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?