10 Splitting Fields. 2. The splitting field for x 3 2 over Q is Q( 3 2,ω), where ω is a primitive third root of 1 in C. Thus, since ω = 1+ 3

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1 10 Splitting Fields We have seen how to construct a field K F such that K contains a root α of a given (irreducible) polynomial p(x) F [x], namely K = F [x]/(p(x)). We can extendthe procedure to build a field that contains all roots of the given polynomial. Definition The extension K/F is a splitting field for the polynomial f(x) F [x] if f(x) splits completely (i.e. factors into linear factors) over K, but does not factor completely over any proper subfield of K containing F.Wewill show later that any two splitting fields for a fixed polynomial are isomorphic, so in fact we may talk about the splitting field for f(x) over F. Theorem Given any f(x) F [x], there exists an extension K F which is a splitting field for f(x) over F. Proof. By iterating the procedure for finding an extension containing one root of the polynomial, we can see there exists an extension E of F containing all roots of f(x). Let K be the intersection of all subfields of E containing F and all roots of f(x). Definition If K/F is an algebraic extension and K is the splitting field over F for a collection of polynomials over F, then K is a normal extension of F. Note: Some people use the term normal for what we will call Galois. Notice that if F E K and K is a splitting field over F for some collection of polynomials, then K is also a splitting field over E for the same collection. Examples: 1. The splitting field for x 2 2 over Q is Q( 2). 2. The splitting field for x 3 2 over Q is Q( 3 2,ω), where ω is a primitive third root of 1 in C. Thus, since ω = 1+ 3, this field is Q( 3 2, 3). 2 We have [Q( 3 2, 3) : Q] =6. The diagram of intermediate field extensions is given below. (We don t yet have the results to prove this is correct.) 3. The field Q( 2, 3) is the splitting field for (x 2 2)(x 2 3) over Q. Again the diagram of intermediate extensions is given below. 52

2 Q( 3 2, 3) Q( 3) Q( 3 2) Q(ω 3 2) Q(ω 2 3 2) Q Q( 2, 3) Q( 3) Q( 6) Q( 2) Q How big is the splitting field of a polynomial of degree n, relative to the original field? Proposition A splitting field of a polynomial of degree n over F is of degree n! over F. Proof. Let F 1 be obtained from F by adjoining one root of f(x) tof. Then if f(x) has degree n, wehave [F 1 : F ] n. The polynomial f(x) has at least one linear factor over F 1,soany other root satisfies a polynomial of degree at most n 1 over F 1. The result follows by induction. 53

3 Example: Cyclotomic Fields The roots of the polynomial x n 1 over Q are the n th roots of 1, e 2πki n, 0 k n 1. The subset of n th roots of 1 form a finite subgroup of the multiplicative group of any field that contains them, thus it is a cyclic group. Definition A generator of the cyclic group of all n th roots of 1 is a primitive n th root ofunity. If ζ n is a primitive n th root of 1, then the others are given by ζn a where 1 a<nis an integer relatively prime to n. All n th roots of unity are powers of ζ n,sothe splitting field of x n 1 over Q is Q(ζ n ). This is called the n th cyclotomic field. (The word cyclotomic means circle-cutting.) Note that if n = p is prime, then x p 1=(x 1)(x p x +1)is a factorization into irreducible polynomials over Q, so[q(ζ p ):Q] =p 1. In general, we will show that [Q(ζ n ):Q] =φ(n) where φ denotes the Euler φ-function. We want to show that any two splitting fields of the same polynomial over F are isomorphic. We need the following result. Theorem Let φ : F F be an isomorphism of fields, f(x) F [x] apolynomial,and f (x) the image of f(x) under the natural extension of φ to an isomorphism F [x] F [x]. LetE be asplitting field for f(x) over F, and E a splitting field for f (x) over F. Then φ extends to an isomorphism σ : E E. Proof. Induct on the degree n of f(x). If f(x) splits completely over F, then f (x) splits completely over F,sotake E = F, E = F,σ = φ. Thus itis true for n =1,and more generally when all irreducible factors of f(x) are of degree 1. Assume the result is true whenever deg f(x) <n. Let deg f(x) =n and let p(x) beanirreducible factor of f(x) ofdegree 2. Let p (x) bethe corresponding factor of f (x). Let α be a root of p(x) and β arootofp (x). Then φ extends to an isomorphism σ : F (α) F (β) aswehave already seen. Write f(x) =(x α)f 1 (x),f (x) =(x β)f 1. Then E is a splitting field for f 1 (x) over F (α), and E is a splitting field for f 1(x) over F (β). The degrees of f 1 (x) and f 1(x) are n 1, so by induction there exists an isomorphism σ : E E extending σ : F (α) F (β), and this σ is the desired extension of φ. Corollary Any two splitting fields for f(x) F [x] over F are isomorphic. 54

4 Proof. Take φ to be the identity, F = F, and E,E the two splitting fields for f(x) in the theorem above. Next we look at the notion of an algebraic closure, when one adds all roots of all polynomials over a field. Definition The field F is an algebraic closure of F if F is algebraic over F and if every polynomial f(x) F [x] splits completely over F.Afield K is algebraically closed if every polynomial with coefficients in K has a root in K. An algebraically closed field is thus its own algebraic closure. (Indeed, K = K if and only if K is algebraicaly closed.) Note that it is not a priori clear that a given field F has an algebraic closure. Then F is alge- Proposition Let F be an algebraic closure of F. braically closed. Proof. Let f(x) F [x],αarootoff(x). Then F (α) isalgebraic over F, and F is algebraic over F,so F (α) isalgebraic over F.Thusα is algebraic over F, and hence α F. Therefore F is algebraically closed. To show that every field F has an algebraic closure, we first construct an algebraically closed field K containing F, which may be much bigger than an algebraic closure of F, and then take the subfield of elements that are algebraic over F. This will be F. Lemma Let K be a field. Then there exists an algebraic extension K 1 of K such that every polynomial in K[x] has a root in K 1. Proof. It suffices to construct a field K 1 such that every monic irreducible polynomial of degree > 1inK[x] has a root in K 1. For each such monic irreducible f(x), define a variable x f, and let S be the set of all such x f. Substituting x f for x, wehave the polynomial f(x f ) K[S]. Let A denote the ideal of K[S] generated by {f(x f ) x f S}. We claim A is a proper ideal of K[S]. For if not, 1 A, and so there exist monic polynomials f 1,...,f r of degree > 1inK[x] and polynomials g 1,...,g r K[S], such that 1=g 1 f 1 (x f1 )+ + g r f r (x fr ). Now there exists a (finite) extension L of K in which each of f 1 (x),...,f r (x) has a root; say α i L is arootoff i (x), 1 i r. Then substituting α i for x i,wehave 1 = g 1 (α 1,...,α r )f 1 (α 1 )+ + g r (α 1,...,α r )f r (α) =0, a contradiction. 55

5 Thus A is a proper ideal of K[S] and there exists a maximal ideal M of K[S] with M A. Let K 1 = K[S]/M. Let β f denote the image in K 1 of x f under the map π : K[s] K 1. Then f(β f )=π(f(x f )) = 0, so each monic irreducible polynomial f(x) K[x] ofdegree > 1hasarootinK 1 as claimed. Each β f is algebraic over K 1, and clearly K K 1. We have K 1 = K[β f x f S] isanalgebraic extension of K. Proposition Let K be a field. Then K has an algebraic closure. Proof. Using the previous lemma, we can construct a sequence K = K 0 K 1 K n... of field extensions such that for each i 0, K i+1 /K i is algebraic, and each polynomial in K i [x] hasarootink i+1. Then K i /K is algebraic for all i. Usning set theory, one can construct a set in which the K i all embed compatibly with their embeddings in each other. (For example, one can take the direct limit lim K i. Let L = i K i in this set. Each element of L lies in some K i, and for any finite set of elements in L, wecan find a K i containing all of them. Therefore L is a field. Each element of L is algebraic over K since it lies in some K I algebraic over K. If f(x) L[x], then there exists K i containing all the coefficients of f. In particular, f(x) K i [x] L[x], so f(x) hasarootink i+1 and hence in L. Thus L is algebraically closed, and hence is an algebraic closure of K. (We will prove uniqueness later probably.) Among algebraic extensions of F we distinguish two types, separable and inseparable. Definition Apolynomial over F is called separable if it has no multiple roots. Otherwise it is called inseparable. We proved already that a polynomial f(x) has a multiple root α if and only if α is also a root of its formal derivative f (x), if and only if both f(x) and f (x) are divisible by m α,f (x). Thus f(x) isseparable if and only if it is relatively prime to its derivative. From this we have the following: Corollary Every irreducible polynomial over a field of characteristic 0 (e.g. Q) isseparable. Proof. If p(x) F [x] isirreducible of degree n, then p (x) isofdegree n 1, and the only factors of p(x) are 1 and p(x), so p(x),p (x) are relatively prime. 56

6 Then the question becomes, where does this proof fail in characteristic p? Ifthe derivative of p(x) 0,then it is relatively prime to p(x), and so we ask when does p (x) =0. Proposition Let char(f )=p. Then for all a, b F,wehave (a + b) p = a p + b p, (ab) p = a p b p,sothe map φ : F F, a a p is an injective field homomorphism. If F is finite, this map is an automorphism of F. (Thus every element of F is a p th power: F = F p. For F either finite or infinite, the elements fixed by φ are precisely the elements of Z/pZ = F p, the prime subfield of F. Proof. The statements are obvious except for the final one. Since φ(x) = x p = x for any element fixed by φ, wesee that φ can only fix the roots of x p x, ofwhich there can be at most p. But every element of F p satisfies this equation, so those are precisely the roots. Definition The map φ is called the Frobenius homomorphism. (It is defined for any commutative ring with 1, of characteristic p.) A field of characteristic p is called perfect if φ : F F is onto, hence an automorphism of F. Inparticular, finite fields are perfect. We also say that any field of characteristic 0 is perfect. Proposition Every irreducible polynomial over a perfect field is separable. Proof. It suffices to show this for a field F of characteristic p such that F p = F. Let p(x) F [x] beirreducible. If p(x) were inseparable, then p(x) =q(x p ) for some polynomial q(x) =a m x m + + a 1 x + a 0. Then since F = F p,wemay write a i = b p i for some b i F. We have p(x) =q(x p )= a m x mp + +a 1 x p +a 0 = b p mx mp + +b p 1x p +b p 0 =(b m x m ) p + +(b 1 x) p +b p 0 = (b m x m + + b 1 x + b 0 ) p.thusp(x) isthep th power ofapolynomial in F [x], contradicting irreducibility. We then ask, what do inseparable polynomials over a field F of characteristic p look like? Proposition Let p(x) be an irreducible polynomial over a field F of characteristic p. Then there exists a unique integer k 0 and a unique separable polynomial p sep (x) F [x] such that p(x) =p sep (x pk ). 57

7 Proof. If p(x) is separable, we are done. If p(x) is inseparable, then p(x) = p 1 (x p ) for some polynomial p 1 (x). If p 1 (x) isseparable, stop. If not, continue. Eventually we get a uniquely determined power p k of p such that p(x) =p k (x pk ), where p k (x) isseparable. Also p k (x) isirreducible, since any factorization of p k (x) gives a factorization of p(x). Definition The degree of p sep (x) is the separable degree of p(x), denoted deg s (p(x)). The integer p k is the inseparable degree of p(x), denoted deg i (p(x)). Wehave deg(p(x)) = deg s (p(x)) deg i (p(x)). Definition The field K is said to be separable over F if every element of K is the root of a separable polynomial over F.Afield which is not separable is inseparable. Corollary Every finite extension of a perfect field is separable. In particular, every finite extension of Q or a finite field is separable. We will return to the topic of separability later. Finite Fields and Cyclotomic Fields We begin with a brief consideration of finite fields. We will have much more to say about them once we have the Fundamental Theorem of Galois Theory at our disposal. Let n>0, and consider the splitting field of x pn x over F p. This polynomial is separable (its derivative is 1 0), and so it has p n distinct roots. Let α, β be two roots of this polynomial. Then α pn = α, β pn = β. This then means (α β) pn = α pn β pn = α β,(α 1 ) pn = α 1, and (α + β) pn = α pn + β pn = α + β, sothe set of roots to this polynomial forms a field, and hence must be the splitting field F. Then F = p n and [F : F p ]=n. This shows that there exist finite fields of degree n over F p for every positive integer n. Now if F is any finite field of characteristic p, say[f : F p ]=n, then F = p n. The multiplicative group F is cyclic of order p n 1, so every nonzero element satisfies x pn 1 1=0, and hence every element satisfies x pn x =0. By cardinality considerations, these must be all roots, so this is the splitting field for x pn x over F p, and hence this field is unique (up to isomorphism). 58

8 Next we turn our attention to the so called cyclotomic fields. The word cyclotomic comes from roots meaning circle cutting, and they are the fields obtained by adjoining n th roots of 1 to Q. Definition Let µ n denote the group of n th roots of 1 over Q. Then the multiplicative group µ n is isomorphic to the additive group Z/nZ where the isomorphism is given by (ζ n ) a a Z/nZ, where ζ n denotes a primitive n th root of 1. There are precisely φ(n) primitive n th roots of 1, corresponding to (ζ n ) a where (a, n) =1. (Here φ denotes the Euler phi-function.) Observe that if d n, then µ d µ n, since a d th root of 1 is also an n th root of 1. Definition The n th cyclotomic polynomial Φ n (x) is the polynomial whose roots are the primitive n th roots of unity: Φ n (x) = (x ζ) = (x ζn), a ζ µ n,ζprimitive which is a polynomial of degree φ(n). 1 a<n,(a,n)=1 Observe that x n 1= ζ µ n (x ζ), and if we group the factors (x ζ) where ζ is a primitive d th root of 1, d n, wehave x n 1= ζ) d n ζ µ d,ζprimitive(x = Φ d (x). d n (Note that n = d n φ(d).) Thus we can compute Φ n(x) recursively for any n. This also shows the following. Lemma Φ n (x) is a monic polynomial of degree φ(n) Z[x]. Proof. Clearly φ(n) isamonic polynomial of degree φ(n) inq(x). One can show inductively that the coefficients lie in Z[x]. It is clearly true for n = 1. Assume Φ d (x) Z[x] for all 1 d < n. Then x n 1 = ( d n,d<n Φ d(x))(φ n (x)). The first factor is a monic polynomial in Z[x], and so by Gauss s Lemma, Φ n (x) isalso a monic polynomial in Z[x]. Theorem Φ n (x) is irreducible over Q. Proof. Suppose Φ n (x) =f(x)g(x),f,g Z[x], monic. We may assume f(x) is irreducible. Let ζ µ n bearootoff(x), so f(x)isthe minimal polynomial for ζ over Q. Let p be aprime such that p n. Then ζ p is a primitive n th root 59

9 of 1, so ζ p is a root of f or g. Suppose g(ζ p )=0. Then ζ is a root ofg(x p ), and since f(x) isthe minimal polynomial for ζ, wehave f(x) g(x p ), say g(x p )=f(x)h(x),h Z[x]. Then consider this equation reduced modulo p: [ḡ(x)] p =ḡ(x p )= f(x) h(x) F p [x]. Now F p [x] isaufd, so f(x) and ḡ(x) have a common factor in F p [x]. Then Φ n (x) = f(x)ḡ(x), so Φ n (x) has a multiple root. Then x n 1 has a multiple root over F p. But we have already seen that x n 1 has n distinct roots over F p if p n. This is a contradiction, so ζ p cannot be a root of g(x), and so ζ p would have to be a root of f9x). This applies to every root ζ of f(x), and so ζ a is a root of f(x) for every integer a relatively prime to n. Thus every primitive n th root of 1 is aroot of f(x), and hence f(x) =Φ n (x) isirreducible. Corollary [Q(ζ n ):Q] =φ(n). We conclude this section with the calculation of some examples. We have [Q(ζ 8 ):Q] =4=φ(8). But 1 Q(ζ 8 ), and also ζ 8 + ζ8 7 = 2, so 2 Q(ζ8 ). Since [Q(i, 2) : Q] =4, we have equality. Note Φ 8 (x) =x 4 + 1=(x 2 +i)(x 2 i), but this also factors over R as (x 2 + 2x+1)(x 2 2x+1). 60

10 Some other Φ n (x): 1. If n = p a prime, Φ p (x) =x p 1 + x p x Φ 6 (x): We have φ(6) = 2, and x 6 1=(x 3 + 1)(x 3 1) =(x 1)(x + 1)(x 2 + x + 1)(x 2 x +1), so Φ 6 (x) =x 2 x + 1). We have ζ 6 = 1+ 3 = ζ Φ 9 (x) =x 6 + x Φ 10 (x) =x 4 x 3 + x 2 x +1,since again ζ 10 = ζ Φ 12 (x) =x 4 x Again, note ζ 12 = ζ 6, since Φ 6 (x 2 )=Φ 12 (x). 61