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

[Zoe-b]

Homework Statement

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, O_K its ring of integers, and Δ(W)² be the discriminant. Write Z for set of integers, Q for set of rationals.

Homework Equations

Dedekind's Theorem on decomposition of (p) for p a rational prime.
Theorem shown previously in the course that Δ ( Z[a] )² = ± Norm_K/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 | [O_K : Z[a]] then p| Δ ( Z[a] ). So suppose p does not divide [O_K : 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] O_K.

It follows that:

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

(and then some more...)

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.

 

[mighty2000]

I can understand your confusion about this step. It may seem counterintuitive to take out a term that is a multiple of p from the norm. However, in this case, we are dealing with a specific situation where p is a prime that ramifies in the ring of integers of a number field K. In this case, the norm is defined as the product of all the conjugates of the element, and since p is a prime that ramifies, it will appear as a factor in all the conjugates. Therefore, we can "take it out" of the norm since it will be present in all the terms. This is a special case and is allowed because of the specific properties of ramified primes in number fields. I hope this helps clarify your understanding. Good luck with your exams!