It is well-known that if $M$ is an algebraically closed field, then the model-theoretic algebraic closure of any set $A \subset dom(M)$ is the same as its algebraic closure in the sense of field theory (roots of polynomials in $A[x]$). Does the same hold if the field is not algebraically closed? I need a proof or a counterexample.

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    $\begingroup$ What have you tried? $\endgroup$ Sep 18, 2020 at 9:31
  • $\begingroup$ @MarkKamsma Frankly, I have no idea. $\endgroup$ Sep 18, 2020 at 11:14
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    $\begingroup$ Maybe I misunderstand the question, but wouldn't any non-algebraically closed field be a counterexample (so concretely e.g. $\mathbb{Q}$)? For a non-algebraically closed field $F$ the model-theoretic algebraic closure of $F \subseteq F$ is just $F$, while the algebraic closure in the field-theoretic sense has to be bigger. $\endgroup$ Sep 18, 2020 at 15:12
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    $\begingroup$ @MarkKamsma I think the most reasonable interpretation of the question is "for an arbitrary field $F$, does model-theoretic algebraic closure of a subset $A$ agree with the relative algebraic closure in $F$ of the subfield generated by $A$? $\endgroup$ Sep 18, 2020 at 15:49
  • $\begingroup$ @MarkKamsma The text of the exercise I'm trying to solve is a bit ambiguous, but I agree with Alex's interpretation. $\endgroup$ Sep 18, 2020 at 16:11

1 Answer 1


There are certainly fields which are not algebraically closed, but in which model-theoretic algebraic closure agrees with relative field-theoretic algebraic closure. For example, this happens in real closed fields, p-adically closed fields, and pseudofinite fields.

On the other hand, this is not true in all fields. Many examples are given / constructed in the following paper: https://www.jstor.org/stable/3648549


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