= Solution
Use the simple-root form of <Hensel lemma>: if $F\in\mathbb Z_p[T]$ and $t_0\in\mathbb Z_p$ satisfy $F(t_0)\equiv0\pmod p$ and $F'(t_0)\not\equiv0\pmod p$, there is a unique $t\in\mathbb Z_p$ with $F(t)=0$ and $t\equiv t_0\pmod p$. 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 expression
$$
x=3^k\epsilon u,\qquad k\in\mathbb Z,\quad\epsilon\in\{1,-1\},\quad u\in U_1:=1+3\mathbb Z_3.
$$
Indeed its normalized <P-adic valuation> fixes $k$, and its unit residue modulo three fixes $\epsilon$. Cubing is an automorphism on $\{1,-1\}$, so it remains to compute the cube map on the <principal units>.
For $t\in\mathbb Z_3$,
$$
(1+3t)^3=1+9\bigl(t+3t^2+3t^3\bigr).
$$
Thus $U_1^3\subseteq U_2:=1+9\mathbb Z_3$. Conversely, given $1+9a\in U_2$, apply <Hensel lemma> to
$$
F(T)=T+3T^2+3T^3-a.
$$
Modulo three it is $T-a$, and $F'(T)=1+6T+9T^2$ is always a unit. The residue root $a\bmod3$ therefore lifts, giving $t$ with $(1+3t)^3=1+9a$. Hence
$$
\boxed{U_1^3=U_2.}
$$
This 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 $1+3t\mapsto t\bmod3$ 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> is
$$
\boxed{\mathbb Q_3^\times/(\mathbb Q_3^\times)^3\cong\mathbb Z/3\mathbb Z\times\mathbb Z/3\mathbb Z.}
$$
Explicit generators are the classes of $3$ and $4=1+3$. Representatives are $3^a4^b$ with $0\leq a,b<3$. 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.
Back to article page
