Please express the first 3 7-adic digits of a root of $x^3-1$ in $\mathbb{Z}_7$ other than 1.

Question

Please express the first 3 p-adic digits of a root of $x^3-1$ in $\mathbb{Z}_7$ other than 1.

Does this just mean find the 7-adic expansion of 2 or 4? Wouldn't their expansions just be 2 and 4?

Answer 1

$2$ is a cube root of unity in $\mathbb{Z}/7\mathbb{Z}$, but not in $\mathbb{Z}_7$, since after all $2^3-1=7\neq 0$ in $\mathbb{Z}_7$. However, the solution $2$ in $\mathbb{Z}/7\mathbb{Z}$ can be lifted to a solution in $\mathbb{Z}_7$ using Hensel's lemma. Most proofs of Hensel's lemma are fairly constructive, so by working through the proof you should be able to find the first three $7$-adic digits of the cube root.

Answer 2

Since you happen to be talking about a root of unity of order dividing $p-1$, there’s a slick way of finding your number. Take the sequence $2,2^7,(2^7)^7=2^{49},\cdots$, and it converges $7$-adically to your root of unity. You don’t need to do your computation in characteristic zero, you can do all your computations modulo $7^3=343$. Don’t forget the trick of repeated squaring, according to which $n^{49}=(n^2\cdot n)^{16}\cdot n$, involving only seven multiplications, if I counted correctly.