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Fix a non-Archimedean local field $K$, its valuation ring $R={\mathcal{O}}_K$, a uniformizer $\pi$, and a residue field with cardinality $q$. The construction below applies both in mixed and equal characteristic. A one-dimensional commutative formal group law is $F(X,Y)\in R[[X,Y]]$ with identity, associativity and commutativity, and $F(X,Y)=X+Y+$ higher-degree terms. It has a unique formal inverse. Its morphisms are series $h(T)\in TR[[T]]$ with $h(F(X,Y))=G(h(X),h(Y))$; a unit linear coefficient makes such a morphism invertible by recursive coefficient comparison. Evaluating its integral series on the maximal ideal of a finite extension of $K$ is legitimate because the series converge there.

Choose a Lubin–Tate series $f(T)$ satisfyingA useful choice is $f(T)=\pi T+T^q$. The Lubin–Tate functional equation lemma constructs a unique series with any prescribed linear form $\ell$ and satisfyingHere is its coefficient argument. If terms below degree $d$ have been found, the unknown homogeneous degree-$d$ term appears with multiplier $\pi -{\pi }^d$. The already known error has every coefficient divisible by $\pi$: after reduction, the equation becomes $H(X{)}^q=H(X_1^q,\dots ,X_r^q)$, valid because the residue coefficients lie in ${\mathbb{F}}_q$. Since $\pi -{\pi }^d$ is $\pi$ times a unit for $d\ge 2$, the next term is uniquely determined and integral. Recursion proves existence and uniqueness.

Use $\ell =X+Y$ to construct $F_f$, and $\ell =aT$ to construct $[a{]}_f$ for every $a\in R$. Uniqueness gives $F_f(X,0)=X$ and $F_f(0,Y)=Y$. Comparing $[a{]}_f(F_f(X,Y))$ with $F_f([a{]}_f(X),[a{]}_f(Y))$, which satisfy the same functional equation and have the same linear form, proves that $[a{]}_f$ is an endomorphism of the law. The same uniqueness applied to three variables proves associativity: $F_f(F_f(X,Y),Z)$ and $F_f(X,F_f(Y,Z))$ have the same linear form and functional equation. It also proves commutativity and the identitiesThus this Lubin–Tate formal group is equipped with an action of the whole ring $R$, not only its integers. For a unit $a$, the inverse of $[a{]}_f$ is $[a^{-1}{]}_f$. The same recursion, with two series for the fixed $\pi$, gives a unique strict Lubin–Tate change of series. Its integral series and inverse converge on torsion points, so it identifies their torsion fields. We may compute with $f(T)=\pi T+T^q$ without losing the general construction for that uniformizer.

Write $f_0(T)=T$ and $f_n=f^{\circ n}$. The Lubin–Tate torsion of level $n$ isFor $n\ge 1$, its primitive points are the roots ofThis is a monic polynomial of degree $D_n=(q-1)q^{n-1}$. Modulo $\pi$ it is $T^{D_n}$, and its constant coefficient is exactly $\pi$. Thus the Eisenstein layers of Lubin–Tate torsion are Eisenstein polynomials. A root ${\lambda }_n$ generates a totally ramified extension of degree $D_n$, and has $v_K({\lambda }_n)=1/D_n$ in the uniquely extended valuation. Also $f_{n-1}({\lambda }_n{)}^{q-1}=-\pi \ne 0$, so it has exact level $n$.

The mapis well-defined and injective. If $a={\pi }^{r}u$ with $r<n$ and $u$ a unit, then $[a]{\lambda }_n=[u]f_r({\lambda }_n)\ne 0$, because $[u]$ is invertible and ${\lambda }_n$ is primitive. Conversely every multiple of ${\pi }^n$ kills it. The image therefore contains $q^n$ distinct roots of the degree-$q^n$ polynomial $f_n$, so it is all of ${\Lambda }_n$. This proves, without merely assuming the torsion cardinality,Its primitive generators are exactly $[a]{\lambda }_n$ with $a\in (R/{\pi }^{n}R{)}^{\times }$.

Define the Lubin–Tate extension $K_{\pi ,n}=K({\Lambda }_n)$. Every integral endomorphism series evaluated at ${\lambda }_n$ converges in the complete field $K({\lambda }_n)$, so all torsion points already lie there. HenceIt is the splitting field of $f_n$, which has $q^n$ distinct roots by the module calculation. Thus it is Galois. Every K-automorphism preserves the uniquely extended absolute value on a field, hence is continuous and commutes with limits of the integral endomorphism series. Its automorphisms therefore commute with the endomorphism action and send a primitive generator to a primitive generator. There is a unique unit class $a_{\sigma }$ such that $\sigma ({\lambda }_n)=[a_{\sigma }]{\lambda }_n$. The Lubin–Tate Galois action on primitive torsion gives an injective homomorphismBoth sides have order $D_n=(q-1)q^{n-1}$, so it is an isomorphism. In particular every finite layer is abelian and totally ramified.

For an example, take $K={\mathbb{Q}}_p$, $\pi =p$, and the equally admissible series $f(T)=(1+T{)}^p-1$. Its formal group law is the formal multiplicative group $F(X,Y)=X+Y+XY$, with $[a](T)=(1+T{)}^a-1$ for $a\in {\mathbb{Z}}_p$. The binomial polynomials make its coefficients integral. Its level-$n$ torsion points are $\zeta -1$ with ${\zeta }^{p^n}=1$. Thus the Lubin–Tate cyclotomic example gives $K_{p,n}={\mathbb{Q}}_p({\mu }_{p^n})$ and Galois action $\zeta \mapsto {\zeta }^a$. The order formula includes $p=2,n=1$, when the first field and its Galois group are trivial over ${\mathbb{Q}}_2$.

Finally choose primitive points compatibly, $[\pi ]{\lambda }_{n+1}={\lambda }_n$. They exist by taking a root of $f(T)-{\lambda }_n$ in the algebraic closure; its roots have positive valuation and exact level $n+1$. Then $K_{\pi ,n}\subseteq K_{\pi ,n+1}$, and restriction of Galois automorphisms corresponds to reduction of unit classes. The Lubin–Tate tower $K_{\pi }={\bigcup }_n{K}_{\pi ,n}$ consequently has the topological Galois groupThe final isomorphism uses completeness of the discrete valuation ring $R$. This computes the finite-layer and infinite-tower Galois groups directly from the formal action and the primitive Eisenstein polynomials.

Use the simple-root form of the Hensel lemma: for $F\in {\mathcal{O}}_K[T]$ and $\bar{a}\in k_K$ satisfying $\overline{F}(\bar{a})=0$ and $\overline{F^{\prime }}(\bar{a})\ne 0$, there is a unique $a\in {\mathcal{O}}_K$ with $\bar{a}$ as its reduction and $F(a)=0$.

Let $d=[\ell :k_K]$. The residue field $k_K$ is finite, so $\ell /k_K$ is separable, and the primitive element theorem gives $\ell =k_K(\bar{\alpha })$. Let $\bar{F}$ be the monic irreducible polynomial of $\bar{\alpha }$, of degree $d$, and choose a monic coefficient lift $F\in {\mathcal{O}}_K[T]$. Irreducible reduction implies that $F$ is irreducible over $K$: any monic factorization over $K$ has integral coefficients, since all its roots are integral over the valuation ring, and would reduce to a factorization of $\bar{F}$.

Adjoin a root $\alpha$ and put $L=K(\alpha )$. Then $[L:K]=d$ and $\alpha \in {\mathcal{O}}_L$. Its reduction satisfies $\bar{F}$, so $k_L$ contains a copy of $\ell$ over $k_K$. For finite extensions of non-Archimedean local fields, the fundamental degree relation isThe residue degree is at least $d$, while the left side equals $d$, henceThis constructs the required unramified extension.

For uniqueness, let $L^{\prime }/K$ be another unramified extension with a chosen identification $k_{L^{\prime }}\cong \ell$. The element corresponding to $\bar{\alpha }$ is a simple residue root of $F$, because finite fields are perfect. The Hensel lemma in $L^{\prime }$ gives a root ${\alpha }^{\prime }\in {\mathcal{O}}_{L^{\prime }}$ with this reduction. Since $F$ is irreducible of degree $d$, $[K({\alpha }^{\prime }):K]=d=[L^{\prime }:K]$, so $L^{\prime }=K({\alpha }^{\prime })$. Sending $\alpha$ to ${\alpha }^{\prime }$ yields a $K$-isomorphism $L\to L^{\prime }$, compatible with the prescribed residue identification. ThusThis proves the existence and uniqueness of unramified local extensions using the stated version of Hensel's lemma, without assuming the classification in advance.

The prime ideal factorization of $pB$ iswhich defines the ramification index of a prime ideal $e_{q/p}=v_q(pB)$. The residue-field degree isLocalizing at $p$ makes $B_p$ a free $R_p$-module of rank $[L:K]$, so $B/pB$ has dimension $[L:K]$ over the residue field $R/p$. The Chinese remainder theorem and the filtration by powers of each $q$ decompose this vector space into $e_{q/p}$ successive quotients isomorphic to $B/q$, each of dimension $f_{q/p}$. Therefore the fundamental identity for prime decomposition gives

Let $\epsilon =1+\sqrt{2}$ and $P=(3+\sqrt{2})$. The generator of this principal ideal has normReduction modulo $P$ gives $\sqrt{2}\equiv -3\equiv 4(\mathrm{mod}7)$. The map below is surjective and has $P$ in its kernel; both the kernel and $P$ have index seven, the latter by the ideal norm, so they are equal. Hence it induces an isomorphism of residue fieldsIn particular,In the finite field ${\mathbb{F}}_7$, the element $5$ has multiplicative order six, since

By part (b), every unit is $(-1{)}^{\delta }{\epsilon }^n$ for $\delta \in \{0,1\}$ and $n\in \mathbb{Z}$. Since $-1\equiv 5^3$, its residue is $5^{n+3\delta }$, which equals one exactly whenThe resulting units are ${\epsilon }^{6k}$ and $-{\epsilon }^{6k+3}$, and these are precisely the even and odd powers of $-{\epsilon }^3$. Thereforewhich is the group of units congruent to one modulo 3 plus square root two.

Let $\mathfrak{m}$ be the maximal ideal and $k=A/\mathfrak{m}$ the finite residue field. The maximal ideal of an Artinian local ring is nilpotent, so ${\mathfrak{m}}^r=0$ for some $r$. Every quotientis both an Artinian $A$-module and a vector space over $k$. An Artinian vector space is finite-dimensional, hence each quotient is a finite set. The finite filtrationtherefore proves that the underlying set of $A$ is finite. This is the Finiteness criterion for an Artinian local ring.

The claim fails for a general extension. If $k$ is an infinite field, all the infinitely many maximal ideals $(x-a)$ of $k[x]$ contract to $(0)$ in $k$.

It still fails for an integral extension. Take a finite field $k={\mathbb{F}}_q$ andEvery $b\in B$ satisfies the monic equation $b^q-b=0$, so $B$ is integral over the diagonal copy of $k$. The coordinate kernels are infinitely many distinct maximal ideals, all lying over $(0)$.

The claim is true for a module-finite ring extension. The primes above $\mathfrak{p}$ correspond to the primes of the fiber ringThis is a finite-dimensional algebra over the residue field $\kappa (\mathfrak{p})$, hence an Artinian ring, and an Artinian ring has only finitely many prime ideals.

For a point $p\in {\mathbb{A}}_k^1$, let $a$ be the image of $t$ in its residue field $\kappa (p)$. The scheme-theoretic fiber isIf $p\ne (t)$, then $a\ne 0$. The quadratic polynomial $xy-a{z}^2$ is irreducible. Indeed, after setting $z=0$, any hypothetical linear factors must restrict, up to nonzero scalars, to $x$ and $y$; comparing the $xz$ and $yz$ coefficients then forces both $z$ coefficients to vanish, contradicting the nonzero $z^2$ coefficient. Hence its homogeneous coordinate ring is an integral domain, so $Y_p$ is an integral scheme.

At the origin $p=(t)$, the fiber is $V_{+}(xy)$, the union of the two distinct projective lines $x=0$ and $y=0$, and is therefore not irreducible. It is nevertheless a reduced scheme because the ideal $(xy)=(x)\cap (y)$ equals its radical. Every other fiber is integral and hence reduced. Thus the fiber is integral exactly away from the origin, and it is reduced at every point of ${\mathbb{A}}_k^1$.

Take $R=\mathbb{Z}$, $M=\mathbb{Q}/\mathbb{Z}$, and $k=2$. The group $M$ is nonzero and divisible. Since every element of $M$ has finite order, $M$ is the filtered union of finite cyclic groups. For every $n\ge 1$,Tensor products commute with filtered colimits, so

Now suppose a nonzero finitely generated $R$-module $M$ satisfied $M^{\otimes k}=0$. Choose a maximal ideal $\mathfrak{m}$ in the support of $M$. The localized module $M_{\mathfrak{m}}$ is nonzero and finitely generated. By Nakayama lemma,This is a nonzero vector space over the residue field $\kappa (\mathfrak{m})$, so its $k$-fold tensor power is nonzero. But it is the reduction modulo $\mathfrak{m}$ of $(M^{\otimes k}{)}_{\mathfrak{m}}$, a contradiction. Thus a nonzero tensor-nilpotent module cannot be finitely generated.

Conversely, assume condition (b), and let $M,N$ be nonzero. The stated property givesThese are nonzero vector spaces over the residue field $k=R/\mathfrak{m}$, so their tensor product over $k$ is nonzero. Associativity and base change giveTherefore $M{\otimes }_{R}N\ne 0$. This proves the reverse implication and the local tensor nonvanishing criterion.

The period is the order of a class in the Brauer group. It divides its index. Over a local field with finite residue field, the period and index are both the reduced denominator of its local Brauer invariant.

A pro-$p$ ring is a topological ring with continuous ring operations whose underlying additive group is a pro-p group. Standard local analytic examples are complete Noetherian local rings with finite residue field of characteristic $p$, such as ${\mathbb{Z}}_p$ and ${\mathbb{F}}_p[[t]]$. The topology is essential to the convergence of power series used for analytic group operations.

For an integral generator $\alpha$ of a number field, let $A=\mathbb{Z}[\alpha ]$ and suppose $p$ is coprime to $[{\mathcal{O}}_K:A]$. Localizing at $p$ makes $A$ and the ring of integers equal, so reduction gives the displayed quotient ring. If $\overline{m}=\prod g_i^{e_i}$ with distinct monic irreducible polynomials, the Chinese remainder theorem gives local factors ${\mathbb{F}}_p[X]/(g_i^{e_i})$. Their residue fields have degrees $\mathrm{deg}{g}_i$ and their maximal ideals have nilpotence indices of an ideal $e_i$. Localizing the Dedekind domain ${\mathcal{O}}_K$ at the corresponding prime ideals gives discrete valuation rings, where this same nilpotence index of an ideal is the ramification index of a prime ideal. Thus $(p)=\prod (p,g_i(\alpha ){)}^{e_i}$ and each residue degree is $\mathrm{deg}{g}_i$.

For a finite-type integer algebra, a nonnilpotent element $r$ gives a nonzero localization of a ring $R[1/r]$. A maximal ideal there has finite residue field by the finite-field theorem for finitely generated integer algebras. The image of $R$ in that field is a finite integral domain, hence a field. Its kernel is therefore maximal in $R$ and avoids $r$. Nilpotents lie in every maximal ideal, proving the radical equality. Applying the same argument to every prime quotient shows that these algebras are Jacobson rings. Merely contracting a localized maximal ideal without the finite-image argument would not prove maximality.

If a finite extension of the P-adic numbers contains a primitive $p$th root of unity $\zeta$, the identity $p={\prod }_{j=1}^{p-1}(1-{\zeta }^j)$ gives $e=(p-1)v_L(1-\zeta )\ge p-1$. The ratios of these factors reduce to the nonzero integers $j$ in the residue field. Thus an extension with $e<p-1$ has only roots of unity of order prime to $p$.

The points of the real affine plane as a scheme are its prime ideals: the zero ideal, the principal ideals of irreducible real polynomials, and its maximal ideals. The latter have residue field $\mathbb{R}$ or $\mathbb{C}$, by the Zariski lemma and the real closed field property. They correspond respectively to real coordinate pairs and conjugate pairs of nonreal complex coordinate pairs. Thus the scheme contains both nonclosed generic points and closed points absent from the real locus. Its topology is the Zariski topology, and its structure sheaf has sections $\mathbb{R}[x,y{]}_h$ on each principal open subscheme $D(h)$.

The residue characteristic of a local ring is the characteristic of a ring of its residue field. A finite extension of ${\mathbb{Q}}_p$ has residue characteristic $p$.

The degree of the residue field extension induced by a finite extension of complete discretely valued fields. Along with the ramification index, it satisfies $[L:K]=ef$. An unramified extension additionally has separable residue extension; a totally ramified extension has residue degree one.

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.

The fibre of a morphism of schemes over $y\in Y$ is its base change to the residue field $\kappa (y)$. For $\mathrm{Spec}B\to \mathrm{Spec}A$, with $y$ corresponding to $\mathfrak{p}$, its fibre ring is $B{\otimes }_A\kappa (\mathfrak{p})$. Its structure sheaf records nilpotents and multiplicities that the set-theoretic inverse image cannot see.

Relative to a nontrivial additive character, the self-dual additive Haar measure makes the twice-applied Fourier transform over a local field equal to reflection. If the annihilator of the valuation ring is ${\pi }^c{\mathcal{O}}_F$ and its residue field has $q$ elements, the normalization is $\mathrm{vol}({\mathcal{O}}_F)=q^{c/2}$. Volumes of a compact open subgroup and its character annihilator then multiply to one.