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A 3-adic number belongs to the completion of $\mathbb{Q}$ for the p-adic absolute value with prime three. Its nonzero elements have the form $3^{k}u$, with $k\in \mathbb{Z}$ and $u\in {\mathbb{Z}}_3^{\times }$. The cube-class group of the 3-adic numbers is generated by the classes of $3$ and $4$.

Use the simple-root form of Hensel lemma: if $F\in {\mathbb{Z}}_p[T]$ and $t_0\in {\mathbb{Z}}_p$ satisfy $F(t_0)\equiv 0(\mathrm{mod}p)$ and $F^{\prime }(t_0)\not\equiv 0(\mathrm{mod}p)$, there is a unique $t\in {\mathbb{Z}}_p$ with $F(t)=0$ and $t\equiv t_0(\mathrm{mod}p)$. More generally the same assertion holds over a complete discrete valuation ring, with its maximal ideal in place of $(p)$.

Every nonzero 3-adic number has a unique expressionIndeed its normalized P-adic valuation fixes $k$, and its unit residue modulo three fixes $ϵ$. Cubing is an automorphism on $\{1,-1\}$, so it remains to compute the cube map on the principal units.

For $t\in {\mathbb{Z}}_3$,Thus $U_1^3\subseteq U_2:=1+9{\mathbb{Z}}_3$. Conversely, given $1+9a\in U_2$, apply Hensel lemma toModulo three it is $T-a$, and $F^{\prime }(T)=1+6T+9T^2$ is always a unit. The residue root $a\mathrm{mod}3$ therefore lifts, giving $t$ with $(1+3t{)}^3=1+9a$. HenceThis is the cube map on 3-adic principal units. Introducing $T$ is essential: applying simple-root lifting directly to $X^3-u$ would fail because its derivative is divisible by three at every unit.

The map $1+3t\mapsto t\mathrm{mod}3$ induces an isomorphism $U_1/U_2\cong (\mathbb{Z}/3\mathbb{Z},+)$, since multiplication adds the first principal-unit digits modulo three. The valuation factor contributes $\mathbb{Z}/3\mathbb{Z}$ independently, so the cube-class group of the 3-adic numbers isExplicit generators are the classes of $3$ and $4=1+3$. Representatives are $3^a{4}^b$ with $0\le a,b<3$. The class of $3$ is detected by valuation modulo three, and that of $4$ by its nonzero first principal-unit digit, so the two generators are independent.

The strong form of Hensel lemma says the following. Let $v$ be a discrete valuation on a complete field $K$. If $f\in {\mathcal{O}}_K[X]$ andthen $f$ has a root $x\in {\mathcal{O}}_K$ satisfying $v(x-x_0)>v(f^{\prime }(x_0))$.

To prove it, apply Newton iteration over a valued field:The initial inequality says that the first correction has valuation greater than $v(f^{\prime }(x_0))$. Taylor expansion then shows inductively that $v(f^{\prime }(x_n))=v(f^{\prime }(x_0))$, while the valuations of the corrections tend to infinity. Hence $(x_n)$ is a Cauchy sequence. Completeness gives a limit $x$, and continuity gives $f(x)=0$.

Now decompose the multiplicative group asThe P-adic valuation givesand cubing is the identity on $\{\pm 1\}$. Put $U_2=1+9{\mathbb{Z}}_3$. Expansion gives $U_1^3\subseteq U_2$. Conversely, for $u=1+9a\in U_2$, choose $b\equiv a(\mathrm{mod}3)$ and put $x_0=1+3b$. For $f(X)=X^3-u$,so the strong form of Hensel lemma produces a cube root in $U_1$. Thus $U_1^3=U_2$. Finally,is an isomorphism. Combining the valuation and principal-unit factors proves the cube-class group of the 3-adic numbers identity