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

Usually Hilbertianity is phrased as saying that some analog of Hilbert's irreducibility theorem holds. One version of this says that {\displaystyle K^{n}} can't be covered by a finite union of sets of the form {\displaystyle f(V(k))} where {\displaystyle V} is a variety over {\displaystyle k}, {\displaystyle f:V\to \mathbb {A} ^{n}} is a {\displaystyle K}-definable morphism of varieties, and where either {\displaystyle \dim V<n} or {\displaystyle \dim V=n} and {\displaystyle 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 {\displaystyle K} is a countable (perfect?) Hilbertian field, and {\displaystyle \sigma } is a random automorphism of {\displaystyle \operatorname {Gal} (K^{alg}/K)}, then the fixed field of {\displaystyle \sigma } will be pseudo-finite for all {\displaystyle \sigma } in a set of Haar measure 1.

Hilbertian fields form an elementary class. (right?) ::: ::: :::