An equation is solvable in $\mathbb Z_p$

Question

Suppose $p$ is an odd prime and $a\in \mathbb Z_p$ with $a\equiv 1 \mod p^2 $. Then $x^p=a$ is solvable over $\mathbb Z_p$.

I want to prove $x^p\equiv a\mod p^v$ is solvable for arbitrary $v\geqslant 1$. It's easy when $v\leqslant 2$. But I don't know how to proceed it.

Answer 1

I think you probably have to use induction somehow to prove the existence of a solution, which is essentially what is used to prove Hensel's Lemma.

Suppose that $a=1+p^2k$, for some $k\in\mathbb{Z}_p$, and also that $x=x_0+py$, where $x_0\in \{0,1,\ldots, p-1\}$ and $y\in \mathbb{Z}_p$. If $x^p=a$, then $(x_0+py)^p=1+p^2y$, that is $$ x_0^p + {p \choose 1} (py)x_0^{p-1} + {p \choose 2} (py)^2x_0^{p-2} + \cdots + {p \choose p} (py)^p = 1+p^2k.$$ So $x_0=1$, and $$ {p \choose 1} py + {p \choose 2} (py)^2 + \cdots + {p \choose p} (py)^p = p^2k.$$ Dividing both sides by $p^2$, we get $$ y+ {p \choose 2} y^2+ \cdots + {p \choose p} p^{p-2}y^p = k. \tag{1}$$ Since $p$ is odd, for every integer $m$ satisfying $2\leq m\leq p-1$, ${ p\choose m}$ is divisible by $p$. This is because ${p \choose m}=\frac{p!}{m!(p-m)!}$ is an integer and the denominator is not divisible by $p$ but the numerator is. So if $p$ is odd, then Equation $(1)$ becomes $\bar{y}=\bar{k}$ in the residue field (if $p=2$, then we would get $\bar{y}+\bar{y}^2=\bar{k}$ which might not have a solution, e.g. $\bar{k}=\bar{1}$). Clearly now one can apply Hensel's Lemma (or the idea of its proof) to $(1)$ to show the existence of $y\in \mathbb{Z}_p$, whence the existence of $x\in \mathbb{Z}_p$.

Answer 2

Perhaps it’s just a personal quirk of mine, but I think that for finding $p$-th roots, Hensel is not the best tool.

Rather, I recommend the Binomial expansion of $(1+p^2t)^{1/p}$, where you’ll be substituting for $t$ any $p$-adic integer. Keep an accurate count of how many $p$’s there are in the denominators as you go along, and you’ll see that you get very good convergence of the series. What complication there is arises from the factorials in the denominators, and you can see how to know how divisible these are by $p$ from various sources such as this.