Showing an irreducible polynomial is a minimal polynomial. We just got into algebraic extensions, and this one threw me for a loop.
Left $f$ be a polynomial irreducible over $F$, and let $E$ be an extension field of $F$ in which $f$ has root $\alpha$. Show that $f$ is a minimal polynomial for $\alpha$ over $F$.
I don't even know where to start so I can't tell you what I've done.
 A: Hint: a minimal polynomial of $\alpha$ divides in $F[X]$ every polynomial vanishing at $\alpha$. Indeed, such polynomials form an ideal and $F[X]$ is principal.
A: This is a proof for finite fields
You want to prove,
If $\mid F \mid = q$ and $ \mid E\mid = q^m $ then the minimal polynomials over $F$ of the elements of $E$ are precisely the monic irreducible polynomials over $F$ whose degrees divide $m$.
Consider monic irreducible polynomials $f_1(x) \in F[x]$, $f_2(x) \in F[x]$ such that $deg(f_1(x))=d_1$ , $deg(f_2(x))=d_2$ , $d_1 \neq d_2$ and $d_1 \mid m\: ,\: d_2 \mid m$. WLOG assume $d1 < d2$. We write $$f_1(x) = \prod_{i=0}^{d_1}(x-\alpha_i) $$ and $$f_2(x) = \prod_{j=0}^{d_2}(x-\beta_j)$$. Note that $f_1$ and $f_2$ may have some common factors  (because they are irreducible over F but not over E.
So we assume $\alpha_i = \beta_j$ for some $i,j$. It is easy to see that through this construction that if there are $s$ such polynomials for which say $\alpha_1$ is a root, then we can consider that polynomial of least degree to obtain the minimal polynomial of $\alpha_1$ and so on.
BONUS: Try to prove that there cannot exist two different monic irreducible polynomials of the same degree $ d_1 = d_2$ and $d_1 \mid m\: ,\: d_2 \mid m$ which share a common root in E
