The question is:
Let $M\models \text{ACF}$. Prove that every definable subset of $M$ is definable in
$M|_\varnothing$ (the reduct of $M$ to the empty language).
Note that this questioned is only concerned with subsets of $M$ not subsets of $M^n$. I've sat on this for a few days and have zero clue where to begin. I figure it has something to do with the fact that $\mathrm{ACF}$ has quantifier elimination. Any hints are appreciated!
Best Answer
For context: This property is called "strong minimality". The question is asking you to show that $\mathrm{ACF}$ is strongly minimal.
You're right that it's a direct consequence of quantifier elimination.
Step 1: Show that every definable (with parameters) subset of $M\models \mathrm{ACF}$ is finite or cofinite. Hint: A non-zero polynomial has only finitely many roots. What does an atomic formula look like? What does a general formula look like (using quantifier elimination)?
Step 2: Show that every finite or cofinite subset of $M$ is definable (with parameters) in the empty language, i.e. using only $=$.