Consider $Q_3$ field which is rational number $Q$ completed at $p=3$.(In other words, $Q_3$ is $3-$adic rational numbers.) Let $f=x^3-17\in Q_3[x]$. Let $O$ be the complete DVR associated to $Q_3$.
Clearly $f\in O[x]$. From Gauss lemma by $O$ PID, I see that if $f=gh,g,h\in Q_3[x]$,then $g,h\in O[x]$. Therefore, I can perform $3-$reduction by considering $\frac{O}{3}[x]$.
Now $\bar{f}\in Z_3[x]$ has $x^3+1=(x+1)^3\in Z_3[x]$. So I cannot apply hensel lemma here to determine factor of $f$ in $Q_3[x]$. The book says it has a degree 2 irreducible factor in $Q_3[x]$.
The other version of Hensel uses $f'(x)=3x^2$. And $x=-1$ yields $3^2\vert f(-1)$ and $3^2\not\vert f'(x)$ but $2=1+1$. The absolute value requires $|f(-1)|<|f'(-1)|^2$ where $|x|$ is the $\frac{1}{3^{v_3(x)}}$ and $v_3(x)$ is $3-$adic valuation of $x\in Q_3$. Both sides yields $\frac{1}{9}$ for absolute value. So I cannot apply this hensel lifting.
$\textbf{Q:}$ Can someone kindly provide hints to methods to determine reducibility for local fields? If I am lucky, I can proceed by hensel lemma but it requires factors after prime reduction being coprime. Have I done something wrong above?
Here’s a method of using Strong Hensel, which you seem to know:
The cubic $X^3-17=f(X)$ has a root if and only if $f(X-1)=X^3-3X^2+3X-18$ has a root. Three-adically, the Newton polygon has a vertex at $(1,1)$, which already tells you that there’s a factorization.
If you’re not comfortable with Newton , go one step further, and substitute $3X$ for $X$, to get $f(3X-1)=27X^3-27X^2+9X-18=9\left(3X^3-3X^2+X-2\right)$. Modulo $3$, we get $3X^3-3X^2+X-2\equiv X-2\pmod3$, and this has the nice factorization $(X-2)\cdot1$ into two relatively prime factors. You can lift one of them to characteristic zero, by Strong Hensel, preserving its degree. Of course you choose the $X-1$. And there you have it.
In case you’re nervous about using $1$ as one of the factors, let me reassure you. This freedom comes in handy surprisingly often.
EDIT: Addendum
You asked about the factorization of $f(X)=X^3-17$ modulo $3$ and why that didn’t tell the splitting of $(3)$ in $\Bbb Q(w)$, where $\text{Irr}(w,\Bbb Q[X])=X^3-17=f(X)$. The theorem should read that the splitting of the irreducible modulo $p$ tells the splitting of $(p)$ only when the root $r$ generates the integers of $\Bbb Q(r)$ over $\Bbb Z$. And of course $r=w$ doesn’t do that.