In the field ℂ with automorphisms fixing ℚ, which pair of numbers shares a Galois type over ℚ, and why?
Aπ and e, because both are transcendental and no algebraic formula can distinguish them over ℚ
B√2 and −√2, because they are roots of the same irreducible polynomial x²−2 over ℚ, related by a ℚ-fixing automorphism
C√2 and √3, because both are irrational square roots with the same degree over ℚ
Di and −1, because they are both non-real complex numbers
√2 and −√2 are conjugate roots of x²−2, and the map √2 ↦ −√2 extends to a field automorphism of ℚ(√2) fixing ℚ pointwise — they share a Galois type over ℚ. While π and e are both transcendental, it is not known whether any ℚ-fixing automorphism of ℂ sends π to e; their orbit structure is far from obvious. √2 and √3 have different minimal polynomials (x²−2 vs x²−3) over ℚ, so no ℚ-fixing automorphism maps one to the other — different Galois types.
Question 2 Multiple Choice
In the dense linear order (ℚ, <), what is the Galois type structure over the empty base set ∅?
AEach rational number has its own unique Galois type, since different rationals occupy different positions
BAll rationals share a single Galois type over ∅, since any rational can be mapped to any other by an order-preserving automorphism
CRationals split into two Galois types: positive and negative
DGalois types are undefined for linear orders because they are not algebraic structures
In (ℚ, <), for any two rationals a and b, the translation x ↦ x + (b−a) is an order-preserving bijection of ℚ to itself sending a to b. Since any element can be mapped to any other by an automorphism, the entire domain is a single orbit under Aut(ℚ, <) over ∅ — one Galois type. This makes (ℚ, <) ω-categorical: maximally symmetric, with no formula over ∅ that can distinguish any two elements by position alone.
Question 3 True / False
Two elements having the same Galois type over A means there is no formula with parameters from A that is satisfied by one but not the other.
TTrue
FFalse
Answer: True
If σ ∈ Aut(M/A) sends a to b, then for any formula φ(x) with parameters from A, M ⊨ φ(a) iff M ⊨ φ(σ(a)) = φ(b) — the automorphism preserves all relations and fixes every parameter. So a and b satisfy exactly the same formulas over A. They are genuinely indistinguishable by the language with A-parameters. This is the model-theoretic formalization of 'symmetric relative to A': the structure cannot, even in principle, tell them apart using A as a reference frame.
Question 4 True / False
In most model, the Galois type of an element over A is the same as its syntactic type — the set of most A-parameter formulas the element satisfies.
TTrue
FFalse
Answer: False
Galois types (orbit-based) and syntactic types (formula-based) agree in saturated and homogeneous models, but can diverge in arbitrary models. Same syntactic type means same set of formulas is satisfied — an outside-in, linguistic description. Same Galois type means an automorphism connects them — an inside-out, structural description. In a model lacking sufficient automorphisms (not saturated or homogeneous), two elements may satisfy exactly the same formulas but no automorphism sends one to the other. Stability theory largely studies when these notions coincide.
Question 5 Short Answer
How does the model-theoretic notion of Galois type generalize the classical notion of algebraic conjugacy from Galois theory?
Think about your answer, then reveal below.
Model answer: In classical Galois theory, two algebraic numbers are conjugate over ℚ iff they are roots of the same irreducible polynomial over ℚ — equivalently, iff a field automorphism of the algebraic closure fixing ℚ maps one to the other. The model-theoretic Galois type over A is the orbit of an element under Aut(M/A), the automorphisms of M fixing A pointwise. Substituting M = ℂ and A = ℚ recovers classical conjugacy exactly: same Galois type over ℚ iff conjugate algebraic numbers. The model-theoretic version applies this idea to any first-order structure and any base set, making 'indistinguishable by symmetry over A' meaningful universally.
The key abstraction is that conjugacy in classical Galois theory is really about orbits under a symmetry group fixing a base — a purely structural idea that model theory extracted and applied to all first-order structures. Stability theory then studies when this orbit-based indistinguishability agrees with formula-based indistinguishability.