Linearly disjoint? Let $f$ and $g$ be irreducible polynomials over a field $K$ with $\deg f=\deg g =3$ and let the discriminant of $f$ be positive and the discriminant of $g$ negative.
Does it follow that the splitting fields of $f$ and $g$ are linearly disjoint? If yes, why?
(Def.: Two intermediate fields $M_1, M_2$ of an algebraic field extension $L|K$ are called linearly disjoint, if every set of elements of $M_1$, that is linearly independent over $K$, is also linearly independent over $M_2$. 
Here, this is equivalent to $M_1 \cap M_2 = K.$)
 A: Let me do the case where $K={\bf Q}$. Let $L_f,L_g$ be the splitting fields. Then $L_f\cap L_g$ must be a subfield of both $L_f$ and $L_g$. So, what are the subfields of those splitting fields? 
Well, $L_f$ has three real cubic subfields and one real quadratic; $L_g$ has one real cubic subfield, two imaginary cubic subfields, and an imaginary quadratic subfield. So if the intersection isn't $\bf Q$, it must be the real cubic subfield. But then we have a single cubic subfield that has three real embeddings and also two complex embeddings, and that's absurd. QED. 
I don't think I made much use of the hypothesis $K={\bf Q}$; this argument ought to work more generally. 
EDIT: More generally, if $L$ and $M$ are splitting fields of irreducible cubics over $K$ and are not equal then their intersection can't be of degree 3 over $K$; it can only be $K$ or the quadratic extension given by adjoining the square root of the discriminant (assuming the discriminant is not a square in $K$). That quadratic case can happen, as non-isomorphic cubic fields can be generated by polynommials with the same discriminant. Over the rationals, there are two non-isomorphic cubics with discriminant -1836, also three with discriminant 22356. 
