MATH2401 Abstract Algebra (6): Galois Theory

This article connects field extensions with groups of automorphisms. It develops fixed fields and the Galois correspondence before applying them to cubic and quartic equations, roots of unity, Kummer extensions, and solvability by radicals.

Collections of MATH2401 Abstract Algebra notes:

1. Group;
2. Ring;
3. Module;
4. Field;
5. Action;
6. Galois Theory (this article).

Symmetric polynomials

Let be a commutative ring. The symmetric group permutes the variables in . A polynomial fixed by all permutations is symmetric. The elementary symmetric polynomials are

Theorem. Fundamental theorem of symmetric polynomials. Every symmetric polynomial has a unique expression as a polynomial in with coefficients in :

Thus the are algebraically independent generators, not a module basis.

Proof outline. Order monomials lexicographically. Symmetry makes the exponents of the leading monomial arrange as . The product has that same leading monomial. Subtracting its appropriate multiple lowers the leading term; induction terminates and proves existence. The distinct leading monomials of these products also show that no nonzero polynomial relation among the exists, which proves uniqueness. ◻

If

Viète’s formulas identify its coefficients with the elementary symmetric functions of its roots. Consequently every symmetric polynomial in the roots is a polynomial in the coefficients. This does not assert that every permutation of the roots extends to an automorphism of their splitting field.

The discriminant

Define the Vandermonde product and discriminant by

For a permutation , let act by permuting the roots (equivalently, the subscripts of the variables). A polynomial is alternating if . Thus is alternating and its square is symmetric.

Proposition. The discriminant is a polynomial in the coefficients of , belongs to the base field, and is nonzero exactly when has no repeated root. If is separable, , is its splitting field, and , then

Proof. is symmetric, hence belongs to , and it is nonzero exactly when the roots are distinct. Moreover,

This implies that is a square in . Conversely, if for , then , so . Here is needed for the sign to distinguish odd permutations. ◻

Fixed fields

For a group , its fixed field is

It is a subfield, and .

If is finite and , the orbit polynomial

has coefficients in . Once the fixed-field theorem below is applied, it is the minimal polynomial of over , so .

An infinite algebraic extension contains finite subextensions of arbitrarily large degree. If the extension is separable, it therefore contains elements of arbitrarily large degree. The separability qualification is essential for the second statement.

Theorem. Artin’s fixed-field theorem. If is a finite group of automorphisms of , then

In particular, is Galois.

Proof outline. Dedekind’s independence lemma says distinct field homomorphisms are linearly independent as functions. Applied to the automorphisms, it gives . In the other direction, for any elements of , the homogeneous system obtained by applying every has a nontrivial solution; choose one of minimal support and compare it with its transforms to show that its coefficient ratios lie in . Hence those elements are -dependent and . Equality follows, and the usual automorphism bound leaves no automorphisms beyond . ◻

Galois extensions

Theorem. Transport of splitting fields. Let be an isomorphism, let , and apply to its coefficients to obtain . If and are splitting fields of and , then extends to an isomorphism .

Proposition. If is finite separable and is an algebraic closure containing an image of , there are exactly distinct -embeddings .

Proof. By the primitive element theorem write . Each embedding is determined by the image of , and every root in of the minimal polynomial gives one. Separability supplies exactly distinct roots. ◻

Definition. A finite extension is Galois if .

Proposition. For every finite extension, divides . A field is Galois over the fixed field of any finite group of automorphisms.

Proof. Put . Since , the tower law and Artin’s theorem give . The second assertion is exactly Artin’s theorem. ◻

Theorem. Characterizations of Galois extensions. For a finite extension , the following are equivalent:

1. is Galois;

2. ;

3. is normal and separable;

4. is the splitting field over of a separable polynomial.

Proof outline. The number of -embeddings of into an algebraic closure is exactly when is separable; all of them have image exactly when the extension is normal. Hence (1) is equivalent to (3). A finite extension is normal exactly when it is a splitting field, and adjoining the product of the minimal polynomials of a basis gives a separable splitting polynomial, proving (3)(4). Artin’s fixed-field theorem applied to gives , so the tower law makes (1) equivalent to (2). ◻

Proposition. Every finite separable extension is contained in a finite Galois extension (its normal closure). If is Galois and , then is Galois and .

Proof. The splitting field of the product of the minimal polynomials of finitely many separable generators is a finite Galois extension containing . For the second claim, a separable polynomial over that splits in remains a separable polynomial over with the same splitting field; the Galois characterization applies. Fixing implies fixing , giving the subgroup. ◻

Fundamental theorem of Galois theory

Theorem. Fundamental theorem of Galois theory. Let be finite Galois and . The maps

are mutually inverse, inclusion-reversing bijections between subgroups and intermediate fields . Moreover,

For , the extension is Galois if and only if ; in that case restriction gives

Proof outline. For , Artin’s theorem gives . For an intermediate , the extension is Galois, so its fixed field is . Thus the two maps are inverse and reverse inclusions. Artin and the tower law give the degree formulas. Conjugation satisfies ; hence is normal exactly when every preserves . In that case restriction has kernel and is surjective by extension of embeddings, so the first isomorphism theorem gives . ◻

Cubic and quartic equations

Assume in this subsection that . Translation by changes a cubic into .

Proposition. Irreducible cubic. Let be irreducible and separable. Its Galois group is or , and

Otherwise the group is .

Proof. Irreducibility makes the action on three roots transitive, so the group is or . The Vandermonde product is fixed exactly by the even permutations. Its square is the cubic discriminant , so the discriminant criterion distinguishes the two groups. ◻

Proposition. Quartic resolvent. An irreducible quartic can be translated to . Its transitive Galois group is, up to conjugacy, one of .

Set

The action of on these three pairings has kernel . Their resolvent cubic is

If is the splitting field of , , and is the splitting field of inside , then

Proof. Transitivity leaves precisely the five listed conjugacy classes of subgroups of . A direct permutation check shows that the action on the three pairings has kernel . Viète’s formulas expand the symmetric product to the displayed polynomial in . Its splitting field is fixed by , so restriction and the first isomorphism theorem give the quotient. The orbit sizes of the give the three factorization patterns in the table; parity and the action over separate the remaining pairs. ◻

For an irreducible quartic, the practical classification is:

Factorization of		Additional test
irreducible	or	discriminant nonsquare/square
three linear factors		—
linear irreducible quadratic	or	stays irreducible/factors over

Roots of unity

Let be a primitive th root of unity. The cyclotomic polynomial is irreducible over , has degree , and is the splitting field of .

Theorem. Cyclotomic Galois group. The map

is an isomorphism. Thus cyclotomic extensions of are abelian.

Theorem. Kronecker–Weber. Every finite abelian Galois extension of is contained in for some .

Kummer extensions

Proposition. Basic Kummer extension. Suppose and contains all th roots of unity. If is irreducible and , then is cyclic Galois of degree , with automorphisms .

Proof. The roots of are , all lying in , and they are distinct because . Thus is a splitting field and is Galois. Irreducibility gives degree ; sending to any of the roots gives all automorphisms, which form the cyclic group of th roots of unity. ◻

For the prime-degree converse, fix a prime , assume , and assume that contains a primitive th root of unity .

Proposition. Prime Kummer dichotomy. For every , the polynomial is either irreducible over or splits completely over .

Proof. Let . Because , every root lies in ; hence is the splitting field of the separable polynomial . Restriction to embeds its Galois group into the cyclic group of order . Thus divides . If the degree is , then and all roots lie in . Otherwise the degree is , so the degree- polynomial is the minimal polynomial of and is irreducible. ◻

Theorem. Prime-degree Kummer converse. If is cyclic Galois of degree , then for some such that .

Proof. Let generate . By the linear independence of distinct field automorphisms, the -linear operator

is not zero. Choose with . Reindexing the sum gives , so and therefore . Since is not fixed by , it is not in . The intermediate degree divides the prime and is greater than , hence equals ; consequently . ◻

Corollary. Prime radical towers. Suppose all required roots of unity have first been adjoined. In a tower

where every is prime, each nontrivial layer is cyclic Galois of degree . In particular, the degree of the tower is a product of the primes occurring in its nontrivial layers; if every , it is a power of .

Proof. Apply the prime Kummer dichotomy to over . The layer is either trivial or, by the basic Kummer proposition, cyclic Galois of degree . The tower law multiplies these layer degrees. ◻

Solvability by radicals

A polynomial with splitting field is solvable by radicals if there is a field and a tower

A finite group is solvable if it has a subnormal series

whose factors are abelian.

Proposition. Refinement of a solvable series. A finite group is solvable if and only if it has a subnormal series whose factors are cyclic groups of prime order.

Proof. A series with cyclic factors has abelian factors, so one implication is immediate. Conversely, suppose the factors in a solvable series are finite abelian. Every finite abelian group has a composition series with prime-order cyclic factors: choose an element of prime order, quotient by the subgroup it generates, and argue by induction on the order. Insert the inverse images of these composition series between each consecutive pair . The resulting refinement has the required prime-order cyclic factors. ◻

Theorem. Galois criterion for solvability. Let have characteristic , let , and let be its splitting field. Then is solvable by radicals if and only if is a solvable group.

Proof outline. First replace arbitrary radical exponents by prime exponents, and adjoin the finitely many roots of unity required by the resulting tower. In a normal closure, adjoin at each stage all conjugates of the next prime radical. The prime Kummer proposition shows that the corresponding Galois group embeds in a product of cyclic groups of prime order and is therefore abelian. Repeated use of the restriction exact sequence shows that the full Galois group is solvable. Subgroups and quotients of solvable groups are solvable, so is solvable.

Conversely, suppose is solvable. First adjoin the finitely many roots of unity needed below; this cyclotomic extension is itself radical. After this base change, the Galois group of the compositum with is a subgroup of and hence remains solvable. Refine a solvable series to cyclic prime-order factors. The fundamental theorem of Galois theory turns it into a tower of cyclic prime-degree extensions. The prime-degree Kummer converse writes every layer as the adjunction of a prime radical. Thus the roots of lie in a radical extension of . ◻

Corollary. Symmetric-group obstruction. If a polynomial has Galois group with , it is not solvable by radicals. In particular, the generic degree- polynomial is not solvable by radicals for .

Proof. For , the subgroup is nonabelian simple. If were solvable, its subgroup would be solvable; but a nontrivial simple solvable group must be cyclic of prime order, a contradiction. Apply the Galois criterion. The polynomial with algebraically independent generic coefficients has Galois group . ◻