I'm trying to show that 4 is not a 3-adic cube.
On the Wikipedia page for Hensel's lemma (https://en.wikipedia.org/wiki/Hensel%27s_lemma) I read that: "4 is not a 3-adic cube since it is not a cube mod 9"
However, I'm not sure why a number not being a cube mod 9 implies it is not a 3-adic cube. Is there a nice proof of this fact?
On
If $4$ were a cube in $\mathbf Z_3$, it would be the cube of a $3$-adic unit, say $4=u^3$. Reduction mod $3$ shows that actually $u$ is a principal unit, i.e. has the form $u=1+w$, with $w\in 3\mathbf Z_3$. Expanding $(1+w)^3$ shows immediately that $v_3 (u^3 -1) \ge 2$, whereas $v_3 (4-1)=1$, a contradiction.
Actually one could tackle similar questions in the general case of $\mathbf Z_p$, where $p$ is any prime, by using the multiplicative structure of the group of units (resp. principal units) $U$ (resp. $U_1$) of $\mathbf Z_p$, see e.g. Serre's booklet "A course in arithmetic", chap. II, §3. In particular, for $p$ odd and $n \ge 1$, denote $U_n =1+p^n\mathbf Z_p$. If $x\in U_n - U_{n+1}$, then $x^p\in U_{n+1} - U_{n+2}$ (§3.2, lemma). Analogous (but more complicated) results are available for any finite extension of $\mathbf Q_p$.
If $4 \in \Bbb Z_3$ was a cube $4=a^3$, then $|a|_3=1$, so $a \in \Bbb Z_3$ and we can reduce the equation $4=a^3$ mod $9$, using that $\Bbb Z_3/9\Bbb Z_3 \cong \Bbb Z/9\Bbb Z$.
To elaborate on the last point, note that any $x \in \Bbb Z_3$ can be written as $x=\sum_{n=0}^\infty a_n 3^n$ for $a_n \in \{0,1,2\}$, pulling out the first two summands gives $x=a_0+3a_1+9 \sum_{n=2}^\infty a_n 3^{n-2}$. If $a_0$ and $a_1$ vary over $\{0,1,2\}$, then $a_0+3a_1$ varies between $\{0, \dots, 8\}$. If we reduce mod $9$, we just get $x \equiv a_0+3a_1 \pmod{9}$. This shows that $\Bbb Z_3/9\Bbb Z_3 \cong \Bbb Z/9\Bbb Z$. The same reasoning can be used to show that $\Bbb Z_p/p^n\Bbb Z_p \cong \Bbb Z/p^n\Bbb Z$ for all primes $p$ and $n \in \Bbb N$.