10. Algebraic Closure in Minimal Structures
Throughout this section let A denote an infinite minimal L-structure with
underlying set A. We will write cl(X) in place of aclA (X) for X ⊆ A.
From Proposition 9.3 we know that cl is a closure operation of finite character on the subsets of A. When A is minimal, cl is actually a (combinatorial)
pregeometry; this means that cl also satisfies the Exchange Property:
10.1. Proposition. Let A be an infinite minimal structure. Let X ⊆ A
and a, b ∈ A. If a 6∈ cl(X) and b 6∈ cl(X), then
a ∈ cl(X ∪ {b}) ⇐⇒ b ∈ cl(X ∪ {a}).
Proof. We argue by contradiction. Suppose a, b 6∈ cl(X), a ∈ cl(X ∪ {b}),
and b 6∈ cl(X ∪ {a}). Let the formula ϕ(x, y, z1 , . . . , zp ) and the parameters
e1 , . . . , ep ∈ X witness the fact that a ∈ cl(X ∪ {b}) (where b is included as
a parameter to be substituted for the variable y). Let N be the cardinality
of the finite set {c ∈ A | A |= ϕ[c, b, e1 , . . . , ep ]}, which contains a as an
element. Let ψ(y, z1 , . . . , zp ) be a formula expressing that there are at
most N values of x for which ϕ(x, y, z1 , . . . , zp ) is true. Note that A |=
ψ[b, e1 , . . . , ep ]. Since b 6∈ cl(X ∪ {a}), the set
{b0 ∈ A | A |= ϕ[a, b0 , e1 , . . . , ep ] and A |= ψ[b0 , e1 , . . . , ep ]}
must be infinite; since A is minimal this set must be cofinite in A. Let M
be the number of elements of A that are not in this set.
Now consider a formula σ(x, z1 , . . . , zp ) that expresses the statement that
ϕ(x, y, z1 , . . . , zp ) ∧ ψ(y, z1 , . . . , zp ) holds for all but M many values of y.
The set
{c ∈ A | A |= σ[c, e1 , . . . , ep ]}
has a as an element; since a 6∈ cl(X) and A is minimal, this set must be
cofinite. Let a0 , . . . , aN be distinct elements of this set. For each j =
0, . . . , N we have that the set
{b0 ∈ A | A |= ϕ[aj , b0 , e1 , . . . , ep ] and A |= ψ[b0 , e1 , . . . , ep ]}
must be cofinite in A, which is infinite. Therefore the intersection of these
sets is also cofinite, hence nonempty. That is, there must exist a single
b0 ∈ A such that for each j = 0, . . . , N we have
A |= ϕ[aj , b0 , e1 , . . . , ep ] and A |= ψ[b0 , e1 , . . . , ep ]
which is a contradiction.
10.2. Definition. Let cl be a pregeometry on the set A; let X, Y ⊆ A.
(1) X is closed if cl(X) = X.
(2) cl(X) is the closure of X.
(3) (Y closed) X spans Y if cl(X) = Y .
(4) X is independent if a 6∈ cl(X \ {a}) for all a ∈ X.
(5) (Y closed) X is a basis for Y if X is independent and X spans Y .
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10.3. Theorem. Let cl be a pregeometry on the set A; let X, Y ⊆ A with
Y closed.
(1) X is independent if and only if each finite subset of X is independent.
(2) X is a basis for Y if and only if X is maximal among independent
subsets of Y . Consequently every closed set has a basis. Indeed, every
independent subset of Y is contained in a basis for Y .
(3) If X spans Y , then there exists Z ⊆ X such that Z is a basis of Y .
(4) X is a basis for Y if and only if X is minimal among subsets of Y that
span Y .
(5) Suppose X is a basis for Y and a ∈ Y . Then there is a smallest finite
set F ⊆ X such that a ∈ cl(F ). We will call F the support of a in X.
(6) Any two bases for Y have the same cardinality.
Proof. (1) Suppose X is independent and let Z be any subset of X. For
each a ∈ Z we have Z \ {a} ⊆ X \ {a} and therefore cl(Z \ {a}) ⊆ cl(X \
{a}). Since X is independent this implies a 6∈ cl(Z \ {a}). Therefore
Z is independent. In particular every finite subset of X is independent.
Conversely, suppose X is dependent, so there exists a ∈ X such that a ∈
cl(X \ {a}). Therefore there is a finite subset Z of X \ {a} such that
a ∈ cl(Z). It follows that Z ∪ {a} is a dependent finite subset of X.
(2) Suppose X is a basis for Y . For each a ∈ Y \ X we have a ∈ cl(X), from
which it follows that X ∪ {a} is dependent. It follows that X is maximal
among independent subsets of Y . Conversely, suppose X is maximal among
independent subsets of Y . Then for each a ∈ Y \ X the set X ∪ {a} is
dependent. If a 6∈ cl(X) then there exists b ∈ X with b ∈ cl((X ∪{a})\{b}).
Since X is independent we have b 6∈ cl(X \ {b}). The Exchange Property
implies a ∈ cl((X \{b})∪{b}) = cl(X). This contradiction proves a ∈ cl(X).
Sinice a ∈ Y was arbitrary, this proves that X spans Y and therefore X is
a basis for Y .
Suppose X is any independent subset of Y . Let Ω be the collection of all
independent subsets of Y that contain X. Part (1) of this Theorem implies
that if C is any subset of Ω that is a chain under ⊆, then ∪C is independent
and thus is a member of Ω. Zorn’s Lemma implies the existence of maximal
elements of Ω under ⊆. Any such set is a basis of Y , by what was proved
in the preceding paragraph.
(3) The proof is similar to the second part of (2). Given X spanning Y ,
let Ω be the collection of all independent subsets of X. By Zorn’s Lemma
and (1) there exists Z ∈ Ω that is maximal with respect to ⊆. By the
argument in the previous paragraph, cl(Z) = cl(X) and therefore Z is an
independent set spanning Y . By (2), Z is a basis for Y .
(4) Suppose X is a basis for Y and Z is a proper subset of X. For each
a ∈ X \ Z we have a 6∈ cl(X \ {a}) ⊇ cl(Z) which shows that Z does
not span Y . Conversely suppose X is minimal among sets that span Y .
We must show X is independent. Otherwise there exists a ∈ X such that
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a ∈ cl(X \ {a}). It follows that cl(X \ {a}) = cl(X), contradicting the
assumption that X is a minimal spanning set.
(5) Since cl has finite character we know there exists a finite F ⊆ X with
a ∈ cl(F ). Let F be such a set of smallest cardinality. We will show
a 6∈ cl(X \ {b}) for each b ∈ F . It follows that F must be contained
in any subset A of X that satisfies a ∈ cl(A). If b ∈ F then we have
a 6∈ cl(F \ {b}) by the minimality of F . The Exchange Property implies
b ∈ cl((F \ {b}) ∪ {a}). Since b 6∈ cl(X \ {b}) we see it is impossible for a
to be in cl(X \ {b}).
(6) Let U and V be bases for Y . The case where one of the bases is
infinite can be proved using a simple counting argument based on the finite
character of cl. Suppose V is infinite and card(U ) ≤ card(V ). For each
a ∈ U there exists a finite set F (a) ⊆ V such that a ∈ cl(F (a)). Let
F = ∪{F (a) | a ∈ U }. Evidently F spans Y , and since V is a basis
for Y it follows from (3) that F = V . Since V is infinite it follows that
U is also infinite and indeed that card(V ) = card(F ) ≤ card(U ). Hence
card(U ) = card(V ).
Now we handle the finite case. Let U be a finite basis for Y and let V be
any independent subset of Y . By what we proved above, V must be finite.
We will show that card(V ) ≤ card(U ), which suffices to complete the proof
of (4). To do this we prove the following statement by induction on the
cardinality of V :
there exists W ⊆ U such that W ∩ V = ∅, W ∪ V is a basis for Y , and
card(W ∪ V ) = card(U ).
As basis step we consider the case card(V ) = 0. Evidently we may take
W = U when V = ∅.
For the induction step consider an independent set V and suppose the
statement is true for all independent sets that are smaller than V . Fix
a ∈ V and let Z = V \{a}. By the induction hypothesis there exists W ⊆ U
such that W ∩ Z = ∅, W ∪ Z is a basis for Y , and card(W ∪ Z) = card(U ).
Let A be the support of a in W ∪ Z. Since V is independent, A must meet
W . Let b be any element of A ∩ W . By (5) we have a 6∈ cl((W \ {b}) ∪ Z).
We complete the proof by showing that W \ {b} is the desired subset of U
for V = Z ∪{a}. The Exchange Property yields b ∈ cl((W \{b})∪(Z ∪{a})).
Therefore (W \ {b}) ∪ (Z ∪ {a}) spans Y . Since b is in the support of a in
W ∪ Z we have a 6∈ cl((W \ {b}) ∪ Z). It follows that (W \ {b}) ∪ (Z ∪ {a})
is independent, that (W \ {b}) ∩ (Z ∪ {a}) = ∅, and that card((W \ {b}) ∪
(Z ∪ {a})) = card(W ∪ Z) = card(U ). This completes the proof.
10.4. Definition. Let cl be a pregeometry on the set A and X ⊆ A. The
dimension of X, which is denoted dim(X), is the unique cardinality of a
basis for the closed set cl(X).
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10.5. Definition. Let A be a minimal L-structure. The dimension of A,
denoted dim(A), is the dimension of the set A with respect to the pregeometry aclA .
10.6. Proposition. Let A and B be L-structures with A ≡ B and A minimal. Suppose X ⊆ A and Y ⊆ B, and let f : X → Y be a function that
is elementary with respect to A, B. For each a ∈ A \ aclA (X) and each
b ∈ B \ aclB (Y ) the extension of f that takes a to b is also elementary with
respect to A, B.
Proof. Otherwise there exists an L-formula ϕ(x, y1 , . . . , yn ) and parameters e1 , . . . , en ∈ X such that A |= ϕ[a, e1 , . . . , en ] and B |=
¬ϕ[b, f (e1 ), . . . , f (en )]. Since a is not algebraic over X and b is not algebraic over Y , the sets {c ∈ A | A |= ϕ[c, e1 , . . . , en ]} and {d ∈ B | B |=
¬ϕ[d, f (e1 ), . . . , f (en )]} are both infinite. Since f is elementary it follows
that {c ∈ A | A |= ¬ϕ[c, e1 , . . . , en ]} is infinite. This contradicts the assumption that A is minimal.
10.7. Corollary. Let A and B be L-structures with A ≡ B and A minimal.
Suppose we have independent sets X ⊆ A (with respect to aclA ) and Y ⊆ B
(with respect to aclB ), and let f : X → Y be a 1-1 function. Then f is
elementary with respect to A, B.
Proof. Let Ω be the collection of subsets S ⊆ X such that the restriction
of f to S is elementary with respect to A, B. We regard ∅ as an element
of Ω (justified since A ≡ B). The partially ordered set (Ω, ⊆) satisfies the
hypothesis of Zorn’s Lemma, so there exists S ∈ Ω that is maximal with
respect to ⊆. We need to show S = X. If not, let a be any element of X \ S
and let b = f (a) ∈ Y \ f (S). By Proposition 10.6, f restricted to S ∪ {a}
is elementary with respect to A, B. This contradicts the maximality of S
and proves S = X.
10.8. Theorem. Let T be a complete strongly minimal theory with infinite
models, and let A, B be models of T .
(1) There is an elementary embedding of A into B if and only if dim(A) ≤
dim(B).
(2) A and B are isomorphic if and only if dim(A) = dim(B).
(3) T is κ-categorical for every cardinal number κ > card(L).
Proof. (1⇐) Since acl defines a pregeometry in each of these structures,
there exist bases X for A (with respect to aclA ) and Y for B (with respect
to aclB ). By hypothesis card(X) ≤ card(Y ) so there is a 1-1 function
f from X into Y . The preceding Corollary yields that f is elementary
with respect to A, B. By Proposition 9.6, f can be extended to a function
g : A → B that is elementary with respect to A, B. It follows easily that g
is an elementary embedding from A into B.
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(1⇒) As above, there is a basis X for A (with respect to aclA ). If f is an
elementary embedding of A into B, then f (X) is independent with respect
to aclB . By Theorem 10.3(2) there is a basis Y for B that contains f (X).
It follows that card(X) ≤ card(Y ) and hence dim(A) ≤ dim(B).
(2) The argument is similar to (1).
(3) By the Löwenheim-Skolem Theorems (Theorems 6.1 and 6.3) there exist
models of T having cardinality κ. Let A and B be two such models of T . As
in the proof of (1), let X be a basis for A (with respect to aclA ) and Y for
B (with respect to aclB ). Because acl is of finite character in each model,
and there are fewer than κ formulas in L, a counting argument shows that
X and Y must each be of cardinality equal to κ. Now use part (2).
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