1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving polynomial congruences modulo a prime power

  1. Apr 29, 2006 #1
    Hey all, I have a problem regarding polynomial congruences. I need to find an iterative procedure for solving a polynomial congruence modulo a prime power. I've been working on the case p^2 and I want to ask my question regarding just this case. Hopefully once I understand the proof here I can extend it generally myself.

    I have seen some proofs of this process, but have been asked to approach the problem in a different way. I need to prove that every solution of f(x)=0 mod p^2 is given in the form x=a0+a1*p, where a0 is a solution of f(x)=0 mod p and a1 is a solution of f'(a0)*a1+f(a0)/p=0 mod p.

    If I call a solution x0, then I have f(x0)=0 mod p^2, and clearly f(x0)=0 mod p as well.. Also, x0=a0+a1*p, so x0=a0 mod p. I know I can factor out a root of f so that f-f(a0)=(x-a0)*g, where g is another polynomial. Now, f'=(x-a0)'*g+(x-a0)*g'=g+(x-a0)*g'.

    From here, I have f'(x0)=g(x0)+(x0-a0)*g'(x0). Since x0=a0 mod p, f'(x0)=f'(a0) mod p and f'(x0)=g(x0) mod p, so f'(a0)=g(x0) mod p.

    Thinking back to the original formula I want, I see how I can get part of it. From f(x0)=0 mod p^2 I have f(xo)/p=0 mod p and thus f(a0)/p=0 mod p. What I am having trouble with is the other part. I need f'(a0)*a1=0 mod p, and so it's enough to have g(x0)*a1=0 mod p. And that's where I'm stuck. Any suggestions?
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?

Similar Discussions: Solving polynomial congruences modulo a prime power
  1. Irreducible polynomial (Replies: 0)