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

  1. Jul 17, 2009 #1
    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 [tex]h_m(x)\in \mathbb{Z}_{p^m}[x][/tex] irreducible over [tex]\mathbb{Z}_{p^m}[/tex] such that [tex]h_m(x)|x^k-1[/tex] in [tex]\mathbb{Z}_{p^m}[x][/tex]. The theorem calls for constructing a unique, irreducible monic polynomial [tex]h_{m+1}(x)\in \mathbb{Z}_{p^{m+1}}[x][/tex] which divides [tex]x^k-1[/tex] in [tex]\mathbb{Z}_{p^{m+1}}[x][/tex].

    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 [tex]\beta^p[/tex] how come [tex]\beta^p\equiv \alpha(mod p^m)[/tex]. This equivalence of roots of the monic polynomial and of [tex]h_m(x)[/tex] 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 [tex]h_{m+1}(x)[/tex].

    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. jcsd
  3. Jul 17, 2009 #2


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

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