I want to determine the number of roots of $f(X) = X^3-5X+20$ in $\mathbb{Z}_p$ using Hensel's lemma (lemma is on the bottom). Unfortunately I am not very well trained to solve this. Take for example $p=2$. Then reducing the polynomial modulo 2 we get
\begin{align*}
f(X) &= X^3 – 5X+20 \equiv 0 \mod 2 \\
f'(X) &= 3X^2-5\equiv X + 1 \mod 2
\end{align*}
That would mean that only $0\in\mathbb{Z}/2\mathbb{Z}$ gives a solution such that $f(0) \equiv 0$ and $f'(0)\not \equiv 0$ modulo 2, which then by Hensel's lemma gives a unique root of $f$ in $\mathbb{Z}_2$, which is $0$ modulo $2$.
The problem is that I don't know the relation between finding a root via Hensel's lemma and there really being one. I have found one root via Hensel's lemma, but what guarantees me there aren't more?
Any help on how to proceed is most welcome. Thank you in advance.
The lemma I am using:
Let $K$ be a complete field with discrete valuation $v$, valuation ring $O$ and residue field $k$. Let $f(X)\in O(X)$. Suppose that the reduction $\bar{f}(X)\in k[X]$ has a simple root $\bar{a}$, ie $\bar{f}(\bar{a}) = 0$ and $\bar{f}'(\bar{a})\neq0$. Then there exists a unique $x\in O$ such that $f(x) = 0$ and $\bar{x}=\bar{a}$.
Best Answer
This polynomial has a double root at $1$ in $\mathbb{F}_2$, so the out of the box Hensel's lemma won't apply. However we can still apply a modified version by going one power of two up. In particular the double root at 1 splits into two simple roots mod 4 (at 1 and 3). Now a modified Hensel's lemma (with essentially the same proof as the usual version) gives us that these each have a unique 2-adic lift.
One thing that is different about this modified Hensel's lemma from what you are used to is that if we want to solve $f(x) \equiv 0 \ \mod 2^n$, the lifts we get will only be unique $\mod 2^{n-1}$. In particular we get that $f(x) \equiv 0 \ \mod 2^n$ has 5 solutions for all $n > 2$.