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

1. May 24, 2013

### Zoe-b

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) $\in$ (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 $\in$ 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.