Count embeddings without assuming every root stays inside

Try

In Q(√2), sending √2 to −√2 preserves every rational relation and gives a nontrivial automorphism. In the real field Q(∛2), the other roots of x³−2 are nonreal and unavailable. Its only Q-automorphism is identity, despite the extension degree being three.

This comparison separates two questions: how many embeddings into a large algebraic closure exist, and how many of them map the field onto itself.

Predict

If σ fixes F and α satisfies its minimal polynomial m, apply σ to m(α)=0. Predict why σ(α) must be another root of m. Then ask whether every such root lies in the original extension. That is a normality issue, not an automatic consequence of irreducibility.

Experiment

Compare three extensions over Q.

Extension	Degree	Q-automorphisms	Normal?
Q(√2)	2	2	Yes
Q(∛2), real root	3	1	No
Splitting field of x³−2	6	6	Yes

All are separable because the characteristic is zero. The middle row shows that separability alone does not make the automorphism count equal the degree. The missing embeddings land in other conjugate copies inside the algebraic closure.

Understand

A finite algebraic extension E/F is normal when every irreducible polynomial over F having a root in E splits completely in E. Equivalently, E is the splitting field of a polynomial over F. For the forward direction, choose finitely many generators and take the product of their minimal polynomials; normality puts all their roots in E, and the roots include the generators. For the reverse direction, any F-embedding of a splitting field into an algebraic closure permutes its generating root set, so its image is E. Extending an embedding of a simple root field then shows every conjugate of any element of E lies in E.

Extend one embedding: transport every coefficient

Let τ:K→Ω be a field embedding into an algebraically closed field Ω, let α be algebraic over K, and let p(X) be its monic minimal polynomial of degree d. Apply τ to each coefficient to obtain p^τ(X) over τ(K). This polynomial is irreducible over τ(K): a factorization would pull back coefficientwise to a factorization of p over K.

Choose a root β of p^τ in Ω. Define the extension on a representative by ∑ⱼ aⱼαʲ↦∑ⱼ τ(aⱼ)βʲ. Here the sum can be taken with 0≤j<d. To prove this is well-defined for arbitrary polynomial representatives, divide their difference by p. A difference vanishing at α is a multiple of p by minimality and polynomial division; its transported polynomial therefore vanishes at β. Conversely, p^τ is the minimal polynomial of β over τ(K), so no polynomial of smaller degree can map to zero. Evaluation thus induces an injective field homomorphism K[X]/(p)→Ω extending τ. It preserves products as well as sums because polynomial evaluation does. Its image is τ(K)(β).

Every extension of τ must send α to a root of p^τ, and the image of α determines all field elements. Thus extensions are in bijection with the distinct roots of p^τ. There are at most d; there are exactly d if p is separable. Transport preserves separability: a polynomial and its formal derivative have greatest common divisor one before transport exactly when they do afterward.

Count a whole tower

For finite E/F, choose generators α₁,…,αᵣ and set K₀=F and Kᵢ=Kᵢ₋₁(αᵢ), ending at Kᵣ=E. Such a finite list exists by choosing a finite F-basis, whose generated field contains its entire F-span E. Start with the inclusion of F into an algebraic closure Ω containing E. For each embedding of Kᵢ₋₁, the preceding construction gives at most dᵢ=[Kᵢ:Kᵢ₋₁] extensions to Kᵢ. Distinct branches cannot merge because their restrictions or generator images differ. Multiplying the bounds gives at most ∏ᵢdᵢ=[E:F] embeddings.

If E/F is separable, the minimal polynomial of αᵢ over Kᵢ₋₁ divides its separable minimal polynomial over F: divide the latter by the former and use minimality to force zero remainder. Thus the step polynomial is separable. Every branch has exactly dᵢ extensions, so the total is exactly [E:F]. This proves equality without assuming normality. The same construction, starting from any embedding of an intermediate field, proves that it extends to E.

If E/F is also normal, every embedding has image E. Finite-dimensional injective maps into E are onto, so these embeddings are automorphisms. Therefore a finite normal separable extension has exactly [E:F] automorphisms. Such an extension is called finite Galois.

The properties remain distinct in positive characteristic. An inseparable polynomial may split in an extension while having only one distinct root. Normality does not repair the missing distinct embeddings. Both hypotheses belong in the finite Galois theorem.

Apply

For Q(√7), construct both automorphisms explicitly on a+b√7 and verify products. For Q(∛2), explain why the image of its generator must be its only real conjugate, forcing identity on the generated field.

For a further practice case, take K=Q(√2), E=K(√3), and let τ negate √2. The polynomial X²−3 is unchanged by τ. Prove it has degree two over K, then write the two extensions on a+b√2+c√3+d√6. The expected sign patterns are (a,−b,c,−d) and (a,−b,−c,d). For the degree claim, if √3=u+v√2 with rational u,v, squaring gives uv=0 and u²+2v²=3; neither rational u²=3 nor rational v²=3/2 is possible. These two extensions are only the branches above this specified τ; there are four embeddings over Q in total.

When counting automorphisms, first prove that generator images satisfy all relations and land inside the field. A list of roots in the algebraic closure gives possible embeddings, not automatically internal automorphisms.

Continue

The finite Galois correspondence next pairs intermediate fields with subgroups of these automorphisms. Its degree formulas depend on the fixed-field theorem and fail if the finite normal separable assumptions are dropped.