Originally Posted by

**chiph588@** How about this:

As you said, $\displaystyle D $ is symmetric in the roots of $\displaystyle f(x) $, so $\displaystyle D $ is fixed by all automorphisms of the Galois group of $\displaystyle f(x) $.

This implies $\displaystyle D \in F $ since $\displaystyle F $ is the fixed field of $\displaystyle \text{Gal}(K/F) $.

(We know $\displaystyle F $ is the fixed field of $\displaystyle \text{Gal}(K/F) $ since $\displaystyle [K:F]=|\text{Gal}(K/F)| $, so by the Fundamental Theorem of Galois Theory we get what we want.)