Pseudo-algebraically closed field

A pseudo-algebraically closed field, or PAC field for short, is a field $$K$$ with the following property: every geometrically integral $$K$$-variety has a $$K$$-point. In terms of rings, this means that if $$A$$ is a finitely generated $$K$$-algebra such that $$A \otimes_K K^{alg}$$ is an integral domain, then there is a morphism of $$K$$-algebras from $$A$$ to $$K$$. (Here $$K^{alg}$$ denotes the algebraic closure of $$K$$.) If $$K$$ is perfect, "geometrically integral" may be replaced with "geometrically irreducible."

The definition of PAC can be rephrased in more model-theoretic terms as follows: a field $$K$$ is PAC if and only if $$K$$ is existentially closed in regular extensions. That is, $$K \le_1 L$$ for every extension $$L/K$$ such that $$L \otimes_K K^{alg}$$ is a domain.

The class of PAC fields turns out to be an elementary class; i.e., one can axiomatize PAC fields. This is somewhat non-trivial, and does not follow directly from the model-theoretic characterization just given.

It turns out that one only needs to check curves, to verify PAC: if $$K$$ is not PAC, then there is a geometrically integral plane curve over $$K$$ without a $$K$$-point. (TODO: double check that this is true.)

Examples of PAC fields
Algebraically closed fields are PAC. (Separably closed fields are also PAC, right?) The other motivating example is probably pseudo-finite fields. Pseudo-finite fields can be described two non-trivially equivalent ways:


 * The perfect PAC fields with a unique field extension of each degree.
 * The infinite fields satisfying the true theory of finite fields.

Aside from algebraically closed fields and separably closed fields, some conrete examples of PAC fields are the following:


 * Any infinite subfield of $$\mathbb{F}_p^{alg}$$, the algebraic closure of $$\mathbb{F}_p$$.
 * The field $$R(i)$$, where $$R$$ is the field of totally real algebraic numbers.

One can obtain pseudo-finite fields by looking at fixed fields of random automorphisms. There is a generalization of this that yields more general PAC fields. Specifically, someone (Jarden?) proved that if $$K$$ is a countable Hilbertian field, $$n$$ is a positive integer, and $$\sigma_1, \ldots, \sigma_n$$ are chosen Haar-randomly from $$\operatorname{Gal}(K^{alg}/K)$$, then the subfield of $$K^{alg}$$ fixed by all the $$\sigma_i$$ is PAC with probability 1. (TODO: check that this theorem is true.)

Quantifier elimination and elementary equivalence
A PAC field $$K$$ is determined up to elementary equivalence by the following data:


 * The isomorphism class of $$Abs(K)$$, the subfield of "absolute numbers." This is the relative algebraic closure in $$K$$ of the prime field.
 * The absolute Galois group $$\operatorname{Gal}(K^{sep}/K)$$, up to elementary equivalence. (One views the profinite group $$\operatorname{Gal}(K^{sep}/K)$$ as a multi-sorted structure through some means…)
 * Something about degree of imperfection, presumably.

(TODO: double check this in the references)

(Warning: the following may be nonsense.) There is also a relative version of this: if $$F$$ is a field, one can describe when two PAC fields extending $$F$$ are elementarily equivalent over $$F$$ (in the language of rings over $$F$$). One probably just replaces $$Abs(K)$$ with the relative algebraic closure of $$F$$. But something also has to be done with the Galois groups.

At any rate, there is some kind of quantifier elimination result involving "Galois stratifications." PAC fields do not have quantifier elimination in the language of rings, however.

Simplicity, or lack thereof
Zoe Chatzidakis showed that a PAC field $$K$$ is simple (in the model-theoretic sense) if and only if it is "bounded," meaning that there are finitely many degree $$d$$ extensions for each $$d$$. Bounded (perfect?) PAC fields are supersimple of finite rank (maybe?).

Omega-free PAC fields
A (perfect?) PAC field is said to be $$\omega$$-free if it is elementarily equivalent to one whose Galois group is the free profinite group $$F_\omega$$ on $$\omega$$ generators. Warning: the free profinite group of rank $$\omega$$ isn't exactly the profinite completion of the free group of rank $$\omega$$, and $$\omega$$-free PAC fields need not have this group as their absolute Galois group.

Jarden once gave a talk in which he may have said the following: the theory of omega-free PAC fields is the model companion of the theory of fields in a language with predicates $$SOL_n(y_1,\ldots,y_n)$$ interpreted as $$\exists x : x^n = y_1 x^{n-1} + y_2 x^{n-2} + \cdots + y_n$$.

Omega-free PAC fields are not bounded, hence not simple. Nevertheless, Zoe Chatzidakis was able to prove some weak forms of the independence theorem for forking and some related independence notions in this setting.