TITLE: Definability of the ring of integer in algebraic extensions of $\mathbb Q$
QUESTION [10 upvotes]: J. Robinson has proved that exists a formula $\psi(x)$ in the language of rings which,applied to the rational numbers, defines the the ring integers (making the theory of $\mathbb{Q}$ undecidable, due to Godel's theorem). However, due to Tarski's results, the theory of the real field is complete and therefore the ring of integers $\mathbb{Z}$ is not definable in $\mathbb{R}$.
Let $K/\mathbb{Q}$ be an algebraic extension of the rational numbers.
I will ask two related questions:

Without any further assumptions on the extension, is $\mathbb{Z}$ necessarily a definable subset in $\langle K,+,\times,0,1\rangle$?
What purely algebraic properties of the extension might yield a more conclusive answer to the above questions? Specifically: separability, dimension, normality, finiteness, simplicity (maybe even Galois group structure). Moreover, is there a deeper side to this? Do there exist two extensions $K_{1}/\mathbb{Q},K_{2}/\mathbb{Q}$ that satisfy exactly the same algebraic properties denoted above, yet in one of them $\mathbb Z$ is a definable subset, and in the other it is not? 

These are not questions on which I have pondered for long, but I am curious whether there exist any interesting examples or logical tools needed in order to tackle these questions.

REPLY [4 votes]: I hope this is not seen as self-promoting, but I just stumbled upon this question when browsing things related to that recent question.
As Tom Scanlon said, it is very difficult if not impossible to get a complete picture of the algebraic extensions of $\mathbb{Q}$ in which $\mathbb{Z}$ is definable. However, one can say that in a certain sense, $\mathbb{Z}$ is not definable in $K$ for most algebraic extensions $K/\mathbb{Q}$: The set of fields $K\subseteq\overline{\mathbb{Q}}$ in which $\mathbb{Z}$ is definable is meager in the set of all such $K$, in the topology induced from $2^{\overline{\mathbb{Q}}}$. This is part of what is shown in this very short note of Philip Dittmann and myself.
The reason for the non-definability used there is a purely algebraic one, as asked for in the question: Namely $\mathbb{Z}$ is not definable in any so-called pseudo-algebraically closed field $K$ (meaning every geometrically integral $K$-variety has a $K$-rational point), and the set of $K\subseteq\overline{\mathbb{Q}}$ that are pseudo-algebraically closed is non-meager.