Is $\frac{10}{459}$ a cube in $\Bbb{Q}_3$, the 3-adic numbers?
My general strategy for showing if something is a cube in $\Bbb{Q}_3$ has been:
- Use Hensel's Lemma to show it is (1)
- If it isn't, suppose for contradiction it is a cube and equate 3-adic valuations to see if we can find a contradiction (2)
- If it isn't, suppose for contradiction it is a cube and reduce modulo $3^k$ for some $k$ and try to arrive at a contradiction. (3)
(2) doesn't help in this instance, and I don't think I can apply either (1) or (3) as the number in question is a fraction.
How can I attempt to answer the question: "Is $x$ a cube in $\Bbb{Q}_3$?" in this specific case, when $x = \frac{10}{459}$, and I'm also interested in a general strategy for when $x$ is a fraction.
Thanks.
First determine the valuation of the fraction by getting it for the numerator and denominator:
$v(10)=0, v(459)=v(17×27)=3, v(10/459)=v(10)-v(459)=-3=\text{a multiple of 3}$
Then since $v(10/459)$ is a multiple of $3$, we will have a cube if the integer obtained by shifting the "decimal point" so that the last nonzero digit is at the units place is a cube. Multiply your fraction by $3^3=27$ to make this adjustment, setting $v$ to zero:
$(10/459)×3^3=10/17$
The resulting fraction will be a 3-adic cube, completing our test, if its residue $\bmod 9$ is $\pm1$. Here $10\equiv1$ and $17\equiv-1$, so $(10/17)\equiv-1$ making it a $3$-adic cube, and then $(10/17)×(1/3^3)=10/459$ will be one also.