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Normalize the discrete valuation by $v_L({\pi }_L)=1$, and put $v_L(0)=+\infty$. For $\sigma \ne 1$, define its ramification numberand set $i_G(1)=+\infty$. The minimum exists: all finite values are nonnegative integers, and a nonidentity automorphism moves some integral element. The lower ramification numbering isThus $G_s=G_{\lceil s\rceil }$, $G_0$ is the inertia group, and $G_1$ is the wild inertia group. This states the convention at integer endpoints as well as between them. Automorphisms preserve $v_L$, so these are normal subgroups of $G$.

Define the Herbrand function and its inverse byThe integrand is positive and piecewise constant. Since $G_t=1$ for sufficiently large $t$, the function is continuous, strictly increasing and unbounded, and its inverse is defined on all nonnegative reals. The upper ramification numbering isIn particular, the denominator in the Herbrand integral involves $G_0$, even when $L/K$ is not totally ramified. These conventions are also given in Milne's Class Field Theory, Chapter I, §4.

Now use the monogenic hypothesis ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$. Since $L/K$ is Galois, its minimal polynomial has the distinct conjugates $\sigma (\alpha )$ as roots, soTaking valuations proves the first equality.

For every polynomial $P\in {\mathcal{O}}_K[T]$, the difference $P(\sigma (\alpha ))-P(\alpha )$ is divisible in ${\mathcal{O}}_L$ by $\sigma (\alpha )-\alpha$. Every element of ${\mathcal{O}}_L$ is such a polynomial value, and taking the element $\alpha$ attains the resulting lower bound. HenceFor an integer $i_G(\sigma )=m\ge 0$, the element $\sigma$ lies in exactly the groups $G_0,\dots ,G_{m-1}$; if $m=0$, it lies in none. Summing this count over the nonidentity elements and interchanging the two finite sums givesThis proves the ramification-group sum for a monogenic integer ring. Equivalently it is the different exponent from ramification groups, since the different ideal is generated by $f^{\prime }(\alpha )$ under the same monogenic hypothesis.

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.