# Intersection of splitting fields of two polynomials

Let $f,g\in \mathbb{Z}[x]$, $(f,g)=h\in \mathbb{Z}[x]$ (if this is not true, can we construct $h$ in another way?). Let $K_1,K_2$ be the splitting fields of $f,g$ over $\mathbb{Q}$ respectively. Is there any relation between $K_1\cap K_2$ and the splitting field over $\mathbb{Q}$ of $h$?

Example: $f(x)=x^7-12$, $g(x)=x^7-11$. Then $(f,g)=1$, $K_1\cap K_2=\mathbb{Q}[\zeta]$, where $\zeta$ is a nontrivial root of $x^7=1$. Is there any theorem in general?

Thanks.

I don't think one can say much in general. Let $f$ be any irreducible polynomial, let $g(x)=f(x+1)$, then you'll have $h=1$ with splitting field the rationals, but $K_1=K_2=K_1\cap K_2$ will just be the splitting field of $f$. At the other extreme, it's easy to find irreducible $f$ and $g$ such that $K_1\cap K_2$ is just the rationals.
As Gerry has pointed out, the two don't need to be equal in general. But there is still a theorem: the splitting field of $(f,g)$ is always contained in the intersection of the splitting fields of $f$ and $g$.
It should be fairly clear why: any root of $(f,g)$ must be a root of $f$ and also a root of $g$.
(Note: I'm assuming that we're working within a specified algebraic closure, say $\overline{\mathbb{Q}}\subset\mathbb{C}$. Otherwise, I would need to say that there is a "natural injection", since there is no a priori notion of containment.)