In the Galois extension ℚ(√2, √3)/ℚ, the Galois group has order 4. If H is a subgroup of order 2, what is the degree [E:ℚ] of the corresponding fixed field E?
A2 — larger subgroup, larger fixed field
B4 — subgroup index equals field degree
C2 — the index [Gal:H] = 4/2 = 2 equals [E:ℚ]
D1 — the entire field is fixed
The Fundamental Theorem states that [E:F] = [Gal(K/F) : H], the index of H in the full Galois group. With |Gal| = 4 and |H| = 2, the index is 2, so [E:ℚ] = 2. Option A names the right number but gives the wrong reason — it inverts the logic. The bijection is order-reversing: a subgroup of order 2 (large relative to index 2) corresponds to a small field of degree 2 over F, not a large one.
Question 2 Multiple Choice
Suppose H is a normal subgroup of Gal(K/F) with fixed field E = K^H. Which of the following is guaranteed by the Fundamental Theorem?
AE/F is an algebraic extension with no proper intermediate fields
BE/F is itself a Galois extension, and Gal(E/F) ≅ Gal(K/F)/H
CH must be abelian for E to be a Galois extension of F
DGal(K/E) is isomorphic to Gal(K/F)
The normal subgroup correspondence is the structural heart of the theorem: H is normal in Gal(K/F) if and only if the fixed field E = K^H is a Galois extension of F (not merely an arbitrary extension). In that case, the Galois group of E over F is the quotient Gal(K/F)/H. This is the link to solvability by radicals — a polynomial is solvable if and only if its Galois group has a chain of normal subgroups with abelian quotients. H need not itself be abelian (option C); it is the quotient that must be.
Question 3 True / False
In the Galois correspondence, a larger subgroup of Gal(K/F) corresponds to a larger fixed field.
TTrue
FFalse
Answer: False
The bijection is order-reversing — this is its most surprising feature. A larger subgroup H imposes more automorphisms on K, and with more symmetries acting, fewer elements of K survive unchanged. The fixed field K^H therefore shrinks: more constraints, smaller fixed set. Conversely, a small subgroup (few automorphisms) fixes many elements, producing a large intermediate field. The correct statement is: larger subgroup ↔ smaller fixed field, smaller subgroup ↔ larger fixed field.
Question 4 True / False
If E is a Galois extension of F and F ⊆ E ⊆ K is an intermediate field in a Galois extension K/F, then the subgroup Gal(K/E) is normal in Gal(K/F).
TTrue
FFalse
Answer: True
This is exactly the normal subgroup correspondence: H is normal in Gal(K/F) if and only if the fixed field K^H is a Galois extension of F. Equivalently, an intermediate field E gives rise to a normal subgroup Gal(K/E) precisely when E/F is itself Galois. The direction stated here is correct — a Galois intermediate extension E/F corresponds to a normal subgroup. The converse also holds.
Question 5 Short Answer
Why is the order-reversing character of the Galois correspondence not a coincidence but a reflection of the relationship between automorphisms and fixed elements?
Think about your answer, then reveal below.
Model answer: A subgroup H consists of automorphisms of K — symmetries that permute elements while fixing F. The fixed field K^H is exactly the set of elements that all of H's automorphisms leave unchanged. The more automorphisms H contains, the more rigidly it acts on K, and the smaller the set of elements that survive every one of H's symmetries. So a larger H means fewer elements escape all automorphisms, giving a smaller K^H. Conversely, a tiny subgroup with only the identity automorphism fixes everything — the full field K. The correspondence reverses order because bigger symmetry group means smaller invariant set.
This is not a formal trick but a direct consequence of what fixed fields measure. The identity subgroup {e} fixes all of K, giving fixed field K itself (maximum). The full Galois group Gal(K/F) fixes precisely F (minimum). Every intermediate subgroup gives an intermediate field, and the containment relationship inverts. Seeing this logic removes the need to memorize the direction — it follows from what 'being fixed by automorphisms' means.