minimal polynomial of $\beta \not\in F(\alpha)$ is same as it in $F$? Give a field $F$, let $\alpha$ is an algebraic element of $F$, let $f(x), g(x)$ are minimal polynomials of $\beta$ over $F$, $\beta$ over $F(\alpha)$, respectively. If $\beta \not\in F(\alpha)$, can I say $f(x)=g(x)$?
I know it is true when $\deg(f(x))=2$:
because $g(x) \mid f(x)$, if $g(x) \neq f(x)$ then $\deg(g(x)) = 1$, so $g(x) = c+x, c \in F(\alpha)$, and $0 = g(\beta) = c+\beta$, so $\beta = -c \in F(\alpha)$, the contradiction.
but for any degree of $f(x)$, is that true? and how to prove that?
 A: In general the minimal polynomial of an algebraic $\beta$ over some field $F$ can change if you adjoin an algebraic $\alpha$ to $F$.
For a simple example, take $\alpha=\sqrt2,\beta=\sqrt[4]2$ over $F=\Bbb Q$.
For a more general load of examples, if the extension $F(\beta)/F$ has any intermediate field, then the minimal polynomial of $\beta$ over that intermediate field will have smaller degree than it does over $F$.
For any natural $n$, there is a polynomial over $\Bbb Q$ of degree $n$ whose splitting field has degree $n!$. If $n\geq3$, then setting $\beta$ to be one root and $\alpha$ to be another, we get that the minimal polynomial of $\beta$ over $\Bbb Q$ has degree $n$ but over $\Bbb Q(\alpha)$ the minimal polynomial of $\beta$ has degree $n-1$. So you can't tell just from the degree, if the degree is greater than $2$, whether the minimal polynomial of $\beta$ will change.
For examples where the minimal polynomial of $\beta$ doesn't change as you extend $F$, you want $F(\beta)$ to be the splitting field of the minimal polynomial of $\beta$. The classic examples of this are the roots of unity over $\Bbb Q$, whose minimal polynomials are the cyclotomic polynomials (the irreducible factors of $x^m-1$).
