Suppose I have a field extension $F\subset E$, with $E$ algebraically closed.
For $f(x),g(x)\in F[x]$, consider $K_{1}$ the splitting field of $f(x)$ and $K_{2}$ the splitting field of $g(x)$.
I'm trying to determine the condition under which $K_{1}\cap K_{2} = F$.
Is this true if $f(x)$ and $g(x)$ have no common roots in $E$?
This is just a guess, but I'm trying to work through a proof of a proposition in Lang's Algebra, Corollary 3.2 in Chapter 5 (snapshot in url):
http://www.math.ualberta.ca/~schlitt/pastelang.png
The problem reduces to isomorphisms $\sigma_{i}:E_{i}\to K_{i}$, and I want to use my above conjecture to prove that this extends to an isomorphism $\sigma:E\to K$ by choosing $E_{i}$'s so that I can write $E$ as a disjoint union, and similarly for the $E_{i}$'s. Without the disjoint union expression for $E$ I don't know how I can guarantee injectivity.
That they have no common roots is not sufficient. Just think of $(x+1)^2 - 2$ and $x^2 - 2$. They have no common roots but they both generate $\mathbb Q(\sqrt 2)$ over $\mathbb Q$.
You might want something more looking like "field-independence" of the roots. This leads you to the notion of resultant, I believe. http://en.wikipedia.org/wiki/Resultant
Hope that helps,