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

1. Jul 17, 2009

### 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:
I have no problems uptil this point in the proof

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

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

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).

Last edited by a moderator: Apr 24, 2017
2. Jul 17, 2009

### Hurkyl

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

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