Contradiction. Theorem 1.9. (Artin) Let G be a finite group of automorphisms of E and F = E G the fixed field of G. Then [E : F ] G.

Size: px

Start display at page:

Download "Contradiction. Theorem 1.9. (Artin) Let G be a finite group of automorphisms of E and F = E G the fixed field of G. Then [E : F ] G."

Ethan Burke
5 years ago
Views:

1 1. Galois Theory 1.1. A homomorphism of fields F F is simply a homomorphism of rings. Such a homomorphism is always injective, because its kernel is a proper ideal (it doesnt contain 1), which must therefore be zero. A field E containing a field F is called an extension field of F. We write [E : F ] = dim F E for the degree of E over F. We say that E is finite over F if [E : F ] <. For E/F extension and α E. Consider F [x] E, f f(α). There are two possibilities: Case 1: ker = (0). We say that α is transcendental over F. We have an isomorphism F (x) F (α). Case 2: ker = (f). Here f is a monic irreducible polynomial. We say that α is algebraic over F and f is a minimal polynomial of α over F. We have an isomorphism F [x]/(f) F (α). Proposition 1.1. E/F is finite iff E is algebraic and finitely generated (as a field) over F. Let E, E be fields containing F. An F -homomorphism is a homomorphism ϕ : E E such that ϕ F = id. Proposition 1.2. Let F (α) be a simple field extension of a field F, and let ϕ 0 : F Ω be a homomorphism of fields. (a) If α is transcendental over F, then the map ϕ ϕ(α) defines a one-to-one correspondence between extensions ϕ : F (α) Ω of ϕ 0 and elements of Ω transcendental over ϕ 0 (F ). (b) If α is algebraic over F, with minimum polynomial f F [x], then the map ϕ ϕ(α) defines a one-to-one correspondence between extensions ϕ : F (α) Ω of ϕ 0 and the roots of ϕ 0 (f) over Ω. In particular, the number of such maps is the number of distinct roots of ϕ 0 (f) in Ω Let f F [x]. We say E/F splits f if f(x) = Π(x α i ) for α i E. If in addition, E is generated by the roots of f, then we say that E is a splitting field of f. Proposition 1.3. Every polynomial f F [x] has a splitting field E f and [E f : E] (deg f)!. Proposition 1.4. Let f F [x], E/F generated by roots of f. Ω/F splits f. Then (1) There exists F -homomorphism E Ω. 1

2 2 (2) {F hom : E Ω} [E : F ] and equals [E : F ] if f has only simple roots in Ω. Corollary 1.5. Let f F [x]. The splitting field E f is unique up to isomorphism. Corollary 1.6. Let E/F be a splitting field of f. Then Aut(E/F ) [E : F ] and equals [E : F ] if f is separable over F (i.e., none of its irreducible factor in F [x] has multiple roots) Here Aut(E/F ) is the group of F -automorphism E E E/F is called separable if minimal polynomial of any element of E is separable (i.e., no multiple roots). E/F is called normal if minimal polynomial of any element of E splits in E[x]. Example 1.7. Q[ 3 2] is separable, but not normal over Q. F p (T ) is normal, but not separable over F p (T p ). Lemma 1.8. (Dedekind) Let F be a field and G be a group. If σ 1,, σ n are distinct characters from G to F. Then σ i are linearly indepdent over F. Proof. Otherwise, a i σ i (g) = 0 with {i; a i 0} as small as possible. So a i σ i (h)σ i (g) = 0 for all g, h G. Thus a i (σ i (h) σ 1 (h))σ i (g) = 0. However, {i; a i (σ i (h) σ 1 (h)) 0} < {i; a i 0}. Contradiction. Theorem 1.9. (Artin) Let G be a finite group of automorphisms of E and F = E G the fixed field of G. Then [E : F ] G. Proof. Assume that G = {1 = σ 1,, σ n } and α 1,, α m E, basis over F. If m > n, then the equations i σ i(α j )x j = 0 for all 1 j m has a nontrivial solution (x 1,, x m ) = (a 1,, a m ) in E. We may also assume that this is a solution with fewest possible nonzero elements. WLOG, x 1 0 and x 1 F after rescalaring. Then for any σ G, (σ(x 1 ),, σ(x m )) and (σ(x 1 ) x 1,, σ(x m ) x m ) are also solutions. By minimality assumption, σ(x i ) = x i for all i and σ G. So x i F for all i. Thus x i a i = 0. Contradiction. Corollary Let G be a finite group of automorphisms of E. Then G = Aut(E/E G ). Proof. G Aut(E/E G ). So G Aut(E/E G ) [E : E G ] G. Theorem For a finite extension E/F, the following statements are equivalent: (a) E is the splitting field of a separable polynomial f F [x]. (b) F = E Aut(E/F ). (c) F = E G for some finite group G of automorphisms of E. (d) E is normal and separable over F. In this case, we call that E is Galois over F and write Gal(E/F ) for Aut(E/F ).

3 Proof. (a) (b). Let G = Aut(E/F ). Then E G F. E is splitting field of f over E G. So [E : E G ] = Aut(E/E G ) = G = Aut(E/F ) = [E : F ] and E G = F. (b) (c). Obvious. (c) (d). Let α E and f minimal polynomial. Let α 1,, α m be orbit of α under G. Set g(x) = Π(x α i ) F [x]. Then f g since g(α) = 0. Also g f since f(α i ) = 0. Thus f = g splits into distinct factors in E. (d) (a). E = F [α 1,, α n ]. Let f i be minimal polynomial of α i over F. Then f i splits in E. Hence f = Πf i splits in E. This is also separable since E/F is separable. Corollary Every finite separable extension is contained in a finite Galois extension. Proof. Let E = F [α 1,, α n ]. Then the splitting field of f = Πf i contains E and is Galois over F. Corollary Let E M F. If E is Galois over F, then it is Galois over M. Proof. Assume that E = E f with f F [x]. Then E is also the splitting field of f M[x]. Theorem (Fundamental theorem) Let E be a finite Galois extension of F and G = Gal(E/F ). Then H E H, M Gal(E/M) gives bijection between subgroup of G and intermediate extension F M E. Moreover, (a) H 1 H 2 iff E H 1 E H 2. (b) (H 1 : H 2 ) = [E H 2 : E H 1 ]. (c) σhσ 1 σ(m). (d) H normal in G iff E H normal over F. In this case, E H is Galois over F and Gal(E H /F ) = G/H. Proof. H = Gal(E/E H ) is proved. Also E is Galois over M. So M = E Gal(E/M). (a) H 1 H 2 E H 1 E H 2 H 1 = Gal(E/E H 1 ) Gal(E/E H 2 ) = H 2. (b) H i = [E : E H i ]. (c) τ G, α E, τα = α στσ 1 (σα) = σα. So Gal(E/σM) = σgal(e/m)σ 1. (d) If H normal in G. Then σe H = E σhσ 1 = E H. Now σ σ E H gives G Aut(E H /F ) with kernel H. So induces G/H Aut(E H /F ). Since (E H ) G/H = F, E H is Galois over F and G/H = Gal(E H /F ). If M = F [α 1,, α m ] normal over F, then σ(α i ) is a root of the minimal polynomial of α i over F, so in M. Thus σ(m) = M for all σ G. Hence σhσ 1 = H. 3

4 4 Proposition Let E, L be field extensions of F contained in some common field. If E/F is Galois, then EL/L and E/E L are Galois, and the map Gal(EL/L) Gal(E/E L), is an isomorphism. In this case [EL : F ] = [E : F ][L : F ]. [E L : F ] σ σ E Proof. E is a splitting field of separable f F [x]. So EL (resp. E) is splitting field of f over L (resp. E L). For σ Gal(EL/L), σ maps roots of f to roots of f. So σ(e) = E. Now σ σ E is defined. If σ Gal(EL/L) fixes the elements in E, then it also fixes the elements in EL. So σ = 1. Thus injective. If α E is fixed by Gal(EL/L), then α E L. So image is Gal(E/E L) by fundamental theorem. Proposition Let E, L be field extensions of F contained in some common field. If E, L are Galois over F, then EL, E L are Galois over F, and Gal(EL/F ) Gal(E/F ) Gal(L/F ), σ (σ E, σ L ) is an isomorphism of Gal(EL/F ) onto the subgroup of Gal(E/F ) Gal(L/F ). H = {(σ 1, σ 2 ); σ 1 E L = σ 2 E L } Proof. E is a splitting field of separable f F [x] and L is a splitting field of separable g F [x]. Then EL is a splitting field of separable fg F [x]. As in the previous Prop., Gal(EL/F ) Gal(E/F ) Gal(L/F ), σ (σ E, σ L ) is defined, injective with image contained in H. Now for α E L with minimal polynomial h F [x]. Then h is separable, and has distinct deg(h) roots in E (resp. L and EL), since E, L, EL are Galois over F. Thus all these roots are in E L. So E L/F is normal and separable, hence Galois. The remaining part follows from comparing the size of Gal(EL/F ) with H For separable f F [x], let E f be the splitting field. Assume that f has only simple roots in E f, f(x) = Π n i=1(x α i ) E f [x]. Set G f = Gal(E f /F ). Then G f S n. Proposition (1) Let (f) = π i<j (α i α j ). Then (f) F iff G f A n. (2) The discriminant D(f) = (f) 2 is always in F. Proposition f is irreducible iff G f acts transitively on the set of roots f.

5 Proof.. If g f and α is a root of g. Then for any root β of f, β = σα for some σ G f. Thus g(σα) = σg(α) = 0. So β is a root of g and f = g.. If α, β are roots of f. Then F (α) = F (β) = F [x]/(f). There exists F -homomorphism φ : F (α) E f, α β. Since E f is splitting field, φ extends to F -homomorphism σ : E f E f. Now σ(α) = β. Example If f(x) irreducible and separable of degree 3. Then G f = A 3 or S 3, according to D(f) is square in F or not. Fact. The discriminant of x 3 + px + q is 4p 3 27q application 2.1. Finite fields. E/F p of degree n. Then E has p n elements. E is a cyclic group of p n 1 elements. Thus x pn 1 = 1 for all nonzero x E. So any element of E is a root of f(x) = x pn x. Since f (x) = 1, f has p n distinct roots and E is the splitting field of f and is Galois over F p. We write it as F p n. Frobenius map F r : x x p is injective, hence surjective on E and E F r = F p. So by fundamental theorem, Gal(E/F p ) =< F r > is a cyclic group of order n. Proposition 2.1. Let F be a field and G a finite subgroup of F. Then G is cyclic. Proof. G is a finite abelian group. By the classification of finite abelian group, there exists r N such that G contains an element of order r, r G and x r = 1 for all x G. Thus every element of G is a root of x r 1, thus G r. Corollary 2.2. (1) For any d n, F p n contains exactly one field with p d elements. (2) F p m F p n iff m n. (3) x pn x is the product of all monic irreducible polynomial over F p whose degree divides n. Proof. (1) Since Gal(F p n/f p ) is cyclic, there is exactly one subgroup of order n/d. (2) Gal(F p m/f p ) is a quotient group of Gal(F p n/f p ). So m n. (3) First, the factors of x pn x are distinct because it has no common factor with its derivative. If g is an irreducible factor of x pn x and α is a root of g in F p n, then F p (α) F p n is a subfield of with p deg g elements. Thus deg g n. If f is an irreducible monic polynomial of degree m with m n, then f has a root α in a field of degree m over F p, thus in F p n. Hence α pn α = 0 and f x pn x. Corollary 2.3. The algebraic closure F of F p is F = F p n!. 5

6 6 Theorem 2.4. (Fundamental theorem of algebra) C is algebraically closed. Proof. C is the splitting field of x R[x]. Let f R[x] and E the splitting field of f(x)(x 2 + 1). Then E is Galois over R. Let G = Gal(E/R) and H a Sylow 2-subgroup. Set M = E H. Then [M : R] = (G : H) is odd. We show that (1) M = R. Let α M and g the minimal polynomial. Then [R(α) : R] is odd and deg(g) is odd. Since g( ) = and g( ) =, g has a root in R. Thus deg(g) = 1 and α R. (2) E = C. Set G = Gal(E/C). Then by (1), G is a 2-group. If not 1, then there is a normal subgroup N of index 2. Thus E N has degree 2 over C. But all the square roots of C are in C. So impossible Constructible numbers. Recall that α R is constructible if it is in a subfield of the form Q[ a 1,, a r ], where a i Q[ a 1,, a i 1 ]. Then in particular, [Q(α) : Q] is a power of 2. Theorem 2.5. If E R is Galois over Q of degree 2 s, then α is constructible for all α E. Remark 2.6. It fails if E is not Galois over Q. For example, f(x) = x 4 4x + 2 has G f = S 4. If all the roots of f are constructible, then so is E f. Let H be a Sylow 2-subgroup of S 4, then [E H : Q] is odd and any element in E H \Q is not constructible. Proof. Let G = Gal(E/Q). Recall that any 2-group has nontrivial center and hence a subgroup in the center of order 2. Now a sequence of groups {1} = G 0 G 1 G s = G with G i+1 /G i = 2. By fundamental theorem, a sequece of fields E = E 0 E 1 E s = Q with [E i : E i+1 ] = 2. Thus E i = E i+1 [ a i ] for some a i E i+1. Corollary 2.7. A regular p-gon with p prime is constructible iff p is a Fermat prime, i.e., p is of the form 2 2r + 1. Proof. [Q(e 2πi/p ) : Q(cos(2π/p))] is a power of 2. Now Q(e 2πi/p ) is Galois over Q with Galois group (Z/pZ) of order p 1. Thus cos(2πi/p) is constructible iff p 1 is a power of 2. Easy to check that such p is of the form 2 2r Primitive element. Finite extension E/F is simple if E = F [α] for some α E. Such α is primitive element.

7 Theorem 2.8. (1) Finite extension E/F is simple iff there are only finitely many intermediate extensions. (2) If E is separable over F, then it is a simple extension. Proof. If F is a finite field, then E is also finite and E is cyclic. So (1) and (2) follows. Now assume that F is infinite. (1). Assume that E = F [α] and f is the minimal polynomial of α. For any intermediate field M, let f M M[x] be the minimal polynomial of α. Then f M f is a monic polynomial in E[x]. So in particular, there are only finitely many possible f M. Now for f M, let M be the subfield of M generated by the coefficient of f M. Then E = M[α] = M [α] and [E : M[α]] = deg f M = [E : M [α]]. So M = M.. Assume that E = F [α, β]. The case with more generators can be proved by induction. Since F is infinite, there exists c c F such that F (α + cβ) = F (α + c β). Then (c c)β F (α + cβ) and β, hence α also in F (α + cβ). So α + cβ is a primitive element. (2) E is contained in a Galois extension E of F. By fundamental theorem, there are only finitely many intermediate extension between F and E. Hence by (1), E/F is simple Cyclotomic extension. Proposition 2.9. Let n N and F a field of char 0 or p with p n. Let E be the splitting field of x n 1. Then (1) There exists a primitive n-th root ξ of 1 in E, i.e., the order of ξ is n. (2) E = F [ξ]. (3) E is Galois over F. The map Gal(E/F ) (Z/nZ), σ i, where σ(ξ) = ξ i, is injective. Proof. (1)&(2) gcd(x n 1, nx n 1 ) = 1. So only simple roots. All the n-th roots of 1 in E form a subgroup of order n, so cyclic and there exists primitive element. (3) Any primitive root is of the form ξ i for some i (Z/nZ). Proposition Let F = Q. Then the cyclotomic polynomial Ψ n (x) = Π ξ primitive (x ξ) is in Z[x] and is irreducible over Q. The above map Gal(E/F ) (Z/nZ) is an isomorphism. Proof. For any n-th root ω of unity, let d be the order of ω. Then ω is a primitive d-th root of unity. Now we have a bijection between the set of n-th root of unity, and the pair (d, ω), where d n and ω is a primitive d-th root of unity. So x n 1 = Π d n Ψ d (x). In particular, by Gauss Lemma, Ψ d (x) Z[x] for all d n. 7

8 8 Now assume that Ψ n (x) = f(x)g(x), where f(x) irreducible and f(x), g(x) are monic polynomials in Z[x]. Assume that ξ is a root of f. We ll show that for any d Z with gcd(d, n) = 1, ξ d is also a root of f. It is suffices to prove the case where d is a prime p. If ξ d is not a root of f, then it is a root of g. Thus f(x), g(x p ) has a common root and g(x p ) = f(x)h(x). Hence ḡ(x) p ḡ(x p ) = f(x) h(x) mod p as elements in F p [x]. Thus f, ḡ has common roots. Therefore x n 1 F p [x] has multiple roots, which is impossible. Now Gal(E/F ) = [E : F ] = deg Ψ n = ϕ(n) = (Z/nZ). So Gal(E/F ) (Z/nZ) is an isomorphism Cyclic and Kummer extesion. Theorem (Hilbert s 90) Let E/F be cyclic extension with Galois group G =< σ >. For β E, define the norm N E/F (β) = Πσ i β. Then N E/F (β) = 1 iff β = α/σα for some α E. Proof. N(β) = N(α)/N(σα) = 1. Assume that G = n. Then by Dedekind s theorem, n 1 i=0 (Πi 1 j=0 σj (β))σ i is not identically zero on E. Hence there exists a E such that α := n 1 i=0 (Πi 1 j=0 σj (β))σ i (a) 0. Now βσ(α) = α. Theorem Let F be a field containing a primitive n-th root of 1. Then (1) For any nonzero a F, the splitting field E for x n a is a cyclic (Galois) extension. (2) If E is a cyclic extension of F of degree n, then E is the splitting field of x n a for some a F. Proof. (1) gcd(x n a, nx n 1 ) = 1. So x n a is separable and E is a Galois extension. Let α be a root. Then all the roots are of the form ξα for some n-th root ξ of unity. Now Gal(E/F ) µ n, σ σ(α)/α is injective since E = F [α]. Here µ n is the group of all n-th root of unity. (2) N(ξ) = ξ n = 1. By Hilbert s 90, ξ = α/σα for some α E. Since G is cyclic, any subgroup of G is normal and F (α) is Galois over F. The map G Gal(F (α)/f ), σ i σ i F (α) is injective. Hence E = F (α). Moreover, σ(α n ) = (ξα) n = α n and α n E G = F. Theorem Let F be a field containing a primitive n-th root ξ of 1. Let E/F be a finite extension. Then E/F is a Kummer extension (i.e., with abelian Galois group) whose Galois group has an exponent dividing n (i.e., any element in G is of order dividing n) iff E is a splitting field of (x n a 1 ) (x n a r ) for some a 1,, a r F. Proof.. We have that E = F (α 1,, α r ), where α i is a root of x n a i. Then for any σ G = Gal(E/F ), σ(α i ) = ξ u i(σ) α i. Hence for τ Gal(E/F ), we have that u i (σ)+u i (τ) = u i (τ)+u i (σ) and στ = τσ. Thus G is abelian. Also for any σ G, σ n = 1.

9 . Since G is a finite abelian group, it is a direct product of cyclic groups C 1,, C r. Set C i = C 1 Ĉi C r G and E i = E C i. Then E i /F is a cyclic extension. Hence E i = F (α i ) for some α i with αi n = a i F. By fundamental theorem of Galois theory, E = E C i is the unique field extension that contains E i for all i. Hence E = F (α 1,, α r ) Galois s solvability theorem. Let F be a field of char 0. A polynomial f F [x] is called solvable if any root of f can be obtained by ±,, / and n. A group G is solvable if there is a sequence 1 = G 0 G s = G, where G i is normal in G i+1 and G i+1 /G i is cyclic for all i. Any subquotient of a solvable group is again solvable. Theorem f F [x] is solvable iff G f is solvable. Proof. Let F = F [ξ], where ξ is a primitive (deg f)!-th root of 1. Set G f = Gal(F f /F ). Then the natural map G f G f is well-defined and injective. Hence G f is solvable. Now there exists a sequence {1} = G 0 G 1 G s = G f, here G i is normal in G i+1 and G i+1 /G i is cyclic. Let F i = (F f )G i. Then F F = F s F 0 = F f and F i /F i+1 is a cyclic extension. There exists F = F 0 F m, where F i+1 = F i (α i ), α n i i F i and F m splits f. Set n = n 1 n m. Let Ω be a field containing F m and a primitive n-th root of unity ξ. Then Ω is contained in a Galois extension of F. We denote this field by Ω. Set G = Gal(Ω/F ) and H = Gal(Ω/F m [ξ]). Set H = σ G σhσ 1. Then H is normal in G and is the maximal subgroup in σhσ 1 for all σ. Thus Ω H = Π σ G Ω σhσi = Π σ G σ(f m [ξ]) is Galois over F and is the minimal subfield containing σω H for all σ G. Now Ω H is generated by ξ, σ(α i ) over F. Consider F F [ξ] F [ξ, α 1 ] Ω H. Here each field is obtained from its predecessor by adding a n-th root of some element, hence cyclic extension. So Gal(Ω H /F ) is solvable. Since Gal(F f /F ) is a quotient of Gal(Ω H /F ), it is also solvable Symmetric polynomial. A polynomial f(x 1,, x n ) F [x 1,, x n ] is symmetric if f(x 1,, x n ) = f(x σ(1),, x σ(n) ) for σ S n. The polynomials p 1 = i x i, 9 p 2 = i<j x i x j,, p n = Π i x i

10 10 are called the elementary symmetric polynomials. Theorem (1) Every symmetric polynomial is a polynomial of elementary symmetric polynomials, i.e., if f F [x 1,, x n ] is symmetric, then f F [p 1,, p n ]. (2) The field F (x 1,, x n ) is the splitting field of F (p 1,, p n ) for the polynomial Π i (X x i ) = X n p 1 X n ( 1) n p n and the Galois group is S n. Proof. (1) For example, x x 2 1x 2 + 4x 1 x x 3 2 = p p 1 p 2. We define the multi-degree on polynomials as follows. The multidegree for x k 1 1 x kn n is (k 1,, k n ) N n. We use lexicographic order on N n. For any polynomial f, let x k 1 1 x kn n be the highest monomial occurring in f with nonzero coefficient and we call (k 1,, k n ) the multi-degree of f. Now for any symmetric polynomial f, its multi-degree is of the form (k 1,, k n ) for k 1 k 2 k n. In other words, there exists (d 1,, d n ) N n such that k 1 = d d n, k 2 = d d n,, k n = d n. Notice that p d 1 1 p dn n has multi-degree (k 1,, k n ). Thus f cp d 1 1 p dn n is again symmetric and has multi-degree strictly less than (k 1,, k n ), here c is the coefficient of x k 1 1 x kn n in f. Now part (1) follows from induction on multi-degree. (2) Let f = g/h with g, h F [x 1,, x n ] be a symmetric function. Set H = Π σ Sn σ(h). Then H and Hf are symmetric polynomials and hence in F [p 1,, p n ]. Thus f = Hf/H F (p 1,, p n ). Therefore F (x 1,, x n ) Sn = F (p 1,, p n ). Corollary We define the general polynomial of degree n to be f(x) = X n t 1 X n ( 1) n t n F (t 1,, t n )[X]. Then the Galois group of f(x) over F (t 1,, t n ) is S n. Proof. The monomials p d 1 1 p dn n for (d 1,, d n ) N n are linearly independent since they have different multi-degree as polynomials of x 1,, x n. Hence p 1,, p n are algebraically independent. Thus the map F [t 1,, t n ] F [p 1,, p n ], t i p i is bijective. So it extends to an isomorphism F (t 1,, t n ) F (p 1,, p n ). Now the corollary follows from part (2) of the previous theorem. References

FIELD THEORY. Contents

FIELD THEORY. Contents FIELD THEORY MATH 552 Contents 1. Algebraic Extensions 1 1.1. Finite and Algebraic Extensions 1 1.2. Algebraic Closure 5 1.3. Splitting Fields 7 1.4. Separable Extensions 8 1.5. Inseparable Extensions