indiscernible sequence?
Morley sequence?
Ramsey theorem?
Erdos-Rado theorem?
Ehrenfeucht-Fraïssé games (back-and-forth games)
Classical Galois theory is about the classification of intermediate field extensions in a (field-theoretic) algebraic closure of a ground field according to the structure of a profinite automorphism group.
This can be adapted to the setting of a universal domain of any first-order theory which eliminates imaginaries: we can analogously classify definably closed subsets of a model-theoretic algebraic closure of some parameter set according to the structure of a profinite automorphism group.
In the language of Grothendieck's Galois theory, we can sketch the next two sections as: the category of finite $A$-definable sets in a monster model $\mathbb{M}$ equipped with the forgetful functor to Set is a Galois category.
It will be instructive to look at the case of finite extensions first, as the case of infinite extensions will be a version of the same argument but souped-up with formalities about profinite groups.
Recall the fundamental theorem of Galois theory for finite extensions of fields:
Theorem. For $L/K$ a finite, separable, normal field extension, there is an order-reversing bijective correspondence
between the subgroups of the group of field automorphisms of $L$ fixing $K$ pointwise, and the intermediate field extensions between $K$ and $L$. The correspondence is given by sending a subgroup to its field of fixed points, and an intermediate extension to its stabilizer subgroup.
This can be translated to general model-theoretic language as follows: we work in a monster model $\mathbb{M} \models T$ for a complete first-order theory which eliminates imaginaries. Let $A$ be a small parameter set in $M$. We have the following
Dictionary, between our general $T$ eliminating imaginaries and the special case $T = [[ACF]]$ the theory of an algebraically closed field.
When $L$ is a normal extension of $K$, $L$ splits into $\operatorname{Aut}(\mathbb{M}/K)$-orbits, and so the latter group acts via restriction on $L/K$. We call the image of the induced group homomorphism $\operatorname{Aut}(\mathbb{M}/K) \to \Sym(L/K)$ $\operatorname{Aut}(L/K)$.
With this in place, we can show:
Theorem. Let $K$ be a definably closed parameter set. Let $A$ be a normal extension of $K$ generated by the finite algebraic tuple $\gamma$. Then there is an order-reversing bijective correspondence between the subgroups of $\operatorname{Aut}(A/K)$ and the definably closed intermediate extensions of $A/K$. The correspondence is given by maps $\mathsf{Fix}$ sending a subgroup to its fixed points and $\mathsf{Stab}$ sending an intermediate definably closed extension to its stabilizer subgroup.
Proof. By saturation, $\mathsf{Fix}$ is well-defined, and $\mathsf{Stab}$ is clearly well-defined.
$\mathsf{Fix}$ is left-inverse to $\mathsf{Stab}$: by saturation in the monster, any fixed points of $\mathsf{Stab}(B)$ for $K \subseteq B \subseteq A$ must be in the definable closure of $B$, so whenever $B$ is definably closed, $\mathsf{Fix} \left( \mathsf{Stab} (B) \right) \subseteq B$, with the reverse inclusion immediate.
$\mathsf{Stab}$ is left-inverse to $\mathsf{Fix}$: for $H$ a subgroup of $\operatorname{Aut}(A/K)$, note that for any $c$ a code for the $H$-orbit of $\gamma$ and for any $\sigma \in \operatorname{Aut}(\mathbb{M}/K)$, $\sigma$ fixes $c$ if and only if the restriction $\sigma \restriction A$ permutes $H.\gamma$. Since $\gamma$ generates $A$, any automorphism in $\operatorname{Aut}(A/K)$ is determined by where it sends $\gamma$, so $\sigma \restriction A \operatorname{.} \gamma \in H.\gamma \iff \sigma \restriction A \in H$.
In particular, since $H.\gamma$ is finite, $c$ is actually $H.\gamma$-definable, hence $A$-definable. Since $A$ is definably closed, $c \in A$, and so $c \in \mathsf{Fix}(H)$. If $g \in \mathsf{Stab}(\mathsf{Fix}(H))$, $g$ in particular fixes $c$, hence $g \in H$, and it’s clear that $H \subseteq \mathsf{Stab}(\mathsf{Fix}(H))$. $\square$
We’ll get the case for infinite extensions by just classifying all subextensions of $\operatorname{acl}(A)/\operatorname{dcl}(A)$ at once.
Let $\mathbb{M} \models T$ be a monster model. Let $A$ be a small parameter set. $\operatorname{acl}(A)$ is a normal extension of $A$, because every finite $A$-definable set splits into $\operatorname{Aut}(\mathbb{M}/A)$-orbits. The absolute Galois group $\operatorname{Gal}(A)$ of $A$ is $\operatorname{Aut}(\operatorname{acl}(A)/\operatorname{dcl}(A)$.
(For example, in $\mathsf{ACF}$, this recovers the usual absolute Galois group.)
Now, $\operatorname{acl}(A)$ is the colimit of the diagram of finite $A$-definable sets. From commutativity of limits and colimits, we know that whenever $F : \mathbf{C} \to G \text{-} \mathbf{Set}$ is a cofiltered diagram of $G$-sets, then taking orbits of $\underset{\longleftarrow}{\lim}F$ is the same as taking a limit of the orbits. Dually, if we take automorphism groups, we get:
So $\operatorname{Gal}(A)$ is profinite.
Theorem. Let $T$ be a first-order theory which eliminates imaginaries, and let $\mathbb{M} \models T$ be a monster. Let $A \subseteq \mathbb{M}$ be a small parameter set. Then there is a bijective order-reversing correspondence
given by taking a subgroup closed in the profinite topology of $\operatorname{Gal}(A)$ to its fixed points and by taking a definably-closed intermediate extension of $\operatorname{acl}(A)/\operatorname{dcl}(A)$ to its stabilizer.
Proof. That $\mathsf{Fix}$ is left-inverse to $\mathsf{Stab}$ again follows from being in a monster.
On the other hand, let $H$ be a closed subgroup. $H$ is an intersection of basic open subgroups $H_i$ which are preimages of $\operatorname{Aut}(B_i/A)$, for $B_i$ finite and $A$-definable. Fixing an ordering on the $B_i$ and treating them as tuples, obtain codes $c_i$ for the orbit of each $B_i$ under $\operatorname{Aut}(B_i/A)$. Since each $B_i$ is finite $A-definable$, $c_i$ is $A$-algebraic. The stabilizer of each $c_i$ is precisely $H_i$, so $H = \bigcap_{i \in I} \mathsf{Stab}(c_i)$. Since each $c_i$ is fixed by $H$, whenever $g \in \mathsf{Stab} \left(\mathsf{Fix}(H) \right)$, then in fact $g \in H$. $\square$
In the motivating examples (see below) it turns out that Galois (i.e. relative automorphism) groups are themselves definable (i.e. arise as interpretations in the model of internal groups in $\mathbf{Def}(T)$.) In this case what you’re taking the Galois group of must formally resemble an internal diagram on this internal group; in model theory these are studied as what model theorists call internal covers. A structure theorem of Hrushovski makes this correspondence explicit: internal covers are torsors of definable groupoids and vice-versa; see here.
All internal groups in $\mathbf{Def}(\mathsf{ACF})$ are in fact algebraic groups, so this is reminiscent of reconstruction results arising from Tannaka duality. As it turns out, one can make this analogy explicit and recover a slightly-weakened version of the Tannakian formalism for algebraic groups using the theory of internal covers; see the paper by Moshe Kamensky here.
If $T = \mathsf{ACF}$ the theory of algebraically closed fields, this recovers the classical fundamental theorem of Galois theory.
If $T = \mathsf{DCF}$ the theory of differentially closed fields?, this recovers differential Galois theory. (In fact, Kolchin’s work was what inspired Poizat to introduce imaginaries and work out classical Galois theory in a model-theoretic setting.)
Any theory $T$ can be conservatively interpreted inside a theory $T^{\operatorname{eq}}$ which eliminates imaginaries and hence “admits a Galois theory.” This is the coherent special case of a result Olivia Caramello spells out in very general terms in her monograph on topological (toposic) Galois theory.
Bruno Poizat, Une Theorie de Galois Imaginaire.
Bruno Poizat, A Course in Model Theory.
Alice Medvedev and Ramin Takloo-Bighash, An invitation to model-theoretic Galois theory
Last revised on September 5, 2016 at 14:03:05. See the history of this page for a list of all contributions to it.