Questions: Definable Closure and Algebraic Closure
5 questions to test your understanding
Score: 0 / 5
Question 1 Multiple Choice
In the real field (ℝ, +, ×, 0, 1) with parameter set A = ℚ, consider the element √2. The formula φ(x) = 'x² = 2' has exactly two solutions: √2 and −√2. Is √2 in dcl(ℚ), in acl(ℚ), or neither?
AIn dcl(ℚ), because √2 satisfies a polynomial formula over ℚ
BIn acl(ℚ) but NOT in dcl(ℚ), because x² = 2 has finitely many solutions but does not uniquely define √2
CIn neither, because √2 is irrational and therefore not definable from rational parameters
DIn both dcl(ℚ) and acl(ℚ), since algebraic elements over ℚ always satisfy both closure conditions
For an element b to be in dcl(A), there must exist a formula φ(x, ā) over A such that b is the UNIQUE solution. The formula x² = 2 has two solutions (√2 and −√2), so it does not uniquely pin down √2 — and no other formula over ℚ uniquely identifies √2 without also using √2 as a parameter. Therefore √2 ∉ dcl(ℚ). However, for acl(A), we only need a formula over A with FINITELY many solutions, and x² = 2 qualifies (two solutions). So √2 ∈ acl(ℚ). This example shows the strict inclusion dcl(A) ⊊ acl(A): algebraically dependent elements that come in pairs or finite groups land in acl but not dcl.
Question 2 Multiple Choice
Which statement correctly distinguishes definable closure from algebraic closure?
Adcl(A) contains elements satisfying some formula over A with finitely many solutions; acl(A) requires the formula to have exactly one solution
Bdcl(A) requires a formula over A with exactly one solution (unique definability); acl(A) requires a formula over A with finitely many solutions
Cdcl(A) and acl(A) are identical in all first-order theories
Dacl(A) contains only elements that appear in no formula with parameters from A
The definitions are: b ∈ dcl(A) iff there exists a formula φ(x, ā) with ā ∈ A such that φ uniquely defines b (the formula has exactly one solution, namely b). b ∈ acl(A) iff there exists a formula φ(x, ā) with ā ∈ A such that b satisfies φ and φ has only finitely many solutions (but not necessarily one). The inclusion dcl(A) ⊆ acl(A) follows immediately: unique definability is a special case of finite definability. Option A has the definitions reversed. Option C is false — in many theories (including real closed fields), there are elements in acl but not dcl, as the √2 example illustrates.
Question 3 True / False
The inclusion dcl(A) ⊆ acl(A) holds in every first-order theory, because unique definability is a special case of finite definability.
TTrue
FFalse
Answer: True
If b ∈ dcl(A), there is a formula φ(x, ā) over A with exactly one solution, namely b. But then φ trivially has finitely many solutions (just one), so b ∈ acl(A) by definition. This is not a non-trivial theorem — it follows directly from comparing the two definitions. The content of model theory is in understanding when acl(A) is strictly larger than dcl(A) (as in real closed fields, where algebraic conjugates like √2 and −√2 both belong to acl but neither belongs to dcl over ℚ) and in the structural role these operators play in stability theory and independence.
Question 4 True / False
If acl(A) = A — meaning A is algebraically closed in the model-theoretic sense — then it also follows that dcl(A) = A.
TTrue
FFalse
Answer: False
acl(A) = A means every element with a finite orbit over A is already in A — no new elements can be added by taking finite definable sets. But this says nothing about dcl(A): it is possible for acl(A) = A while dcl(A) is a proper subset of A. An element b ∈ A might not be uniquely definable from A\{b} even though it belongs to A. The condition dcl(A) = A (every element of A is uniquely definable from other elements) is independent of acl(A) = A. In stable theories, having both dcl(A) = A and acl(A) = A is what makes A a 'sufficiently closed' set for independence theory, but the two conditions are not equivalent to each other.
Question 5 Short Answer
Explain the difference between an element b being in dcl(A) versus acl(A), using the analogy between these model-theoretic closures and classical algebraic dependence in field theory.
Think about your answer, then reveal below.
Model answer: b ∈ dcl(A) means b is uniquely determined by A — there is a formula over A with b as its only solution, analogous to an element being definable by a formula with a unique root. b ∈ acl(A) means b lives in a finite 'orbit' over A — it satisfies a formula over A that has only finitely many solutions (b may be one of several conjugates), analogous to being a root of a polynomial over a base field. In classical field theory, √2 is algebraic over ℚ (satisfies x²−2=0, which has two roots) but is not uniquely defined by ℚ because its conjugate −√2 is equally valid. Model-theoretic acl(ℚ) captures this: √2 ∈ acl(ℚ) because it has a finite orbit {√2, −√2}, but √2 ∉ dcl(ℚ) because no ℚ-formula uniquely identifies it.
This analogy makes the model-theoretic definitions intuitive for students with algebraic background. The key shift is that model theory generalizes these notions to any first-order structure, not just fields. In a general structure, 'being algebraic over A' means living in a finite set definable from A, which captures the essential feature of algebraic elements (bounded degree over the base) without requiring a field structure.