Minimal polynomial of an element over a subfield when the degrees of extensions are coprime This is a question from D.J.H. Garling's "A course in Galois Theory", and I am struggling to find an answer. Can anyone help me?

Suppose $K\leq L\leq L(\alpha)$ are field extensions, and that
  $|L(\alpha):L|$ and $|L:K|$ are relatively prime. Show that the
  minimal polynomial of $\alpha$ over $L$ has its coefficients in $K$.

Let  $|L(\alpha):L| = n$ and $|L:K| = m$ where $(m, n) = 1$. I can prove that the minimal polynomial of $\alpha$ over $K$ must have degree which is a multiple of $n$. Hence, $|K(\alpha):K|=nt$ for some integer $t$. Let $|L(\alpha):K(\alpha)|=s$. 
Now by the tower law, we have $nts = mn \implies ts =m$. 
If I can show that $t=1$, then we are done. But why can't we have $ts=m$ where $t$ and $s$ are proper factors of $m$? 
I know I need to use the fact that $(m,n) = 1$ somewhere, but I can't see where.
It's easy if we are given two quantities on 'different sides of the ladder' as coprime, but here $m$ and $n$ are on the same side of the ladder. If this is a typo in the book, is there a counterexample?
 A: Try $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{2})$, $\alpha=\sqrt[6]{2}$.
A: Another counterexample to the exercise as written is to take $\alpha \in L \setminus K$.  Then $|L(\alpha):L|=1$ is relatively prime to everything, but the minimal polynomial for $\alpha$ over $L$ is simply $x - \alpha$ which clearly does not have coefficients in $K$.
I think it is a typo and that Garling meant $|K(\alpha):K|$ and not $|L(\alpha):L|$.  Another statement for the exercise, which is probably more to the point and has essentially the same proof, is:

Suppose that $K \subset L \subset L(\alpha)$ are field extensions, and that $|K(\alpha):K|$ and $|L:K|$ are relatively prime, then $|L(\alpha):L| = |K(\alpha):K|$.

A proof would be:
Since $|K(\alpha):K|$ and $|L:K|$ are relatively prime, and both divide $|L(\alpha):K|$, their product must divide $|L(\alpha):K|$, implying $|L(\alpha):K| \geq |K(\alpha):K| \cdot |L:K|$.
But then, the general fact that $|K(\alpha):K| \geq |L(\alpha):L|$ implies that $|K(\alpha):K| \cdot |L:K| \geq |L(\alpha):L| \cdot |L:K| = |L(\alpha):K|$.
So the inequalities collapse, forcing $|L(\alpha):L| = |K(\alpha):K|$.
The exercise in this form can be used, for example, in exercise 5.9 to show that the positive $p$-th roots of $2$, as $p$ varies over the primes, are linearly independent over $\mathbb Q$.
