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Let $ K$ be a finite extension of $F$ and assume $K$ is a splitting field over $F$. Is it given that for any element $ \alpha \in K, \alpha \not\in F $ that there exists a polynomial $ f(x) \in F[x] $ in which $ \alpha $ is a root?

It relates to the following exercise: Let $ K_1 $ and $ K_2 $ be finite extensions of $ F $ contained in the field $ K $, and assume both are splitting fields over $ F $. Prove that $ K_1 \cap K_2 $ is a splitting field over $ F $.

I don't want an answer to the exercise, I'm just posting it for reference. The thing is, how can I be certain that $ K_1 $ and $ K_2 $ do not share only a set of elements which are not algebraic (a part from the obviosly shared $ F $)? Because in that case $ K_1 \cap K_2 $ would only split the same polynomials as $ F $ and since $ F $ is a proper subset, $ K_1\cap K_2 $ cannot be a splitting field.

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Suppose $K$ has dimension $n$ over $F$, then the $n+1$ elements $1,a,a^2 \dots a^n$ are linearly dependent over $F$. An explicit dependence gives a polynomial over $F$ which is satisfied by $a$.

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  • $\begingroup$ I'm not sure I grasp that completely, is it a "yes"? $\endgroup$ – zo0x Sep 14 '14 at 13:06
  • $\begingroup$ What I mean, when I say I don't grasp it, is: What is your criteria for picking $ a $? Is it any just any element not in $ F $? I know that if $ a $ is algebraic over $ F $ then what you state is satisfied in $ F(a) $, but then I also have a polynomial by choice of $ a $, which proves nothing. $\endgroup$ – zo0x Sep 14 '14 at 13:15
  • $\begingroup$ @zo0x The point is that $K$ has finite dimension over $F$. Whatever $a$ is, list its powers until you get a dependent set. This will happen because of the finite dimension. And this is inherent in "Let $K$ be a finite extension of $F$". You could argue by contradiction too to say that if $a$ is not algebraic, the extension is not finite, because you can find an infinite linearly independent set. $\endgroup$ – Mark Bennet Sep 14 '14 at 13:23

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