By Hyman Bass
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Contradiction. Hence, β p is a root of ϕ(t). Let now α be the root generating the multiplicative subgroup of roots of degree n of unity in Q(tn − 1), and Φn (t) be the minimal polynomial of α, then, for arbitrary k ∈ (Z/n)∗ , k can be presented as a product k i pi of primes relatively prime to n. It follows from Lemma 122 that α is also a root of Φn (t). Thus, we k have an automorphism σk : Q(α) → Q(α) sending α to α . This shows that the map ρ is surjective. And the above considerations also give us the description of the minimal polynomial Φn (t) for the primitive n-th root of 1.
T + 1, for p-prime. 19. Radical extensions Consider polynomials of slightly more general form: tn − c, where c ∈ K is arbitrary element now. These are exactly the polynomials producing radical extensions. We will assume that char(K) = 0. The polynomial tn − c is still separable (like tn − 1), and so it has no multiple roots. If β1 , . . , βn are all roots of tn − c in its splitting field, then α1 = 1 = ββ11 , α2 = ββ21 , . . , αn = ββn1 are different roots of tn − 1. So, these are all roots of the latter polynomial.
1) Φ1 (t) = t − 1; (2) Φ2 (t) = t + 1; (3) Φ3 (t) = t2 + t + 1; (4) Φ4 (t) = t2 + 1; (5) Φ5 (t) = t4 + t3 + t2 + t + 1; (6) Φ6 (t) = t2 − t + 1; (7) Φp (t) = tp−1 + tp−2 + . . + t + 1, for p-prime. 19. Radical extensions Consider polynomials of slightly more general form: tn − c, where c ∈ K is arbitrary element now. These are exactly the polynomials producing radical extensions. We will assume that char(K) = 0. The polynomial tn − c is still separable (like tn − 1), and so it has no multiple roots.