A-ManESL
Jul17-09, 06:33 AM
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 here (http://books.google.co.in/books?id=H0mAM-Zr0HAC&pg=PA25&lpg=PA25&dq=Hence+the+monic+polynomial,+whose+roots+are+the +p-th+powers&source=bl&ots=phoiqeZbaW&sig=cvx-dyfYArKEW6qOdN1DNw7abJM&hl=en&ei=jFVgSryrKYiVkAW2k9jZDA&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 lets \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).
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 here (http://books.google.co.in/books?id=H0mAM-Zr0HAC&pg=PA25&lpg=PA25&dq=Hence+the+monic+polynomial,+whose+roots+are+the +p-th+powers&source=bl&ots=phoiqeZbaW&sig=cvx-dyfYArKEW6qOdN1DNw7abJM&hl=en&ei=jFVgSryrKYiVkAW2k9jZDA&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 lets \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).