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Algebraic Number Theory, don't understand one step in a given proof

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi, sorry to be a pain, if anyone could help me understand this I'd be very grateful (exams next week, no more revision classes and no tutors I can easily ask...)

    Let K be a number field, OK its ring of integers, and Δ(W)2 be the discriminant. Write Z for set of integers, Q for set of rationals.


    2. Relevant equations
    Dedekind's Theorem on decomposition of (p) for p a rational prime.
    Theorem shown previously in the course that Δ ( Z[a] )2 = ± NormK/Q(m'(a)) where ' denotes derivative, m is the minimal polynomial of a over Z.

    Corollary:
    If p ramifies then p| Δ ( Z[a] )2

    Proof:
    If p | [OK : Z[a]] then p| Δ ( Z[a] ). So suppose p does not divide [OK : Z[a]] and so Dedekind's Theorem applies. By Dedekind, is p ramifies, with a factor p2, then n(x) := m(x) (mod p) has a multiple irreducible factor mod h(x) over the finite field with p elements, for which h(a) [itex]\in[/itex] (p, h(a)) = (p). We then have:

    m(x) = g(x)2s(x) + pk(x) for some polynomials s, k, so that

    m'(x) = g(x) (2g'(x)h(x) + g(x)h'(x)) + pk'(x) = g(x)j(x) + pl(x) say for polynomials j,l.

    Thus m'(a) = g(a)j(a) + pb for b [itex]\in[/itex] OK.

    It follows that:

    NormK/Q(m'(a)) = ∏σ σ(m'(a)) = ∏σ σ(g(a)j(a)) + pβ for some algebraic integer β. ***

    (and then some more...)
    3. The attempt at a solution

    I understand most of this, and I understand the bit following, I am stumped at how on the line *** we can take the term that's a multiple of p 'out of' the norm. Norm is multiplicative but not additive, in general, so what allows this step? Thank you in advance.
     
  2. jcsd
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