Exercise 5.3.4

Proof. Write $\mathrm{\Delta }=\mathrm{\Delta }({\sigma }_1,\cdots ,{\sigma }_n)\in F({\sigma }_1,\cdots ,{\sigma }_n)\subset F(x_1,\cdots ,x_n)$ the discriminant.

Let $f=x^n+a_1{x}^{n-1}+\cdots +a_0\in \mathbb{Z}[x]$ be a monic nonconstant polynomial.

In the section 2.4, $\mathrm{\Delta }(f)$ is defined by

obtained by applying to $\mathrm{\Delta }({\sigma }_1,\cdots ,{\sigma }_n)$ the evaluation homomorphism defined by ${\sigma }_i\mapsto {(-1)}^i{a}_i$, which sends $\tilde{f}=x^n-{\sigma }_1{x}^{n-1}+\cdots +{(-1)}^n{\sigma }_n$ on $f$, and $\mathrm{\Delta }$ over $\mathrm{\Delta }(f)$.

Write $f_p$ the reduction of $f$ modulo $p$ :

$f_p=x^n+{\bar{a}}_1{x}^{n-1}+\cdots +{\bar{a}}_0$, where we write $\bar{k}={[k]}_p$ the class of $k\in \mathbb{Z}$ modulo $p$.

By definition,

$\mathrm{\Delta }$ is a polynomial with coefficients in $\mathbb{Z}$ of ${\sigma }_1,\cdots ,{\sigma }_n$, thus $\mathrm{\Delta }(-{\bar{a}}_1,\cdots ,{(-1)}^i{\bar{a}}_i,\cdots ,{(-1)}^n{\bar{a}}_n)$ is the reduction modulo $p$ of $\mathrm{\Delta }(-a_1,\cdots ,{(-1)}^i{a}_i,\cdots ,{(-1)}^n{a}_n)$, so $\mathrm{\Delta }(f_p)$ is the reduction modulo $p$ of $\mathrm{\Delta }(f)$ :

□