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

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.

• 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. Feb 15 '20 at 10:33
• Do you mean $\Delta(f) = (-1)^{n(n-1)/2} R(f,f')$? I don't know any formulas involving a determinant Feb 15 '20 at 10:36
• Yes, I mean that formula. Keep in mind that $R\left(f,g\right)$ is defined as a determinant. Feb 15 '20 at 10:37
• Using this formula, how would it help solving the task? Feb 15 '20 at 10:37

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.