 Solution

Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. In lower numbering,These are the ramification groups; $G_0$ is the inertia group and $G_1$ is the wild inertia group.

Because ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$, every $x\in {\mathcal{O}}_L$ is $P(\alpha )$ for some $P\in {\mathcal{O}}_K[X]$. The polynomial identity $P(Y)-P(X)=(Y-X)Q(X,Y)$ has integral coefficients. Consequently the inequality for $\alpha$ implies it for every $x$, and the converse follows by taking $x=\alpha$. Therefore

Since $L/K$ is a Finite Galois extension, the minimal polynomial factors asDifferentiating and evaluating at $\alpha$ givesand henceFor a fixed nonidentity $\sigma$, its valuation is exactly the number of integers $s\ge 0$ for which $\sigma \in G_s$. Interchanging the two finite sums proves the ramification-group sum for a monogenic integer ring:

The extension is unramified exactly when $G_0=1$, which by this nonnegative sum is equivalent to $v_L(f^{\prime }(\alpha ))=0$, or $f^{\prime }(\alpha )\in {\mathcal{O}}_L^{\times }$. Moreover $|G_0|=e(L/K)$, so the $s=0$ term is $e(L/K)-1$. Equalityholds exactly when $G_1=1$, which is exactly tame ramification.