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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.

Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. The lower ramification groups arewith $G_{-1}=G$. If $K\subseteq M\subseteq L$, the same inequality defines $G_s(L/M)$ inside $\mathrm{Gal}(L/M)$, so directly

Now suppose $M/K$ is Finite Galois extension. The inertia group is the kernel of the action on the residue field. Restriction sends $G_0(L/K)$ into $G_0(M/K)$. Conversely, the maximal unramified subextension of $M/K$ is the intersection of $M$ with the maximal unramified subextension of $L/K$. The Galois correspondence therefore shows that the restriction image is all of $G_0(M/K)$.

For the explicit extension, take ${\alpha }^3=3$ and a primitive cube root of unity $\zeta$. The polynomial $X^3-3$ is Eisenstein over ${\mathbb{Q}}_3$, while ${\mathbb{Q}}_3(\zeta )/{\mathbb{Q}}_3$ is a ramified quadratic extension. Its splitting fieldis therefore a totally ramified extension of degree six with Galois group $S_3$. With $v_L(3)=6$, one has $v_L(\alpha )=2$ and $v_L(\zeta -1)=3$, sois a uniformizer. Let $\tau (\alpha )=\zeta \alpha$, $\tau (\zeta )=\zeta$, and let $\sigma (\alpha )=\alpha$, $\sigma (\zeta )={\zeta }^{-1}$. ThenThe two nonidentity elements of $⟨\tau ⟩=A_3$ have ramification number four, whereas each transposition has ramification number one. HenceIn particular, $A_3$ is the wild inertia group.

For an odd prime $p$, this Eisenstein polynomial generates a totally ramified cubic root field. Its discriminant is $-p^2(4p+27)$. For $p>3$ the unit part is a square exactly when $p\equiv 1(\mathrm{mod}3)$. The splitting field then has cyclic Galois group of order three; otherwise it has symmetric group $S_3$, with an unramified extension of degree two and inertia group $C_3$. Both cases are tamely ramified, so all positive ramification groups are trivial.

At $p=3$, adjoining $\beta =\sqrt{-39}$ gives a ramified quadratic subfield, and the splitting field has total ramification index six. A root $\alpha$ has normalized valuation two, while $\beta$ has valuation three, so $\pi =\beta /\alpha$ is a uniformizer. The derivative $3{\alpha }^2+3$ has valuation six, and all distinct root differences have valuation three. A three-cycle fixes $\beta$, giving $v_L(\sigma \pi -\pi )=2$. A transposition fixing $\alpha$ negates $\beta$, giving valuation one; conjugation handles the other transpositions. Hence $G_0=S_3$, $G_1=C_3$, and $G_i=1$ for $i\ge 2$. The different exponent is seven.

For a finite Galois extension of non-Archimedean local fields, this is a homomorphism from the inertia group to $k_L^{\times }$ with kernel the wild inertia group. Inertia acts trivially on residue units, which proves multiplicativity and independence of the chosen uniformizer. Passing to the maximal unramified subextension reduces the kernel calculation to a totally ramified extension. Hence it embeds the tame inertia quotient into the cyclic multiplicative group of the residue field.

Let $R$ be an integrally closed domain and $f\in R[X]$ monic, with discriminant outside a prime ideal $P$. The roots stay distinct after reduction. In the splitting algebra choose a prime above $P$. Its decomposition group maps onto the Galois group of the residue splitting field; its inertia group fixes every reduced root and is trivial because reduction bijects the distinct root sets. Thus the residue Galois group embeds in the generic group as a permutation group. For finite residue fields, irreducible factor degrees give the Frobenius cycle type.

The Weil-group topology makes the inertia group open with its profinite topology and makes the quotient by inertia discrete. It is generally finer than the topology induced from the ambient Galois group.

The wild inertia group is the first ramification group $G_1$. It is the unique Sylow subgroup of the inertia group for the residue characteristic.