Let $n\geq 1$, $$D=\prod_{i<j}(x_i-x_j)^2, B=\prod_{i<j}(x_i+x_j), C=(B^2-D)/4$$ be polynomials in $\mathbb{Z}[x_1, ..., x_n]$
Then if $f(x)\in F[x]$ has different roots $\alpha_1, ... \alpha_n$ in its splitting field $E$, it has a discriminant equation $$x^2+bx+c$$ where $$b=B(\alpha_1, ... \alpha_n), c=C(\alpha_1, ... \alpha_n), d=D(\alpha_1, ... \alpha_n)$$ This equation is called a discriminant equation, because $d$ is the ordinary discriminant, and when $char(F) \neq 2$, $F(\sqrt{d})=F(\beta)$, where $\beta$ is any root of $x^2+bx+c$, and $Gal(E/F)\subset A_n$ iff $\beta\in F$. When $char(F)=2$, it might happen that $F(\sqrt{d})\neq F(\beta)$, but still $Gal(E/F)\subset A_n$ iff $\beta\in F$.
Show that $x^2+bx+c$ splits over $E$ even when $char(F) = 2$
Define $R$ and $S$ in $\mathbb{Z}[x_1, ..., x_n]$ by means of the following equations: $$ \begin{align} R-S &= -\prod_{i<j} (x_i-x_j) \\ R+S &= -\prod_{i<j} (x_i+x_j) \\ \end{align} $$
The latter product equals $-B$ and $(R-S)^2=D$. Thus $r=R(\alpha_1,\alpha_2,\dots,\alpha_n)$ and $s=S(\alpha_1,\alpha_2,\dots,\alpha_n)$ are roots of the quadratic equation $x^2+bx+c$, since $R+S=-B$ and $$RS = \frac{(R+S)^2-(R-S)^2}{4}=\frac{B^2-D}{4}=C$$ It's then obvious that $x^2+bx+c$ must split over $E=F[\alpha_1,\alpha_2,\dots,\alpha_n]$.