My attempt at this has been with induction on $m$; for instance, the case $m=0$ is vacuous. For the case $m=1$, we first consider the case where $i_1$ and $j_1$ are both odd. Then $$M\models \psi(\bar{b}^{i_1})\iff\psi(\bar{v})\in p\upharpoonright_{\emptyset}\iff M\models \psi(\bar{b}^{j_1}),$$ as needed. Next we might consider the case where $i_1$ and $j_1$ are both even, but even here I seem to get stuck. We know that $M\models \psi(\bar{b}^{i_1})$ forces $\psi(\bar{v})$ to be non-algebraic, but this alone is certainly not enough to conclude that $M\models\psi(\bar{b}^{j_1})$. I imagine that we want to use strong minimality of $\varphi$ here somehow, to show that if $M\models\neg\psi(\bar{b}^{j_1})$ then $\bar{b}^{j_1}$ would be algebraic, but how can we exploit this when $\psi$ has more than one free variable? Indeed, $\bigwedge_{i=1}^n\varphi(v_i)$ will not be a strongly minimal formula when $n>1$ (for instance, consider its intersection with the formula $v_1=c$ for any $c\in \varphi(M)$, which is infinite and coinfinite), and I don't see an obvious way to "extract" a strongly minimal formula in $n$ free variables from a strongly minimal formula in $1$ free variable. So I'm not sure how to proceed even in the case $m=1$, let alone cases where $m>1$. I get even further stuck when $i_m$ and $j_m$ have different parities, as I can't see a way to link the construction of odd entries in the sequence with even entries in the sequence.

Furthermore, even if the $(\bar{b}^i)$ were indiscernible, why would this mean that $\pi(q)$ is a type? The only thing I can imagine they might mean by $\pi(q)$ is $$\{\psi(\bar{v},\bar{b}^{\pi(i_0)},\dots,\bar{b}^{\pi(i_n)}):\psi(\bar{v},\bar{b}^{i_0},\dots,\bar{b}^{i_n})\in q\},$$ but for general sequences of indiscernibles I can't see why this construction would possibly yield types. For instance, consider $(\mathbb{Q},<)$ and any infinite sequence of distinct elements $B\subseteq\mathbb{Q}$. If $b_1<b_2\in B$, then any type over $B$ will contain the sentence $b_1<b_2$, so the "type" induced by any permutation $\pi:B\rightarrow B$ that swaps $b_1$ and $b_2$ will contain the sentence $b_2<b_1$ and hence not be consistent. I'm clearly missing something here; what do they actually mean by $\pi(q)$?


1 Answer 1


Q1: Yes, your sketch of why we may assume $M$ is $\aleph_0$-saturated is correct.

Q2: Yes, your construction of $\bar{b}^i$ when $i$ is even is correct. But there is a typo here in Tent and Ziegler: $\bar{b}^i$ should be chosen to be algebraically independent over $\bar{a}, \bar{b}^1,\dots,\bar{b}^{i-1}$, not just $\bar{b}^1,\dots,\bar{b}^{i-1}$ (so that $\text{tp}(\bar{a}/\bar{b}^i)$ contains a formula witnessing algebraic dependence of $\bar{b}^i$ over $\bar{a}$ when $i$ is odd, but no such formula when $i$ is even).

Q3: Why is the sequence $(\bar{b}^i)_{i\in \omega}$ indiscernible? Well, it comes down to the fact that in a strongly minimal set, for each $m$, there is a unique type satisfied by any algebraically independent $m$-tuple (this is Corollary 5.7.4 in Tent and Ziegler). Now for any $i_1<\dots<i_m$, we have that $\bar{b}^{i_1}\dots\bar{b}^{i_m}$ is an algebraically independent $(mn)$-tuple, so its type does not depend on the choice of $i_1<\dots<i_m$.

Note that the distinction between the odd and even cases has disappeared, since we're just asking for indiscernibility (over $\varnothing$), not indiscernibility over $\bar{a}$. If we forget about $\bar{a}$, each new $\bar{b}^i$ is just a tuple which is algebraically independent over $\bar{b}^1,\dots,\bar{b}^{i-1}$. In the even case, we explicitly chose it that way, and in the odd case, that's part of the type $p\restriction \bar{b}^1,\dots,\bar{b}^{i-1}$, since $\bar{b}^1,\dots,\bar{b}^{i-1}\in \varphi(M)$ and $\bar{b}$ is algebraically independent over $\varphi(M)$.

Some superfluous information: This kind of indiscernible sequence, where each element of the sequence is chosen to be independent over the previous elements, is called a Morley sequence. The concept of Morley sequence, and especially its generalizations outside the strongly minimal context (where we need to use more general notions of independence than algebraic independence - most commonly non-forking independence) are hugely important in model theory.

Q4: Your understanding of what is meant by $\pi(q)$ is correct. So why is $\pi(q)$ a type? The crucial thing that Tent and Ziegler have neglected to tell you here is that $(\bar{b}^i)_{i\in \omega}$ is not just an indiscernible sequence, it's an indiscernible set. This means that for any tuple $i_1,\dots,i_m$ of pairwise distinct elements of $\omega$, and any other tuple $j_1,\dots,j_m$ of pairwise distinct elements of $\omega$, $\text{tp}(\bar{b}^{i_1},\dots,\bar{b}^{i_m}) = \text{tp}(\bar{b}^{j_1},\dots,\bar{b}^{j_m})$. This is just the same as the definition of indiscernible sequence, but we no longer require $i_1,\dots,i_m$ and $j_1,\dots,j_m$ to be increasing tuples, i.e., we forget about the ordering of the sequence. An indiscernible set is also called a totally indiscernible sequence - see the definition at the beginning of Section 9.1 of Tent and Ziegler.

It follows that for any permutation $\pi$ of $\omega$, $\text{tp}((\bar{b}^i)_{i\in \omega})) = \text{tp}((\bar{b}^{\pi(i)})_{i\in \omega})$, and thus if $p$ is a consistent type over $(\bar{b}^i)_{i\in \omega}$, then $\pi(p)$ (which I would denote by $\pi_*p$) is also consistent (by the lemma on pushforward types, which we've already discussed here).

Ok, so why is $(\bar{b}^i)_{i\in \omega}$ an indiscernible set? The key point here is that algebraic independence is a symmetric notion: if $a_1,\dots,a_k$ is a tuple of algebraically independent elements, then so is $a_{\sigma(1)},\dots,a_{\sigma(k)}$ for any permutation $\sigma$ of $\{1,\dots,k\}$. So for any tuple $i_1,\dots,i_m$ of pairwise distinct elements of $\omega$, we have that $\bar{b}^{i_1}\dots\bar{b}^{i_m}$ is an algebraically independent $(mn)$-tuple, and this does not depend on the choice of tuple $i_1,\dots,i_m$ (and in particular not on its ordering).

Some more superfluous information:

  1. There's a higher-level reason why $(\bar{b}^i)_{i\in \omega}$ is an indiscernible set: We're working in the context of an $\omega$-stable theory. And in any stable theory (a theory $T$ is stable if there is some cardinal $\kappa$ such that $T$ is $\kappa$-stable), every indiscernible sequence is an indiscernible set. In fact, this condition characterizes stability! See Lemma 9.1.1 and Exercise 9.1.5 in Tent and Ziegler.
  2. Let $\varphi(\bar{a},\bar{y})$ be the formula witnessing algebraic dependence of $\bar{b}$ over $\bar{a}$. Note that because $(\bar{b}^i)_{i\in \omega}$ is an indiscernible set, it doesn't actually matter that the truth value of $\varphi(\bar{a},\bar{y})$ alternates on the sequence (true on odds, false on evens) - what matters is just that there are infinitely many $i$ such that $\varphi(\bar{a},\bar{b}^i)$ is true and infinitely many such that it is false. In fact, this is another characterization of stability: A theory $T$ is stable if and only if for every indiscernible sequence $(\bar{b}^i)_{i\in \omega}$ and every formula $\varphi(\bar{a},\bar{y})$, the set $\{i\in \omega\mid \varphi(\bar{a},\bar{b}^i)\}$ is finite or cofinite. So stability is characterized by a kind of minimality condition on indiscernible sequences.
  3. Let's say we didn't know that $(\bar{b}^i)_{i\in \omega}$ was an indiscernible set. We could still get continuum-many types, we would just have to restrict ourselves to using order embeddings instead of permutations. Let $X\subseteq \omega$ be an arbitrary subset. Then there is an order embedding $\sigma_X\colon \omega\to\omega$ such that $\sigma^{-1}_X(\text{Odds}) = X$. We build $\sigma_X$ by induction: define $\sigma_X(n)$ to be the least odd number greater than $\text{ran}(\sigma_X\restriction \{0,\dots,n-1\})$ if $n\in X$ and the least even number greater than $\text{ran}(\sigma_X\restriction \{0,\dots,n-1\})$ if $n\notin X$. Now we have an inverse order-isomorphism $\sigma_X^{-1}\colon \text{ran}(\sigma_X)\to \omega$, which induces a partial elementary map $(\bar{b}^i)_{i\in \text{ran}(\sigma_X)}\to (\bar{b}^i)_{i\in \omega}$. Letting $p_X = \text{tp}(\bar{a}/(\bar{b}^i)_{i\in \text{ran}(\sigma_X)}$ and $q_X = (\sigma_X^{-1})_* p_X$, we have $\{i\in \omega\mid \varphi(\bar{x},\bar{b}^i)\in q_X\} = X$, where again $\varphi(\bar{a},\bar{y})$ is the formula witnessing algebraic dependence of $\bar{b}$ over $\bar{a}$. So if $X\neq X'$, then $q_X \neq q_{X'}$, and we get continuum-many types. Note that this time (in the construction of $\sigma_X$) it was actually important that the truth value of $\varphi(\bar{a},\bar{y})$ alternated on the sequence. It turns out that for a complete theory $T$, the existence of an indiscernible sequence $(\bar{b}^i)_{i\in \omega}$ and a formula $\varphi(\bar{a},\bar{y})$ such that $\varphi(\bar{a},\bar{b}^i)$ is true if and only if $i$ is odd characterizes $T$ having the independence property. Theories without the independence property (called NIP theories for "Not the Independence Property") are a class of theories which contain the stable theories but also theories with definable orders, like the theory DLO of dense linear orders without endpoints or the theory RCF (real closed fields) of $\mathbb{R}$ as a field.
  • $\begingroup$ thank you so much as always for the very thorough answer, will read through it later tonight. I feel bad for taking up so much of your time on this website!! truly I greatly greatly appreciate your help and the time you put in; thank you $\endgroup$ Dec 5, 2020 at 23:31
  • 1
    $\begingroup$ You're welcome! I wouldn't spend so much time answering questions here if I didn't enjoy it. And your questions in particular are always very good and well-considered, so it's a pleasure to help. $\endgroup$ Dec 6, 2020 at 1:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.