p-adic unit (${\mathbb{Z}}_p^{\times }$)

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.

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 $C_2\times C_2$.

A 2-adic unit is an odd element of ${\mathbb{Z}}_2$. It is a square in ${\mathbb{Z}}_2^{\times }$ exactly when it is congruent to $1$ modulo $8$.

An odd square is congruent to one modulo eight. Conversely, for $u\equiv 1(\mathrm{mod}8)$, begin with $x_3=1$ and $x_n^2\equiv u(\mathrm{mod}{2}^n)$. For $n\ge 3$, adding ${\epsilon }_n{2}^{n-1}$ changes the square by ${\epsilon }_n{2}^n{x}_n$ modulo $2^{n+1}$. Since $x_n$ is odd, one choice of ${\epsilon }_n\in \{0,1\}$ gives the next congruence. The corrections tend to zero 2-adically, so completeness produces an odd square root. This directly proves the unit square criterion without needing a unit derivative in Hensel lemma.

An odd $2$-adic unit is a square exactly when it is congruent to $1$ modulo $8$. Valuation parity and the four odd residue classes modulo $8$ therefore givewith generators represented by $-1$, $2$, and $5$.

The three generators $-1,2,5$ of the square-class group of the 2-adic numbers give these seven distinct extensions. In characteristic different from two, completing the square shows every quadratic extension is of this form, and comparing coefficients in $\sqrt{e}=a+b\sqrt{d}$ proves that two such fields coincide precisely when $e/d$ is a square. The only unramified member is ${\mathbb{Q}}_2(\sqrt{5})$: $(1+\sqrt{5})/2$ is integral with irreducible reduction $X^2+X+1$. Conversely any unramified quadratic extension has that residue field, and Hensel lemma lifts a residue root of $X^2-X-1$, recovering the same field.

In $L={\mathbb{Q}}_2(\sqrt{2})$, set $\pi =\sqrt{2}$ and normalize $v_L(\pi )=1$, so $v_L(2)=2$. The two terms of $a+b\pi$ have valuations of opposite parity, giving ${\mathcal{O}}_L={\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\pi$. The nonidentity automorphism sends $\pi$ to $-\pi$, and $v_L(-2\pi )=3$. The lower ramification numbering therefore gives the displayed groups. The different exponent is three.

For an odd prime $p$ and $\alpha \in {\mathbb{Z}}_p^{\times }$,This follows from ${\mathbb{Z}}_p^{\times }\cong {\mu }_{p-1}\times (1+p{\mathbb{Z}}_p)$ and $(1+p{\mathbb{Z}}_p{)}^p=1+p^2{\mathbb{Z}}_p$.

For a non-Archimedean local field $K$ with maximal ideal ${\mathfrak{m}}_K$, a principal unit is an element of $1+{\mathfrak{m}}_K$. The higher principal-unit groups are $1+{\mathfrak{m}}_K^n$.

In a complete discrete valuation ring of residue characteristic $p$, the power map $u\mapsto u^m$ is a bijection on $1+\mathfrak{m}$ when $p\nmid m$. For a principal unit $a$, the Hensel lemma lifts the simple residue root one of $T^m-a$ uniquely. This can absorb principal-unit factors in tame radical descriptions of local fields.

For a non-Archimedean local field $K$, the $n$th higher principal-unit group is $1+{\mathfrak{m}}_K^n$. These groups form a decreasing filtration of the principal units.

For $r\ge 1$, the map $1+{\pi }^{r}a\mapsto \bar{a}$ is a surjective group homomorphism to the additive residue field with kernel $U_{r+1}$. The extra product term is divisible by ${\pi }^{r+1}$. Thus each quotient has cardinality $|k|$, a useful count when splitting the unit group by roots of unity.

The cube of $1+3t$ is $1+9(t+3t^2+3t^3)$. Conversely the polynomial $T+3T^2+3T^3-a$ has a unique simple residue root modulo three for each $a\in {\mathbb{Z}}_3$, so Hensel lemma produces a cube root of every $1+9a$. Thus the principal units modulo their cubes have order three. The change of variable removes the derivative obstruction that prevents direct simple-root lifting for $X^3-u$.

For a finite extension $K/{\mathbb{Q}}_p$ and all sufficiently large $r$, the p-adic logarithm and p-adic exponential are inverse group isomorphisms

The valuation and first principal-unit digit give