Reduction of polynomials mod p

1-Let $f_1,f_2\in\mathbb{Z}[X]$ be two different irreducible monic polynomials. Is it true that for almost all primes $p$ (that is, for all but a finite number of primes), the polynomials $\bar{f}_1$ and $\bar{f}_2$ have no common roots in $\mathbb{F}_p$? (here $\bar{f}$ means reduction modulo $p$).

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By Gauss' lemma, $f_1$ and $f_2$ remain irreducible over $\mathbb{Q}[x]$. The Euclidean algorithm then gives polynomials $g_1,g_2\in\mathbb{Q}[x]$ such that $g_1(x)f_1(x)+g_2(x)f_2(x)=1$. For all primes $p$ which do not divide the denominator of any coefficient of $g_1$ or $g_2$, we can reduce the above equation mod $p$, and conclude that $\overline{f}_1$ and $\overline{f}_2$ are coprime, so don't share a root.

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How can you reduce it when the coefficients arn't integers ? – Belgi Sep 16 '12 at 15:27
Either you multiply away all denominators and obtain an equation $\ldots=N$ which remains nonzero after reduction mod $p$ unless $p|N$. Or you note that $\frac1n$ "exists" mod $p$ in the sense that $\bar n$ has an inverse mod $p$. – Hagen von Eitzen Sep 16 '12 at 15:38

The resultant of two polynomials over a field is an element of the field which is zero if and only if the polynomials have a common factor over the field. The resultant of your two polynomials is an integer, which is zero modulo~$p$ for only finitely many primes $p$, unless the integer is zero, in which case the polynomials have a common factor over the integers. But that is ruled out by the hypotheses.

The resultant can be calculated as the determinant of a certain matrx whose entries are taken from the coefficients of the polynomials --- it's called the Sylvester matrix.

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