Strong minimality

A one-sorted infinite structure $$M$$ is said to be minimal if every definable subset is finite or cofinite. Here, "definable" means "definable with parameters." If this remains true in elementary extensions of $$M$$, $$M$$ is called strongly minimal. A theory $$T$$ is strongly minimal if every model is minimal, or equivalently, every model is strongly minimal.

More generally, an infinite definable set $$S$$ in a structure $$M$$ is said to be minimal if for every definable set $$D$$, either $$S \cap D$$ or $$S \setminus D$$ is finite. If this remains true in elementary extensions, $$S$$ is said to be a strongly minimal set. In terms of Morley rank, $$S$$ is strongly minimal exactly when $$S$$ has Morley rank and degree 1.

Example
The canonical examples of strongly minimal theories:


 * The theory of an infinite set with no structure.
 * The theory of infinite $$k$$-vector spaces, for $$k$$ a field (or skew field).
 * ACF

In each case, strong minimality follows from quantifier elimination results. For example, in ACF, any definable subset of the home sort must be a constructible subset of the affine line, hence a boolean combination of Zariski closed subsets of the affine line. But Zariski closed subsets of the affine line are either finite or everything.

The simplest example of a structure which is minimal, but not strongly minimal, is $$(\mathbb{N},<)$$.

Pregeometries from Minimal Sets
'''Lemma. ''' If $$M$$ is a minimal structure, then the closure operator acl(-) on $$M$$ satisfies the exchange principle: if $$a$$ and $$b$$ are singletons and $$C \subset M$$, and if $$a \in \operatorname{acl}(Cb) \setminus \operatorname{acl}(C)$$, then $$b \in \operatorname{acl}(Ca)$$.

Proof. By definition of $$\operatorname{acl}$$, there must be some $$C$$-formula $$\phi(x;y)$$ such that $$\phi(a;b)$$ holds and $$\phi(M;b)$$ is finite, of size $$n$$. Let $$\psi(x;y)$$ be the $$C$$-formula $$\phi(x;y) \wedge \exists_{\le n} x' \phi(x';y)$$, the formula which says that $$\phi(x;y)$$ holds and $$|\phi(M;y)| \le n$$. Then $$\psi(a;b)$$ holds, and $$\psi(M;c)$$ has size at most $$n$$ for each $$c \in M$$.

If $$\psi(a;M)$$ is finite, then $$b \in \operatorname{acl}(Ca)$$, and we are done. Otherwise, $$\psi(a;M)$$ is cofinite. Let $$m = |M \setminus \psi(a;M)|$$. Let $$D = \{d \in M : |M \setminus \psi(d;M)| = m\}.$$ Then $$D$$ is $$C$$-definable, and contains $$a$$. If $$D$$ is finite, then $$a \in \operatorname{acl}(C)$$, contradicting the hypotheses. So $$D$$ is infinite. Let $$d_1, \ldots, d_{n+1}$$ be distinct elements of $$D$$. Then $$M \setminus \psi(d_i;M)$$ has size $$m$$ for each $$i$$. So $$\bigcup_{i = 1}^{n+1} M \setminus \psi(d_i;M) = M \setminus \bigcap_{i = 1}^{n+1} \psi(d_i;M)$$ has size at most $$m(n+1)$$, and in particular, is finite. This ensures that $$\bigcap_{i = 1}^{n+1} \psi(d_i;M)$$ is non-empty. If $$c \in \bigcap_{i = 1}^{n+1} \psi(d_i;M)$$, however, then $$\{d_1,\ldots,d_{n+1}\} \subset \psi(M;c)$$, contradicting the fact that $$\psi(M;c)$$ has size at most $$n$$ for every $$c$$. QED

Since $$\operatorname{acl}$$ is always finitary closure operator, it follows that on minimal structures (and more generally, minimal sets), $$\operatorname{acl}$$ is the closure operator of some pre-geometry.

For ACF, this recovers the algebraic independence pregeometry. For the theory of $$k$$-vector spaces, this recovers the pregeometry of linear independence in an infinite dimensional $$k$$-vector space. For the theory of equality, this recovers the trivial pregeometry in which every set is closed.

The (false) Zilber Trichotomy Conjecture said (among other things) that these were essentially the only possible geometries for a strongly minimal set. Namely, if $$S$$ was a strongly minimal set, then the associated geometry had to be either


 * Trivial, so every set is its own closure.
 * The geometry associated to linear independence, i.e., the geometry associated to infinite dimensional projective space over a skew field $$k$$.
 * The geometry associated to algebraic independence.

Hrushovski found counterexamples to the Zilber Trichotomy Conjecture. However, there are a number of settings in which the Zilber trichotomy conjecture is known to hold. For example, it holds of the strongly minimal sets occurring in DCF and CCM. Moreover, it is known to hold in a totally categorical setting. In fact, if $$S$$ is $$\aleph_0$$-categorical, then the associated geometry must either be trivial, or must be the geometry associated to some finite field.

Dimension theory in strongly minimal sets
Assume $$M$$ is strongly minimal. If $$C$$ is a set of parameters, then the operation $$\operatorname{acl}_C(-) : S \mapsto \operatorname{acl}(C \cup S)$$ yields a pregeometry on $$M$$. A finite set $$\{a_1, \ldots, a_n\}$$ is independent (over $$C$$) if $$a_i \notin \operatorname{acl}_C(a_1, \ldots, a_{i-1})$$ for each $$i$$. The rank of a finite tuple $$a$$ over $$C$$ can be defined to be the size of a maximal independent subset of $$a$$. The rank of a $$C$$-definable set $$D$$ is defined to be the maximum of $$R(a/C)$$ for $$a \in D$$. It turns out that $$R(D)$$ does not depend on the choice of defining parameters $$C$$.

In the case of ACF, $$R(a/C)$$ is the transcendence degree of $$a$$ over $$C$$, and $$R(D)$$ is the dimension of $$D$$.

This notion of rank agrees with Lascar rank and Morley rank, and has many intuitive properties that one expects of dimension. Some of these are listed below:


 * If $$D_1 \subset D_2$$, then $$R(D_1) \subset R(D_2)$$.
 * If $$D_1 \to D_2$$ is a definable surjection, then $$R(D_1) \ge R(D_2)$$. If the fibers of this surjection are finite, then $$R(D_1) = R(D_2)$$.
 * $$R(D) = 0$$ if and only if $$D$$ is finite.
 * $$R(D_1 \cup D_2) = \max(R(D_1),R(D_2))$$.
 * $$R(D_1 \times D_2) = R(D_1) + R(D_2)$$.
 * Rank is definable in families: if $$\phi(x;y)$$ is a formula, then the set of $$b$$ such that $$R(\phi(M;b)) = n$$ is definable for each $$n$$.
 * If $$f : D_1 \to D_2$$ is a definable surjection and every fiber has rank $$k$$, then $$R(D_1) = k + R(D_2)$$, unless $$D_2$$ is empty.
 * $$R(ab/C) = R(a/bC) + R(b/C)$$.
 * $$R(a/C) = 0$$ if and only if $$a \in \operatorname{acl}(C)$$.
 * If $$R(a/C) = n$$, then there is some $$C$$-definable set $$D$$ containing $$a$$ with dimension $$n$$.
 * The rank function can also be extended to interpretable sets (to $$M^{eq}$$) in a way which maintains all the above properties.

Strong minimality and uncountable categoricity
Strongly minimal (complete, countable) theories are always uncountably categorical. This happens for essentially the same reason that ACF is uncountably categorical. If $$T$$ is a strongly minimal theory, then it can be shown that a model $$M$$ of $$T$$ is determined up to isomorphism by the cardinality of a basis of the associated pregeometry.

In more detail, here is a proof:

Proof. Let $$\mathbb{U}$$ be a monster model of $$T$$. If $$C$$ is any subset of $$\mathbb{U}$$, let $$p_C(x)$$ be the type over $$C$$ asserting that $$x$$ is not in any finite $$C$$-definable sets, i.e., that $$x$$ is not algebraic over $$C$$. Then $$p_C(x)$$ is consistent (because in an elementary extension of $$\mathbb{U}$$, all the new elements will not be algebraic over $$C$$). Also, $$p_C(x)$$ is a complete type over $$C$$. Indeed, if $$D$$ is a $$C$$-definable set, then either $$D$$ is finite or cofinite. In the first case, $$x \notin D$$ is in $$p(x)$$, and in the second case, $$x \in D$$ is in $$p(x)$$.

If $$C \subset C'$$, then $$p_{C'} | C = p_C$$, because if an element in an elementary extension of $$\mathbb{U}$$ were not algebraic over $$C'$$, it would certainly not be algebraic over the smaller set $$C$$. Consequently, the union of the $$p_C$$’s, as $$C$$ ranges over all sets, is $$p_\mathbb{U}$$, and $$p_C = p_\mathbb{U}|C$$ for each $$C$$. Let $$p = p_\mathbb{U}$$. The type $$p$$ is a global type, invariant under $$\operatorname{Aut}(\mathbb{U})$$. From generalities on global invariant types, it follows that if $$a_1, \ldots, a_n$$ and $$b_1, \ldots, b_n$$ are two sequences such that $$a_i \models p | \{a_1, \ldots, a_{i-1}\}$$ $$b_i \models p | \{a_1, \ldots, a_{i-1}\}$$ for each $$i$$, then $$(a_1,\ldots,a_n)$$ and $$(b_1,\ldots,b_n)$$ have the same type over the empty set. Equivalently, the map sending $$a_i \mapsto b_i$$ is a partial elementary map.

But the condition that $$a_i \models p | \{a_1, \ldots, a_{i-1}\}$$ for each $$i$$ just means that $$a_i \notin \operatorname{acl}(a_1, \ldots, a_{i-1}),$$ for each $$i$$, or equivalently, that the set $$\{a_1, \ldots, a_n\}$$ is independent.

So if $$I$$ and $$J$$ are two finite independent sets and $$f$$ is a bijection from $$I$$ to $$J$$, then $$f$$ is a partial elementary map. More generally, if $$I$$ and $$J$$ are two independent sets of arbitrary size, and $$f$$ is any bijection from $$I$$ to $$J$$, then $$f$$ is a partial elementary map. (It suffices to check that $$f$$ is a partial elementary map on finite subsets $$I_0$$ of $$I$$.)

Now suppose $$M$$ and $$M'$$ are two models of the same uncountable cardinality. We may assume that $$M$$ and $$M'$$ sit inside $$\mathbb{U}$$. Let $$I$$ be a maximal independent subset of $$M$$, and let $$I'$$ be a maximal independent subset of $$M'$$. Then $$M = \operatorname{acl}(I)$$ and $$M' = \operatorname{acl}(I')$$. Since $$T$$ is countable, $$|M| = |I| + \aleph_0$$ and similarly for $$M'$$. Since $$|M|$$ is uncountable, $$|M| = |I|$$, and similarly, $$|M'| = |I'|$$. Finally, since $$M$$ and $$M'$$ have the same cardinality, $$|I| = |I'|$$. Let $$f$$ be a bijection from $$I$$ to $$I'$$. Then $$f$$ is a partial elementary map, so it can be extended to an automorphism $$\sigma$$ of $$\mathbb{U}$$. Then $$\sigma(M) = \sigma(\operatorname{acl}(I)) = \operatorname{acl}(\sigma(I)) = \operatorname{acl}(I') = M'.$$ So $$\sigma$$ gives an isomorphism between $$M$$ and $$M'$$. QED

The converse is far from true. However, some variant of it is still true:

'''Fact. ''' Suppose $$T$$ is a countable theory which is $$\kappa$$-categorical for some uncountable cardinal $$\kappa$$. Then any model $$M$$ of $$T$$ contains a strongly minimal set $$D$$, with $$M$$ constructible over $$D$$. The model $$M$$ is determined up to isomorphism by the size of a pregeometry basis in $$D$$. Moreover, if $$D$$ and $$D'$$ are two strongly minimal sets definable in $$M$$, the underlying geometries of $$D$$ and $$D'$$ are the same.

This fact is a key step in the proof of Morley's Theorem and the Baldwin-Lachlan theorem.