= Solution
The <strong form of Hensel lemma> says the following. Let $v$ be a <discrete valuation> on a complete field $K$. If $f\in\mathcal O_K[X]$ and
$$
v(f(x_0))>2v(f'(x_0)),
$$
then $f$ has a root $x\in\mathcal O_K$ satisfying $v(x-x_0)>v(f'(x_0))$.
To prove it, apply <Newton iteration over a valued field>:
$$
x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.
$$
The initial inequality says that the first correction has valuation greater than $v(f'(x_0))$. Taylor expansion then shows inductively that $v(f'(x_n))=v(f'(x_0))$, 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 as
$$
\mathbb Q_3^\times
=3^{\mathbb Z}\times\{\pm1\}\times U_1,
\qquad U_1=1+3\mathbb Z_3.
$$
The <P-adic valuation> gives
$$
3^{\mathbb Z}/3^{3\mathbb Z}\cong\mathbb Z/3\mathbb Z,
$$
and cubing is the identity on $\{\pm1\}$. 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 $b\equiv a\pmod3$ and put $x_0=1+3b$. For $f(X)=X^3-u$,
$$
v_3(f(x_0))\geq3>2=2v_3(f'(x_0)),
$$
so the <strong form of Hensel lemma> produces a cube root in $U_1$. Thus $U_1^3=U_2$. Finally,
$$
U_1/U_2\longrightarrow\mathbb Z/3\mathbb Z,
\qquad 1+3b\longmapsto b\pmod3,
$$
is an isomorphism. Combining the valuation and principal-unit factors proves the <cube-class group of the 3-adic numbers> identity
$$
\mathbb Q_3^\times/(\mathbb Q_3^\times)^3
\cong\mathbb Z/3\mathbb Z\times\mathbb Z/3\mathbb Z.
$$
Back to article page
