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For odd $p$ and p-adic units $u,v$, the quadratic Hilbert symbol isThe Legendre symbols record the two unit square classes. For $p=2$, put $ϵ(u)=(u-1)/2\mathrm{mod}2$ and $\omega (u)=(u^2-1)/8\mathrm{mod}2$. ThenThe basis $-1,2,5$ of the square-class group of the 2-adic numbers gives this formula: its only negative basis pairings are $(-1,-1{)}_2$ and $(2,5{)}_2=(5,2{)}_2$. The first is obstructed by sums of squares modulo four, the second by $x^2-2y^2$ modulo eight; the remaining pairings have explicit field norm representatives.

Let $a$ be an integer not divisible by $p$. Its modular inverse modulo $p^n$ has a unique least nonnegative representative $r_n$. Reducing the inverse equation modulo $p^n$ shows that the representatives are compatible. Since $a$ is a p-adic unit,so the sequence converges to the inverse $a^{-1}$ in ${\mathbb{Z}}_p$. This computes a fraction through finite integer operations.

For $a=3$ and $p=2$, the first eight representatives are $1,3,3,11,11,43,43,171$. Each satisfies $3r_n\equiv 1(\mathrm{mod}{2}^n)$. The resulting expansion isThe geometric series has ratio $4$, whose 2-adic absolute value is $1/4$, and its sum is $1+2/(1-4)=1/3$. In the ordinary real number absolute value, the representatives grow. Different metrics impose different meanings of a good approximation. The periodic digits here are a specific instance of the rational expansions studied in Conrad's authored note.^[ref]

A p-adic unit is an invertible element of the ring ${\mathbb{Z}}_p$. For a p-adic integer $x$, this is equivalent to $v_p(x)=0$, $|x{|}_p=1$, or a nonzero initial p-adic digit $b_0$. A rational number $a/b$ is a p-adic unit when neither its reduced numerator nor its denominator is divisible by $p$.

Thus $3$ and $1/3$ are p-adic units at $p=2$, while $2$ is a p-adic integer with no inverse inside ${\mathbb{Z}}_2$. The modular inversion converges to a p-adic inverse construction computes the inverse of an integer unit to arbitrary finite precision.

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.

False. Use the polynomial $X(X^2-p)$, with splitting field $K={\mathbb{Q}}_p(\theta )$, ${\theta }^2=p$, and label the polynomial roots $0,\theta ,-\theta$. Take $\beta =p\theta$. ThenBoth $p-1$ and $p+1$ are p-adic units, also when $p=2$. Hence zero is the uniquely closest polynomial root, but $\beta \notin {\mathbb{Q}}_p(0)={\mathbb{Q}}_p$, because $\theta \notin {\mathbb{Q}}_p$. The polynomial need not be irreducible in the given setup. The inclusion established by Krasner's lemma goes in the direction in part (iii), not its converse.

A strong lifting lemma. Let $R$ be a complete discrete valuation ring, with fraction field $F$ and normalized discrete valuation $v$. A useful strong form of Hensel lemma says that for $f\in R[T]$ and $a\in R$ satisfyingthere is a root $b\in R$ withIf $f(a)=0$, take $b=a$. Otherwise the hypothesis ensures that $f^{\prime }(a)\ne 0$. The root is unique among $b$ satisfying $v(b-a)>v(f^{\prime }(a))$. In particular, a simple root of the reduction of $f$ lifts uniquely with its specified residue.

Here is a proof by Newton iteration over a valued field. Put $a_0=a$ andWrite $A_j=v(f(a_j))$ and $B=v(f^{\prime }(a_0))$. Taylor expansion over $R$ shows, inductively, that the increment has valuation $A_j-B>B$, that $v(f^{\prime }(a_{j+1}))=B$, and thatIndeed, the constant and linear terms of $f(a_{j+1})$ cancel, leaving terms of valuation at least twice that of the increment. The change in $f^{\prime }$ has valuation greater than $B$, so its valuation stays $B$. Thus all iterates lie in $R$ and form a Cauchy sequence. Completeness and continuity give a root $b$. The later increments have strictly larger valuations than the first, proving the displayed distance formula. For uniqueness, if $b,c$ are in the stated ball, Taylor expansion factorsThe second factor is nonzero, so two roots must coincide. This proves Hensel's lemma in the form needed here.

Odd residue characteristic. For odd $p$, a p-adic unit $u$ is a square exactly when its reduction is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction. For sufficiency, a nonzero residue root of $T^2-u$ has derivative $2T$ invertible modulo $p$, so the Hensel lemma lifts it. The multiplicative group of a finite field is cyclic, and its subgroup of squares has index two. ThereforeRepresentatives are $1$ and any unit whose reduction is a quadratic nonresidue.

Residue characteristic two. An odd integer has square congruent to one modulo eight, so a square 2-adic unit must lie in $1+8{\mathbb{Z}}_2$. Conversely, if $u\equiv 1(\mathrm{mod}8)$, take $f(T)=T^2-u$ and $a=1$. Then $v_2(f(1))\ge 3>2v_2(f^{\prime }(1))=2$, and the strong form of Hensel lemma produces a square root. HenceThe four unit square classes of the p-adic integers have representatives $1,3,5,7$.

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.

Every element of $A_{\neg p}$ can be written $m/n$ with $p\nmid n$. Since $n$ is a p-adic unit, defineThis is the restriction of the standard embedding $\mathbb{Q}\hookrightarrow {\mathbb{Q}}_p$, so it is an injective group homomorphism. Equivalently, if $m{n}^{-1}=0$ in ${\mathbb{Z}}_p$, then $m=0$ because the characteristic is zero.

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.

An element $\alpha \in {\mathbb{Z}}_p$ topologically generates the additive group of ${\mathbb{Z}}_p$ exactly when $\alpha \not\equiv 0(\mathrm{mod}p)$, equivalently when $\alpha$ is a p-adic unit. In that case its image generates every cyclic quotient $\mathbb{Z}/p^n\mathbb{Z}$.

For odd $p$, a p-adic unit is a square exactly when its residue is a square, by the Hensel lemma. Its unit square-class group is $C_2$. For $p=2$, a unit is a square exactly when it is congruent to one modulo eight, by the strong form of Hensel lemma. The four square classes are represented by $1,3,5,7$, giving $C_2\times C_2$.