# Extension fields II. Sergei Silvestrov. Spring term 2011, Lecture 13

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1 Extension fields II Sergei Silvestrov Spring term 2011, Lecture 13 Abstract Contents of the lecture. Algebraic extensions. Finite fields. Automorphisms of fields. The isomorphism extension theorem. Splitting fields. Separable extensions. Typeset by FoilTEX

2 Finite extensions Definition 1. Let E be an extension field of F. E is called an algebraic extension of F if every element of E is algebraic over F. Definition 2. Let E be an extension field of F. E is called a finite extension of F if E is of finite dimension n as a vector space over F. We denote [E : F] = n. Theorem 1. If E is a finite extension field of F, then E is an algebraic extension of F. Proof. Let [E : F] = n, and let u E. The set of n + 1 elements {1,u,u 2,...,u n } is linearly dependent. Hence there are a i F, not all zero, such that a 0 + a 1 u + a 2 u a n u n = 0, which implies that u is algebraic over F. Since u was arbitrary, E is algebraic over F. Typeset by FoilTEX 1

3 The theorem of Lagrange for fields Theorem 2. Let E be a finite extension field of a field F, and let K be a finite extension field of a field E. Then [K : F] = [K : E][E : F]. Proof. Let k 1,..., k n be a basis for K as a vector space over E, and let e 1,..., e m be a basis for E as a vector space over F. If u K, then u = space over F, each α i may be written as α i = n m i=1( j=1 over F. Then 0 = f i j e j )k i = Suppose that n i=1 n i=1 n i=1 m f i j (e j k i ) = j=1 n i=1 α i k i (α i E, k i K). Since E is a vector m f i j e j ( f i j F, e j E). Thus u = j=1 n i=1 α i k i = m f i j e j k i. Therefore, {e j k i : e j E,k i K } span K as a vector space j=1 m f i j (e j k i ) = 0 ( f i j F, e j E, k i K). For each i, let α i = j=1 n m i=1( j=1 f i j e j )k i = n i=1 m f i j e j E. j=1 α i k i. The linear independence of e 1,..., e m over F implies that f i j = 0 for all i, j. Therefore, {e j k i : e j E,k i K } is linearly independent over F and hence a basis. Consequently, we may conclude that [K : F] = [K : E][E : F]. Typeset by FoilTEX 2

4 Adjoining the elements Let α, β be algebraic over F. We suppose α, β are contained in some extension E of F. Then, β is algebraic over F(α). We can form the field F(α)(β). Any field which contains F and α, β, will contain F(α)(β). Hence F(α)(β) is the smallest field containing F and both α, β. Furthermore, by Theorem 2, F(α)(β) is finite over F, being decomposed in the extensions F < F(α) < F(α)(β). Hence by Theorem 1, the field F(α)(β) is algebraic over F and in particular αβ and α + β are algebraic over F. Furthermore, it does not matter whether we write F(α)(β) or F(β)(α). Thus we shall denote this field by F(α,β). Inductively, if α 1,..., α n are algebraic over F and contained in some extension of F, we let F(α 1,...,α n ) be the smallest field containing F and α 1,..., α n. It can be expressed as F(α 1 )(α 2 )...(α n ). It is algebraic over F by repeated applications of Theorem 2. We call it the field obtained by adjoining α 1,..., α n to F. Typeset by FoilTEX 3

5 Example of adjoining Example 1. Consider the field Q( 2, 3) generated over Q by 2 and 3. Since 3 is of degree 2 over Q, the degree of the extension Q( 2, 3) over Q( 2) is at most 2 and is precisely 2 if and only if x 2 3 is irreducible over Q( 2). Since this polynomial is of degree 2, it is reducible only if it has a root, i.e., only if 3 Q( 2). Suppose 3 = a + b 2 with a, b Q. Squaring this we obtain 3 = (a 2 + 2b 2 ) + 2ab 2. If ab = 0, then we can solve this equation for 2 in terms of a and b which implies that 2 is rational, which is not. If b = 0, then we would have that 3 = a is rational, a contradiction. Finally, if a = 0, we have 3 = b 2 and multiplying both sides by 2 we see that 6 would be rational, again a contradiction. This shows 3 / Q( 2), proving [Q( 2, 3) : Q] = 4. Elements in this field (by closing" 1, 2, 3) include 1, 2, 3, 6 and by the computations above, these form a basis for this field: Q( 2, 3) = {a + b 2 + c 3 + d 6: a,b,c,d Q}. Typeset by FoilTEX 4

6 Characterising the finite extensions Theorem 3. An algebraic extension E of F is finite if and only if E is generated by a finite number of algebraic elements over F. Proof. Let E be a finite extension of degree n over F. Let α 1, α 2,..., α n be a basis for E as a vector space over F. Each α i is algebraic over F. Since E is obviously generated over F by α 1, α 2,..., α n, we see that E is generated by a finite number of algebraic elements over F. The converse follows from Theorem 2 by induction. Typeset by FoilTEX 5

7 The algebraic closure Theorem 4. Let F be a field, and E an extension of F. Let F E be the set of all elements from E which are algebraic over F. Then F E is a field. Proof. Whenever α, β are algebraic, then α + β and αβ are algebraic, being contained in the finite extension F(α,β) of F. Definition 3. The field F E is called the algebraic closure of F in E. In particular, the set of all algebraic numbers is the algebraic closure of Q in C, hence a field. Typeset by FoilTEX 6

8 Algebraically closed fields Definition 4. A field F is algebraically closed if every polynomial of degree 1 with coefficients in F has a root in F. Definition 5. An algebraic closure F of a field F is an algebraic extension of F that is algebraically closed. The fundamental theorem of algebra says that C is algebraically closed; moreover, C is an algebraic closure of R. Perhaps the simplest proof of this theorem is by Liouville s theorem in complex variables: Every bounded entire function is constant. If f (x) C[x] had no roots, then 1/ f (x) would be a bounded entire function that is not constant. To prove that every field F has an algebraic closure, we require some appropriate tools of set theory. Typeset by FoilTEX 7

9 Tools of set theory Definition 6. Let A be a set, and let P(A) { } be the family of all its nonempty subsets. The axiom of choice states that if A is a nonempty set, then there exists a function β : P(A) { } A with β(s) S for every nonempty subset S of A. Such a function β is called a choice function. Informally, the axiom of choice is a harmless looking statement; it says that we can simultaneously choose one element from each nonempty subset of a set. The axiom of choice is easy to accept, and it is one of the standard axioms of set theory. However, the axiom is not convenient to use as it stands. There are various equivalent forms of it that are more useful, the most popular of which is Zorn s lemma. Definition 7. A set X is a partially ordered set if there is a relation x y defined on X that is reflexive, transitive and antisymmetric: if x y and y x, then x = y. As usual, we write x < y if x y and x = y. Definition 8. An element m in a partially ordered set X is a maximal element if there is no x X for which m < x. Definition 9. An upper bound of a nonempty subset Y of a partially ordered set X is an element x 0 X, not necessarily in Y, with y x 0 for every y Y. Definition 10. A partially ordered set X is a chain if, for all x, y X, either x y or y x. Lemma 1 (Zorn). If X is a nonempty partially ordered set in which every chain has an upper bound in X, then X has a maximal element. Theorem 5. Zorn s lemma and the axiom of choice are equivalent statements. Typeset by FoilTEX 8

10 The existence of an algebraic closure Theorem 6. Every field F has an algebraic closure. Sketch of proof. The basic idea is to apply Zorn s lemma to a suitably chosen set of algebraic extension fields of F. Let A be a set that contains an element for every possible zero ω f, j of any f (x) F[x]. Consider the set Ω 1 = {F(ω): ω A} and, for any chain T = {F i } Ω 1, add the field W = i T F i to Ω 1. Denote the resulting set by Ω and partially order it by inclusion. By construction, any chain T in Ω has an upper bound W. By Zorn s lemma, there is a maximal element F Ω. Typeset by FoilTEX 9

11 General structure of finite fields Let F be a finite field with q elements and prime field Z p. Theorem 7. The number of elements of F is equal to a power of p. Proof. We may view F as a vector space over Z p. Since F has only a finite number of elements, it follows that the dimension of F over Z p is finite. Let this dimension be n. If w 1, w 2,..., w n is a basis, then every element of F has a unique expression x = a 1 w a n w n, with elements a i Z p. Since the choice of these a i is arbitrary, it follows that there are p n possible elements in F, thus proving that q = p n, as desired. Definition 11. An element α of a field is an nth root of unity if α n = 1. It is a primitive nth root of unity if α n = 1 and α m = 1 for 0 < m < n. Theorem 8. The nonzero elements of a finite field of p n elements are all (p n 1)th roots of unity. They form a cyclic group. Theorem 9. A finite field GF(p n ) of p n elements exist and is unique for every prime power p n. Typeset by FoilTEX 10

12 Conjugation Definition 12. Let E be an algebraic extension of a field F. Two elements α, β E are conjugate over F if irr(α,f) = irr(β,f). Example 2. A polynomial x is irreducible over F = R. Let E = C, α = i, and β = i. Then α and β are conjugate complex numbers over R. Theorem 10 (Conjugation Isomorphism Theorem). Let F be a field, let α and β be algebraic over F, and let deg(α,f) = n. The conjugation homomorphism ψ α,β : F(α) F(β) defined by ψ α,β (c 0 + c 1 α + + c n 1 α n 1 ) = c 0 + c 1 β + + c n 1 β n 1 is an isomorphism of F(α) onto F(β) if and only if α and β are conjugate over F. Typeset by FoilTEX 11

13 Algebra course FMA190/FMA190F An explanation to the Conjugation Isomorphism Theorem F(α) ϕ α ψ α γ F[x] F[x]/p(x) ϕβ ψ β F(β) Here p(x) = irr(f,α) = irr(f,β), γ : F[x] F[x]/p(x) is the canonical residue class map, ϕ α and ϕ β are evaluation homomorphisms, ϕ α = γ ψ α and ϕ β = γ ψ β by the Fundamental Homomorphism Theorem. If both ϕ α and ϕ β are one-to-one, then ψ α and ψ β must be one-to-one, and ψ α,β = ψ β (ψ α ) 1. Corollary to the Conjugation Isomorphism Theorem Corollary 1. Let α be algebraic over a field F. The correspondence β ψ α,β is one-to-one correspondence between the set of numbers β conjugate to α and the set of isomorphisms between F(α) and a subfield of F fixing F element-wise. Example 3. Let F = Q and α = 2. Then irr( 2,Q) = x 2 2. The set of its zeros is { 2, 2}. According to Corollary 1, there exist exactly two isomorphisms between Q( 2) and a subfield of Q = C leaving Q fixed: ψ 2, 2 (a + b 2) = a + b 2, for a, b Q. ψ 2, 2 (a + b 2) = a b 2, Typeset by FoilTEX 12

14 Automorphisms and fixed fields Definition 13. Let F be a field. An isomorphism σ between F and itself is an automorphism of F. The word automorphism is made up of two Greek roots: auto, meaning self", and morph, meaning shape" or form". Just as an isomorphism carries one field onto an identical replica, an automorphism carries a field onto itself. Definition 14. Let σ be an automorphism of a field F. An element a E is left fixed by σ if σ(a) = a. Theorem 11. Let S = {σ i : i I } be a collection of automorphisms of a field E. The set is a subfield of E. E {σi } = {a E : σ i (a) = a,i I } The field E {σi } is called the fixed field of S. Example 4. An identical map σ(x) = x is an automorphism of any field. Example 5. The complex conjugation z z is an automorphism of C. Moreover, z = z if and only if z R. In other words, the complex conjugation leaves R fixed. Example 6. The map ψ 2, 2 of Example 3 is an automorphism of Q( 2) leaving Q fixed. Typeset by FoilTEX 13

15 The group of E over F Theorem 12. Let E and F be fields with F E. The set G(E/F) of all automorphisms of E leaving F fixed forms a subgroup of the group of all automorphisms of E. Moreover, F E G(E/F). The group G(E/F) is called the group of E over F. Example 7. C = R(i) and ±i are the roots of x The group G(C/R) has order 2 and hence is isomorphic to Z 2. Example 8. Let E = Q( 2, 3). We have seen that {1, 2, 3, 6} is a basis of F over Q. Thus if σ G(E/Q), then σ is completely determined by σ( 2) and σ( 3). The possible values are σ( 2) = ± 2 and σ( 3) = ± 3 and this means that there are at most four distinct Q-automorphisms of F. It is easily verified that each of the four possibilities is indeed a Q- automorphism of F and that G(F/Q) is the Klein 4-group. Typeset by FoilTEX 14

16 Algebra course FMA190/FMA190F The Frobenius automorphism Theorem 13. Let F = GF(p n ). The map σ p : F F defined by σ p (a) = a p is an automorphism of F (the Frobenius automorphism). The field F σp is isomorphic to Z p. The isomorphism extension theorem Theorem 14. Let E be an algebraic extension of a field F. Let σ be an isomorphism of F onto a field F. Let F be an algebraic closure of F. Then σ can be extended to an isomorphism τ of E into a subfield of F such that τ(a) = σ(a) for all a F. F E τ τ[e] F σ F Proof. Let S consists of all triples (L,M,λ), where F L E, F M F and λ : L M is an isomorphism that extends σ. Define (L 1,M 1,λ 1 ) (L 2,M 2,λ 2 ) if L 1 L 2, M 1 M 2 and λ 2 (l 1 ) = λ 1 (l 1 ) for all l 1 L 1. Verify that S is a nonempty partially ordered set in which every chain has an upper bound in S. By Zorn s lemma there is a maximal element (L 0,M 0,λ 0 ) of S. We claim that L 0 = E and λ 0 : L 0 M 0 is the desired extension of σ. If L 0 = E, let α E L 0. Now α is algebraic over F, so α is algebraic over L 0. Let p(x) = irr(α,l 0 ). By the Fundamental Homomorphism Theorem, the evaluation homomorphism ϕ α : L 0 [x] L 0 (α) may be represented as ϕ α = γ ψ α, where γ is a canonical homomorphism from L 0 [x] onto L 0 [x]/p(x), and ψ α is the isomorphism between L 0 [x]/p(x) and L 0 (α). Typeset by FoilTEX 15

17 Algebra course FMA190/FMA190F L 0 (α) λ 0,x L 0 [x] M 0 [x] ϕ α λ 0 ϕ α L 0 M 0 γ γ ψ α L 0 (x)/p(x) λ 0 λ 0 =γ λ 0,x γ 1 M 0 [x]/q(x) ψ α M 0 (α ) Let p(x) = a 0 + a 1 x + + a n x n, q(x) = λ 0 (a 0 ) + λ 0 (a 1 )x + + λ 0 (a n )x n. Since λ 0 is an isomorphism, q(x) is irreducible in L 0 [x]. Since L 0 F, there is a zero α of q(x) in F. By the Fundamental Homomorphism Theorem, the evaluation homomorphism ϕ α : M 0 [x] M 0 (α) may be represented as ϕ α = γ ψ α, where γ is a canonical homomorphism from M 0 [x] onto M 0 [x]/q(x), and ψ α is the isomorphism between M 0 [x]/q(x) and M 0 (α ). Let λ 0 = γ λ 0,x γ 1 be the isomorphism extending λ 0 on L 0 and mapping x + p(x) onto x + q(x). Then the composition of maps ψ α λ 0 ψ 1 α : L 0 (α) M 0 (α ) is an isomorphism of L 0 (α) onto M 0 (α ) F. Clearly, (L 0,λ 0 ) < (L 0 (α),ψ α λ 0 ψα 1 ), which contradicts that (L 0,λ 0 ) is maximal. Therefore, we must have had L 0 = E. Typeset by FoilTEX 16

18 Corollaries to the isomorphism extension theorem Corollary 2. Let a field E be an algebraic extension of a field F and a subfield of the algebraic closure F. Let α, β E be conjugate over F. The conjugation isomorphism ψ α,β : F(α) F(β) can be extended to the isomorphism of E onto a subfield of F. Corollary 3. Let F and F be two algebraic closures of F. between F and F leaving each element of F fixed. Then there exists an isomorphism Typeset by FoilTEX 17

19 The index of a field extension Theorem 15. Let E be a finite extension of a field F. Let σ be an isomorphism of F onto a field F, and let F be an algebraic closure of F. Then the number of extensions of σ to an isomorphism τ of E onto a subfield of F is finite and independent on F, F, and σ. Definition 15. Let E be a finite extension of a field F. The index {E : F} of E over F is the number of isomorphism τ of E onto a subfield of F. Splitting fields Definition 16. Let E be an extension field of a field F. A polynomial f (x) F[x] splits in E if it factors into a product of linear factors in E[x]. Definition 17. Let { f i (x): i I } be a collection of polynomials in F[x]. A field E F is the splitting field of { f i (x): i I } over F if all f i (x) split in E, but do not split over any proper subfield of E. Example 9. Consider f (x) = x Q(x). The roots of f (x) are ±i, and so f (x) splits in C; that is, f (x) = (x + i)(x i) is a product of linear polynomials in C[x]. However, C is not a splitting field of f (x) over Q, for C is not the smallest field containing Q and all the roots of f (x). The splitting field of f (x) F(x) depends on F as well as on f (x). Here, the splitting field of f (x) over Q is Q(i), the splitting field over R is R(i) = C. Example 10. Let f (x) = x 4 10x Q[x]. If β is the root of f (x), then the quadratic formula gives β 2 = 5±2 6. But the identity a+2 ab+b = ( a+ b) 2 gives β = ±( 2+ 3). Similarly, = ( 2 3) 2, so that the roots of f (x) are 2 + 3, 2 3, 2 3, We claim that f (x) is irreducible in Q[x]. If g(x) is a quadratic factor of f (x) in Q[x], then g(x) = (x a 2 b 3)(x c 2 d 3), where a, b, c, d {1, 1}. Multiplying, g(x) = x 2 ((a + c) 2 + (b + d) 3)x + 2ac + 3bd + (ad + bc) 6. Typeset by FoilTEX 18

20 We check easily that (a + c) 2 + (b + d) 3 is rational if and only if a + c = b + d = 0, but these equations force ad +bc = 0, and so the constant term of g(x) is not rational. Therefore, g(x) / Q[x], and so f (x) is irreducible in Q[x]. If β = 2 + 3, then f (x) = irr(β,q). Consider the field F = Q(β) = Q( 2 + 3). There is a chain of fields Q F E, where E = Q( 2, 3), and so [E : Q] = [E : F][F : Q]. Since F = Q(β) and β is a root of an irreducible polynomial of degree 4, namely, f (x), we have [F : Q] = 4. On the other hand, [E : Q] = [E : Q( 2)][Q( 2) : Q]. Now [Q( 2) : Q] = 2, because 2 is a root of the irreducible quadratic x 2 2 in Q[x]. We claim that [E : Q( 2)] 2. The field E arises by adjoining 3 to Q( 2); either 3 Q( 2), in which case the degree is 1, or x 2 3 is irreducible in Q( 2)[x], in which case the degree is 2 (in fact, the degree is 2). It follows that [E : Q] 4, and so the equation [E : Q] = [E : F][F : Q] gives [E : F] = 1, that is, E = F. It follows that F = Q( 2+ 3) is a splitting field of f (x) = x 4 10x 2 +1, as well as a splitting field of (x 2 2)(x 2 3). We have the following diagram of known subfields. Q( 2) 2 2 Q( 2, 3) 2 2 Q( 6) Q 2 2 Q( 3) Example 11. The splitting field of x 3 2 over Q is not just Q( 3 2), since the three roots of this polynomial in C are 3 ( 3 2, i ) 3, 2 3 ( 1 i ) 3 2, 2 and the latter two roots are not elements of Q( 3 2) since the elements of this field are of the form a + b c 3 4 with rational a, b, c and all such numbers are real. Typeset by FoilTEX 19

21 The splitting field E of this polynomial is obtained by adjoining all three of these roots to Q. Note that since E contains the first two roots above, then it contains their quotient hence 2 E contains the element 3. On the other hand, any field containing 3 2 and 3 contains all three of the roots above. It follows that E = Q( 3 2, 3) is the splitting field of x 3 2 over Q. Since 3 satisfies the equation x = 0, the degree of this extension over Q( 3 2) is at most 2, hence must be 2 since we observed above that Q( 3 2) is not the splitting field. It follows that [Q( 3 2, 3) : Q] = 6. This gives us the diagram of known subfields. Q( 3 2, 3) Q(θ 1 ) Q(θ 2 ) Q( 3) Q Q(θ 3 ) where θ = 3 2, θ 2 = 3 2 ( 1 + i ) ( 3, θ 3 = 3 1 i ) Example 12. One must be careful in computing splitting fields. The splitting field for the polynomial x over Q is smaller than one might at first suspect. In fact this polynomial factors over Q: x = x 4 + 4x x 2 = (x 2 + 2) 2 4x 2 = (x 2 + 2x + 2)(x 2 2x + 2) Typeset by FoilTEX 20

22 where these two factors are irreducible (Eisenstein again). Solving for the roots of the two factors by the quadratic formula, we find the four roots ±1 ± i so that the splitting field of this polynomial is just the field Q(i), an extension of degree 2 of Q. Typeset by FoilTEX 21

23 Splitting fields and automorphisms Theorem 16. Let E be any intermediate field between F and F. E is a splitting field over F if and only if every automorphism of F leaving F fixed maps E onto itself. Corollary 4. If E is a splitting field of finite degree over F, then {E : F} = G(E/F). Multiplicity of zeros of a polynomial Definition 18. Let F be a field and f (x) F[x]. An element α F is a zero of f(x) of multiplicity ν, if ν is the greatest integer such that (x α) ν is a factor of f (x) in F[x]. Theorem 17. Let f (x) be irreducible in F[x]. Then all zeroes of f (x) in F have the same multiplicity. Corollary 5. If E is a finite extension of F, then {E : F} divides [E : F]. Typeset by FoilTEX 22

24 Separable extensions Definition 19. Let E be a finite extension of F. E is a separable extension of F if {E : F} = [E : F]. An element α F is separable over F if F(α) is a separable extension of F. An irreducible polynomial f (x) F[x] is separable over F if every zero of f (x) in F is separable over F. Theorem 18. Let E be a finite extension of F. 1. E is a separable extension of F if and only if each α E is separable over F. 2. An element α F is separable over F if and only if all zeroes of irr(α,f) in F have multiplicity An irreducible polynomial f (x) F[x] is separable over F if and only if all zeroes of f (x) in F have multiplicity 1. Typeset by FoilTEX 23

25 Perfect fields Definition 20. A field F is perfect if every finite extension is separable. Theorem 19. A field F is perfect if and only if either F has characteristic 0 or F has characteristic p and every element if F is a pth power in F, i.e., F = F p. The primitive element theorem Theorem 20. Let E be a finite separable extension of a field F. There is α E (a primitive element) with E = F(α). In particular, if F has characteristic 0, then every finite extension E is a simple extension. The theorem of the primitive element was known by Lagrange, and Galois used a modification of it in order to construct the original version of the Galois group. Typeset by FoilTEX 24

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