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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.

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Sorry, I'm thick. It's assumed in the hypothesis that $\sigma$ is an embedding. –  Kyle Schlitt Nov 25 '12 at 4:10
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up vote 1 down vote accepted

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,

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Thanks. Fortunately the problem I was facing had a simpler solution than the one I was thinking. But this answers my question nonetheless. –  Kyle Schlitt Nov 25 '12 at 4:11
    
@lovinglifein : I guessed that, so I just gave you the short and simple answer telling you "no, think of something else, this approach is complicated". :P –  Patrick Da Silva Nov 25 '12 at 4:33
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