Math 429/581 (Advanced) Group Theory. Summary of Definitions, Examples, and Theorems by Stefan Gille

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1 Math 429/581 (Advanced) Group Theory Summary of Definitions, Examples, and Theorems by Stefan Gille 1

2 2 0. Group Operations 0.1. Definition. Let G be a group and X a set. A (left) operation of G on X is a map G X X, (g, x) g.x, such that (i) e.x = x for all x X, where e denotes the neutral element of G, and (ii) (g 1 g 2 ).x = g 1.(g 2.x) for all x X and g 1, g 2 G. We say then also that X is a G-set. A right operation is defined analogous. If not otherwise specified operation means here always left operation. Let Y be another G-set. A homomorphism of G-sets between X and Y is a map f : X Y, such that f(g.x) = g.f(x) for all x X and g G. If f is bijective than we say that f is an isomorphism of G-sets. Note that then also the inverse of f is a homomorphism and so an isomorphism of G-sets Example. Let X be a set and Bij(X) be the set of bijective maps X X. This is a group with composition as multiplication, and operates on X by setting f.x := f(x) for f Bij(X) and x X. Note an operation of the group G defines a group homomorphism G Bij(X) be mapping g G to the bijective map x g.x, and vice versa Orbits and stabilizer. Let G be a group and X a G-set. Then G.x := { g.x g G } is called the orbit, and Stab(x) := { g G g.x = x } is called the stabilizer of x X (under the operation of G). Note that Stab(x) is a subgroup of G. We have G.x G.y if and only if G.x and G.y coincide. As a consequence X is the disjoint union of the different orbits under the action of G: X = i I G.x i, where G.x i, i I, are all different orbits. If X is a finite set then also I is finite, say I = {1, 2,..., n} for some integer n 1, and so we have the so called Bahnensatz n n X = G.x i = (G : Stab(x i )). i=1 The last equation since for all x X the homomorphism of G-sets is an isomorphism of G-sets. i=1 G/Stab(x) G.x, g Stab(x) g.x 0.4. Three operations. Let G be group with subgroups H, K. Then H operates (i) on G by conjugation: H G G, (h, g) h g h 1 ;

3 3 (ii) on G by left translation H G G, (h, g) h g, and similar by right translation which is then a right operation of H on G; and (iii) on the left cosets G/K also be left translation H G/K G, (h, gk) (h g)k. Again there is a similar operation on the right cosets.

4 4 1. p-groups and Sylow groups Convention. Throughout this section p denotes a prime number Definition. A group G, such that G = p l for some l 1 is called a p-group Example. As the order of an element divides the order of a group, a group of order exactly p is cyclic generated by any element except the neutral element Lemma. Let G be a finite group whose order is divisible by p. Then G contains a subgroup of order p. The proof (given in class) of this lemma has the following 1.4. Corollary. Let G be a p-group. Then the center of G: Z(G) := { g G x g = g x for all x G } is not trivial, i.e. Z(G) {e} Example. The above corollary can be used to show that a group of order p 2 is either cyclic or a product of two cyclic groups of order p. In the proof (given in class) of the Sylow Theorems the following easy but very useful lemma plays some role Lemma. Let G be a p-group which acts on the finite set X. Denote by X G the set of fix points of this action, i.e. X G := { x X g.x = x for all g G }. Then we have X X G mod p Definition. Let G be a group of order p l m with l 1 and m not divisible by p. A subgroup H of G is called a p-sylow subgroup of G if H = p l The Sylow theorems. Let G be a group of order p l m with l 1 and m not divisible by p. Then: (i) There exists a p-sylow subgroup of G. (ii) If H is a subgroup of G which is a p-group. Then there exists a p-sylow subgroup S of G, such that H S. (iii) Let S 1 and S 2 be two p-sylow subgroups of G. Then there is an element x G, such that S 2 = x S 1 x 1, i.e. two p-sylow subgroups are conjugate by an inner automorphism of G. There is also some information about the number of p-sylow subgroups. In the following results G is a group of order p l m with l 1 and m prime to p, i.e. p does not divide m Theorem. Let G be a group of order p l m with l 1 and p not dividing m. Then the number of different p-sylow subgroups if congruent 1 modulo p Theorem. Let H be a subgroup of G which is a p-group of order p t with 0 t l 1. Then there exists a subgroup K H of G of order p t+1. The proof (given in class) of this result has the following corollary Corollary. Let H be a subgroup of G of order p t with 0 t l 1. Then N G (H) H.

5 Note that by assumption H is not a Sylow subgroup of G as the order is to small. If H is a p-sylow subgroup the corollary appears to be wrong in general as there groups with several p-sylow subgroups. 5

6 6 2. Solvable and nilpotent groups 2.1. Reminder. Recall that a group G is called simple if G and {e} are the only normal subgroups. Note that an abelian group is simple if and only if it is cyclic group of prime order Definition. Let G be a finite group and {H i } = {H i } m i=0 : {e} = H 0 H 1 H 2... H m 1 H m = G a chain of subgroups of G. (i) The chain {H i } is called a subinvariant or subnormal series if H i 1 is a normal subgroup of H i for all i = 1,..., m. In this case we call the factor groups H i /H i 1 the factors of the series. If all the factors in the subnormal series {H i } are simple we call it a composition series. Note that H i /H i 1 is simple if and only if H i 1 is a maximal proper normal subgroup of H i. (ii) The chain {H i } is called a normal series if H i is a normal subgroup of G for all i = 0,..., m. Note that a normal series is also a subnormal series. (iii) The group G is called solvable if it has a composition series with all factors abelian. This implies that all factors are cyclic of prime order. (iv) The group G is called nilpotent if it has a normal series {H i } m i=1, such that the factor groups H i /H i 1 are in the center of G/H i 1 for all i = 1,..., m Lemma. Let G be a group with normal subgroup H. Then G is solvable if and only if H and G/H are solvable Definition. Let G be a group and {H i } m i=0 and {K j} n j=0 be two subnormal (respectively normal series) of G. Then: (i) The series {H i } is called a refinement of {K j } if there is an injective map α : {1,..., n} {1,..., m}, such that K j = H α(j) for all 0 j n. (ii) The series {H i } and {K j } are called isomorphic if there is a bijection α : {1,..., m} {1,..., n} (in particular m = n), such that for all 1 i m. H i /H i 1 K α(i) /K α(i) The Zassenhausen Lemma. Let G be a group with subgroups H, K. Let further H H and K K be normal subgroups of H and K, respectively. (Note that that none of these subgroups is assumed to be normal in G.) Then we have: (i) K (K H) is a normal subgroup of K (K H), and H (H K) is a normal subgroup of H (K H). (ii) The factor groups K (K H)/ K (K H) and H (K H)/ H (H K) are isomorphic Schreier s Refinement Theorem. Let {H i } m i=0 and {K j} n j=0 be two subnormal (respectively normal) series of a group G. Then there exists subnormal (respectively normal) series {H i }m i=0 and {K j }n j=0 which are isomorphic and refinements of {H i } m i=0 and {K j} n j=0, respectively Corollary (Jordan-Hölder). Two composition series of a group are isomorphic.

7 7 Another consequence of Schreier s refinement theorem is the following fact Corollary. Let H be a normal subgroup of the group G. If G has a composition series then there exists a composition series {H i } m i=0 of G with the property that H = H i0 for some 0 i 0 m Definition. Let G be group. The commutator of g, h G is the element [g, h] := g h g 1 h 1. (Note that [g, h] = e if and only if g h = h g, i.e. if and only if g and h commute with each other.) Let further A, B be two subsets of G. We denote by [A, B] the subgroup generated by all commutators [a, b] with a A and b B. The commutator subgroup G is defined as G := [G, G] and the higher commutator subgroups G (i), i 0, are defined inductively by G (0) := G and G (i+1) := [G (i), G (i) ] for i 0. In particular we have then G (1) = G Theorem. Let G be a group. Then: (i) G is a normal subgroup of G. (ii) The factor group G/G is abelian, and if f : G A is a homomorphism of groups with A abelian then Ker f G. Hence there is a commutative diagram of morphisms of groups: f G A q f G/G, where q : G G/G is the canonical quotient map and f : G/G maps the coset xg to f(x). The higher commutator groups can be used for a characterization of solvable groups. Actually this is the correct definition of solvability for groups which are not finite Theorem. A finite group G is solvable if and only if for some integer i 0. G (i) = {e} Example. Let S n be the permutation group on n letters, i.e. S n = Bij ( {1,..., n} ). Recall that the j-cycle (i 1 i 2... i j ) in S n is the bijection of {1,..., n} which maps m to m if m {1,..., n} \ {i 1,..., i j }, i k to i k+1 for 1 k j 1 and i j to i 1. A straightforward calculation shows that if γ S n then γ (i 1 i 2... i j ) γ 1 = (γ(i 1 )γ(i 2 )... γ(i j )). The 2-cycles (ij) for 1 i j n generate S n, and there is a well defined homomorphism of groups sign : S n Z/2Z sending γ to 0 mod 2 if γ is a product of an even number of 2-cycles and to 1 mod 2 otherwise. The kernel

8 8 of sign is the alternating group A n. It is a normal subgroup of index 2 in S n. The groups S n are solvable for 2 n 4 but not for n 5. As for solvable groups using the higher commutator groups one can characterize nilpotent groups using the lower central series Definition. Let G be a group. The lower central series of G: Γ 0 (G) Γ 1 (G) Γ 2 (G) Γ 3 (G)... is defined inductively by Γ 0 (G) := G and Γ i+1 (G) := [Γ i (G), G] for i 0. Note that Γ i (G) is a normal subgroup of G for all i 0, and that by the very definition of the commutator the factor group Γ i (G)/Γ i+1 (G) is in the center of G/Γ i+1 (G) Theorem. Let G be a finite group. Then G is nilpotent if and only if for some i 0. Γ i (G) = {e} Example. A direct product of p-groups (for different or equal prime number p s) is nilpotent Theorem. Let G be a finite nilpotent group. Then for any prime p dividing the order of G there is exactly one p-sylow subgroup, and G is isomorphic to the direct product of its Sylow subgroups. The proof uses the following lemma which of interest on its own Lemma. Let G be a finite nilpotent group and H a proper subgroup. Then N G (H) H.

9 9 3. Reminder on ring theory 3.1. Rings. Recall that a ring R is a non empty set with two operations: An addition: R R R, (r, s) r + s, and a multiplication: R R R, (r, s) r s. With respect to the addition R is an abelian group. The neutral element is usually denoted 0 and called zero. The multiplication is associative and there is also a neutral element for the multiplication denoted 1 and called one: 1 r = r 1 = r for all r R. Multiplication and addition satisfy the distributive laws: r (s + t) = r s + r t and (s + t) r = s r + t r for all r, s, t R. The ring R is called commutative if r s = s r for all r, s R. Most rings appearing in these lectures are commutative. An element r R is called a unit or invertible if there is s R, such that s r = r s = 1. The element s is unique and called the inverse of R, denoted r 1. The set of units of R is denoted by R. This is a group with respect to the multiplication, which is abelian if R is a commutative ring. A field is a commutative ring R, such that every r 0 in R is a unit, i.e. R = R \ {0}. A subset I of a commutative ring R is called an ideal if r s I for all r, s I, and x r I for all r I and x R. The ideal I is called a maximal ideal if I R and there is no (two sided) ideal J R with J I except J = I. If R is a commutative ring this is equivalent to the assertion that R/I is a field. Let R and S be commutative rings. A homomorphism of rings is a map f : R S, such that f(r + r ) = f(r) + f(r ), f(r r ) = f(r) f(r ) and f(1 R ) = 1 S, where 1 R and 1 S denote the ones in R and S, respectively. Note that the kernel of a homomorphism of rings is an ideal. A subring S of a ring R is a subset S R, such that 1 = 1 R S, s t S, and s t S for all s, t S. Note that the only subring of a commutative R which is also an ideal is R itself A commutative ring R is called an integral domain if it has no zero divisors, i.e. if r s = 0 for r, s R is only possible if either r = 0 or s = 0. An integral domain R has a quotient field. This is the set of fractions { r } Q = Q(R) := r, R, s R \ {0}, s which is a field with respect to addition r s + u rv + su := v sv and multiplication r s u ru := v sv. The injection R Q(R), r r 1 is a homomorphism of rings. This is a generalization of the construction of the rational numbers out of the integers Z. Let R be an integral domain and f : R S a homomorphism of rings, such that f(r) is invertible for all r 0 in R. Then there is a unique extension of f to the quotient field Q(R) of R: Q(R) S, r s f(r) f(s) 1.

10 Polynomial rings. A important example of rings here are the so called polynomial rings. Let K be a field. The polynomial ring K[T ] in one variable T over K is the set of all formal expressions n f(t ) = a i T i, a i K. i=0 Assume that a n 0 the n is the degree of f, denote deg f. The polynomial f(t ) is called monic if the highest coefficient a n is equal 1. The set K[T ] is a commutative ring with addition m a i T i + i=0 n b j T j := j=0 max(n,m) l=0 (a l + b l ) T l, where a l = 0 if l > m and b l = 0 if l > n is understood, and multiplication m n n+m ( ) a i T i b j T j := a i b j T l. i=0 j=0 l=0 i+j=l The units of K[T ] are exactly the units of K, i.e. the constants, and K[T ] is a an integral domain. Moreover, every ideal I K[T ] is generated by one element: I = K[T ] f(t ) for some polynomial f(t ), i.e. K[T ] is a principal ideal domain. A polynomial p(t ) 0 in K[T ] is called irreducible if p(t ) is not a unit, and an equation p(t ) = f(t ) g(t ) implies that either f(t ) or g(t ) is a unit in K[T ]. As K[T ] is a principal ideal domain it is factorial and so every polynomial f(t ) m can be written as a product f(t ) = u(t ) p i (T ) ki, where u(t ) is a unit and p 1 (T ),..., p m (T ) are different irreducible polynomials. This expression is unique if we require that the irreducible polynomials are monic: If m n u(t ) p i (T ) ki = f(t ) = v(t ) q i (T ) li i=1 with u(t), v(t) units and p i (T ), q i (T ) monic irreducible polynomials then u(t ) = v(t ), n = m, and after (if necessary) reordering p i (T ) = q i (T ) and k i = l i for all i = 1,..., m. The polynomials p 1 (T ),..., p m (T ) are called the monic irreducible factors of f(t ). Two polynomials f(t ) and g(t ) are called prime if they do not have common irreducible factors. Then there exist polynomials a(t ) and b(t ), such that i=1 i=1 1 = a(t ) f(t ) + b(t ) g(t ). This follows from the Euclidean algorithm: Given polynomials f(t ) and g(t ) there exists polynomials h(t ) and r(t ) with deg r(t ) < deg g(t ), such that f(t ) = h(t ) g(t ) + r(t ). The polynomials h(t ) and r(t ) with this property are unique. A particular example of an irreducible polynomial in K[T ] is the linear polynomial T a, where a K. If f(t ) is any non constant polynomial then either T a divides f(t ) are f(t ) and T a are prime to each other.

11 11 An irreducible polynomial p(t ) of K[T ] generates a maximal ideal of K[T ] and so K[T ]/(p(t )) is a field. (p(t )) := K[T ] p(t ) 3.3. Algebras. Let K be a field. A commutative K-algebra is a commutative ring A together with a homomorphism of rings ι : K A, which is automatically injective as K is a field. The homomorphism ι is called the structure map. One identifies then K with its image in A, i.e. K is identified with the subring ι(k) of A. Note that the polynomial ring K[T ] is a K-algebra. The structure morphism sends x K to the constant polynomial x. A homomorphism of commutative K-algebras f : A B is a homomorphism of rings, such that f restricted to K is the identity. More precisely, this means that if ι : K A and j : K B are the respective structure maps then f(ι(x)) = j(x) for all x K. A K-subalgebra of the K-algebra A is a subring B of A, such that the inclusion B A is a homomorphism of K-algebras. Let A be a commutative K-algebra. Then for every r R there is a unique homomorphism of K-algebras α r : K[T ] R, f(t ) f(r), where f(r) = n a i r i for f(t ) = n a i T i. If f(r) = 0 then r R is called a root i=0 i=0 or zero of the polynomial f(t ). Assume that K = R. Then r K is a zero of f(t ) K[T ] if and only if T r divides f(t ) in K[T ], i.e. if and only if there is g(t ) K[T ], such that f(t ) = (T r) g(t ). This implies in particular, that an irreducible polynomial in K[T ] of degree 2 can not have a zero in K.

12 12 4. Field extensions 4.1. Prime fields. Let F be field. A subfield of F is a subring L of F which is a field. An arbitrary intersection of subfields of F is a field. The intersection P of all subfields of F is called the prime field of F. The image of the homomorphism of rings ψ F : Z F, n n 1 F is in P as P contains 1 = 1 F. There are two cases. (i) Ker ψ F = {0}: Then P Q the field of rational numbers. In this case the characteristic of F is 0; notation char F = 0. (ii) Ker ψ F (0): Then P Z/Zp for some prime number p. In this case the characteristic of F is p; notation char F = p, and one says that F has positive characteristic Field extensions. Let F E be a field extension, i.e. F is a subfield of E. Let further S E a subset. There is a smallest field F (S) E which contains F and S, the field generated by S over F. The field F (S) is an extension field of F. It is called a finitely generated extension if S is a finite set, and a simple extension if S contains only one element. If S = {s 1,..., s n } on writes also F (s 1,..., s n ) instead of F (S) Algebraic field extensions. Let E F be a field extension. Given x E there is a unique homomorphism α x : F [T ] E which maps T to x. If α x is injective one says that x is transcendental or algebraic independent over F. Otherwise x is called algebraic over F. If all x E are algebraic over F then E is called an algebraic field extension of F. In this case the kernel of α x is a maximal ideal of K[T ] which is generated by an irreducible polynomial. There is a unique monic polynomial which generated the kernel. This polynomial p(t ) is called the minimal polynomial of x over F. The map α x induces then an isomorphism of fields and of F -algebras ᾱ x : F [T ]/(p(t )) F (x) E. In particular dim F F (x) = deg p(t ). On the other hand, given an irreducible (monic) polynomial p(t ) in F [T ] then K := F [T ]/(p(t )) is a field extension of F, such p(t ) has a zero in K, the class of T modulo (p(t )) Lemma. The field extension E F is finitely generated and algebraic if and only if dim F E <. In this case E is called a finite field extension of F, and the dimension of E over F is called the degree of the field extension; notation: [E : F ] : dim F E Lemma. Let K E F be field extension. (i) The extension K F is finite if and only if K E and E F are finite. In this case [K : F ] = [K : E] [E : F ]. (ii) The extension K F is algebraic if and only if K E and E F are algebraic.

13 Definition. A field E is called algebraically closed if every polynomial in E[T ] has a root in E. In this case every polynomial in E[T ] is a product of linear, i.e. degree 1, polynomials. The field E F is called an algebraic closure of F if E is algebraically closed and the extension E F is algebraic. For instance, using some complex analysis one can show that C is algebraically closed. But C is not the algebraic closure of Q, why? 4.7. Theorem. Every field has an algebraic closure. To show that an algebraic closure is unique up to isomorphism one can use the following lemma which is of interest on its own Lemma. Let F be a field and K E F algebraic field extensions. Let further Ω F be another field extension which is assumed to be algebraically closed (but not necessarily algebraic over F ). If there is a F -algebra homomorphism σ : E Ω then there is an extension of σ to K, i.e. then there is a F -algebra homomorphism τ : K Ω, such that τ E = σ. If K E is moreover a finite extension then there are at most [K : E] such extensions of σ to K Corollary. If E 1 and E 2 are two algebraic closures of the field F then there is an F -algebra isomorphism E 1 E Spitting fields. Let K be a field and f i, i I, a family of polynomials in K[T ]. The field extension L K is called a splitting field of the set {f i } i I if (i) Every f i is a product of linear polynomials in L[T ], and (ii) L = K(S), where S is the set of the roots of the polynomials f i, i I, in L Theorem-Definition. Let L K be an algebraic extension of fields. Then the following is equivalent: (i) L is a splitting field of a set of polynomials in K[T ]. (ii) If Ω K is an algebraically closed extension field and τ 1, τ 2 : L Ω are K-algebra homomorphisms then Image τ 1 = Image τ 2. (iii) If f(t ) K[T ] is an irreducible polynomial which has a root in L then f(t ) is a product of linear polynomials in L[T ]. A field extension L K satisfying these three equivalent properties is called normal Separable polynomials. Let K be a field and f(t ) K[T ] a non constant polynomial. The polynomial f(t ) is called separable if it does not has multiple roots in an algebraic closure of K, i.e. if K is an algebraic closure of K and deg f f(t ) = u (T a i ) in K[T ] for a unit u K and a i K then a i a j for all 1 i j deg f. (This does not depend on the choice of the algebraic closure.) Lemma. Let f(t ) K[T ] be an irreducible polynomial. Then f(t ) is separable if and only if f (T ) 0. In particular if char K = 0 then all irreducible polynomials are separable. i=1

14 14 Recall here that the derivation f (T ) of the polynomial f(t ) = m a i T i is defined by the formula known from analysis: f (T ) := The usual rules for differentiation hold: and m ia i T i 1. i=1 [f(t ) + g(t )] = f (T ) + g (T ) [f(t ) g(t )] = f (T ) g(t ) + f(t ) g (T ) Definition. Let E F be an algebraic field extension. An element a E is called separable over F if the minimal polynomial of a in F [T ] is separable. The field extension E F is called a separable field extension if all elements of E are separable over F Theorem. Let E F be a finite extension and σ : F Ω a field homomorphism with Ω algebraically closed. The extension E F is separable if and only if there are [E : F ] different field homomorphisms τ : E F, such that τ F = σ. This has the following consequence Corollary. Let F be a field and K E F field extensions. (i) Let S E. The elements of the set S are separable over F is and only if F (S) F is a separable field extension. (ii) The extension K F is separable if and only if both extensions K E and E F are separable. (Note that the analogous statement does not hold for normal extensions.) Let E F be an algebraic field extension. It follows from the corollary above that the set K := { a E a is separable over F } is a subfield of E containing F which is called the separable closure of F in E. If E = F is an algebraic closure then this field is called the separable closure of F and is denoted by F s. It is unique up to F -algebra isomorphism. Note also that F s is a normal extension of F Definition. The field F is called perfect if all algebraic field extensions are separable. By the lemma in 4.12 all fields of characteristic 0 are perfect. This lemma has also the following consequence. Corollary. A field F of characteristic p > 0 is perfect if and only if F = F p := { x p x F } Finite fields. Let F be a finite field. Then char F = p > 0 is a prime number, i.e. F p := Z/Zp is the prime field. As F is finite n := dim Fp F <, and F has p n elements. For every n 1 there is up to isomorphism exactly one field with p n elements which is denoted F p n, i.e. F F p n. The finite field F of characteristic p > 0 is perfect since x x p is an isomorphism of F onto itself and so F = F p. i=0

15 If m divides n then F p m F p n, and this is a normal and separable extension of degree n m. It is even a simple extension as the following lemma implies Lemma. Let K be a field and G K a finite subgroup of the multiplicative group of K. Then G is a cyclic group. Note that all elements x of a finite subgroup G K satisfy x n = 1, where n is the order of G. Such an element of K is called a n-th root of unity. If ξ K has (exact) order n, i.e. if ξ n = 1, but ξ l 1 for all 1 l n 1, the ξ is called a primitive n-th root of unity. Such a primitive n-th root of unity generates the subgroup of K consisting of all n-th roots of unity Theorem. Let E F be a finite separable extension. Then this is a simple extension, i.e. there is x E, such that E = F (x) Definition. Let E F be an algebraic field extension with char F = p > 0. An element x E \F is called purely inseparable over F if there is an integer n 1, such that x pn F. One shows then that the minimal polynomial of x over F is equal T pl a for some integer 1 l n and some a F. More precisely, a = x pl, where l is the smallest integer, such that x pl F. Note that x is then the only root of its minimal polynomial, and so a homomorphism of fields σ : F Ω, where Ω is an algebraically closed field, has exactly one extension to F (x). The extension E F is called purely inseparable if all elements of E \ F are purely inseparable over F Theorem. Let F be a field of characteristic p > 0 and E F an algebraic extension. Let E K F be the separable closure of F in E. Then either E = K, or E K is purely inseparable. If [E : F ] < then [E : K] is a power of p, and every homomorphism of fields σ : F Ω, where Ω is algebraically closed, has exactly [K : F ] extensions to E Norms and traces. Let L K be a finite field extension. Then L is a finite dimensional K-vector space. Let a L. The map l a : L L, x a x is K-linear. (i) The trace of a (over K) is defined as the trace of the K-linear map l a and is denoted by Tr L/K (a). The trace is a K-linear map Tr L/K : L K, such that Tr L/K (a) = [L : K] a if a K. If K E is another finite extension then Tr K/E (Tr L/K (a)) = Tr L/E (a). (ii) The norm of a (over K) is defined as the determinant of the K-linear map l a and is denoted by N L/K (a). The norm N L/K is a map L K, and restricting N L/K to the group of units L it is a homomorphism of groups N L/K : L K, such that N L/K (a) = a [L:K] if a K. If K E is another finite extension then N K/E (N L/K (a)) = N L/E (a). 15

16 16 If the finite extension L K is separable then the trace and norm can be compute as follows. Let Ω K be an algebraic closure of K and Γ the set of all K-algebra homomorphisms L Ω. Then for all a L. Tr L/K (a) = σ Γ σ(a) and N L/K (a) = σ Γ σ(a) If the finite extension L K is not separable then Tr L/K (a) = 0 for all a L. This gives one direction of the following characterization of separable extensions Theorem. The finite extension L K is separable if and only if L L K, (x, y) Tr L/K (x y) is a non degenerate symmetric bilinear form. (Recall from linear algebra that a symmetric bilinear form b : V V K, where V is a finite dimensional K-vector space, is called non degenerate if b(v, w) = 0 for all w V if and only if v = 0.) The proof that this symmetric bilinear form is non degenerate if the extension is separable uses the following fact called linear independence of characters. Let G be a monoid,i.e. G is a non empty set with a associative multiplication G G G, (x, y) x y. A character of G with values in a field K is a map χ : G K, such that χ(x y) = χ(x) χ(y). Then the following lemma holds Lemma. Let G be a monoid, K a field, and χ 1,..., χ n different characters of G with values in K. Then these characters are linear independent, i.e. a relation n a i χ i (x) = 0 for all x G with a 1,..., a n K implies a 1 = a 2 =... = a n = 0. i=1

17 17 5. Galois theory 5.1. Definition. The algebraic field extension E F is called a Galois extension or Galois if it is normal and separable Theorem. Let L be a field. Then L is a finite Galois extension of a field K if and only if there is a finite group G of automorphisms of L, such that K = L G := { x L g(x) = x for all g G }. In this case the order of G is equal the degree [L : K]. Recall that an automorphism of the field L is a homomorphism σ : L L of fields. The set of all automorphisms of L is a group with composition of maps as multiplication: σ τ : L L, x σ(τ(x)). Denote this group by Aut(L). A group of automorphisms of L is a subgroup of Aut(L). Note that if G is a subgroup of Aut(L) then every σ G is a L G - algebra homomorphism of L onto itself Definition. The group G in the theorem above is called the Galois group of the Galois extension L K, and is denoted Gal(L/K). The Galois extension L K is called abelian respectively cyclic if the Galois group Gal(L/K) is abelian respectively cyclic. Note that if L = K(α 1,..., α n ) is the splitting field of a separable and irreducible polynomial f(t ) K[T ], where α 1,..., α n are all different roots of f(t ), then the Galois group Gal(L/K) permutes the roots α 1,..., α n and so is isomorphic to a subgroup of the permutation group S n on n letters The fundamental theorem of (finite) Galois theory. Let L K be a finite Galois extension with group G = Gal(L/K). Then there is a one-to-one (inclusion reversing) correspondence between the subgroups of G and the intermediate extension fields L F K. The correspondence is given by G H L H = { x L h(x) = x for all h H } with inverse map F Gal(L/F ) := { g G g(x) = x for all x F }. The extension L F = L H is also Galois with group H = Gal(L/F ), and the extension F = L H K is Galois if and only if H is a normal subgroup of G, in which case Gal(F/K) G/H Examples. (i) Let K be a field with char K 3, and f(t ) K[T ] an irreducible polynomial of degree 3. Since char K 3 the polynomial f(t ) is separable, and one can assume after a linear change of the variable T that f(t ) = T 3 + bt + c for some b, c K. Let L = K(α 1, α 2, α 3 ) be a splitting field of f(t ), where α 1, α 2, and α 3 are the three distinct roots of f(t ). The Galois

18 18 group G := Gal(L/K) is isomorphic to a subgroup of S 3 and has at least 3 elements. Hence there are two possible cases: G S 3 or G A 3, where A 3 S 3 is the subgroup of even permutations which is cyclic of order 3. Consider now δ := (α 1 α 2 ) (α 1 α 3 ) (α 2 α 3 ) and the square := δ 2, the discriminant of f(t ). The Galois group G acts by permutation on the roots of f(t ). Therefore G fixes δ if and only if it consists only of even permutations, i.e. if and only if G A 3. (Note that the square = δ 2 is invariant under any permutation of the roots of f(t ).) Hence G A 3 and so [L : K] = 3 if and only if is a square in K. A long but straightforward computation (or a clever argument) shows = 4b 3 27c 2. From this it follows for instance that if b = 0 and K = Q is the field of rational numbers then the splitting field of f(t ) = T 3 + c is always of degree 6 over Q as 27 is not square in Z. 1 (ii) Let K = Q and ξ n := e 2π n C a primitive nth root of unity. The field Q(ξ n ) is a Galois extension of Q of degree exactly the number of primitive nth roots of unity, which is the same as the number of units in the ring Z/Zn. The Galois group Gal(Q(ξ n )/Q) is isomorphic to the group of units in Z/Zn. Denote this isomorphism by ι : G (Z/Zn), σ ι(σ). Then the action of Gal(Q(ξ n )/Q) is given by σ(ξ n ) = ξ ι(σ) n. Note that if n = p is a prime number this implies that Q(ξ p ) Q is a cyclic extension of degree p 1, and the minimal polynomial of ξ p is given by f(t ) = T p 1 + T p T Hilbert s Theorem 90. Let L K be a cyclic extension, x L, and σ a generator of the Galois group. Then N L/K (x) = 1 if and only if there is α L, such that x = α σ(α). The following description of cyclic extensions is essentially a corollary of this theorem. Note the special (and already known) case n = 2: Quadratic extensions of a field of characteristic 2 are always cyclic Galois extensions Theorem. Let K be a field which contains a primitive nth root of unity, where n is an integer 2. (i) If L K is a cyclic extension of degree n then there is α L, such that L = K(α), and the minimal polynomial of α is given by T n a, where a = α n K. (ii) Let α be a root of T n a K[T ] in some algebraic closure of K. Then the extension K(α) K is cyclic of degree d dividing n, and the minimal polynomial of α is T d a, where a = α d. The integer d 1 is the smallest integer i, such that α i K. *****************