Let $\mathbb F$ be a finite Galois extension of $\mathbb K$.
a) For $\alpha \in \mathbb F$ show that $f(x) = \prod_{\sigma \in Gal(\mathbb F / \mathbb K)}(x - \sigma(\alpha))$, belongs to $\mathbb K[x]$.
b)Show that $f(x) = \prod_{\sigma \in Gal(\mathbb F/\mathbb K)}(x - \sigma(\alpha))$ is a power of $m(\alpha, \mathbb K)$, and $f(x) = m(\alpha, \mathbb K)$ if and only if $\mathbb F = \mathbb K(\alpha)$.
a) For $\tau \in G(K/F)$, we have a natural extension of to an automorphism $\bar \tau of K[x]$ where $\bar \tau (\alpha_0+\alpha_1x +...+\alpha_nx^n)= \tau(\alpha_0)+\tau(\alpha_1)x+...+\tau(\alpha_n)x^n$. Clearly the polynomials left fixed by $\bar \tau$ for all $ \tau \in G(K/F)$ are precisely those in $F(x)$. For $f(x) = \prod_{\sigma \in G(K/F)}(x-(\sigma)(\alpha))$. we have $f(x) =\bar \tau (f(x)) = \prod_{\sigma \in G(K/F)}(x-(\tau \sigma)(\alpha))$. Now as $\sigma$ runs through all elements of $G(K/F)$, we see that $\tau \sigma$ also runs through all elements because $G(K/F)$ is a group. Thus $\bar \tau (f(x)) = f(x)$ for each $\tau \in G(K/F)$, so $f(x) \in F[x]$.
b) Because $\sigma(\alpha)$ is a conjugate of $\alpha$ over $F$ for all $\sigma \in G(K/F)$, we see that $f(x)$ has precisely the conjugates of as zeros. Because $f(\alpha) = 0$,and, knowing that $p(x) = irr(\alpha, F)$ divides $f(x)$. Let $f(x) = p(x)q_1(x)$. If $q_1(x) \ne 0$, then it has as zero some conjugate of $\alpha$ whose irreducible polynomial over $F$ is again $p(x)$, so $p(x)$ divides $q_1(x)$ and we have $f(x) = p(x)^2 q_2(x)$. We continue this process until we finally obtain $f(x) = p(x)^rc$ for some $c \in F$. Because $p(x)$ and $f(x)$ are both monic, we must have $f(x) = p(x)^r$. Now $f(x) = p(x)$ if and only if $deg(\alpha, F) = |G(K/F)| = [K : F]$. Because $deg(\alpha, F) = [F(\alpha) : F]$,we see that this occurs if and only if $[F(\alpha) : F] = [K : F]$ so that $[K : F(\alpha)] = 1$ and $K = F(\alpha).$
That's what I managed to do.
Is it complete enough? Can I improve on something? is not acceptable?