Let M be a structure. Definable closure and algebraic closure are two closure operators on M. If S is a subset of M, and a is an element (or finite tuple) from M, then

• a is in the definable closure of S if the singleton set {a} is definable over S. An equivalent condition is that a = f(b) for some 0-definable function f and some bS.
• a is in the algebraic closure of S if a is an element of some finite S-definable set.

The definable closure of S is denoted dcl(S) and the algebraic closure is denoted acl(S). Sometimes we want dcl(S) and acl(S) to be subsets of M, in which case we take dcl(S) and acl(S) to be the sets of singletons in the definable closure and algebraic closure of S, respectively. There is not a very big distinction, because it turns out that a tuple a is in dcl(S) or acl(S) if and only if each coordinate of a is in dcl(S) or acl(S), respectively.

Both dcl(-) and acl(-) are finitary closure operators on M, meaning that

• S ⊆ dcl(S) and S ⊆ acl(S)
• If ST, then dcl(S) ⊆ dcl(T) and acl(S) ⊆ acl(T)
• dcl(dcl(S)) = dcl(S) and acl(acl(S)) = acl(S)
• If a is in dcl(S) or acl(S), then a is in dcl(S0) or acl(S0) for some finite subset S0 of S.

Also, in general one has dcl(S) ⊆ acl(S).

A set S is algebraically closed if S = acl(S), and definably closed if S = dcl(S).

## Preservation in Elementary Extensions

Like most good concepts in model theory, these notions behave well under elementary extensions. Specifically, if is an elementary extension of M, and SM, then dcl(S) computed inside N is the same as dcl(S) computed inside M. The same holds for acl(S).

## Conjugates

If a ∈ acl(S), then a is in some finite S-definable set D. The intersection of all such D is still a finite S-definable set containing a, so there is a unique minimal finite S-definable set D containing a. The set D consists exactly of the realizations of tp(a/S). The elements of D are called the conjugates of a over S.

## Interpretation in Saturated Models

If M is |S|+-saturated, then

• a ∈ dcl(S) if and only if tp(a/S) has a unique realization in M (namely a).
• a ∈ acl(S) if and only if tp(a/S) has finitely many realizations in M.

If furthermore M is |S|+-strongly homogeneous, then one gets a nice description of dcl(S) and acl(S) in terms of automorphisms:

• a ∈ dcl(S) if and only if a is fixed by Aut(M/S)
• a ∈ acl(S) if and only if a has a finite orbit under Aut(M/S). This orbit consists exactly of the conjugates of a over S.

## Examples

ACF: Let K be an algebraically closed field, and S be a subset of K. Then

• dcl(S) is the perfect hull of the field generated by S
• acl(S) is the algebraic closure of the field generated by S.

In particular, in characteristic zero, dcl(S) is just the field generated by S.

DLO: If M is a dense linear order, and S is a subset of M, then acl(S) = dcl(S) = S. In other words, all sets are algebraically closed and definably closed.

RCF: If K is a real closed field, and S is a subset of K, then acl(S) and dcl(S) are both equal to the relative algebraic closure in K of the field generated by S.

ACVF: If K is an algebraically closed valued field, non-trivially valued, and S is a subset of K, then

• acl(S) is the algebraic closure of the field generated by S
• dcl(S) is the smallest perfect henselian subfield of K containing S.

Linearly ordered structures: In general, if M is a structure with a 0-definable total ordering, then acl(-) and dcl(-) agree on M.

## Proofs

This section proves the unproven claims made above.

### The definitions

We asserted that a ∈ dcl(S) if and only if a = f(b) for some 0-definable function f and some tuple b from S.

Proof: If a ∈ dcl(S), then {a} is S-definable. So there is some formula and some tuple b from S such that

Let be the formula

In other words, asserts that not only does hold, but that no other value of x works.

Now holds. And for every tuple c from M, the set is either a singleton or empty. Therefore is the graph of (the transpose of) a 0-definable (partial) function f such that a = f(b).

(If we wanted a total definable function, we could set f(b) = b when f(b) is currently undefined, or something similar.) QED.

### Tuples vs Singletons

Lemma: An n-tuple a is in dcl(S) if and only if each coordinate of a is in dcl(S). Similarly, a is in acl(S) if and only if each coordinate of a is in acl(S).

As a consequence of this, the notions of algebraic and definable closure are determined by what they do to singletons. Proof: For 1 ≤ in, let πi : Mn -> M be the projection onto the ith coordinate. Let ai be the ith coordinate of a.

Suppose first that a is in dcl(S). Then {a} is S-definable. So is πi({a}) = {ai} for each coordinate i. Consequently, ai is in the definable closure of S for each i. Conversely, if {ai} is S-definable for each i, then so is

,

so a is in dcl(S).

Next suppose that a is in acl(S). Thus a is in some finite S-definable set DMn. For each i, ai is in the finite S-definable set πi(D) so ai ∈ acl(S).

Conversely, suppose that ai ∈ acl(S) for each i. Then for each i there is an S-definable finite set DiM containing ai. Then a is in a finite S-definable set:

so a ∈ acl(S). QED.

### Dcl and acl are finitary closure operators

Finitariness: Suppose a is in dcl(S). Then {a} is S-definable. The definition can only use a finite subset S0S. So {a} is S0-definable, so a ∈ dcl(S0). Similarly, if a is in acl(S), then a is in some finite S-definable set D. But the definition of D can only use finitely many elements of S, so D is S0-definable for some S0. Then a ∈ acl(S0).

Increasingness: Suppose aS. Then the set {a} is S-definable, by the formula x = a. So a ∈ dcl(S). And {a} is finite, so a ∈ acl(S). Therefore S ⊆ dcl(S) and S ⊆ acl(S).

Monotonicity: Suppose ST. If a ∈ dcl(S), then {a} is S-definable, hence T-definable, and a ∈ dcl(T). Similarly, if a ∈ acl(S), then a is in a finite S-definable set D. Of course D is T-definable, so a ∈ acl(T). Thus dcl(S) ⊆ dcl(T) and acl(S) ⊆ acl(T).

Idempotence of dcl(-): As dcl(-) is increasing, we have dcl(S) ⊆ dcl(dcl(S)). We need to show the reverse inclusion. Suppose a ∈ dcl(dcl(S)). Then for some tuple b from dcl(S) and some formula . By the Lemma above, b ∈ dcl(S). Consequently, for some tuple c from S and some formula .

Let assert that there is a unique y such that holds and that for this y, holds as well, so is the formula:

Then holds. Also, if holds for some tuple a' then holds, and consequently a' = a. So . But is 0-definable, and c is a tuple from S, so a ∈ dcl(S). Consequently, dcl(dcl(S)) ⊆ dcl(S).

Idempotence of acl(-): This one is hardest. Because acl(-) is increasing, we have acl(S) ⊆ acl(acl(S)). We need the reverse inclusion. Suppose that a ∈ acl(acl(S)). Then a is in a finite acl(S)-definable set D. Write where b ∈ acl(S). If k is the cardinality of D, we can modify the formula so that for every value of y, at most k values of x satisfy . Specifically, we replace with the formula

Similarly, since b ∈ acl(S), we can find a formula , a number j, and a tuple c from S such that holds, and such that for every value of z, at most j values of y satisfy .

Now let be the formula

For each value of z, there at at most kj values of x satisfying . So . But , so a ∈ acl(S), because c is from S.

### Behavior in elementary extensions

If is an elementary extension of M, and S ⊆ M, then dcl(S) is the same when computed in N or M. The same holds for acl(S).

Proof: Both of these claims follow easily from the fact that if is finite, then , a general fact about elementary extensions. From the definition of elementary extension, one knows that

However, we can write a statement asserting that there are exactly values of x such that holds. Then , and hence . Thus . Two finite sets of the same size must be equal, if one includes the other. QED.

### Conjugates and Types

Suppose a ∈ acl(S), and D is the set of conjugates of a over S. Then D is exactly the set of realizations of tp(a/S).

Proof: Suppose b realizes tp(a/S). The type tp(a/S) includes the statement that xD, so bD. Conversely, suppose bD, but b does not realize tp(a/S). Then there is some S-definable set E containing a but not b. Then DE is a finite S-definable set containing a. By choice of D, D = DE, so DE. But bD and b is not in E, a contradiction. QED.

### The case of saturated models

Claim: Suppose M is |S|+-saturated, and a is a finite tuple. Then a ∈ dcl(S) if and only if tp(a/S) has a unique realization in M.

Proof: If a ∈ dcl(S), then D := {a} is S-definable. The type of a over S includes the statement that xD. Consequently, any other realization of tp(a/S) must be in D, i.e., must equal a.

Conversely, suppose a is the unique realization of tp(a/S). Let p(x) be the partial type over S ∪ {a} consisting of tp(a/S) together with the statement x ≠ a. By assumption, there are no realizations of p(x) in M. By the saturation of M, it follows that p(x) must be inconsistent, i.e., not finitely satisfiable. So there is some formula such that

or equivalently

,

so . Since we know that . Thus . Since b is a tuple from S, we conclude that a ∈ dcl(S). QED.

Claim: Suppose M is |S|+-saturated, and a is a finite tuple. Then a ∈ acl(S) if and only if tp(a/S) has finitely many realizations.

Proof: If a ∈ acl(S), then a is in a finite S-definable set D. The type of a over S includes the statement that xD. So any realization of tp(a/S) must be in D. Therefore tp(a/S) has finitely many realizations.

Conversely, suppose that tp(a/S) has finitely many realizations b1, ..., bn. Let p(x) be the partial type over S ∪ {b1, ..., bn} asserting that x satisfies tp(a/S) and x does not equal any bi. By assumption, p(x) is not realized in M. By saturation, p(x) must be inconsistent. It follows that there is some formula such that

or equivalently

So is finite. As , we know that . Therefore a is in a finite S-definable set, and consequently a ∈ acl(S). QED.

Claim: Suppose that M is |S|+-saturated and |S|+-strongly homogeneous. Then a ∈ dcl(S) if and only if a is fixed by Aut(M/S), and a ∈ acl(S) if and only if a has a finite orbit under Aut(M/S).

Proof: By strong homogeneity, the orbit of a under Aut(M/S) is exactly the set of realizations of tp(a/S). Now use the previous claims. QED. ::: ::: :::