Questions: Algebraically Closed Fields: Model-Theoretic Analysis

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A logician wants to decide algorithmically whether the first-order sentence 'every element has a square root' holds in all algebraically closed fields of characteristic 0. Which property of ACF makes such algorithmic decision possible?

AThe axiom of choice guarantees root existence in any algebraically closed field
BQuantifier elimination reduces every sentence to a quantifier-free statement, which is evaluable as true or false — giving a decision procedure
CGödel's completeness theorem implies every sentence is provable or refutable in ACF₀
DCategoricity in ℵ₁ means all models agree on truth values of all sentences
Question 2 Multiple Choice

A student argues: 'Since there are many algebraically closed fields of characteristic 0 — the complex numbers, their elementary extensions of larger cardinality, etc. — the theory ACF₀ cannot be complete: different models might disagree on some sentence.' What is wrong with this reasoning?

AACF₀ is not complete — the student is correct
BCompleteness requires that all models be isomorphic, which fails here, so the argument has a false premise
CCompleteness means every sentence has the same truth value across all models; QE guarantees this even when models are non-isomorphic
DAll algebraically closed fields of characteristic 0 are isomorphic regardless of cardinality, so the premise is false
Question 3 True / False

ACF is categorical in most infinite cardinalities: for each infinite cardinal κ and each characteristic, there is exactly one algebraically closed field of cardinality κ up to isomorphism.

TTrue
FFalse
Question 4 True / False

The key step in proving quantifier elimination for ACF — eliminating an existential quantifier ∃y from a formula involving polynomials in x and y — uses the fact that every polynomial of degree ≥ 1 over an algebraically closed field has a root.

TTrue
FFalse
Question 5 Short Answer

Why does quantifier elimination imply decidability for ACF, and why would QE alone not give decidability for a theory with infinitely many complete extensions?

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