Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In my lecture notes of algebraic number theory they are dealing with the polynomial $$f=X^3+X+1, $$ and they say that

If f has multiple factors modulo a prime $p > 3$, then $f$ and $f' = 3X^2+1$ have a common factor modulo this prime $p$, and this is the linear factor $f − (X/3)f'$.

Please could you help me to see why this works? And moreover, how far can this be generalized?

share|improve this question

2 Answers 2

up vote 2 down vote accepted
  1. If $f$ has a multiple factor, say $h$ (in any field containing the current base field), then with appropriate $g$, we have $$f(x)=h(x)^2\cdot g(x)$$ If you take its derivative, it will be still a multiple of $h(x)$, so it is a common factor of $f$ and $f'$.

If polynomials $u$ and $v$ have common factors, then all of their linear combinations will have that as common factor. Now in your particular example, note that the written $f-(X/3)f'$ is already linear (hence surely irreducible), so, if there is a common factor, it must be this one.

share|improve this answer
    
linear and monic. –  lhf Nov 23 '12 at 10:49
1  
Note that this works for formal derivatives, which can be defined without any consideration of limits or continuity. –  Mark Bennet Nov 23 '12 at 11:42

By the rules of differentiation, if $f=g^2h$, then $f'=2gg'h+g^2h'=g\cdot (2g'h+gh')$.

share|improve this answer
1  
Ok, this is clear, but that was not the point of my question. It doesn't explain why that particular factor is the common one –  Abramo Nov 23 '12 at 10:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.