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Let $R$ be a discrete valuation ring with uniformizer $\pi$, fraction field $K$, and normalized discrete valuation $\nu :K^{\times }\to \mathbb{Z}$. A map $\mathrm{Spec}K\to {\mathbb{P}}_{\mathbb{Z}}^n$ is a projective pointwith at least one nonzero coordinate. Put $m={\min }_{a_i\ne 0}\nu (a_i)$ and set $b_i={\pi }^{-m}{a}_i$. Then every $b_i$ lies in $R$, and at least one $b_j$ is a unit.

On the standard affine chart $U_j$ of projective space, the ratios $b_i/b_j$ all lie in $R$. They therefore define a map $\mathrm{Spec}R\to U_j\subseteq {\mathbb{P}}_{\mathbb{Z}}^n$ whose generic restriction is the original point. This proves existence in the DVR case. The assumed separatedness of ${\mathbb{P}}_{\mathbb{Z}}^n\to \mathrm{Spec}\mathbb{Z}$, through the valuative criterion for separatedness, gives uniqueness. Thus the morphism satisfies the requested DVR form of the valuative criterion for properness.

Let $K$ be an equal-characteristic-p local field and let $c\in K$ have $v_K(c)=-m<0$, with $p\nmid m$. If ${\alpha }^p-\alpha =c$, the Artin–Schreier extension $L=K(\alpha )$ has degree $p$ and is totally ramified: a root in $K$ would force the pole order to be divisible by $p$, and $p{v}_L(\alpha )=-me(L/K)$ forces $e=p$. Choose $1\le s<p$ and an integer $r$ with $rp-sm=1$. For a base uniformizer $t$, the element ${\pi }_L=t^r{\alpha }^s$ has valuation one. Under ${\sigma }_a(\alpha )=\alpha +a$, $a\in {\mathbb{F}}_p^{\times }$, the leading term in ${\sigma }_a({\pi }_L)-{\pi }_L$ has valuation $rp-(s-1)m=m+1$. The uniformizer criterion for lower ramification groups therefore gives the single lower break $m$. The Herbrand function is the identity up to that break, so the upper break is also $m$.

For ${\mathbb{Q}}_2({\zeta }_8)/{\mathbb{Q}}_2$, the shifted polynomial $(1+T{)}^4+1$ is Eisenstein and ${\zeta }_8-1$ is a uniformizer. The three nonidentity automorphisms have displacement valuations two, four and two. Thus the lower breaks are one and three, the upper breaks one and two, and the different exponent from ramification groups is eight.

For the splitting field $M/{\mathbb{Q}}_3$ of $T^3-3T+3$, the cubic is Eisenstein and its discriminant is $-135$. The quadratic subfield is ${\mathbb{Q}}_3(\sqrt{3})$, and the degree-six extension is totally ramified. If $\alpha$ is a root and $s=\sqrt{3}$, then $\pi ={\alpha }^2/s$ is a uniformizer. Transpositions have $v_M(\sigma \pi -\pi )=1$, while three-cycles have value $2$, giving the displayed ramification groups. Its different exponent from ramification groups is $7$.

For a nonconstant morphism between nonsingular curves and a local target uniformizer $t$ at the image of $P$, the index is the order of vanishing of its pullback. Changing $t$ multiplies it by a local unit in a ring and does not change that order. This is a valuation definition and remains meaningful for inseparable morphisms.

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.

The ramification polynomial records the normalised differences of conjugates of a uniformizer, including the zero difference from the identity automorphism. Its quotient by $X$ is the reduced ramification polynomial. In the usual coefficient-index convention, the finite slopes of its lower Newton polygon are negatives of the valuations of its nonzero roots; the coefficient at the zero root is treated as having infinite valuation.

For a degree-$n$ totally ramified Galois extension and a uniformizer $\pi$, let $f$ be its minimal polynomial. Dividing the ramification polynomial $f(\pi (1+X))/{\pi }^n$ by its known factor $X$ gives the reduced ramification polynomial. It is monic of degree $n-1$, integral over the extension's valuation ring, and has roots $(\sigma (\pi )-\pi )/\pi$ for nonidentity automorphisms. Its constant coefficient is nonzero in a separable extension. Removing the zero root allows the usual Newton polygon root valuation theorem to apply directly.

The p-adic integers form a complete metric space and a discrete valuation ring. Every nonzero $x$ is uniquely $p^{k}u$ with $k=v_p(x)\ge 0$ and $u$ a p-adic unit. The prime $p$ is a uniformizer, and reduction modulo $p$ gives the residue field ${\mathbb{F}}_p$. The element $p$ has no inverse inside ${\mathbb{Z}}_p$, so the ring is not a field.

For a short ideal-classification outline, take a nonzero ideal and choose an element of smallest P-adic valuation $k$, which exists by the well-ordering principle. Its unit factor is invertible, so the ideal contains $p^k$. Every other element is divisible by $p^k$, proving that the ideal is $p^k{\mathbb{Z}}_p$. Completeness follows because ${\mathbb{Z}}_p$ is a closed subset of the complete metric space ${\mathbb{Q}}_p$. The zero ideal is separate from this nonzero-ideal classification.

For a finite unramified extension of non-Archimedean local fields, the field norm on residue units is surjective: on a cyclic group of order $q^n-1$ it is exponentiation by $(q^n-1)/(q-1)$ and has image of order $q-1$. For a common uniformizer $\pi$ and $r\ge 1$, expansion of the conjugate product givesThe field trace of a finite finite field extension is onto: its polynomial $X+X^q+\cdots +X^{q^{n-1}}$ is nonzero of degree below $q^n$, so defines a nonzero linear map to the base field. Lift a residue norm and correct successively with elements of $1+{\pi }^r{\mathcal{O}}_L$. The infinite product converges by completeness, and its norm is the prescribed unit. This argument also works in positive characteristic and when $p$ divides $n$.

If $K$ is a complete discretely valued field whose residue field $k$ is a perfect field, and $\pi$ is a uniformizer, the multiplicative Teichmuller lift $[\cdot ]:k\to {\mathcal{O}}_K$ gives each $x\in {\mathcal{O}}_K$ a unique expansion

For an integral element of a finite extension of non-Archimedean local fields, reduce the multiplication endomorphism on ${\mathcal{O}}_L$ modulo a base uniformizer. Its filtration by the powers of the maximal ideal has $e$ successive quotients, each the residue field ${\kappa }_L$. Multiplication by $a$ induces multiplication by $\bar{a}$ on each quotient. Additivity of the matrix trace through the filtration proves the formula. If the residue characteristic divides the ramification index, the trace of every integral element lies in the base maximal ideal; a Galois extension is not required.

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.

If $L/K$ is a finite totally ramified Galois extension and ${\pi }_L$ is a uniformizer, thenfor every $i\ge 0$.

For an unramified extension of non-Archimedean local fields, let $\varphi$ generate its cyclic Galois group. The ordinary Hilbert theorem 90 supplies $b\in L^{\times }$ with $x=\varphi (b)/b$. A common uniformizer $\pi \in K$ gives $y={\pi }^{-v_L(b)}b\in {\mathcal{O}}_L^{\times }$, retaining the same quotient because $\varphi$ fixes $\pi$. The reverse implication follows by telescoping the field norm.

For $K={\mathbb{Q}}_3({\zeta }_3)$, the element $\pi ={\zeta }_3-1$ is a uniformizer and

If $L/K$ is an unramified extension and $M/K$ is a totally ramified extension of non-Archimedean local fields inside a fixed algebraic closure, then $LM/L$ is totally ramified of degree $[M:K]$. A uniformizer of $M$ has an Eisenstein polynomial over $K$. Since a uniformizer of $K$ remains a uniformizer in $L$, that polynomial remains Eisenstein over $L$ and generates $LM$.

If $t$ is a uniformizer at a point $p$ of a smooth projective curve and a nonzero rational differential is $\omega =f(t)dt$, then ${\nu }_p(\omega )={\nu }_p(f)$. This integer is independent of the chosen uniformizer. Its positive and negative values are respectively the orders of the zero and pole of $\omega$ at $p$.