Let $\alpha_1, \ldots, \alpha_n$ be the roots of $f$ in some splitting field $K$. Then define $disc_K(f) = \prod_{i<j} (\alpha_i - \alpha_j)^2$. Is it true $disc_K(f) \in F$?
I used to prove $disc(f)\in F$ when $f$ is irreducible and (wlog) separable by some Galois Theory.
I think it is true: Let $A:=V(\alpha_1, \ldots, \alpha_n)$ denote the Vandermonde matrix. Then $disk_K(f) = \det A^2$. But $\det(A)^2=\det (AA^T)$ where $AA^T$ has $ij$-th entry $t_{i+j-2} = \sum \alpha_k^{i+j-2}$. There is recursion formula for $t_{i+j-2}$ in terms of coefficients of $f$, beginning with $t_0 = n$. Hence $disc_K(f) \in F$
Is my proof correct? Or is there a counter example?
Edit, A proof with Galois: Let $\sigma \in Gal(K/F)$, and suppose $f = m_1^{r_1} \cdots m_k ^{r_k}$, where $m_i$ are irreducibles over $F$. May suppose $f$ is separable, so each $m_i$ is separable and $r_i=1$. $K/F$ is a Galois extension, being splitting field of separable polynomial.
Each $\sigma $ permutes the roots of individual $m_i$, thus must send $\alpha_i$ to some $\alpha_j$ for each $i$. So a $\sigma$ is a permutation of $\{ \alpha _i \}$. So $\sigma(disc(f)) =disc(f)$ and $disc(f) \in F$.
The easiest proof is to note that the discriminant is a symmetric polynomial function of the roots and so can be expressed as a symmetric polynomial function over $F$ on the coefficients of $F$.