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!

Irreducibility problem help

  1. Dec 3, 2005 #1
    We are given Z_p, where p is prime. It is known that there is an 'a' such that a^3 = 2 (mod p) and that 3 does not divide p-1. Does this imply that f(x) in (x^3-2) = (x-a)f(x) is irreducible over Z_p?

    I think this should be trivially easy to answer, but my mind is so muddled up at the moment, that i cant even think straight currently.

    -- AI
  2. jcsd
  3. Dec 3, 2005 #2
    The answer is usually.
    Exceptions are p=2 and p=3
    For p=2, a=0 and eqn is x^3-2=x^3=(x-a)x^2 so f=x^2=x.x
    For p=3 a=-1 and eqn is
    so f=(x+1)^2=(x+1)(x+1)
    Otherwise p must be (3n + or - 1)+1, but except when p=3 this must be 3n+2.
    r.h.s. is then (x-a)(x^2+ax+a^2) so f is x^2+ax+a^2
    which clearly has a root iff (-3/p)=1 (meaning -3 is a quadratic residue mod p).
    If p is 4k+3 then
    (-3/p)=-(3/p)=(p/3)=((3n+2)/3)=(2/3)=-1 so no soap.
    If p is 4k+1 then
    (-3/p)=(3/p)=(p/3)=-1 so also no soap.
    So 2 & 3 are the only exceptions.
  4. Dec 3, 2005 #3
    Aha! (One of those aha moments!)
    p = 2,3 case were being evident but i had no idea how to proceed for the general case.

    Very Neat Solution Martin!!

    -- AI
    P.S -> LoL! For a while i was wondering why we were considering the cases p = 4k+3 and 4k+1 above. Geez it didnt hit me completely for a while :P
    Last edited: Dec 3, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Irreducibility problem help
  1. Is this irreducible? (Replies: 4)

  2. Irreducibility in Zp (Replies: 3)