For the discriminant of a polynomial, how to prove $\Delta(f \bmod p) = \Delta(f) \bmod p$?

Question

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.

Answer 1

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.