What happened in 13.4. Let F be a field and f(x) 2 F [x]. An extension K ⊃ F is a splitting field for f(x) if it is the minimal extension of F over which f(x) splits into linear factors. More precisely, K is a splitting field for f(x) if it splits over K but does not split over any subfield of K. Theorem 1. For any field F and any f(x) 2 F [x], there exists a splitting field K for f(x). Idea of proof. We shall show that there exists a field over which f(x) splits, then take the intersection of all such fields, and call it K. Now, we decompose f(x) into irreducible factors 0 0 p1(x) : : : pk(x), and if deg(pj) > 1 consider extension F = f[x]=(pj(x)). Since F contains a 0 root of pj(x), we can decompose pj(x) over F . Then we continue by induction. Proposition 2. If f(x) 2 F [x], deg(f) = n, and an extension K ⊃ F is the splitting field for f(x), then [K : F ] 6 n! Proof. It is a nice and rather easy exercise. Theorem 3. Any two splitting fields for f(x) 2 F [x] are isomorphic. Idea of proof. This result can be proven by induction, using the fact that if α, β are roots of an irreducible polynomial p(x) 2 F [x] then F (α) ' F (β). A good example of splitting fields are the cyclotomic fields | they are splitting fields for polynomials xn − 1. Roots of xn − 1 are n-th roots of unity, they form a group isomorphic to Z=nZ. This group is cyclic and has φ(n) elements that can serve as the generator, here φ(n) is the Euler function: φ(n) is equal to the number of positive integers k < n, such that (k; n) = 1. Every such generator is called a primitive root of unity. Thus, if ζn is a primitive n n-th root of unity, then Q(ζn) is a cyclotomic field, i.e. the splitting field for x − 1. It is natural to ask, what the degree of Q(ζn) over Q is. For n = p prime, we can answer right away: [Q(ζp): Q] = p − 1. Indeed, xp − 1 = (x − 1)(xp−1 + ··· + x + 1); p−1 and it is enough to show that Φp(x) := x + ··· + x + 1 is irreducible over Q. The latter is not immediately obvious, but we could use a trick and show instead that Φp(x + 1) is irreducible. We have (x + 1)p − 1 Φ (x + 1) = ; p x (x+1)p−1 and one can notice that every coefficient of except for the leading one Φp(x + 1) = x is divisible by p. Then Eisenstein's criterion applies, Φp(x + 1) is irreducible, hence so is Φp(x), and we have [Q(ζp): Q] = deg(Φp) = p − 1. Definition 4. A field F is an algebraic closure of F if it is an algebraic extension of F , and every polynomial in F [x] splits completely over F . A field K is called algebraically closed if every polynomial in K[x] splits completely over K. Proposition 5. For any field F its algebraic closure F is algebraically closed (that is if one allows polynomials with coefficients in F instead of F , they still split over F ). Idea of proof. One shall use that F is algebraic over F and hence if α is a root of f¯(x) 2 F [x] it is algebraic over F , and hence belongs to F . Theorem 6. For any field F , there exists an algebraically closed field containing F . Theorem 7. For field C of complex numbers is algebraically closed (we will prove this later). 1 2 Section 13.5. Since we have just discussed splitting fields, i.e. fields over which a particular polynomial decomposes into linear factors, it makes sense to ask whether all of these factors are distinct, or if the polynomial has repeated roots. Definition 8. A polynomial is called separable if all of its roots are distinct (in the corre- sponding splitting field), and inseparable otherwise. 2 Example. A polynomial x − 2 is separable over Q, but inseparable over any field of char- acteristic 2. In order to check whether polynomial has repeated roots it is convenient to use its deriv- ative. We will not define the latter as a limit, because they do not make sense over finite fields, but rather we will define it by the following explicit formula: for n f(x) = anx + : : : a1x + a0 2 F [x] we define its derivative to be n−1 Dxf(x) = nanx + ::: 2a2x + a1 2 F [x]: Clearly, Dx : F [x] ! F [x] is still a linear operation. Proposition 9. A polynomial f(x) has a multiple root α if and only if α is a root of Dxf(x). Proof. Suppose α is a multiple root of f(x), then over a splitting field we have f(x) = (x − α)ng(x); where n > 1. Then n−1 n Dxf(x) = n(x − α) g(x) + (x − α) Dxg(x); and therefore Dxf(α) = 0. Now, let F (α) = Dxf(α) = 0. Write f(x) = (x − α)g(x), then Dxf(x) = g(x) + (x − α)Dxg(x); 2 which implies g(α) = 0. But then g(x) = (x − α)h(x) and f(x) = (x − α) h(x). Corollary 10. A polynomial f(x) is separable of and only if it is relatively prime with its derivative, that is (f(x);Dxf(x)) = 1. Corollary 11. Every irreducible polynomial over the field of characteristic 0 is separable. A polynomial over such field is separable if and only if it is a product of distinct irreducible polynomials. Proof. Let p(x) 2 K[x] be irreducible, deg(p) = n, and char(K) = 0. Then the derivative Dxp(x) has degree n − 1, p(x) and Dxp(x) are co-prime since p(x) is irreducible, and the first statement follows. For the second statement, one only need to notice that distinct irreducible polynomials can't have common roots, otherwise, they would be divisible by the minimal polynomial for that root. Example. pn n pn−1 pn (1) Let f(x) = x − x 2 Fp, then Dxf(x) = p x − 1 = −1 in Fp. Hence x − 1 is separable over Fp, as Dxf(x) has no roots at all. n n−1 (2) For Let f(x) = x − 1 2 F [x], we have Dxf(x) = nx unless char(F ) j n. Since nxn−1 has only root 0 of multiplicity n − 1 and 0 is not a root of f(x), we derive that xn − 1 is separable. 3 We would now like to prove Corollary 11 in greater generality, that is for fields of finite characteristic. Let us note however, that in the proof of the Corollary we used the fact that char(K) = 0 to make sure that Dxf 6≡ 0. Indeed, if char(K) = p and f(x) 2 K[x], we have Dxf ≡ 0 if and only if all exponents of x present in f(x) are divisible by p, that is mp p f(x) = amx + ··· + a1x + a0: Definition 12. A field K of characteristic p is perfect if K = Kp, that is for any α 2 K there exists β 2 K such that α = βp. Every field of characteristic 0 is also considered perfect. Proposition 13. Every irreducible polynomial over a perfect field is separable. A polynomial over a perfect field is separable if and only if it is a product of distinct irreducible polynomials. Proof. Suppose K is a perfect field cf characteristic p, and f(x) 2 K[x] is an irreducible polynomial. If Dxf 6≡ 0, then f(x) is separable exactly as in the proof of Corollary 11. Otherwise, Dxf ≡ 0, which in turn implies mp p f(x) = amx + ··· + a1x + a0; but since K is perfect, we can write m p p p f(x) = (bmx ) + ··· + (b1x) + b0; p where aj = bj . As we show in the following lemma, we can rewrite the above equality as m p f(x) = (bmx + ··· + b1x + b0) ; which contradicts the irreducibility of f(x) and finishes the proof. Lemma 14. Let K be a field of characteristic p, then the Frobenius map φ: K ! K; φ(α) = αp is an injective endomorphism. Proof. It is clearly injective, since K has no nilpotent elements. To prove that φ is an endomorphism, it is enough to show equalities (a + b)p = ap + bp and (ab)p = apbp for any a; b 2 K. The second equality is obvious. For the first one, one only needs to use the p binomial formula, and recall that k is divisible by p for any prime p and 0 < k < p. Corollary 15. Every finite field is perfect. Proof. The fact the Frobenius endomorphism is injective implies that it is an automorphism of a finite field. Corollary 16. For any prime integer p and any integer n > 1, there exists a unique up to isomorphism finite field with pn elements. pn Proof. Let K be a splitting field for the polynomial x − x over Fp. We already know that this polynomial is separable, therefore has precisely pn roots. Moreover, its roots form a field: indeed if αpn = α and βpn = β one has (αβ)pn = αβ,(α−1)pn = α−1, and by Lemma 14 (α ± β)pn = αpn ± βpn .