Compute a correspondence and test its boundaries

Try

Let E=Q(√2,√3). The tower argument gives basis 1,√2,√3,√6 and degree four. It is the splitting field of (x²−2)(x²−3), and characteristic zero gives separability. Thus E/Q is finite Galois, so the correspondence applies with all its hypotheses checked.

Predict the automorphisms by deciding where each square root can go. Their images must square to two and three respectively, allowing independent sign choices.

Predict

If an automorphism negates both square roots, what happens to √6? Predict its fixed field before solving for coefficients. The product of two changed signs remains unchanged, suggesting a quadratic field different from either original root field.

Experiment

Write σ for negating √2 alone and τ for negating √3 alone. The four automorphisms form C₂×C₂.

Subgroup	Fixed field
Identity subgroup	E
Generated by σ	Q(√3)
Generated by τ	Q(√2)
Generated by στ	Q(√6)
Full group	Q

Check the middle nonobvious row directly. A general element is a+b√2+c√3+d√6. Applying στ gives a−b√2−c√3+d√6. Equality and basis independence force b=c=0, leaving exactly Q(√6).

Understand

Each sign assignment defines a well-defined automorphism. One route is to use the four-dimensional quotient presentation with relations u²=2,v²=3 and uv=vu. Changing either sign preserves these relations and therefore multiplication. Alternatively expand products in the established basis and check the substitutions. Each map is its own inverse, and their distinct effects on generators distinguish the four maps. Since the extension degree is four, the automorphism bound leaves no additional possibilities.

The subgroup list is complete: C₂×C₂ has the trivial subgroup, three subgroups generated by its nonidentity involutions, and the whole group. The Galois correspondence therefore proves the displayed field list is complete. Finding three attractive quadratic subfields alone would not rule out others; the subgroup theorem supplies that conclusion.

All subgroups here are normal because the Galois group is abelian. Each intermediate quadratic extension over Q is therefore Galois, and its automorphism group is the appropriate quotient of order two. The degree identities can be checked independently with the tower law: each quadratic field has degree two, and E has degree two over it.

Now compare the real cubic Q(∛2). Its degree is three by Eisenstein, but its only Q-automorphism is identity because its nonreal conjugates are absent. The fixed field of the identity subgroup is the entire cubic field, not Q. Applying the finite Galois correspondence without normality would give a false equality. The counterexample identifies precisely the failed hypothesis rather than suggesting the correspondence is unreliable.

Finite fields supply another complete family. F_(pⁿ)/F_p is the splitting extension of x^(pⁿ)−x and is separable because the derivative is −1. Frobenius has order n, so its n powers already supply every automorphism by the degree bound. The Galois group is cyclic of order n. Its subgroups correspond exactly to the unique subfields of sizes pᵈ for d dividing n, agreeing with the earlier root-set construction.

A complete nonabelian example

Now reset notation: let a be the positive real cube root of 2, let ω be a nonreal root of X²+X+1, and let M=Q(a,ω). Thus ω³=1 and ω≠1. The roots of X³−2 are a, aω and aω². We will find every field between Q and M, including three fields that are not normal over Q.

Eisenstein at 2 gives [Q(a):Q]=3. Because Q(a) is real and ω is nonreal, X²+X+1 has no root in Q(a), so adjoining ω has degree two. Therefore [M:Q]=6. The field M is the splitting field of a separable polynomial, hence is Galois. Its automorphisms permute the three roots faithfully: fixing all roots fixes their generated field M. There are six automorphisms by the embedding theorem and six permutations, so this faithful map onto a subgroup of S₃ is onto S₃.

In particular there is an automorphism r cycling a→aω→aω²→a, and there is complex conjugation s fixing a and interchanging aω,aω². Since ω=(aω)/a, the cycle r fixes ω, whereas s sends ω to ω². These are actual field automorphisms by the preceding count, not merely substitutions asserted to respect every relation. They satisfy r³=s²=1 and srs=r⁻¹. Products act from right to left.

Before reading the table, use r(a)=aω to predict the image of the field Q(a). Does conjugating its fixing subgroup make that field larger, or give a different field of the same degree?

Subgroup H of Gal(M/Q)	Explicit fixed field Mᴴ	[Mᴴ:Q]	[M:Mᴴ]	Is Mᴴ/Q normal?
Identity subgroup	M	6	1	Yes
⟨s⟩	Q(a)	3	2	No
⟨rsr⁻¹⟩	Q(aω)	3	2	No
⟨r²sr⁻²⟩	Q(aω²)	3	2	No
⟨r⟩, the alternating subgroup	Q(ω)=Q(√−3)	2	3	Yes
Whole S₃	Q	1	6	Yes

This table is also a field-lattice description: Q lies below each of the four middle fields, each lies below M, and there are no inclusions between distinct middle fields. Draw those eight edges, reversing every subgroup inclusion. Field degree and subgroup order run in opposite directions.

Certify each row and prove the list exhaustive

The element a is fixed by s, so Q(a)⊆M^⟨s⟩. Artin gives the fixed field degree 6/2=3 over Q, equal to the degree of Q(a); the tower law forces equality. For j=1,2, the automorphism rʲsr⁻ʲ fixes rʲ(a)=aωʲ. Each of these elements still has irreducible polynomial X³−2 and degree three. The same containment and degree comparison prove the two conjugate rows. Likewise r fixes ω; since Q(ω) has degree two and the fixed field of ⟨r⟩ has degree 6/3=2, they coincide. The identity and whole-group rows follow from the correspondence.

A subgroup of S₃ has order 1,2,3 or 6 by Lagrange. Order-two subgroups are generated by transpositions, of which there are three. The two elements of order three generate the same unique subgroup ⟨r⟩. Orders one and six give the trivial and whole groups. This exhausts all possibilities and therefore all intermediate fields; exhibiting generators alone would not establish completeness.

The three order-two subgroups are conjugate and distinct, so are not normal. Their fields are distinct by the inverse correspondence. There is also a direct field check: if Q(aωʲ) contained a different root aωᵏ, their quotient would give ω or ω², hence Q(ω) would be contained in a degree-three field. The tower law would then make 2 divide 3, impossible. Thus each cubic field lacks the other conjugates and is not normal. Its only Q-automorphism is identity, although it has three embeddings into M. In contrast, ⟨r⟩ has index two and is normal; Q(ω)/Q has Galois group S₃/⟨r⟩≅C₂.

Combine fields instead of combining degrees blindly

Two distinct cubic fields have intersection Q: its degree over Q divides three, and degree three would force the two fields equal. Their compositum is M, because it contains two different roots, hence their ratio ω or ω², and then also a. Thus the compositum has degree six, not nine. A cubic field and Q(ω) also intersect in Q, now by coprime degrees; their compositum is again M. These arguments explain every missing edge between the middle fields and warn against multiplying degrees without checking the relevant hypotheses.

Apply

Reconstruct the biquadratic diagram from the subgroup side without looking at the table, then verify one fixed field by coefficient comparison. Use both a general theorem and a direct calculation; each can reveal a different error.

For a changed case, replace a by the positive real cube root b of 5. Derive the degree of N=Q(b,ω), construct the same six-row correspondence and justify every fixed generator. Use Eisenstein at 5 instead of silently reusing the irreducibility claim for X³−2. Then compare Q(b)∩Q(bω) and Q(b)Q(bω): predict why an intersection of degree one does not force a compositum of degree nine.

Check the changed cubic and its limitation

Eisenstein gives degree three over Q; reality and the nonreal quadratic root give total degree six. The faithful action on the three roots identifies the full group with S₃. The quadratic fixed field is Q(ω); the three cubic fields are Q(b), Q(bω), Q(bω²), each fixed by its corresponding transposition. Their intersection in pairs is Q and their pairwise compositum is N, of degree six. Multiplication of extension degrees always applies to a tower, but the two cubic fields are not a tower. The subgroup enumeration proves the list complete. This changed example exercises the same theorem in a new polynomial; it is not an independent examination.

Explain why these examples do not by themselves prove the general correspondence or the solvability-by-radicals theorem. Those results require separate proof machinery. A correct small field diagram is valuable practice, not a substitute for quantifying over all finite Galois extensions.

Continue

This unit supplies a ring-to-field proof sequence and a bounded introduction to finite Galois theory. Continue to the primitive-element theorem to combine generators and prove when this is possible. Then study solvability by radicals for the full characteristic-zero criterion. Deeper module methods and broader algebraic extensions remain further study. Mixed proof assessment and independent review are needed before treating these methods as demonstrated proficiency.