---
title: "The Galois Correspondence"
description: "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 it"
canonical: https://lightmysky.com/learn/mathematics/the-galois-correspondence-mt_30kyliQR_L
source: https://lightmysky.com/learn/mathematics/the-galois-correspondence-mt_30kyliQR_L.md
retrieved: 2026-11-12
---

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# 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.

Subject: Mathematics · Area: Abstract Algebra · Ages 22 to 24
Page: https://lightmysky.com/learn/mathematics/the-galois-correspondence-mt_30kyliQR_L

## Ready when they can

- 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

## Lesson: Symmetries of fields and the groups that mirror them

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.

*(drawing: Here K=Q(√2,√3), and s flips only √2. More maps impose more conditions on fixed elements.)*

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.

**Recap.** 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.*

## Practice

17 questions on this page, each with its working shown.

## Needs first

- [Field Extensions and Their Degrees](https://lightmysky.com/learn/mathematics/field-extensions-and-their-degrees-mt_CjOMNt5_i0)

## Optional connections

- [Normal Subgroups and Quotient Groups](https://lightmysky.com/learn/mathematics/normal-subgroups-and-quotient-groups-mt_RU7A-t7wQ5)

## Opens up

- [The Fundamental Group and Loops on a Circle](https://lightmysky.com/learn/mathematics/the-fundamental-group-and-loops-on-a-circle-mt_zH50_L0nu4)
