Constructing an irreducible polynomial in Z_{p^(m+1)}

[A-ManESL]

Hello PF members
This is my first post. It is rather complicated to understand but I request you to bear with me.

The Problem: I have a theorem in my book, the proof of which I do not understand fully. The theorem may be viewed http://books.google.co.in/books?id=...9jZDA&sa=X&oi=book_result&ct=result&resnum=1" (The book is: Finite commutative rings and their applications By Gilberto Bini, Flaminio Flamini The theorem is on Page 24).

My specific problem is as follows:
We are given a monic polynomial h_m(x)\in \mathbb{Z}_{p^m}[x] irreducible over \mathbb{Z}_{p^m} such that h_m(x)|x^k-1 in \mathbb{Z}_{p^m}[x]. The theorem calls for constructing a unique, irreducible monic polynomial h_{m+1}(x)\in \mathbb{Z}_{p^{m+1}}[x] which divides x^k-1 in \mathbb{Z}_{p^{m+1}}[x].

The proof in the book runs as follows:

By Hensel's Lemma, (something already proved) the proof starts off with taking a polynomial h(x)\in \mathbb{Z}_{p^{m+1}}[x] of the form h(x)=h_m(x)+p^mg(x). It then let's \alpha be a root of h_m(x) and \beta a corresponding root of h(x) of the form \beta=\alpha+p^m\delta. Then it states that \alpha^k=1+p^m\epsilon, since h_m(x) divides x^k-1 in \mathbb{Z}_{p^m}[x].

I have no problems uptil this point in the proof

Moreover \beta^p=(\alpha+p^m\delta)^p=\alpha^p and \beta^{kp}=(\alpha+p^m\delta)^{kp}=(1+p^m\epsilon)^p=1. (Here the book doesn't say so but I assume that the equalitites hold modulo p^{m+1})

My major problem is with the next two lines (Underlined portion specially):

Hence the monic polynomial, whose roots are the p-th powers of the roots of h(x), divides x^k-1 and these roots coincide modulo p^m with those of h_m(x).

I don't understand what the monic polynomial referred to is? If it is the polynomial with roots all of the type \beta^p how come \beta^p\equiv \alpha(mod p^m). This equivalence of roots of the monic polynomial and of h_m(x) is very crucial as the next line also seems to be related to it

This polynomial is the required polynomial h_{m+1}(x)\in \mathbb{Z}_{p^{m+1}}[x]; in fact it is irreducible, by construction.

For the life of me I can't understand why this polynomial is irreducible.

The proof then goes on to establish the uniqueness of such an h_{m+1}(x).

I'll be very extremely grateful if someone points me in the right direction. Thank you for your time (all those who have read the whole post).

 

[Hurkyl]

Didn't Hensel's lemma say h was monic?

Have you pondered what would happen if if h was reducible?