# Let $f \in F[x]$, then discriminant of $f$ lies in $F$.

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 $disc(f) \in F$.

• If you know the rather easy proof with Galois theopry, what is then the problem here? Take the splitting field of $\;f\;$ over whatever field you're working on and apply the automorphisms, check what remains fixed and etc. It is the same! – DonAntonio Apr 2 '18 at 17:47
• I think I was afraid with working with general $f$, is the proof I written correct using Galois? – Bryan Shih Apr 2 '18 at 17:56
• (1) Your proof with matrices is fine...as long as you can use, or prove, that thing about that recursion formula in terms of coefficients of $\;f\;$ , and (2) The proof with Galois is very fishy imo. First, you can assume $\;f\;$ is irreducible so simplify. Then, as you did, you use the fact that any Galois autom. permutes this irred. pol.'s roots. Finally, in the expression of the discriminant, one observes that this means $\;\sigma(disk f)=disk f\;$ for any autom. $\;\sigma\;$ , which of course means the discriminant belongs to the fixed field of the whole Galois group, namely: $\;F\;$ . – DonAntonio Apr 2 '18 at 21:47
• I probably wasn't clear, I edited my proof now, which added what you wrote. Is this still fishy? – Bryan Shih Apr 2 '18 at 22:52

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 $F$.