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?)