New PDF release: Algebraic Number Theory: summary of notes [Lecture notes]

New PDF release: Algebraic Number Theory: summary of notes [Lecture notes]

By Robin Chapman

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Extra info for Algebraic Number Theory: summary of notes [Lecture notes]

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J j This must be an integer, yet it cannot be as p bj . This contradiction shows that Ok = Z[ζ]. Example Let K = Q(ζ) where ζ = exp(2πi/5). Then OK = Z[ζ] and ζ has minimum polynomial f (X) = X 4 + X 3 + X 2 + X + 1. For each prime number q we aim to factorize the ideal q by factorizing the polynomial f modulo q. Consider the case q = 5. Then (X − 1)4 = X 4 − 4X 3 + 6X 2 − 4X + 1 ≡ X 4 + X 3 + X 2 + X + 1 (mod 5). It follows that 5 = P54 where P5 = 5, ζ − 1 . For λ = ζ − 1 we have seen that λ4 | 5 so that P5 = λ .

The norm of bj λp−2 /p is bp−1 pp−2 /pp−1 = bp−1 /p. j j This must be an integer, yet it cannot be as p bj . This contradiction shows that Ok = Z[ζ]. Example Let K = Q(ζ) where ζ = exp(2πi/5). Then OK = Z[ζ] and ζ has minimum polynomial f (X) = X 4 + X 3 + X 2 + X + 1. For each prime number q we aim to factorize the ideal q by factorizing the polynomial f modulo q. Consider the case q = 5. Then (X − 1)4 = X 4 − 4X 3 + 6X 2 − 4X + 1 ≡ X 4 + X 3 + X 2 + X + 1 (mod 5). It follows that 5 = P54 where P5 = 5, ζ − 1 .

Let K be a field and f ∈ K[X] have positive degree. We say that f is irreducible over K if there are no g, h ∈ K[X] with f = gh and deg(g), deg(h) < deg(f ). 1 (Unique factorization) Let K be a field and let f ∈ K[X] be a monic polynomial of positive degree. Then there are monic polynomials p1 , p2 , . . , pk ∈ K[X], each irreducible over K, such that f = p1 p2 · · · pk . Furthermore the pj are determined, up to order, uniquely by f . 2 Symmetric polynomials Let R be a ring. Then R[T1 , . . , Tk ] denotes the ring of polynomials in the n indeterminates T1 , .

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Algebraic Number Theory: summary of notes [Lecture notes] by Robin Chapman


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