# Are there any 1-decidable algebraic extensions of $\mathbb{Q}$ which are not decidable?

A model $$M$$ is decidable if the set of all first-order formulas which are true in $$M$$ is a recursive set. And a model is $$1$$-decidable if the set of all existential formulas which are true in $$M$$ is a recursive set. This paper gives an example of a linear order which is $$1$$-decidable (in fact $$n$$-decidable for all $$n$$) but not decidable. As pointed out by Laurent Moret-Bailly in the comments, by this MO answer, the field $$\mathbb{R}(t)$$ is 1-decidable but not decidable.

My question is, are there any known examples of algebraic extensions of $$\mathbb{Q}$$ which are $$1$$-decidable but not decidable?

• See this post. – Laurent Moret-Bailly Mar 14 at 16:41
• Seems indeed to be a duplicate of Laurent's question: First order decidability of rings vs Diophantine decidability. The question asked for any ring with this property, but the accepted answer yields a field. – YCor Mar 14 at 17:51
• @YCor Do you know of any infinite algebraic extensions of Q with that property? – Keshav Srinivasan Mar 14 at 18:17
• I guess to make the question not formally a duplicate you could edit it to ask only about infinite algebraic extensions of $\mathbb Q$ (or fields whose elements can be enumerated, or something else that avoids the answers to Laurent's question). – Will Sawin Mar 15 at 2:03
• @WillSawin OK. I edited my question as you suggested. – Keshav Srinivasan Mar 15 at 2:17