Hilbertian field

A (pure) field $$K$$ is Hilbertian if there is some elementary extension $$K^* \succeq K$$ and an element $$t \in K^* \setminus K$$ such that $$K(t)$$ is relatively algebraically closed in $$K^*$$. (Note that $$t$$ must be transcendental over $$K$$.)

Usually Hilbertianity is phrased as saying that some analog of Hilbert's irreducibility theorem holds. One version of this says that $$K^n$$ can't be covered by a finite union of sets of the form $$f(V(k))$$ where $$V$$ is a variety over $$k$$, $$f : V \to \mathbb{A}^n$$ is a $$K$$-definable morphism of varieties, and where either $$\dim V < n$$ or $$\dim V = n$$ and $$V \to \mathbb{A}^n$$ is a dominant rational map of degree greater than 1.

Global fields are Hilbertian, as are function fields over arbitrary fields. In Fried and Jarden's book on Field Arithmetic, there is a theorem to the effect that any field having a suitable product formula is Hilbertian. Hilbertian fields play an important role in field arithmetic. There is some theorem, for example, which says that if $$K$$ is a countable (perfect?) Hilbertian field, and $$\sigma$$ is a random automorphism of $$\operatorname{Gal}(K^{alg}/K)$$, then the fixed field of $$\sigma$$ will be pseudo-finite for all $$\sigma$$ in a set of Haar measure 1.

Hilbertian fields form an elementary class. (right?)