Questions: Fundamental Theorem of Galois Theory

Question 1

Let Gal(F/K) have a large subgroup H containing most of the group's automorphisms. What does the Galois correspondence predict about the fixed field F^H?

• A. F^H is large, close to F itself, because a large subgroup fixes many elements
• B. F^H is small, close to K, because many automorphisms must simultaneously fix an element for it to be in F^H
• C. F^H has the same cardinality as H, reflecting the bijective nature of the correspondence
• D. The size of F^H cannot be predicted without knowing which specific automorphisms H contains

Answer: (B)

The Galois correspondence reverses inclusion: a larger subgroup corresponds to a smaller fixed field. For an element to be in F^H, it must be fixed by every automorphism in H. The more automorphisms in H, the fewer elements can satisfy all their constraints simultaneously, making F^H smaller. Conversely, a small H places fewer constraints, allowing more elements to be fixed. This reversal is confirmed by the degree-index relation: [Gal(F/K):H] = [F^H:K].

Question 2

An intermediate field E (with K ⊆ E ⊆ F) is not itself a Galois extension of K. What does the Fundamental Theorem tell you about the subgroup of Gal(F/K) corresponding to E?

• A. The corresponding subgroup is trivial — only the identity fixes E
• B. The corresponding subgroup is not normal in Gal(F/K)
• C. The corresponding subgroup equals all of Gal(F/K)
• D. The corresponding subgroup does not exist — non-Galois intermediate fields fall outside the bijection

Answer: (B)

The Fundamental Theorem establishes a bijection between ALL intermediate fields and ALL subgroups — including non-Galois intermediate fields and non-normal subgroups. The theorem also states that E is a Galois extension of K if and only if its corresponding subgroup is normal in Gal(F/K). Contrapositively: if E/K is not Galois, the corresponding subgroup is not normal. Every intermediate field corresponds to exactly one subgroup; the Galois/non-Galois distinction tracks the normal/non-normal distinction.

Question 3

The Galois correspondence reverses inclusion: if H₁ ⊆ H₂ are subgroups of Gal(F/K), then F^(H₂) ⊆ F^(H₁).

• T. True
• F. False

Answer: True

If H₂ contains H₁ (H₁ ⊆ H₂), then H₂ imposes more constraints — every element of F^(H₂) must be fixed by all of H₂, which includes all of H₁ and more. So F^(H₂) ⊆ F^(H₁): the fixed field of the larger subgroup is contained in the fixed field of the smaller subgroup. Inclusion reverses. This is consistent with the degree-index relation: [F^(H₁):K] = [Gal(F/K):H₁] ≥ [Gal(F/K):H₂] = [F^(H₂):K].

Question 4

The question of whether a polynomial is solvable by radicals reduces, via the Galois correspondence, to whether its Galois group is abelian.

• T. True
• F. False

Answer: False

Solvability by radicals corresponds to whether the Galois group is *solvable* — a broader class than abelian. A group is solvable if it has a composition series where each successive quotient is abelian, but the group itself need not be abelian. Abelian groups are solvable (since every subgroup of an abelian group is normal and quotients are abelian), but solvability extends further. The quintic has Galois group S₅, which is not solvable — not merely because it is non-abelian, but specifically because it has no such composition series.

Question 5

Explain why the Galois correspondence reverses inclusion — why does a larger subgroup of Gal(F/K) correspond to a smaller fixed field?

Model answer: For an element x ∈ F to belong to the fixed field F^H, it must satisfy σ(x) = x for every automorphism σ in H. The larger H is, the more simultaneous constraints x must satisfy, and the fewer elements can satisfy all of them — so a large H leaves few elements fixed, giving a small F^H. Conversely, a small H places fewer constraints, and more elements can satisfy them all, yielding a large F^H. The correspondence reverses inclusion because 'fixed by everything in H' becomes a stricter condition as H grows.

This reversal is confirmed algebraically: [Gal(F/K):H] = [F^H:K]. A large subgroup H has small index in Gal(F/K), corresponding to a small extension degree [F^H:K], meaning F^H is close to K. The index-degree equality is the quantitative version of the qualitative inclusion-reversal argument, and it underlies the correspondence's power as a tool for translating field questions into group questions.