Definition 17.2.1 (Field extension and degree). A field extension is a field containing a field as a subfield (Chapter 4). Then is a vector space over , and the degree of the extension is . The extension is finite if .
Proposition 17.2.2 (Tower law). If are fields and and are finite, then .
Proof. Let be a basis of over and a basis of over . We show that the products form a basis of over . Every is with , and each with ; so . If with , then , so each by independence of the over , and then each by independence of the over .∎
Definition 17.2.3 (Algebraic and transcendental elements). Let be a field extension and . Then is algebraic over if for some nonzero , and transcendental over otherwise.
For , evaluation ,, is a ring homomorphism, and is algebraic exactly when its kernel is nonzero. For , we write for the smallest subfield of containing and (the intersection of all such subfields).
Proposition 17.2.4 (Minimal polynomial). Let be algebraic over . There is a unique monic polynomial with , the minimal polynomial of over . It is the monic polynomial of least degree with , and it is irreducible.
Proof. The kernel of is a nonzero ideal of , which is a PID (Example 17.1.10 and Theorem 17.1.12); a generator is a nonzero element of least degree, and exactly one generator is monic, since two monic polynomials that divide each other are equal. If with , then in the field , so or , contradicting minimality. And is not a unit, since . So is irreducible. ∎
Proposition 17.2.5 (Simple algebraic extensions). If is algebraic over with minimal polynomial of degree , then and , with basis .
Proof. By Proposition 17.1.2, the image of is isomorphic to . Since is irreducible, is maximal (Corollary 17.1.14), so is a field. It contains and , and it is contained in every subfield of containing them; so . Division with remainder by shows that every element of is with or , and this is unique: if with , then , so . Hence is a basis. ∎
Example 17.2.6 (Two simple extensions of ). The minimal polynomial of over is , so has degree 2 over , with basis . The minimal polynomial of over is , which is irreducible by Eisenstein’s criterion (Proposition 17.1.25) with ; so .
17.2.2 Splitting Fields and Separability
Definition 17.2.7 (Splitting field). Let be nonconstant. A splitting field of over is an extension such that factors into linear factors in and , where are the roots of in .
Theorem 17.2.8 (Existence and uniqueness of splitting fields). Every nonconstant has a splitting field over . Any two splitting fields of over are isomorphic by an isomorphism that is the identity on .
Proof of existence. We use induction on . Let be an irreducible factor of in , which exists because is a UFD. By Corollary 17.1.14, is a field. The composite is a field homomorphism, hence injective (Proposition 4.2.3), so we may regard as an extension of ; and the class of satisfies , hence . In we can write . If is constant, is a splitting field. Otherwise, by induction has a splitting field over . Then factors into linear factors in , and is generated over by and the roots of , because ; so is a splitting field of over .∎
We omit the proof of uniqueness, which extends an isomorphism of fields one root at a time; see Artin, Algebra (2nd edition), Chapters 15 and 16.
Definition 17.2.9 (Normal extension). A finite extension is normal if every irreducible polynomial in that has a root in factors into linear factors in .
Definition 17.2.10 (Separable polynomials and extensions). A nonconstant polynomial is separable if it has no repeated roots in a splitting field. An algebraic element of an extension of is separable over if its minimal polynomial is separable, and an extension is separable if every element of is algebraic and separable over .
The formal derivative of is . The product rule holds: it is easily checked for monomials and extends by bilinearity.
Proposition 17.2.11 (Separability in characteristic 0). If , every irreducible polynomial in is separable. Hence every algebraic extension of is separable.
Proof. If is a repeated root of in some extension , then in , so and . Now let be irreducible of degree , with leading coefficient . Since , the coefficient of in is nonzero, so and . The ideal of the PID is for some ; as does not divide , is not an associate of , so is a unit because is irreducible. Thus for some . If were a repeated root of , evaluating at would give . The second statement follows because minimal polynomials are irreducible (Proposition 17.2.4). ∎
17.2.3 The Galois Group
For an extension , let be the group of field automorphisms with .
Lemma 17.2.12 (Automorphisms permute roots). Let be an extension and .
If and , then .
If , where are the distinct roots of in , then sending to the permutation of that it induces is an injective homomorphism .
Proof. (i) Since is a ring homomorphism fixing the coefficients of , we have .
(ii) By (i), maps the finite set injectively to itself, hence permutes it, and composition of automorphisms corresponds to composition of permutations. If fixes every , then is a subfield of containing and all the , so it is , and .∎
The following theorem characterizes the extensions with as many automorphisms as possible. We state it without proof; see Artin, Algebra, Chapter 16.
Theorem 17.2.13 (Galois extensions). Let be a finite extension. Then , and the following are equivalent:
;
is normal and separable;
is a splitting field over of a separable polynomial;
the elements of fixed by every element of are exactly the elements of .
Moreover, is normal if and only if is a splitting field over of some polynomial.
Definition 17.2.14 (Galois extension and Galois group). A finite extension satisfying the equivalent conditions of Theorem 17.2.13 is Galois, and its Galois group is . Thus .
In characteristic 0 the splitting field of any nonconstant polynomial is a Galois extension of : it is finite by Proposition 17.2.5 and the tower law (each is algebraic over ), normal by the last statement of Theorem 17.2.13, and separable by Proposition 17.2.11.
Example 17.2.15 (Quadratic extensions). Let be a non-square. Then is the splitting field of over , so it is Galois of degree 2. An automorphism must send to a root of , that is, to , so with , and .
Example 17.2.16 (The splitting field of ). Let . The roots of in are and they generate , since ; so is the splitting field of over , and is Galois. We have , and is a root of but does not lie in , so and by the tower law. By Lemma 17.2.12, embeds in through its action on ; since both groups have order 6, .
17.2.4 The Fundamental Theorem of Galois Theory
Theorem 17.2.17 (Fundamental theorem of Galois theory). Let be a finite Galois extension with Galois group . There is an inclusion-reversing bijection given by and . If corresponds to , then:
and ;
is Galois if and only if is a normal subgroup of , and in that case ;
the intersection of two intermediate fields corresponds to the subgroup generated by the two subgroups, and the smallest field containing both corresponds to the intersection of the subgroups.
Note. We do not prove the theorem. Besides Theorem 17.2.13, the proof uses the primitive element theorem (every finite separable extension is of the form ) and Artin’s lemma ( for every finite group of automorphisms of a field ).
Example 17.2.18 (Galois correspondence for ). Let as in Example 17.2.16, and identify with through the numbering of the roots.
The subgroup has index 2 and is normal. The 3-cycle sends , so ; hence fixes , which has degree , and corresponds to . Accordingly is Galois with group .
The subgroup fixes , so it corresponds to , which has degree . Likewise corresponds to and to . These subgroups are not normal, and the three cubic fields are not Galois over ; for instance has the root in but does not split there.
17.2.5 Cyclotomic Extensions
Let and . The th roots of unity are for , and is a primitiveth root of unity (one of order exactly ) if and only if . The th cyclotomic polynomial is of degree , Euler’s totient. The field is the th cyclotomic field. We state the following theorem without proof.
Theorem 17.2.19 (Irreducibility of cyclotomic polynomials). The polynomial has integer coefficients and is irreducible in . Hence it is the minimal polynomial of over , and .
Proposition 17.2.20 (Galois group of a cyclotomic field). The extension is Galois, and : the class of corresponds to the automorphism with .
Proof. The roots of are the powers of , so is the splitting field of and is Galois over , of degree by Theorem 17.2.19. An automorphism sends to a root of its minimal polynomial (Lemma 17.2.12), so for a unique , and is determined by because generates the field. Since , the map is an injective homomorphism . Both groups have order , so it is an isomorphism. ∎
Note. Since is abelian, every subgroup of it is normal, so by the fundamental theorem every intermediate field of is Galois over with abelian Galois group. The Kronecker–Weber theorem, a deep result of algebraic number theory, says conversely that every finite Galois extension of with abelian Galois group is contained in some cyclotomic field.
17.2.6 Solvability by Radicals
Definition 17.2.21 (Solvable group). A group is solvable if it has a chain of subgroups in which each is normal in and each quotient is abelian.
Proposition 17.2.22 (Subgroups of solvable groups). Every subgroup of a solvable group is solvable.
Proof. Let be as in the definition, and . The homomorphism ,, has kernel , which is therefore normal in , and by the First Isomorphism Theorem (Theorem 3.3.3) the quotient is isomorphic to a subgroup of the abelian group . So the chain shows that is solvable. ∎
Example 17.2.23 (Symmetric groups).
For , is solvable. The groups and are abelian; has quotients and ; and , with , has quotients of orders 2, 3, 4, namely ,, and . Each subgroup is normal in the previous one: is the kernel of the sign, and consists of and a full conjugacy class of .
is not solvable. If is a normal subgroup of a group with abelian, then contains the commutator subgroup (Definition 3.3.9), since every commutator maps to the identity of the abelian group . The commutator subgroup is normal in and nontrivial, as is not abelian; since is simple (Theorem 11.4.9), . Now let be a chain as in the definition. Then every contains : this holds for , and if , then . So the chain never reaches .
Definition 17.2.24 (Radical extension). An extension is radical if there is a chain of fields with and for some . A polynomial is solvable by radicals if its splitting field over is contained in a radical extension of .
So is solvable by radicals when its roots can be obtained from elements of by field operations and the extraction of th roots. The following theorem, which we state without proof (see Artin, Algebra, Chapter 16), translates this into group theory.
Theorem 17.2.25 (Galois’s criterion). Let be a field of characteristic 0 and nonconstant, with splitting field . Then is solvable by radicals if and only if is a solvable group.
Lemma 17.2.26 (Generating ). Let be a prime. A transposition and a -cycle together generate .
Proof. Let be a transposition and a -cycle. Since moves every point through a single cycle, for some . The power has order (as is prime), so it is again a -cycle. Renumber the points so that ,, and . Conjugating by the powers of gives , so the subgroup generated contains all adjacent transpositions , which generate (Proposition 3.4.9). ∎
Example 17.2.27 (A quintic with Galois group ). Let . It is irreducible by Eisenstein’s criterion with . Let be its splitting field and . The five roots of are distinct (Proposition 17.2.11), and by Lemma 17.2.12 is isomorphic to a subgroup of .
Since ,,, and , the polynomial has at least three real roots. Its derivative has only two real zeros, so by Rolle’s theorem has at most three real roots. So has exactly three real roots and two nonreal, complex conjugate roots.
Complex conjugation permutes the roots of , which generate , so it maps to itself and restricts to an element of . It fixes the three real roots and swaps the other two: it is a transposition.
For a root , divides by the tower law. So contains an element of order 5, by Cauchy’s theorem (Corollary 11.5.4); and the elements of order 5 in are the 5-cycles.
Corollary 17.2.28 (The quintic is not solvable by radicals). The polynomial is not solvable by radicals over . Consequently there is no formula expressing the roots of a general polynomial of degree 5 in terms of its coefficients using only the field operations and radicals.
Proof. By Example 17.2.27, its Galois group is , which is not solvable (Example 17.2.23); apply Theorem 17.2.25. ∎
Note. The impossibility of a general formula is the Abel–Ruffini theorem (Abel, 1824); the group-theoretic criterion is due to Galois (1832). For polynomials of degree at most 4, the Galois group is a subgroup of with , hence solvable (Proposition 17.2.22), which is why the quadratic formula and the formulas of Cardano and Ferrari exist.
Exercise 17.2.29 (Exercises).
Compute , and draw its lattice of subgroups alongside the lattice of intermediate fields.
Show that the Galois group of over is isomorphic to the dihedral group of order 8 (Definition 2.2.14).
Show solution
(1) Let . First, : if with , then , so ; but would make rational, and would give . So is irreducible over , and by the tower law. The field is the splitting field of , so it is Galois over and . An automorphism sends and and is determined by these signs; as there are four automorphisms, all four sign choices occur. Let change the sign of only and that of only. Then and , so .
The three subgroups of order 2 are ,,; they fix ,, and respectively, so they correspond to ,,. The two lattices, with corresponding positions matched (the correspondence reverses inclusions):
(In the field lattice, larger fields are drawn lower, so that the two pictures correspond position by position.)
(2) The polynomial is irreducible by Eisenstein’s criterion with , and its roots are and , so its splitting field is . We have , and is a root of , so and has order 8. An automorphism sends to one of the four roots and to , and is determined by these choices; since there are eight automorphisms, all eight choices occur. Let ,, and let be complex conjugation: ,. Then has order 4, has order 2, and : both sides send and . Since (it moves ),, and is an isomorphism onto . Acting on the four roots, is a subgroup of of order 8, not all of .