The Galois Correspondence
For a finite Galois extension, subgroups of its automorphism group correspond to intermediate fields, with inclusions reversed. Over Q, a polynomial's roots are expressible by radicals exactly when its Galois group is solvable.
What a learner can do afterwards
- List the automorphisms of a small splitting field
- Match subgroups to intermediate fields in a concrete degree-four example
- State what solvability of the group says about a general quintic
1 · Read
Write Q for the rationals. A splitting field over Q is the smallest field containing Q and every root of a given polynomial. An automorphism over Q is a bijection of the field that fixes Q and preserves addition and multiplication. It sends each root to another root of the same polynomial. Finite splitting fields over Q are Galois extensions. Their automorphism group order equals their extension degree.
A quadratic over Q with distinct roots has two cases. If its roots are rational, its splitting field is Q and its group has order 1. If it is irreducible, adjoining one root r gives degree 2 and includes the other root: for ax²+bx+c, that root is -b/a-r. Its group has order 2. For prime p, one map on Q(√p) sends a+b√p to itself; the other sends it to a-b√p, with a,b rational. Both preserve (√p)²=p.
For distinct primes p and q, adjoining √p and then √q gives degree 2×2=4 over Q. More generally, the second step has degree 2 whenever √q is outside Q(√p). The four automorphisms choose the two root signs independently. In K=Q(√2,√3), call them e, s, t and st: e fixes both roots, s flips only √2, t flips only √3, and st flips both. Each nonidentity map has order 2. This is the Klein four group.
For a finite Galois extension K/Q, let G be its automorphism group. The fixed field K^H contains the elements fixed by every map in a subgroup H. Every intermediate field occurs exactly once this way. If H is contained in J, then K^J is contained in K^H. Also [K:K^H]=|H| and [K^H:Q]=|G|/|H|. Here [E:F] is extension degree and |H| is subgroup order. The identity subgroup fixes K; the whole group fixes Q.
For K=Q(√2,√3), the subgroups {e,s}, {e,t} and {e,st} fix Q(√3), Q(√2) and Q(√6). For example, write an element as a+b√2+c√3+d√6 with rational coefficients. The map st negates the √2 and √3 terms but fixes √6, so its fixed elements have b=c=0. These are the three proper intermediate fields. For distinct primes p,q, the same sign argument gives Q(√p), Q(√q) and Q(√(pq)).
For prime p, x³-p is irreducible by Eisenstein at p. Let α be its real root and ω a nonreal cube root of 1. The roots are α, αω and αω². The field Q(α) has degree 3 and is real, so ω is outside it. Since ω²+ω+1=0, adjoining ω adds degree 2. Multiply the degrees to find the splitting-field degree, which equals the automorphism count. For x³-2, write L=Q(α,ω); its degree is 3×2=6 and its group is S3.
In S3, (12) swaps roots 1 and 2. The cycle (123) sends 1 to 2, 2 to 3 and 3 to 1. The subgroups are {e}, three groups generated by individual swaps, {e,(123),(132)}, and S3 itself. There are six, giving six intermediate fields of L/Q when Q and L are included. To count fields strictly between the endpoints, omit the identity subgroup and the whole group.
Let G be finite. A subgroup N is normal if gNg^-1=N for every g in G. For normal N, G/N has cosets gN={gn:n in N}, product (gN)(hN)=ghN and order |G|/|N|. Abelian means all elements commute. G is solvable if it has a chain down to {e} with each subgroup normal in the preceding group and each quotient abelian. S3 is nonabelian but solvable: N={e,(123),(132)} is normal and cyclic of order 3. The chain S3, N, {e} has abelian quotients of orders 2 and 3.
For a polynomial over Q, all its roots are expressible by radicals if and only if the Galois group of its splitting field is solvable. Such expressions use arithmetic and repeated root extractions, starting from rational numbers. Complex numbers may occur along the way. The criterion does not require the group to be abelian, and it does not say the roots are rational.
S5 is not solvable. Quintics over Q with Galois group S5 exist, so no formula using radicals can solve every quintic. This is the obstruction to a radical formula for the general quintic. A particular quintic may have a solvable Galois group and radical expressions for its roots. Degree 5 alone does not decide the group or rule out radicals.
For finite Galois extensions, subgroup and field inclusions reverse. Over Q, solvable groups give radical expressions; S5 rules out a formula for all quintics.
This lesson, question bank and flashcards were written for this dataset.
These readings, additional questions and flashcards were written for this dataset.
More to read
When a quadratic gives one symmetry or two
The base field matters when counting automorphisms. Over Q, every rational number must stay fixed. An automorphism must also preserve addition and multiplication. If r satisfies a polynomial with rational coefficients, its image must satisfy that same polynomial.
A quadratic with distinct roots has two cases. When both roots are rational, no extension is needed: its splitting field is Q. Fixing Q then fixes every element of the splitting field. The group contains only the identity.
When the quadratic is irreducible over Q, adjoining one root r gives an extension of degree 2. For ax²+bx+c, the other root is -b/a-r. It lies in the same extension, so this extension is already the splitting field. Its Galois group has order 2.
For a prime p, the elements of Q(√p) have the form a+b√p, where a and b are rational. The two automorphisms keep √p or replace it with -√p. Both choices preserve the relation (√p)²=p. The resulting maps are the identity and the map sending a+b√p to a-b√p.
An arbitrary permutation of named roots need not preserve their equations. For instance, √2 cannot be sent to √3 by a map fixing Q: the image would square to 3, although the preserved equation requires it to square to 2.
Finding a field fixed by sign changes
Set K=Q(√2,√3). Its degree over Q is 4, and its automorphisms choose the two signs independently. Let s change only the sign of √2, and let t change only the sign of √3. Their product st changes both signs. Together with the identity e, these maps form the Klein four group.
Write an element of K as a+b√2+c√3+d√6, with rational coefficients. Under s, the √2 and √6 terms change sign. The fixed elements have b=d=0, so the fixed field of {e,s} is Q(√3). Under t, the fixed field is Q(√2).
The map st changes the signs of √2 and √3 but preserves their product √6. Its fixed elements have b=c=0, giving Q(√6). These three fields correspond to the three subgroups of order 2. There are no other fields strictly between Q and K.
For a subgroup H, the degree [K:K^H] equals |H|. The degree [K^H:Q] equals 4/|H|. These formulas measure different parts of the extension tower.
The endpoints also correspond. The identity subgroup fixes all of K because the identity changes nothing. The whole group fixes Q. A larger subgroup imposes more fixed-element conditions, so its fixed field is smaller.
From cubic root permutations to radical formulas
Let α be the real cube root of 2 and ω a nonreal cube root of 1. The roots of x³-2 are α, αω and αω². The real field Q(α) has degree 3 over Q. It cannot contain ω, which satisfies a quadratic equation. Adjoining ω adds degree 2, making the splitting field L=Q(α,ω) an extension of degree 6.
The group is S3, acting on the three roots. Its subgroups are the identity subgroup, three subgroups generated by individual swaps, the subgroup consisting of the identity and the two three-cycles, and S3 itself. The correspondence gives six intermediate fields when Q and L are included. The subgroup of order 3 fixes a field of degree 2 over Q.
S3 is nonabelian but solvable. Its cyclic subgroup of order 3 is normal, and the quotient has order 2. A chain through that subgroup to the identity has abelian quotients of orders 2 and 3.
For polynomials over Q, solvability of the Galois group is equivalent to expressing all roots by radicals. S5 is not solvable, and some quintics over Q have that group. They rule out a radical formula that works for every quintic. Other particular quintics may have solvable groups and radical expressions.
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