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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?

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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.

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linear and monic. – lhf Nov 23 '12 at 10:49
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 $f'=2gg'h+g^2h'=g\cdot (2g'h+gh')$.

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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

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