By Jürgen Müller

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**Extra resources for Algebra**

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Latter case occurs if and only if ∂X Thus for f irreducible, if char(K) = 0 then f has only simple roots in L, while if char(K) = p > 0 then f has a multiple root in L if and only if there is ∂f g ∈ K[X] such that f = g(X p ) ∈ K[X]: If f = i≥0 ai X i such that ∂X = ip i−1 a X , ia X = 0, then we have a = 0 for all p | i ∈ N , thus f = ip i i 0 i≥0 i≥1 33 and if conversely f = i≥0 aip X ip then we have ∂f ∂X = i≥1 ipaip X ip−1 = 0. 13) Separability. Let K be a field. An irreducible polynomial f ∈ K[X] is called separable, if it has only simple roots in a splitting field for f ; otherwise f is called inseparable.

38 Conversely let M/K be normal, then the minimum polynomial µa ∈ K[X] of a over K, having the root a ∈ M , splits in M [X]. Thus M contains all roots of µa in L. Since for ϕ ∈ G we have µa (aϕ ) = (µa (a))ϕ = 0, we conclude aϕ ∈ M , and thus M ϕ ⊆ M , by K-linearity and M/K finite implying M ϕ = M . If M/K is Galois, since M ϕ = M for all ϕ ∈ G, we have the group homomorphism resL M : G → Aut(M/K) : ϕ → ϕ|M . Since L/M is a splitting field for some polynomial in M [X] \ M , any element of Aut(M/K) extends to an L element of G, thus resL M is surjective.

Let n := dimZ(K) (K) ∈ N and na := dimZ(K) (CK (a)) ∈ N, letting q := |Z(K)| ≥ 2 we get |K| = q n and |CK (a)| = q na . From CK (a)∗ ≤ K ∗ we get q na − 1 | q n − 1, which implies na | n: For m ∈ N let k ∈ Z and l ∈ {0, . . , m − 1} such that n = km + l; hence we have k gcd(n, m) = gcd(m, l) ⊆ Z. Then X n − 1 = (X m − 1) · i=1 X m(k−i)+l + (X l − 1) ∈ Z[X] implies gcd(q n − 1, q m − 1) = gcd(q m − 1, q l − 1) ⊆ Z and thus gcd(q n − 1, q m − 1) = q d − 1, where 0 < d ∈ gcd(n, m). Writing K ∗ as the disjoint union of its conjugacy classes, where T ⊆ K ∗ is a set of representatives, we from Z(K)∗ = {a ∈ T ; na = n} get |K ∗ | = ∗ n | n |Z(K)∗ | + a∈T ,na