Consider the field extension $[F:\mathbb{C}(t)]=d$. Now consider $A$, the integral closure of $\mathbb{C}[t]$ in $F$. Why is it true that every prime ideal $\mathfrak{p}\subset \mathbb{C}[t]$ has inertia index one in $A$ (for every prime $\mathfrak{q}|\mathfrak{p}$)?
This result has a nice interpretation (and proof) once we analyze $F$ as the function field of a $d-$branched covering of the Riemann sphere, but I do not see how to prove the result algebraically.
It is trivial if the ideal in $A$ is the zero ideal, so let $\mathfrak{P}$ be a non-zero prime ideal in $A$ and let $\mathfrak{p}=\mathfrak{P}\cap\mathbb{C}[t]$. Since $\mathbb{C}[t]$ is a PID then we have $\mathfrak{p}=g(t)\mathbb{C}[t]$ for some $g(t)\in\mathbb{C}[t]$. By the fundamental theorem of algebra we see that there is a factorisation $$g(t)=(t-c)f(t)$$ for some $c\in\mathbb{C}$ and $f(t)\in\mathbb{C}$. Therefore we have an inclusion of ideals $$\mathfrak{p}=(g(t))\subseteq(t-c).$$ But $\mathfrak{p}$ and $(t-c)$ are both maximal ideals and $\mathfrak{p}\neq 0$ whence $\mathfrak{p}=(t-c)$. So $\mathbb{C}[t]/\mathfrak{p}=\mathbb{C}$.
Recall that the inertia degere $[A/\mathfrak{P}:\mathbb{C}[t]/\mathfrak{p}]<\infty$ which implies that $[A/\mathfrak{P}:\mathbb{C}]<\infty$ whence we conclude that $A/\mathfrak{P}=\mathbb{C}$. Thus the inertia degree is 1, as desired.