Algebra 2 [Lecture notes] by Jan Nekovar

By Jan Nekovar

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17) Proposition. Let H ⊂ Sn be a subgroup. (i) If (Sn : H) = 2, then H = An . (ii) If n ≥ 5 and H Sn is a normal subgroup of Sn , then H = {e}, An or Sn . (iii) If n ≥ 5 and H = An , Sn , then (Sn : H) ≥ n. (iv) If n ≥ 5 and H An , then (An : H) ≥ n. Proof. (i) Exercise. (ii) We use the fact that An is a simple group for n ≥ 5 ([De 1], Thm. 1), which ∼ implies that H ∩ An is equal to An (=⇒ H = An , Sn ) or to {e} (=⇒ H → Sn −→ Sn /An −→ {±1} is injective =⇒ |H| ≤ 2; a subgroup of order two is never normal in Sn , hence H = {e}).

N−1 form a basis of K[α] as a K-vector space. (4) The polynomial f is irreducible in K[X] (conversely, if g(α) = 0 for an irreducible monic polynomial g ∈ K[X], then g = f ). (5) The K-algebra K[α] is a field; thus K[α] = K(α) and [K(α) : K] = n. (6) Conversely, if f ∈ K[X] is an irreducible monic polynomial of degree n ≥ 1, then the K-algebra L = K[X]/(f ) is a field extension of K, the element α = X (mod f ) ∈ L is algebraic over K, its minimal polynomial over K is equal to f and L = K[α] = K(α) (in particular, [L : K] = n).

10) Exercise (Newton’s formulas). The polynomials sk = xk1 + · · · + xkn satisfy recursive relations sk − σ1 sk−1 + · · · + (−1)k−1 σk−1 s1 + (−1)k kσk = 0 (k ≥ 1) (of course, σk = 0 for k > n). 11) Discriminant. Let n ≥ 2. The polynomial (xi − xj ) ∈ Z[x1 , . . , xn ] ∆ := i

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