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Let $f \in \mathbf{Z}[X]$ be a monic polynomial. I am trying to prove that $\Delta(f \bmod p) = \Delta(f) \bmod p$, where $\Delta$ is the discriminant of $f$, defined by $$ \Delta(f) = \prod_{1 \leq i < j \leq n} (\alpha_i - \alpha_j)^2 $$ with $\alpha_i$ de roots of $f$ in some field that contains them.

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    $\begingroup$ If $f$ is monic of degree $N$, then the discriminant of $f$ is the determinant of a certain matrix whose entries are certain coefficients of $f$ (sometimes with pre-factors). Ring homomorphisms (such as the projection from $\mathbb{Z}$ to $\mathbb{Z}/p$) clearly preserve this matrix. $\endgroup$ Feb 15 '20 at 10:33
  • $\begingroup$ Do you mean $\Delta(f) = (-1)^{n(n-1)/2} R(f,f')$? I don't know any formulas involving a determinant $\endgroup$
    – Sigurd
    Feb 15 '20 at 10:36
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    $\begingroup$ Yes, I mean that formula. Keep in mind that $R\left(f,g\right)$ is defined as a determinant. $\endgroup$ Feb 15 '20 at 10:37
  • $\begingroup$ Using this formula, how would it help solving the task? $\endgroup$
    – Sigurd
    Feb 15 '20 at 10:37
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Notice that the discriminant can be seen as a symmetric polynomial with variables $\alpha_1, \dots, \alpha_n$. By the fundamental theorem of symmetric polynomials, we can therefore write it as a polynomial with as variables the elementary symmetric polynomials $$ s_k = \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \alpha_{i_1} \alpha_{i_2} \dots \alpha_{i_k}, $$ and these are the coefficients of $f = \prod_{i = 1}^n (X - \alpha_i)$ (possibly with a different sign). So, the discriminant is a polynomial expression in the coefficients of $f$, and since $x \mapsto x$ mod $p$ is a homomorphism, we see that reducing the coefficients of polynomial mod $p$ and then computing the discriminant is the same as doing it the other way around.

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