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Take rings and modules to be unital. Let $J_{\ell }$ denote the intersection of the maximal left ideals. For each nonzero vector $v$ of a simple module $M$ on the left, the map $R\to M$, $r\mapsto rv$, is onto and its kernel is a maximal left ideal. Consequently $J_{\ell }$ annihilates every simple module on the left. Conversely, annihilating $R/L$ forces an element to belong to the maximal left ideal $L$. Thuswhich shows that $J_{\ell }$ is a two-sided ideal.

For $j\in J_{\ell }$ and $r\in R$, the left ideal $R(1-rj)$ cannot be proper: a containing maximal left ideal would also contain $rj$ and hence $1$. Therefore $u(1-rj)=1$ for some $u\in R$. Here $u=1+urj$ also differs from $1$ by an element of $J_{\ell }$, so it too has a left inverse $v$. Multiplying the first equality by $v$ gives $1-rj=v$; hence $(1-rj)u=1$. We have proved that $1-rj$ is a unit, not just left invertible.

The identityis verified by multiplying on either side, using $(1-ab{)}^{-1}(1-ab)=(1-ab)(1-ab{)}^{-1}=1$. It implies that $1-jr$ is a unit as well. If $j$ were outside a maximal right ideal $I$, then $I+jR=R$, so $1-jr\in I$ for some $r$. A proper right ideal cannot contain a unit, giving a contradiction. Thus $J_{\ell }$ belongs to every maximal right ideal. Applying the same argument to the opposite ring proves the reverse inclusion. This proves the Jacobson radical is independent of handedness:For the zero ring the intersections are interpreted as the whole ring, so the equality remains valid.

If a two-sided ideal $N$ satisfies $N^m=0$, then for $x\in N$ and $r\in R$, $(rx{)}^m=0$. The finite geometric series ${\sum }_{i=0}^{m-1}(rx{)}^i$ inverts $1-rx$. The unit criterion for the Jacobson radical, also proved above by the maximal-ideal argument, gives $N\subseteq J(R)$. This assertion concerns a nilpotent ideal; arbitrary nilpotent elements of a noncommutative ring need not lie in its Jacobson radical.

For the formal power series ring over the p-adic integers, a series $f={\sum }_{i\ge 0}{a}_i{t}^i$ is a unit exactly when $a_0$ is a unit of ${\mathbb{Z}}_p$. Necessity follows by comparing constant coefficients. If $a_0$ is invertible, construct the inverse recursively by $b_0=a_0^{-1}$ and $b_m=-a_0^{-1}{\sum }_{i=1}^m{a}_i{b}_{m-i}$. Therefore the nonunits are exactly the series whose constant coefficient lies in $p{\mathbb{Z}}_p$. They form the unique maximal ideal $(p,t)$, and the quotient is ${\mathbb{F}}_p$. Hence

A one-dimensional commutative formal group law over the p-adic integers is a power series $\mathcal{F}(X,Y)\in {\mathbb{Z}}_p[[X,Y]]$ withIts linear part is $X+Y$, and it has a unique formal inverse $i(T)=-T+O(T^2)$ satisfying $\mathcal{F}(T,i(T))=0$. A formal-group homomorphism is a series $h(T)\in T{\mathbb{Z}}_p[[T]]$ satisfying $h(\mathcal{F}(X,Y))=\mathcal{F}(h(X),h(Y))$.

Define $[n](T)$ by adding $T$ to itself $n$ times with the formal group law. Induction on the linear part givesAssociativity and commutativity make $[n]$ a formal-group homomorphism. Since $p\nmid n$, its linear coefficient is a unit in ${\mathbb{Z}}_p$. We now construct its compositional inverse of a formal power series, rather than merely assert that the nonzero coefficient is sufficient.

Write $[n](T)=nT+a_2{T}^2+\cdots$ and seek $g(T)=b_{1}T+b_2{T}^2+\cdots$ with $[n](g(T))=T$. The degree-one coefficient gives $b_1=n^{-1}$. At degree $k\ge 2$, the coefficient is $n{b}_k$ plus a polynomial in the already chosen $b_1,\dots ,b_{k-1}$ and $a_2,\dots ,a_k$. Division by the unit $n$ uniquely chooses $b_k\in {\mathbb{Z}}_p$. This recursively constructs $g$ over the same ring. A right inverse is constructed in the same way; associativity of composition identifies it with $g$, so $g\circ [n]=[n]\circ g=T$.

To prove that $g$ respects the group law, apply $[n]$ to the two candidate series:Cancel by composing with $g$. The result is $\mathcal{F}(g(X),g(Y))=g(\mathcal{F}(X,Y))$. ThusThis is the multiplication isomorphism of a formal group law, a special case of the invertible morphism criterion for formal group laws. If $p\mid n$, its linear coefficient is not a unit, so the same isomorphism assertion would fail.

A basis of open sets in the p-adic integers isIndeed this is the set of $x$ with $v_p(x-a)\ge n$, equivalently $|x-a{|}_p<p^{-(n-1)}$; these neighborhoods become arbitrarily small. Reduction identifies ${\mathbb{Z}}_p/p^n{\mathbb{Z}}_p$ with $\mathbb{Z}/p^n\mathbb{Z}$, so each has index $p^n$.

If a subgroup $H\le ({\mathbb{Z}}_p,+)$ is open, a neighborhood of zero of the displayed form lies inside $H$. Thus $p^n{\mathbb{Z}}_p\subseteq H$ for some $n$, and $[{\mathbb{Z}}_p:H]\le p^n$ is finite.

Conversely suppose the abstract index is the finite integer $m$. The quotient is a finite abelian group of order $m$, so Lagrange's theorem gives $mx\in H$ for every $x\in {\mathbb{Z}}_p$. Write $m=p^{r}u$ with $p\nmid u$. Multiplication by the p-adic unit $u$ is a bijection of ${\mathbb{Z}}_p$, its inverse being multiplication by $u^{-1}\in {\mathbb{Z}}_p$. ConsequentlyThe subgroup $H$ is a union of open cosets of $p^r{\mathbb{Z}}_p$, and is therefore open. This proves the characterization of open subgroups of the p-adic integers:No prior assumption of closedness is required for the finite-index direction.

The correspondence concerns uniform pro-p groups and powerful Lie lattices, with their natural morphisms. Put $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. A powerful pro-p group is a pro-p group whose closed commutator subgroup lies in its closed $p^{ϵ}$-power subgroup. It is uniform when it is topologically finitely generated and all successive indices in its lower p-series are equal. Equivalently, a finitely generated powerful pro-p group is uniform precisely when it is a torsion-free group. A powerful Lie algebra over p-adic integers means a Lie algebra over ${\mathbb{Z}}_p$ that is free of finite rank as a module and satisfiesHere a Lie algebra over a ring is a module with an alternating bilinear Lie bracket satisfying the Jacobi identity. Finite freeness is an essential part of the lattice convention. The zero-bracket module ${\mathbb{F}}_p$ satisfies the bracket inclusion but has torsion, and its additive group is not uniform. Thus allowing arbitrary torsion modules would make the categorical assertion false.

The basic structural facts behind the construction can be seen from the power layers. Write $P_n=G^{p^n}$ with $P_0=G$. Collection for a powerful pro-p group givesFor odd $p$, the power lifting is exactly the successive-correction argument of Question 3, followed by compactness; at $p=2$ the stronger fourth-power commutator condition makes the corresponding collection corrections work. The maps between successive layers are surjective linear maps. Uniformity makes their finite dimensions equal, so these maps are isomorphisms. Comparing the first layer in which two elements differ now shows that the power map $P_n\to P_{n+1}$ is injective, while the power-set formula makes it surjective. Thus all the roots used below exist uniquely in the indicated layer. If $g_1,\dots ,g_d$ lift a basis of $G/P_1$, the images of $g_i^{p^n}$ form a basis of $P_n/P_{n+1}$. Successively choosing their coefficients modulo $p$ gives the unique ordered expressionExistence follows by lifting one layer at a time and taking a limit; uniqueness follows by comparing the first nonzero coefficient layer. In particular $[G:P_n]=p^{nd}$ and these layers form a neighbourhood basis. These are the power-layer and coordinate facts needed in the proof, rather than an assumption that an arbitrary pro-p group has unique roots.

On the underlying set of $G$, set the additive zero to $1$, the additive negative of $x$ to $x^{-1}$, and defineThe p-adic integers act through limits of integer powers in the closed cyclic subgroup generated by $x$. In the second formula the product belongs to $P_n$; in the third the group commutator belongs to $P_{2n+ϵ}$. The symbols $p^{-n}$ and $p^{-2n}$ denote the unique group roots, not unrestricted division in a group. We next justify convergence and the Lie algebra laws together.

A useful concrete tool is a complete normed group algebra containing $G$. In its completed group algebra, put $b_i=g_i-1$. Ordered-coordinate uniqueness gives unique expansions ${\sum }_{\alpha \in {\mathbb{N}}^d}{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To see why, first work modulo $p$ and $P_n$: the ordered products of the $g_i$ with exponents below $p^n$ form the regular basis, and the change to the corresponding products of $g_i-1$ is triangular with diagonal entries one. Take inverse limits and then lift the coefficient digits successively in powers of $p$. Define the weightIt is a genuine separated, submultiplicative weight. Indeed reordering $b_j{b}_i$ produces a group-commutator correction in $P_{ϵ}$; the binomial expansion of a $p^{ϵ}$th power gives that correction weight at least $2ϵ$, so collection never decreases total weight. Also $v(pc)=1+v(c)$. Inverting $p$ and completing therefore gives a Banach algebra $B$ over ${\mathbb{Q}}_p$. For $g\in G$, ordered coordinates showwith value infinity at $g=1$. Thus the induced topology on $G$ is exactly its original topology. This construction supplies the ambient algebra without presupposing the uniform pro-p group and powerful Lie lattice correspondence. The weighted construction, including the case $p=2$, is described in Equivariant D-modules on rigid analytic spaces, Section 6.5.

Since $ϵ>1/(p-1)$, the p-adic logarithm and p-adic exponential converge on the relevant balls in $B$:The estimates $v_p(k!)\le k/(p-1)$ and $v_p(k)\le {\log }_{p}k$ show that term weights tend to infinity. Formal substitution is legitimate there and proves that the two maps are inverse. They preserve the leading weight. In particular $\log$ embeds the compact set $G$ homeomorphically into $B$, and $\log (g^{p^n})=p^n\log g$.

Put $a=\log x$ and $b=\log y$. Expanding the convergent Baker--Campbell--Hausdorff formula in $B$ givesThe first expansion has linear part $p^n(a+b)$ and all other terms have degree at least two in $p^{n}a,p^{n}b$. The second has leading term $p^{2n}(ab-ba)$ and all remaining terms have degree at least three. After division by the displayed powers of $p$, the extra factor of $p^n$ forces those errors to zero. The normalized expressions are logarithms of the unique roots already specified. Since $\log G$ is compact and closed in $B$, both limits belong to $\log G$. Scalar multiplication also remains in $\log G$ by continuity of powers. Consequently $\log G$ is a module over ${\mathbb{Z}}_p$, closed under the associative-algebra Lie bracket $ab-ba$. Bilinearity, alternation and the Jacobi identity follow by expanding products in the associative algebra $B$. Transporting these operations back gives exactly $L_G$ above.

The commutator-layer inclusion places every normalized bracket root in $P_{ϵ}$. Since $\log P_{ϵ}=p^{ϵ}\log G$ and this set is compact, passage to the limit proves $[L_G,L_G]\subseteq p^{ϵ}{L}_G$. Addition modulo $p{L}_G$ is the original multiplication modulo $P_1$: collection shows $(x^{p^n}{y}^{p^n}{)}^{p^{-n}}\equiv xy(\mathrm{mod}{P}_1)$, as the nonlinear errors lie at least one layer deeper. Thus $L_G/p{L}_G$ has dimension $d$, with basis $e_i=\log g_i$. Given $a\in \log G$, write its residue as ${\sum }_i{\lambda }_{0i}{e}_i$ and repeat on $(a-{\sum }_i{\lambda }_{0i}{e}_i)/p$. This iterative digit lifting converges because the remainders stay in the bounded compact set $\log G$. It gives $a={\sum }_i{\lambda }_i{e}_i$ with ${\lambda }_i\in {\mathbb{Z}}_p$. A nontrivial relation among the $e_i$ could be divided by the smallest power of $p$ among its coefficients, contradicting their independence modulo $p$. Hence $L_G$ is free of rank $d$, proving that it is a powerful Lie lattice.

Conversely, let $L$ be a powerful Lie lattice. In $L{\otimes }_{{\mathbb{Z}}_p}{\mathbb{Q}}_p$, defineThe formal Baker--Campbell--Hausdorff formula and its coefficient estimate are the algebraic ingredients used here. More precisely, the degree-$m$ component can be expressed as an integral Lie polynomial divided by a number of $p$-adic valuation at most $\lfloor (m-1)/(p-1)\rfloor$. One way to obtain this estimate is to bound the associative coefficients of $\log (\exp X\exp Y)$ and use an integral ordered basis of the free Lie algebra. The denominator bound $v_p(m!)+\lfloor {\log }_p{s}_p(m)\rfloor$, where $s_p(m)$ is the base-$p$ digit sum, is at most $\lfloor (m-1)/(p-1)\rfloor$. A precise primary reference for that denominator estimate is Smallest common denominators for the homogeneous components of the Baker-Campbell-Hausdorff series. A length-$m$ Lie bracket in $L$ lies in $p^{ϵ(m-1)}L$. Thus the degree-$m$ term belongs toThese exponents are nonnegative and tend to infinity; for $m\ge 2$ they are at least one. This proves integral convergence, continuity and $x\ast y\equiv x+y(\mathrm{mod}pL)$. Associativity follows from the formal identity $\exp (\mathrm{BCH}(X,Y))=\exp X\exp Y$ in the completed universal enveloping algebra. Substitution of three elements is justified by the same valuation estimates, so the two associative expressions agree. The identity is $0$, the inverse is $-x$, and $x^{\ast k}=kx$ because every nonlinear term involving only $x$ vanishes.

For powerfulness, formal conjugation givesThe difference $\delta =y-e^{-\mathrm{ad}x}y$ lies in $p^{ϵ}L$: its $k$th term has valuation at least $kϵ-v_p(k!)\ge ϵ$ for $k\ge 1$. The group commutator is therefore $(-y+\delta )\ast y$. Its linear term is $\delta$. Every nonlinear Lie term contains $\delta$, since it vanishes when $\delta =0$, and the extra divisibility makes all those terms lie in $p^{ϵ}L$ too. Hence $[(L,\ast ),(L,\ast )]\subseteq p^{ϵ}L$. The ideals $p^{n}L$ are open normal subgroups for $\ast$, and their finite quotients have order $p^{nd}$. The group is consequently a pro-p group. Its power sets are exactly $p^{n}L$, its lower-series indices are constantly $p^d$, and a basis of $L$ topologically generates it by the same successive digit lifting. The BCH group $(L,\ast )$ is uniform.

To check the inverse constructions, powers in the Baker--Campbell--Hausdorff formula group are ordinary multiples in $L$. Thereforeand the similarly normalized group commutator tends to $[x,y]$. The limit constructions recover the original Lie algebra exactly. In the other direction, the ambient Banach algebra construction gives $\log (xy)=\mathrm{BCH}(\log x,\log y)$, so the recovered operation is the original multiplication of $G$.

Finally a continuous group homomorphism between uniform pro-p groups preserves integer powers and powers with exponents in the p-adic integers, preserves the unique roots in the power layers, and commutes with their limits. It therefore becomes a linear Lie algebra homomorphism. Conversely a linear Lie algebra homomorphism between finite free modules is continuous and preserves every term of the convergent Baker--Campbell--Hausdorff formula, hence is a continuous group homomorphism for $\ast$. The constructions preserve identity maps and composition, and the inverse identifications are natural. They are mutually inverse functors, giving the stated category correspondence with the finite-free lattice convention.

Put $C={\bigcup }_{r\ge 1}\mathbb{Q}({\mu }_{p^r})$, where ${\mu }_{p^r}$ is the group of $p^r$th roots of unity. The compatible Galois groups of the cyclotomic fields giveFor odd $p$, the Teichmüller character givesThe p-adic logarithm identifies $1+p{\mathbb{Z}}_p$ with $p{\mathbb{Z}}_p$, which is topologically isomorphic to ${\mathbb{Z}}_p$. More explicitly, $a\mapsto (1+p{)}^a$ is an isomorphism from the additive p-adic integers onto $1+p{\mathbb{Z}}_p$: the logarithm of $1+p$ has valuation one and generates $p{\mathbb{Z}}_p$. Take the fixed field of ${\mu }_{p-1}$ in $C$.

For $p=2$, use insteadThe p-adic logarithm identifies the second factor with $4{\mathbb{Z}}_2$, and $a\mapsto 5^a$ identifies it with ${\mathbb{Z}}_2$. The fixed field of $\{\pm 1\}$ in $C$ is the union of the maximal real subfields of the cyclotomic fields of $2$-power conductor. Thus in both cases the cyclotomic Zp-extension exists:where ${\Delta }_p={\mu }_{p-1}$ for odd $p$ and ${\Delta }_2=\{\pm 1\}$.

The closed subgroup $p^n{\mathbb{Z}}_p$ is the unique subgroup of index $p^n$, so the Galois correspondence supplies the unique degree-$p^n$ field $\mathbb{Q}[p^n]$. In the odd case it is the ${\Delta }_p$-fixed field in $\mathbb{Q}({\mu }_{p^{n+1}})$; in the even case it is the real subfield of $\mathbb{Q}({\mu }_{2^{n+2}})$. These finite layers are all totally real number fields.

Now set $A=\mathbb{Q}[p^n]$, $B=\mathbb{Q}[p^m]$, and let $H_A$ be the ordinary Hilbert class field of $A$. Thus $H_A/A$ is abelian, unramified at finite primes, and split at every real place, with degree $h(A)$. If $m=n$ there is nothing to prove. Otherwise the unique prime over $p$ is totally ramified in $B/A$, by the permitted total-ramification fact and multiplicativity of the ramification index. Any nontrivial intermediate field of $B/A$ is ramified at that prime. HenceIt follows that the two extensions are linearly disjoint field extensions, andUnramified extensions remain unramified after base change. The real places also remain split: $H_A$ and $B$ are totally real, so their compositum is totally real. Therefore $B{H}_A/B$ lies inside the ordinary Hilbert class field $H_B/B$. Its degree divides $[H_B:B]=h(B)$, provingThis is class number divisibility with split real places; using a class-field statement restricted to imaginary fields would unnecessarily omit the present real layers.

In the one-dimensional commutative convention, a formal group law over $R$ is a formal power series $F(X,Y)=X+Y+\text{terms of degree at least two}$ satisfying $F(X,0)=X$, $F(0,Y)=Y$, $F(X,Y)=F(Y,X)$ and $F(F(X,Y),Z)=F(X,F(Y,Z))$. The formal inverse is a series $I(T)=-T+O(T^2)$ with $F(T,I(T))=0$. For $R={\mathbb{Z}}_p$ all these series converge on $p{\mathbb{Z}}_p$, which becomes a group with operation $F$.

Repeated formal addition gives the multiplication isomorphism of a formal group law $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the coefficient $n$ is a unit in the p-adic integers. We can construct an inverse $H(T)=n^{-1}T+{\sum }_{j\ge 2}{h}_j{T}^j$ by coefficient recursion. At degree $j$, the equation $[n{]}_F(H(T))=T$ has the form $n{h}_j+\text{known integral terms}=0$; dividing by the unit $n$ keeps $h_j$ integral. The inverse is two-sided by uniqueness of formal composition inverses. Since integral coefficients multiplied by successive powers of $t\in p{\mathbb{Z}}_p$ tend to zero, both series converge on that set and their formal identities hold there. This proves prime-to-residue-characteristic multiplication on a formal group:The argument also applies at $p=2$ for odd $n$; it does not require a logarithm or torsion-freeness.

On continuous functions on the p-adic integers, define the forward difference operator and the Mahler coefficients byWriting $Sf(x)=f(x+1)$, the explicit iterate follows from ${\Delta }^n=(S-I{)}^n$. The ultrametric inequality and integral binomial coefficients imply $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$ and $|a_n(f){|}_p\le \Vert f{\Vert }_{\infty }$, where the supremum norm is taken over ${\mathbb{Z}}_p$.

The Mahler theorem states that every such $f$ has a unique expansion with uniform convergenceConversely, every sequence in ${\mathbb{Q}}_p$ tending to zero yields a continuous function by this expansion. The binomial polynomials form an orthonormal expansion in the non-Archimedean sense: $\Vert f{\Vert }_{\infty }={\sup }_n|a_n(f){|}_p$.

Here is the requested proof under the permitted coefficient-decay assumption. For $n\ge 1$, the binomial polynomial $B_n(x)=x(x-1)\cdots (x-n+1)/n!$ is a continuous function on ${\mathbb{Z}}_p$. Its values on nonnegative integers are integral, and those integers are dense in the p-adic integers. Since ${\mathbb{Z}}_p$ is a closed set in ${\mathbb{Q}}_p$, $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$. Also $B_n(n)=1$, so $\Vert B_n{\Vert }_{\infty }=1$, including $B_0=1$.

If $a_n\to 0$, the ultrametric inequality gives the uniform tail boundCompleteness of ${\mathbb{Q}}_p$ gives a uniform limit $F$, and the uniform limit theorem makes $F$ continuous. At any nonnegative integer $m$, all $B_n(m)$ with $n>m$ vanish. Finite binomial inversion givessince the inner sum is $(1-1{)}^{m-j}$. Hence $F=f$ on a dense subset and therefore on all of ${\mathbb{Z}}_p$. The values at $m=0,1,2,\dots$ recover each coefficient recursively because $B_m(m)=1$; this proves uniqueness. The expansion bounds $\Vert f{\Vert }_{\infty }$ by $\sup |a_n{|}_p$, and the earlier coefficient inequality proves equality. This also proves the converse statement.

Although the problem allows us to assume decay, it can be established independently. By compactness and uniform continuity, approximate $f$ uniformly by $h$ constant on residue classes modulo $p^r$. On this finite-dimensional space, $S^{p^r}=I$ andThe matrix of ${\Delta }^{p^r}$ has entries divisible by $p$, so its operator norm is at most $p^{-1}$; consequently $\Vert {\Delta }^{j{p}^r+s}h{\Vert }_{\infty }\le p^{-j}\Vert h{\Vert }_{\infty }$ for $0\le s<p^r$. Thus $a_n(h)\to 0$. Since $|a_n(f)-a_n(h){|}_p\le \Vert f-h{\Vert }_{\infty }$, arbitrary uniform approximation proves automatic decay of Mahler coefficients.

For the last claim, construct the discrete antidifferentiation on the p-adic integersIts coefficient sequence is $0,a_0,a_1,\dots$, still tending to zero, so it is continuous. By Pascal's identity, $\Delta B_{n+1}=B_n$. The boundedness of $\Delta$ permits applying it to the uniform limit, giving $\Delta G=f$. For the stated linear map, invariance under translation by one now givesThus translation-invariant linear forms on p-adic continuous functions vanish:No continuity of $T$ has been assumed or used. In particular, one must not justify this by applying $T$ termwise to an infinite Mahler expansion; it is the existence of a continuous discrete antiderivative that makes the conclusion valid.

Put $f(X)=X^3+5X-12$. Any root in ${\mathbb{Q}}_p$ is a p-adic integer: if its valuation were negative, $X^3$ would be the unique term of least valuation.

For $p=2$, reduction modulo two has roots zero and one. The root zero is simple because $f^{\prime }(0)=5$ is odd, so it lifts uniquely. An odd integer $x$ satisfies $x^2\equiv 1(\mathrm{mod}8)$, and henceThus there is no odd $2$-adic root and the number of roots is one.

For $p=3$,All three roots are simple because $f^{\prime }(X)=3X^2+5\equiv 2(\mathrm{mod}3)$. Each lifts uniquely, giving three roots in ${\mathbb{Q}}_3$.

For $p=5$, reduction gives $X^3-2$, whose unique root is $X=3$; it is simple because $3X^2\not\equiv 0(\mathrm{mod}5)$. It lifts uniquely, so there is one root in ${\mathbb{Q}}_5$.

An element of the p-adic integers is a unit in a ring exactly when its reduction modulo $p$ is nonzero. More explicitly, if $\alpha \not\equiv 0(\mathrm{mod}p)$, its inverses modulo $p^n$ are unique and compatible, so they define $\beta \in {\mathbb{Z}}_p$ with $\alpha \beta =1$. The converse follows by reducing $\alpha \beta =1$ modulo $p$. Part ii therefore proves that $\alpha$ topologically generates the additive group if and only if $\alpha$ is a p-adic unit.

The ring of p-adic integers is the inverse limitEquivalently, each element has a unique convergent expansion ${\sum }_{j\ge 0}{b}_j{p}^j$ with digits $0\le b_j<p$. It is a complete discrete valuation ring with maximal ideal $(p)$ and residue field ${\mathbb{F}}_p$.

A Lie algebra over the p-adic integers whose underlying module is free of finite rank and whose Lie bracket satisfies $[L,L]\subseteq p^{ϵ}L$, where $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. Finite freeness is part of this convention: merely requiring a small Lie bracket on a torsion module would not correspond to uniform pro-p groups.

A pro-p group whose closed commutator subgroup lies in its closed power subgroup $G^{p^{ϵ}}$, with $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. This is the profinite analogue of a powerful p-group. Finitely generated torsion-free examples are exactly the uniform pro-p groups. The additive group of a finite product of p-adic integers is an abelian example.

A profinite group $G$ is procyclic if every quotient by an open normal subgroup is cyclic. Equivalently $G=\overline{⟨g⟩}$ for one element $g$. For the forward implication, the sets of lifts of generators of $G/N$ are nonempty clopen sets. Any finite intersection contains a lift of a generator of the cyclic quotient by the intersection of those open normal subgroups. Compactness supplies one element generating every finite quotient, and hence generating densely. The reverse implication follows by projecting that generator. For a pro-p group, two distinct maximal open subgroups would give a quotient $C_p^2$, so a nontrivial procyclic group has a unique maximal open subgroup $M$. Its profinite Frattini subgroup is $M$, and any $g\notin M$ generates topologically: otherwise the proper closed subgroup it generates lies in a maximal open subgroup. The additive group of p-adic integers is an infinite example, generated topologically by $1$.

A profinite abelian group is an inverse limit of finite abelian groups. Its topology is compact and totally disconnected. An abelian pro-p group acquires scalar multiplication by p-adic integers by taking limits of integer multiples.

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.

Every abstract subgroup of finite index in a finitely generated pro-p group is open. Fixed-generator commutator collection in nilpotent groups makes the abstract commutator subgroup closed. The resulting abelian quotient is a quotient of a finite power of the p-adic integers, so its subgroup of $p$th powers is closed and of finite index. Thus the algebraic $G^p[G,G]$ is open and every abstract index-$p$ subgroup is open. An arbitrary finite quotient is a finite p-group; induction on its order, using a normal subgroup of index $p$ and the Schreier lemma, gives the assertion. The normal core reduces nonnormal finite-index subgroups to this case.

For an abelian variety $A$ and a prime number $\ell$ different from the characteristic, the inverse limit of its geometric ${\ell }^n$-torsion, with transition multiplication by $\ell$, is its Tate module. For an elliptic curve, it is free of rank two over the p-adic integers and satisfies $T_{\ell }(E)/{\ell }^n{T}_{\ell }(E)\simeq E[{\ell }^n]$. The absolute Galois group acts continuously and commutes with all base-defined endomorphisms.

Let $T:\mathcal{C}({\mathbb{Z}}_p,{\mathbb{Q}}_p)\to {\mathbb{Q}}_p$ be linear and invariant under translation by one. For any $f$, choose its continuous discrete antiderivative $Jf$. Then $T(f)=T(\Delta Jf)=T(Jf(\cdot +1))-T(Jf)=0$. No continuity or boundedness of $T$ is assumed. In particular this applies to a form invariant under every translation by a p-adic integer.

An ultrametric space is a metric space whose metric satisfies the ultrametric inequality. The rational numbers with $d_p(x,y)=|x-y{|}_p$, and their P-adic number completion, are examples. Their geometry organizes points into nested clusters rather than measuring separation along a number line.

For integers $a,b$ and $k\ge 0$, the definition of the P-adic valuation givesThus a closed ball of radius $p^{-k}$ in the integers is a congruence class modulo $p^k$. In the p-adic integers the same closed ball is $a+p^k{\mathbb{Z}}_p$. More shared low-order digits mean a smaller p-adic absolute value of the difference.

The binary tree first splits the sixteen integers into even and odd congruence classes, then refines modulo $4$, $8$, and $16$. Each node fixes one more low-order digit. The leaves $5$ and $13$ share the class $5$ modulo $8$, then separate modulo $16$, so their distance is exactly $1/8$. The leaves are deliberately ordered by their low-order digits; horizontal spacing and edge lengths are layout choices, not the metric. The singleton leaves describe this finite sample; a congruence class modulo $16$ in all p-adic integers remains infinite.

The ultrametric inequality also explains the unusual behavior of balls. Every point of an open ball can serve as its center: if $d(a,b)<r$, then $d(a,x)<r$ if and only if $d(b,x)<r$. Two open balls that intersect are nested, since a point of their intersection serves as a common center and their radii can be compared. The same nesting rule holds for closed balls and for a pair consisting of one open ball and one closed ball.

Every open ball is a clopen set. For a point $y$ outside $B(a,r)$, the open ball $B(y,d(a,y))$ stays outside $B(a,r)$: if $d(y,z)<d(a,y)$, the ultrametric inequality forces $d(a,z)=d(a,y)\ge r$. Hence the complement is an open set. Also every closed ball of positive radius is an open set, because $B(b,r)$ lies in the closed ball of radius $r$ around $a$ whenever $d(a,b)\le r$. A radius-zero closed ball is a singleton and need not be an open set. Distinct points can therefore be separated by a clopen set, proving that an ultrametric space is totally disconnected. These ball properties turn distance thresholds into a hierarchy of compatible clusters. ^[ref]