# Irreducibility problem help

1. Dec 3, 2005

### TenaliRaman

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. Dec 3, 2005

### Martin Rattigan

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
x^3-2=x^3+1=x^3+3x^2+3^x+1=(x+1)^3=(x-a)(x+1)^2
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
=(x+a/2)^2+3(a/2)^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.

3. Dec 3, 2005

### TenaliRaman

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