# Is every element in a finite splitting field K over F a root in a polynomial?

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.

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 $a$.
• 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. – zo0x Sep 14 '14 at 13:15
• @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. – Mark Bennet Sep 14 '14 at 13:23