Galois Theory III Splitting fields.


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1 Galois Theory III Splitting fields. We know how to construct a field extension L of a given field K where a given irreducible polynomial P (X) K[X] has a root. We need a field extension of K where P (X) has all its roots. Definition. Let E be a field and f(x) E[X]. A field extension F E is called a splitting field for f(x) over E if F = E(α 1,..., α n ) with f(x) = c(x α 1 )... (X α n ), where c K is the (nonzero) leading coefficient of F (X). Notice that F must contain all roots of f(x) and it is a minimal field with this property. Examples: (1) C is a spliting field for X over R; (2) C is a splitting field for X over R; (3) C is not a splitting field for X over Q (C is too large); (4) Q(i) is a splitting field for X over Q; (5) Q is a splitting field for X 2 4 over Q; (6) Q( 3 2, 3) is a splitting field for X 3 2 over Q. Theorem (Existence of splitting fields). For any field E and any polynomial f(x) E(X) there is a splitting field F E for f(x) over E and [F : E] n!, where deg f(x) = n. Proof. Use induction on n = deg f(x). If n = 1 then f(x) is linear and F = E. Let n > 1 and assume our theorem is proved for polynomials of degree < n. Choose an irreducible factor m(x) for f(x) in E[X]. Consider the field extension E 1 = E(α 1 ) E, where α 1 is a root of m(x). Notice that [E(α 1 ) : E] = deg m(x) and f(x) = (X α 1 )f 1 (X), where f 1 (X) E 1 [X] and deg f 1 (X) = n 1 < n. By inductive assumption, there is a splitting field for f 1 (X) over E 1. In other words, there is a field extension F E 1 such that f 1 (X) = c(x α 2 )... (X α n ) and F = E 1 (α 2,..., α n ). But E 1 = E(α 1 ) implies that F = E(α 1, α 2,..., α n ) and f(x) = (X α 1 )f 1 (X) = c(x α 1 )(X α 2 )... (X α n ). So, F is a splitting field for f(x) over E. It remains to notice that [F : E] = [F : E 1 ][E 1 : E] (n 1)! deg m(x) (n 1)!n = n!. 1
2 Theorem (Uniqueness of splitting fields). Let E and Ẽ be fields and let σ : E Ẽ be a field isomorphism. Let f(x) E[X] and σ(f(x)) = f(x) Ẽ[X]. Let K be a splitting field for f(x) over E. Let K be a splitting field for f(x) over Ẽ. Then there is a field isomorphism τ : K K such that τ E = σ : E Ẽ. Proof. Use induction on n = deg f(x) = deg f(x). If n = 1 then K = E and K = Ẽ and we can take τ = σ. Suppose n > 1 and choose an irreducible factor m(x) of f(x) in E[X]. Then m(x) = σ(m(x)) is an irreducible factor of f(x). Choose a root α K of m(x) and a root α K of m(x). Because the polynomial m(x) is obtained from m(x) via the field isomorphism σ there is a field isomorphism τ 1 : E(α) Ẽ( α) such that τ 1 E coincides with σ and τ 1 (α) = α. (Use the uniqueness of a minimal field extension where a given polynomial has a root.) But then notice that K became a splitting field of f 1 (X) over E(α), where f(x) = (X α)f 1 (X). Similarly, K is a splitting field of f 1 (X) over E( α), where f(x) = (X α) f 1 (X). It remains to notice that the degree of f 1 (X) is n 1 < n and f 1 (X) is obtained from f 1 (X) via the field isomorphism τ 1 : E(α) Ẽ( α). Therefore, by inductive assumption τ 1 can be extended to required field isomorphism τ from K to K Normal extensions. Definition. Let F/E be a field extension. It is normal if the following condition is satisfied: whenever P (X) E[X] is irreducible and has a root in F then P (X) splits completely in F [X]. In other words, a field F is normal over its subfield E if any irreducible polynomial P (X) from E[X] with a root in F has all its roots in F. Examples; a) If [F : E] = 1 then F is normal over E; b) Prove that any quadratic extension F/E is normal. Indeed, suppose P (X) E[X] is irreducible and let θ F is such that P (θ) = 0. Then E E(θ) F and because [F : e] = 2 we have [E(θ) : E] = deg P (X) is either 1, or 2. If deg P (X) = 1 there is nothing to prove. If deg P (X) = 2, then P (X) = c(x θ)(x θ ) in F [X]. This implies that c(θ + θ ) F and, therefore, θ F. So, P (X) splits completely in F [X]. c) Not every cubic extension is normal. Indeed, take E = Q and F = Q(θ), where θ is a root of X 3 2 Q[X]. Then two remaining roots of X 3 2 are θζ 3 and θζ3, 2 where ζ 3 = ( 1 + 3)/2. If θζ 3 Q(θ) then ζ 3 Q(θ). Therefore, Q Q(ζ 3 ) Q(θ). This is impossible because this would imply that 2 = [Q(ζ 3 ) : Q] divides 3 = [Q(θ) : Q]. d) Give an example of cubic normal extension of Q. Let f(x) = X 3 3X 1 Q[X]. It is irreducible and if θ 1 C is its root then F = Q(θ 1 ) is cubic extension of Q. Apply Cardano formulas to find explicitly all roots of f(x) in C. Any root of f(x) appears in the form u + v, where uv = 1 and u 3 + v 3 = 1. Therefore, u 3 and v 3 are roots exp(±iπ/3) = cos(π/3) ± i sin(π/3) of the quadratic polynomial 2
3 T 2 T + 1 = 0. This gives the following three pairs of solutions (u, v): (exp(iπ/9), exp( iπ/9), (exp(7iπ/9), exp( 7iπ/9)) and (exp(13iπ/9), exp( 13iπ/9)). Therefore, the roots of our cubic polynomial f(x) are θ 1 = 2 cos(π/9), θ 2 = cos(7π/9) and θ 3 = cos(13π/9). This implies that θ 2 = 2 cos(2π/9) = 2(1 2θ 2 1) Q(θ 1 ). And θ 3 Q(θ 1 ) because θ 1 + θ 2 + θ 3 = 0. Theorem (Criterion of normality). A finite field extension F/E is normal if and only if there is a polynomial G(X) E[X] such that F is a splitting field for G(X) over E. Proof. Suppose, first, that F is normal over E. Choose a finite number of elements α 1,..., α r F such that F = E(α 1,..., α r ). (Explain, why such finite set of elements exists.) For 1 i r, let P i (X) E[X] be the minimal polynomial for α i over E. Each polynomial P i (X) is irreducible and has a root α i in F, therefore, P i (X) splits completely in F [X]. (Use that F/E is normal.) Let G(X) be the product P 1 (X)... P r (X). Then G(X) splits completely in F [X], therefore, F contains the splitting field E G for G(X) over E. On the other side, F = E(α 1,..., α r ) contains E G because all α i are roots of G(X) and E G is generated over E by all roots of G(X). So, F = E G. Prove our theorem in the opposite direction. Suppose F = E G is the splitting field for some polynomial G(X) E[X]. Take any irreducible P (X) E[X] such that P (a) = 0 for some a F. We must prove that P (X) splits completely over F. Let q(x) be an irreducible factor of P (X) in F [X]. Consider the minimal field extension F (b) of F, where b is a root of q(x). WE must prove that b F, i.e. F = F (b). Step 1. a and b are roots of the same irreducible polynomial P (X) E[X]. Therefore, there is a field isomorphism σ : E(a) E(b) such that σ(a) = b and σ E = id. Notice that [E(a) : E] = [E(b) : E] = deg P (X). Step 2. We know that F is generated by all roots of G(X) E[X] over E. This iomplies that the roots of G(X) generate the field extensions F (a)/e(a) and F (b)/e(b). In particular, F (a) is the splitting field for G(X) over E(a) and F (b) is the splitting field for F (b) over E(b). Step 3. By the uniqueness property of splitting fields, the field isomorphism σ from step 1 can be extended to a field isomorphism τ : F (a) F (b). In particular, [F (a) : E(a)] = [F (b) : E(b). Remind that a F and, therefore, F = F (a). It remains to count the degrees: [F : E] = [F (a) : E] = [F (a) : E(a)][E(a) : E] = [F (b) : E(b)][E(b) : E] = [F (b) : E] = [F (b) : F ][F : E]. Therefore, [F (b) : F ] = 1, i.e. F = F (b) and b F. Our Theorem is completely proved. Examples. 1) Q( 2, 3, 5, 7) is normal over Q. (It is a splitting field of (X 2 2)(X 2 3)(X 2 5)(X 2 7) Q[X].); 2) Q( 3 2, 3) is the splitting field of X 3 2 Q[X] and is, therefore, normal over Q; 3) Q( 3 2) is not normal over Q. (Consider X 3 2 Q[X] and prove that it has only one root in Q( 3 2).); 3
4 4) Q( 4 2) is not normal but Q( 4 2, i) is normal over Q. 5) If α is a root of X 3 + X + 1 F 2 [X] then F 2 (α) is a normal extension of F 2 of degree 3; (Later we shall prove that any finite field extension of F p, where p is a prime number, is automatically normal.) 6) Let K = F p (T ) = Frac F p [T ] be a field of rational functions in one variable T over prime field F p. Then the polynomial F (X) = X p T K[X] is irreducible and L = K(θ), where θ is a root of P (X), is normal over K Why normal extensiona are good but not good enough? Consider a finite field extension. The Galois Theory deals with symmetries of L over K. These symmetries are field automorphisms σ : L L such that σ K = id. For a given field extension L/K we shall denote the set of such symmetries by Aut K L. Examples: a) Aut R C contains (at least) two elements: the identity map id : C C and the complex conjugation σ : C C given by the correspondence a + bi a bi; b) Aut Q Q( 2) again contains (at least) two elements a + b 2 a ± b 2. The above examples give as a matter of fact precise information about the corresponding sets of symmetries due to the following result. Proposition. Suppose L = K(θ), where θ is a root of an irreducible polynomial P (X) K[X]. Then Aut K L deg P (X). Proof. We know that any element of α L can be written (uniquely) as a linear combination α = b 0 + b 1 θ + + b n 1 θ n 1, where n = deg P (X) and all coefficients b i K. If σ Aut K L then σ(α) = b 0 + b 1 σ(θ) + + b n 1 σ(θ) n 1 (Use that σ is compatible with addition and multiplication and σ K = id.) In other words, the knowledge of the whole map σ is equivalent to the knowledge of just the image σ(θ) of θ. But σ(θ) must be again a root of our polynomial P (X). Indeed, if P (X) = X n + b 1 X n b n 1 X + b n, then P (σ(θ))σ(θ) n + b 1 σ(θ) n b n 1 σ(θ) + b n = σ(p (θ)) = 0 (use again that σ is compatible with operations and acts as identity on K.) This implies our proposition because the number of distinct roots of P (X) is n. Taking into account results of section 2 we obtain Corollary. Aut K L = n is maximal if and only if P (X) splits completely in L[X] (i.e. L is normal over K) and all roots of P (X) are different (i.e. each root of P (X) appears with multiplicity 1). The above properties can be generalized as follows: 4
5 Theorem. For any finite field extension L/K, Aut K L [L : K] and this inequality becomes equality if and only if L/K is normal and the minimal polynomials of all elements of L over K have no multiple roots Separable extensions. Definition. Let F E be an algebraic field extension. An element θ F is separable over E if its minimal polynomial over E has no multiple roots (in its splitting field). Definition. F/E is separable if any element of F is separable over E. So, the last theorem of n.3.3 states that if L/K is a finite field extension then it has a maximal possible number of field automorphisms if and only if it is normal and separable. Inseparable extensions do not appear very often. More precisely, we have the following theorem. Theorem. A finite field extension F/E is separable if it satisfies to one of the following conditions (where p is a prime number): a) char E = 0; b) char E = p and [F : E] is not divisible by p; c) char E = p and E = E p (i.e. any element of E is pth power of element of E) Proof. We must prove that under one of above conditions a)c) the minimal polynomial P (X) of any element θ F has no multiple roots. Let F be a splitting field of P (X) over E. Then P (X) = c 1 i n (X θ i ), with c E and θ 1 = θ. Notice, for any i, P (X) is the minimal polynomial for θ i over E. Indeed, P (θ i ) = 0 and deg P (X) = [E(θ 1 ) : E] = [E(θ i ) : E]. For any polynomial f(x) = a n X n + a n 1 X n a 1 X + a 0 F [X] we can introduce its formal derivative by explicit formula f (X) = a n nx n 1 + a n 1 (n 1)X n a 1. For any polynomials f(x), g(x) F [X] and a F, one can verify the usual rules: (f(x) ± g(x)) = f (X) ± g (X) (af(x)) = af (X) (f(x)g(x)) = f (X)g(X) + f(x)g (X). 5
6 Suppose P (X) has multiple roots, i.e. for some different indices i j, θ i = θ j, i.e. P (X) = (X θ i ) 2 P 1 (X) in F [X]. Then P (X) = 2(X θ i )P 1 (X) + (X θ i ) 2 P 1(X) and, therefore, P (θ i ) = 0. But P (X) is the minimal polynomial for θ i and P (X) E[X] has degree < deg P (X). Therefore, P (X) = 0 is the zero polynomial, i.e. the polynomial with zero coefficients. So, if P (X) = X n + a n 1 X n a 1 X + a 0 then n = 0, a n 1 (n 1) = 0,..., a 1 = 0. If char E = 0 then it is impossible. This proves the case a) of our theorem. If char E = p then n 0 mod p. This proves the case b) of our theorem. Even more, P (X) can have nonzero coefficients a i only if i is divisible by p. In other words, P (X) = X mp + b m 1 X (m 1)p + + b 1 X p + b 0. (Verify that P (X) = 0!) Under assumption c) there are c 0, c 1,..., c m 1 E such that b 0 = c p 0, b 1 = c p 1,..., b m 1 = c p m 1. Therefore, P (X) = X mp + c p m 1 X(m 1)p + + c p 0 = (Xm + c m 1 X m c 0 ) p (Use that char E = p.) This means that P (X) is not irreducible in E[X]. Contradiction. The theorem is completely proved. The above theorem explains why it is not easy to construct an example of inseparable field extension. In the case of fields of characteristic 0 any algebraic field extension is separable. The simplest field extensions in characteristic p are extensions of the finite field F p, but any element of a F p satisfies tha condition a p = a and, therefore any finite field extension of F p is separable. (This also implies that any algebraic extension of any finite field is automatically separable.) So, the first chance to get an example of inseparable extension is to take an infinite field of characteristic p. Example of inseparable extension. Suppose E = F 2 (t) = Frac F 2 [t] is the field of rational functions in one variable t with coefficients in F 2. Let P (X) = X 2 t E[X]. Then P (X) is irreducible (use that E[X] is a unique factorisation domain). Let F = E(θ), where θ is a root of P (X). Then in F [X] we have that P (X) = (X θ) 2 (use that F is a field of characteristic 2). So, P (X) has θ as a root with multiplicity 2. Therefore, F is an inseparable extension of E of degree 2. Notice that replacing everywhere 2 by a prime number p we shall obtain an example of an inseparable extension of degree p. There is also the following general criterion. Theorem. A finite field extension F = E(α), where α is a root of an irreducible polynomial f(x) E[X], is separable if and only if f(x) has no multiple roots (in its splitting field over E). Proof. The only if part is easy: f(x) is the minimal polynomial for α over E and, therefore, if our extension is separable then f(x) can t have multiple roots. Prove the if part of the theorem. (It is very far from to be straightforward!) 6
7 Suppose θ F and P (X) has multiple roots. As earlier, this implies that for some prime number p, char E = p and P (X) = Q(X p ) for some polynomial Q(X) E(X). Let K = E(θ) and K 1 = E(θ p ). Then K and K 1 are subfields in F, K K 1, [K : E] = deg P (X), [K 1 : E] = deg Q(X) and, therefore, [K : K 1 ] = p. Notice that K p K 1, i.e. pth power of any element of K is an element of the smaller field K 1. Now take our α F and consider its minimal polynomials f K (X) over K and f K1 (X) over K 1. Then both f K (X) and f K1 (X) are factors of f(x) and, therefore, have no multiple roots. In addition, f K1 (X) is divisible by f K (X), deg f K (X) = [F : K] and deg f K (X) = [F : K 1 ] = p[f : K]. Finally, consider the pth power g(x) = f K (X) p of f K (X). Then g(x) has coefficients in K p K 1, it has α as a root and its degree is p[f : K] = [F : K 1 ]. Therefore, g(x) satisfies all conditions for the minimal polynomial for α over K 1. Therefore, g(x) = f K1 (X). But this is impossible because then f K1 (X) is a pth power of f K (X) and, therefore, has multiple roots. The theorem is completely proved. Finally notice that above methods allow to prove the following general property: Suppose K L F are finite field extensions. Then F/K is separable if and only if both F/L and L/K are separable Galois extensions and their Galois groups. Definition. A finite field extension E/K is Galois if Aut K E = [E : K]. Theorem. A finite field extension is Galois if and only if it is normal and separable. We do not prove this theorem, but notice that it was completely proved in the previous section in the case of simple extensions E/K, i.e. such that E = K(θ) for some θ E. Let E be any field. Consider the set Aut E of all field automorphismsm of the field E. By definition, Aut E consists of all injective and surjective maps ψ : E E such that for any α, β E, it holds ψ(α + β) = ψ(α) + ψ(β) and ψ(αβ) = ψ(α)ψ(β). Notice that automatically, ψ(0) = 0, ψ(1) = 1, ψ( α) = ψ(α) and for α 0, ψ(α 1 ) = ψ(α) 1. One can prove that: a) Aut Q = {id}; b) Aut Q( 2) = {id, σ}, where for any a, b Q, σ(a + b 2) = a b 2; c) Aut R = {id}. Remark. The properties a) and b) can be easily proved. In order to prove c), prove that for any ψ Aut R, if x, y R and x < y then ψ(x) < ψ(y). Proposition. Aut E is a group, where the operation is the composition of automorphisms. Proof. Suppose ψ, ϕ Aut E. Then for any a E, (ψϕ)(a) = ϕ(ψ(a)). Then standard settheoretic arguments prove that ψϕ is injective and surjective. In addition, for any α, β E, (ψϕ)(α + β) = ψ(ϕ(α + β)) = ψ(ϕ(α)) + ψ(ϕ(β)) = (ψϕ)(α) + (ψϕ)(β) 7
8 and, similarly, (ψϕ)(αβ) = (ψϕ)(α)(ψϕ)(β). This proves that the set Aut E is closed under operation given by the composition of morphisms. Then we must verify group axioms: associativity Indeed, if ψ, ϕ, χ Aut E then for any a E, ((ψϕ)χ)(a) = χ(ϕ(ψ(a)) = (ψ(ϕχ))(a) ; existence of identity element Indeed, the identity map id E Aut E and for any a E, (ψ id E )(a) = ψ(a) = (id E ψ)(a). existence of inverse map One can easily see (do this!) that for any ψ Aut E, one can define the map ψ : E E by the following rule: if α, β E and ψ(α) = β then ψ (β) = α. Also then (check it!) ψ Aut E and ψψ = ψ ψ = id E. The proposition is proved. Examples: a) Aut Q and Aut R are trivial groups, i.e. groups which consist of only one (identity!) element; b) Aut Q( 2) is the cyclic group of order 2, i.e. it equals {id, σ} where σ 2 = id. Suppose now that K is a subfield of E. Then (as earlier) Aut K E is the subset of all ψ Aut E such that ψ K = id, i.e. for any a K, ψ(a) = a. Clearly (prove this!), Aut K E is a subgroup in Aut E. Definition. If E is a finite Galois extension of K then Aut K E is the Galois group of the extension E/K which will be denoted by Gal(E/K). Example. Gal(K/K) is the trivial group, Gal(Q( 2)/Q) is the cyclic group of order Basic results of Galois theory. Suppose E/K is a finite Galois extension and G = Gal(E/K) is its Galois group. Let E G = {a E σ G, σ(a) = a}. In other words, E G is the subset of all invariant elements of E with respect to the action of G on E. Clearly, E G K. Example. If E = C and K = R then C/R is Galois, Gal(C/R) = G = {id, σ}, where σ is the complex conjugation and C G = R. 8
9 Lemma. E G is a subfield in E. Proof. Suppose α, β E G. Then for any σ G, σ(α) = α and σ(β) = β. Because σ is a field automorphism of E, σ(α ± β) = σ(α) ± σ(β) = α ± β and σ(αβ) = σ(α)σ(β) = αβ. Therefore, α ± β and αβ belong to E G. In other words, E G is closed with respect to operations on E and, therefore, is a subfield in E. Notice that the subfield E G of E contains K, i.e. K E G E. The following statement is he first basic result of Galois theory. (We are not going to prove it.) Theorem A. With above notation, E G = K. Suppose H is any subgroup in G. Consider E H = {a E σ H, σ(a) = a}. As earlier, E H is a subfield in E and E H K. Therefore, the correspondence H E H gives a map {subgroups in G} {fields L such that K L E}. This map is a very important component of Galois Theory, it is called the Galois correspondence. Theorem B. The Galois correspondence is a oneone correspondence between the sets of all subgroups of G = Gal(E/K) and all subfields of E which contain K. Notice that if a subgroup is bigger then the corresponding subfield is smaller : for subgroups {e} H 1 H 2 G we have E = E {e} E H 2 E H 1 E G = K. Proposition. If E/K is Galois and H is a subgroup in Gal(E/K) then E is Galois over E H and Gal(E/E H ) = H. Example. If H = {e} then E H = E and Gal(E/E) = {e}. Proof of the proposition. Let E H = L, then K L E and: E/K is normal implies that E/L is normal; E/K is separable impolies that E/L is separable. So, E/L is Galois and we can introduce H 1 = Gal(E/L). Then by theorem A, E H 1 = L = E H and by theorem B, H = H 1, i.e. Gal(E/E H ) = H. The proposition is proved. Corollary. The correspondence L Gal(E/L) gives the inverse to the Galois correspondence and, therefore, defines a oneone map from the set of all fields between K and E to the set of all subgroups of G = Gal(E/K). Corollary. If, as earlier, E/K is finite Galois, G = Gal(E/K) and H G is a subgroup then [E : E H ] = H and [E H [E : K] : K] = [E : E H ] = G = (G : H) the H index of H in G. Problem. If H is a subgroup of Gal(E/K) then E/E H is Galois but, generally, E H /K is not Galois. (Give an example!) How we can characterize the subgroups H such that E H /K is Galois? What will be then Gal(E H /K)? The answer to this question is given by third main result of Galois theory. 9
10 Theorem C. In the above notation, E H /K is Galois if and only if H is a normal subgroup of G. In this case, Gal(E H /K) is the quotient group G/H. Reminder. A subgroup H of G is normal if for any g G, g 1 Hg = H. Equivalently, for any g G, the left coset gh must coincide with the right coset Hg. (In particular, if G is abelian group then any its subgroup is automatically normal.) Then the quotient group G/H appears as a set of all cosets gh with group operation induced by the group operation on G. This means that the natural projection from G to G/H given by the correspondence g gh is a group epimorphism. Remark. There is a natural map p H from G = Gal(E/K) to Gal(E H /K). It is given by the restriction g g E H of any g G to E H. (One can verify that if H is a normal subgroup in G then g(e H ) = E H.) Finally, with respect to the identification Gal(E H /K) = G/H from above Theorem C, the map p H is just the natural projection from G to G/H. 10
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