Reverse the relation between subgroups and intermediate fields

Try

An automorphism fixing more elements has less freedom. In a finite Galois extension E/F, a larger subgroup of automorphisms therefore has a smaller fixed field. This reverses inclusion: if H₁⊆H₂, every element fixed by all of H₂ is certainly fixed by all of H₁.

Define Eᴴ as the elements of E fixed by every member of H. It is a field because automorphisms preserve sums, products and inverses. Normality of H is not needed for this definition.

Predict

If G is the full Galois group, predict the fixed fields of the trivial subgroup and of G. The trivial subgroup fixes all of E. Under the finite Galois hypotheses, G fixes exactly F. The second assertion is a theorem, not part of the definition of an automorphism fixing F.

Experiment

For a finite Galois extension of degree twelve, compare subgroup sizes and the corresponding degrees.

Subgroup order	Degree from fixed field up to E	Degree of fixed field over F
1	1	12
3	3	4
12	12	1

The table is conditional on those subgroups existing; not every divisor must occur in every Galois group. Its purpose is to keep the two degree directions separate. The subgroup order measures [E:Eᴴ], while its index measures [Eᴴ:F].

Understand

The finite Galois correspondence assumes E/F is finite, normal and separable. It pairs intermediate fields K with subgroups H=Gal(E/K), and the inverse sends H to Eᴴ. The maps reverse inclusion, [E:K]=|H| and [K:F]=[G:H]. Furthermore K/F is Galois exactly when H is normal in G; then Gal(K/F)≅G/H.

The central dependency is Artin’s fixed-field theorem: if H is a finite group of distinct automorphisms of a field E, and K=Eᴴ, then [E:K]=|H|. No initial assumption that E is finite over another field is needed. Write m=|H|. We prove the degree is at most m before using finite-dimensional arguments to prove it is at least m.

Worked example: descend coefficients to the fixed field

Take E=Q(√2) and H consisting of identity and the sign change σ. For x₁=1, x₂=√2, x₃=2, the two equations for coefficients c₁,c₂,c₃ are

Automorphism row	Equation
Identity	c₁+√2 c₂+2c₃=0
Sign change	c₁−√2 c₂+2c₃=0

Subtracting gives c₂=0; setting c₃=1 gives c₁=−2. Thus the same relation has coefficients in the fixed field Q. One row alone would also permit the relation (−√2,1,0), whose coefficients are not fixed. The whole automorphism group supplies the additional constraints. Try replacing x₃=2 with x₃=1+√2: predict and check the fixed-coefficient relation (−1,−1,1).

Upper bound: minimal support forces fixed coefficients

Choose any m+1 elements x₁,…,xₘ₊₁ of E. Form the m by m+1 matrix A whose σ,j entry is σ(xⱼ). More columns than rows gives a nonzero vector c in the kernel over E. Among all such vectors choose one with the fewest nonzero coordinates; this finite positive number is its support size. Choose an index k with cₖ≠0 and divide every coefficient by cₖ, so cₖ=1.

For τ∈H, apply τ to the equation for each σ-row. We obtain ∑ⱼ τ(cⱼ)(τσ)(xⱼ)=0. Since σ↦τσ permutes H, the vector τ(c), obtained by applying τ coordinatewise, is also in the kernel of A. Its nonzero coordinates are exactly those of c, and its kth coordinate is still one. Therefore d=τ(c)−c is a kernel vector supported inside the support of c, with its kth coordinate zero. If d were nonzero, its support would be smaller, contradicting the choice of c. Hence τ(c)=c for every τ∈H, so every coefficient belongs to K.

The identity row now gives ∑ⱼ cⱼxⱼ=0, a nonzero relation over K. Thus every m+1 elements of E are K-linearly dependent. In particular, a maximal independent list can be obtained by adding elements outside its span until this bound prevents further additions; it is a basis of size at most m. Consequently n=[E:K] is finite and n≤m. This is where finite-dimensionality is established, not an assumption used at the start.

Lower bound: distinct automorphisms are independent functions

Suppose distinct field automorphisms σ₁,…,σᵣ had a nontrivial relation ∑ᵢ aᵢσᵢ(x)=0 for every x∈E, with coefficients aᵢ∈E. Choose such a relation with the smallest number r of nonzero coefficients, discarding zero coefficients. A one-term relation is impossible by evaluating at x=1. For r≥2, distinctness gives b∈E with σ₁(b)≠σᵣ(b).

Evaluate the relation at bx and subtract σᵣ(b) times the relation at x. For every x the result is ∑ᵢ aᵢ(σᵢ(b)−σᵣ(b))σᵢ(x)=0. The last coefficient is zero, while the first is nonzero because E is a field. This is a nontrivial relation with fewer terms, a contradiction. Hence distinct automorphisms are linearly independent as functions over E.

Each σ∈H is K-linear. The space Hom_K(E,E) of K-linear maps is an E-vector space: multiply map outputs by any element of E. Evaluating a map on a K-basis of n elements identifies this space with Eⁿ, because basis images may be chosen arbitrarily and determine the map uniquely. Its dimension over E is therefore n (its dimension over K is n²). The m independent automorphisms lie in this space, so m≤n. Together with n≤m, this proves [E:Eᴴ]=m.

Recover both correspondence maps

For finite Galois E/F, the embedding theorem gives |G|=[E:F]. Applying Artin to G gives [E:Eᴳ]=|G|. Since F⊆Eᴳ, the tower law forces [Eᴳ:F]=1 and Eᴳ=F.

For an intermediate field L, E/L remains finite Galois: a separable polynomial over F whose roots generate E is still separable over L and its roots still generate E over L. Applying the preceding argument over L gives E^Gal(E/L)=L.

Conversely, take H≤G and put K=Eᴴ. Every member of H fixes K, so H⊆Gal(E/K). By the embedding bound, |Gal(E/K)|≤[E:K]; by Artin this degree is |H|. Containment and equal finite sizes force Gal(E/K)=H. This direction does not assume that K/F is normal. The subgroup order gives [E:K]; the tower law gives [K:F]=[G:H].

Apply

Conjugation sends Eᴴ to E^(σHσ⁻¹). Thus normal H makes its fixed field invariant under G. Every F-embedding of K into an algebraic closure extends to E, and normality of E/F makes that extension an element of G. Invariance of K therefore puts every conjugate of each element of K inside K, proving K/F normal. Separability is inherited from E/F. Restriction gives G→Gal(K/F), with kernel H; the same extension argument supplies surjectivity. This yields the quotient-group conclusion. Without normality, a fixed field still exists but G/H need not be a group.

Conversely, if K/F is normal, every restriction of an element of G preserves K. Its fixing subgroup is therefore invariant under conjugation, so H is normal. Separability of K/F is inherited from E/F, completing the stated Galois equivalence.

Reconstruct the proof before reading the checks

1. In the upper-bound proof, why is an arbitrary nonzero null vector insufficient? Identify the two roles of normalization and minimal support.
2. If H has three elements and four chosen elements of E are given, explain exactly which matrix and which field of coefficients establish dependence. Do not assume [E:Eᴴ] is already finite.
3. Repair this claim: “The automorphisms lie in an n²-dimensional space, so there can be at most n² of them; this proves Artin.”
4. Starting only from Artin, the embedding bound and H⊆Gal(E/Eᴴ), recover H. State why no normality assumption on H is needed.

Check the proof obligations

Even for the full two-row matrix, (−2√2,0,√2) is a nonzero null vector with nonfixed coefficients; divide by √2 to obtain the fixed-coefficient relation. Normalization gives a common coordinate equal to one after applying every automorphism; minimal support forces the difference to vanish. For a three-element H, use all three automorphism rows and four columns over E, then descend the selected coefficients to Eᴴ using the row permutation argument. The identity row is the resulting dependence. For the lower bound, independence is over E, and Hom_K(E,E) has dimension n over E, not n²; comparing dimensions over different fields is invalid. Finally, H is contained in the fixing group and both sizes are squeezed by [E:Eᴴ]=|H|. Fixed fields and this equality require H to be a subgroup, not a normal subgroup. These are written proof checks; opening the feedback does not demonstrate independent understanding.

Continue

The final lesson computes every subgroup and fixed field in a biquadratic example. It also revisits a nonnormal cubic to show exactly how the correspondence fails outside its hypotheses.