FANDOM

78 Pages

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?)