indiscernible sequence?
Morley sequence?
Ramsey theorem?
Erdos-Rado theorem?
Ehrenfeucht-Fraïssé games (back-and-forth games)
Hrushovski construction?
generic predicate?
Fields are finitely first-order axiomatizable. An algebraically closed field is further axiomatized by infinitely many sentences which state that all non-constant polynomials have a root. Once we additionally specify a characteristic $p$, $\mathsf{ACF}_p$ turns out to be complete, eliminates imaginaries, is stable, and admits quantifier elimination.
$\mathsf{ACF}$ is the countable collection of sentences in the language $\mathcal{L}_{\operatorname{ring}}$ of rings given by:
where $n = 1, 2, \ldots$.
We can additionally specify a characteristic $p$ to obtain $\mathsf{ACF}_p$ by either adding the axioms $\{ 1 \ne 0 , 1+1 \ne 0 , \cdots \}$ to get $\mathsf{ACF}_0$ or adding the axiom $\underset{p\; terms}{\underbrace{1 + \cdots + 1}} = 0$ to get $\mathsf{ACF}_p$ (where $p$ is prime).
$\mathsf{ACF}$ has quantifier elimination. This amounts to a special case of Chevalley’s direct image theorem from algebraic geometry.
$\mathsf{ACF}$ is stable.
$\mathsf{ACF}$ is totally transcendental: Morley rank? is defined everywhere. In this setting Morley rank subsumes the usual Krull dimension of an algebraic variety.
$\mathsf{ACF}$ is uncountably categorical: for each uncountable $\kappa$, models of $\mathsf{ACF}$ of size $\kappa$ must all be isomorphic. This fails at $\aleph_0$: $\mathbb{Q}^{\operatorname{alg}}$ has transcendence degree zero while there exist countable algebraically closed overfields of $\mathbb{Q}$ with infinite transcendence degree.
$\mathsf{ACF}$ eliminates imaginaries. This means that its syntactic category $\mathbf{Def}(\mathsf{ACF})$ has effective internal congruences, and has a good Galois theory.
It’s easy to see that $\mathsf{ACF}$ codes all finite sets: if $R$ is a finite set of points inside a monster $\mathbb{M}$, its code is the tuple $c$ of coefficients for the polynomial
so that $c$ is fixed by an automorphism of $\mathbb{M}$ if and only if that automorphism permutes $R$.
Last revised on February 19, 2018 at 18:40:39. See the history of this page for a list of all contributions to it.