Consider the field extension ℚ(∛2)/ℚ. A student claims this extension has a Galois group of order 3 because ∛2 has three cube roots. What is wrong with this reasoning?
ANothing — the Galois group of ℚ(∛2)/ℚ does have order 3
BThe Galois group has order 6, not 3, because you must count all permutations of the root
Cℚ(∛2)/ℚ is not a Galois extension because the other cube roots of 2 are complex and not in ℚ(∛2); the automorphism group has order 1
DThe Galois group has order 2 because only real automorphisms are allowed
ℚ(∛2)/ℚ is not a Galois extension — it is not a normal extension because the minimal polynomial x³−2 has three roots (∛2, ∛2·ω, ∛2·ω²) but only one lies in ℚ(∛2). An automorphism of ℚ(∛2) fixing ℚ must send ∛2 to a root of x³−2 that lies in ℚ(∛2), and the only such root is ∛2 itself. So the only automorphism is the identity, giving |Gal(ℚ(∛2)/ℚ)| = 1, not [ℚ(∛2):ℚ] = 3. The Galois correspondence holds only for Galois (normal and separable) extensions.
Question 2 Multiple Choice
What determines which elements an automorphism φ ∈ Gal(K/F) is allowed to send an extension element α to?
Aφ(α) can be any element of K, since automorphisms are bijections of K
Bφ(α) must equal α, since automorphisms preserve all algebraic relationships
Cφ(α) must be another root of the same minimal polynomial of α over F
Dφ(α) must be an element of F, since the automorphism fixes the base field
An automorphism φ fixes F and preserves field operations. If α satisfies an irreducible polynomial p(x) over F, then p(φ(α)) = φ(p(α)) = φ(0) = 0. So φ(α) must also be a root of the same minimal polynomial p(x). This constraint — that automorphisms permute roots of irreducible polynomials — is what limits the size of the Galois group and connects group structure to the root relationships. For ℚ(√2)/ℚ, √2 has minimal polynomial x²−2 with roots ±√2, so φ(√2) ∈ {√2, −√2} — exactly two choices, giving |Gal| = 2.
Question 3 True / False
The Galois group Gal(K/F) can include automorphisms that move elements of the base field F.
TTrue
FFalse
Answer: False
By definition, every element of Gal(K/F) fixes F element-wise: φ(f) = f for all f ∈ F. This constraint is the 'relative' part of the Galois group — it captures symmetries of the extension K that respect the structure of F as a rigid backbone. An automorphism of K that moved elements of F would not belong to Gal(K/F), even if it were a valid field automorphism of K in isolation.
Question 4 True / False
For a Galois extension K/F of degree n, the Galois group Gal(K/F) has exactly n elements.
TTrue
FFalse
Answer: True
This is the fundamental counting theorem of Galois theory: for separable (and hence Galois) extensions, |Gal(K/F)| = [K:F]. The example ℚ(√2)/ℚ illustrates this — the extension has degree 2 (a basis is {1, √2}), and the Galois group {identity, conjugation} has exactly 2 elements. This equality is non-trivial: there could in principle be fewer automorphisms (if the polynomial has repeated roots) or more (if we allow automorphisms not fixing F). Separability ensures neither pathology occurs.
Question 5 Short Answer
Why must any automorphism φ in Gal(ℚ(√2)/ℚ) send √2 to either √2 or −√2, and not to some other value like √3 or 2?
Think about your answer, then reveal below.
Model answer: The automorphism φ must fix ℚ (all rational numbers) and preserve field operations. Since (√2)² = 2 and 2 is rational (fixed by φ), we get φ(√2)² = φ((√2)²) = φ(2) = 2. So φ(√2) must be a square root of 2, which means φ(√2) ∈ {√2, −√2}. Both lie in ℚ(√2), so both give valid automorphisms. Neither √3 (which satisfies x²−3, not x²−2) nor 2 (which satisfies x²−2 only if 4=2, which is false) are roots of x²−2.
This reasoning generalizes: every Galois automorphism permutes the roots of each irreducible polynomial over F among themselves. The minimal polynomial of α over F encodes all the algebraic constraints on α, and any automorphism fixing F must respect those constraints. This is why computing Gal(K/F) reduces to counting how many ways the roots of the generating minimal polynomial can be permuted while remaining within K.