Proving Hensel's Lemma in a special case According to a proof sketched in nlab, the general form of Hensel's Lemma (concerning polynomials, not power series) can eventually be reduced to the following special case:

Let $R$ be a commutative ring, $I$ an ideal satisfying $I^2 = 0$, and let $f \in R[X]$ be a monic polynomial. Then any factorization $\overline{f} = gh$ into strongly coprime (*) monic polynomials $g,h \in (R/I)[X]$ can be lifted to a factorization $f = g'h'$ of $f$ into monic polynomials $g',h' \in R[X]$ with $\overline{g'} = g$ and $\overline{h'} = h$.

(*) Two polynomials are strongly coprime, if they generate the unit ideal (i.e. Bézout's identity holds).
I was trying to prove that statement, and I already managed to find polynomials $g'', h''$ lifting $g,h$ such that $f = g''h''$. However, I cannot guarantee that $g''$ and $h''$ are monic.
incomplete proof:

Let $g',h' \in R[X]$ be monic polynomials lifting $g,h$. Then $f-g'h' =: r \in I[X]$. Since $g$ and $h$ are strongly coprime and $I$ is nilpotent, $g'$ and $h'$ are strongly coprime as well, i.e. we have $1 = g's + h't$ for certain $s,t \in R[X]$. Now set $g'' := g' + tr$ and $h'' := h' + sr$. Then we have $\overline{g''} = g$, $\overline{h''} = h$, and
  $$ g''h'' = (g' + tr)(h' + sr) = g'h' + r(g's + h't) + str^2 = g'h'+r = f. $$

I realize that that many quantities in this proof are chosen fairly arbitrary. By choosing $g', h', s$, and $t$ more carefully, we get bounds on the degrees of $r,s$ and $t$. However, I wasn't able yet to get $\deg(tr) < \deg(g)$ and $\deg(sr) < \deg(h)$ which I need to ensure that $g''$ and $h''$ are monic.
Any ideas are greatly appreciated!
 A: I've looked up Bourbaki's proof. They use the following lemma, which I wasn't aware of:

Lemma: Let $R$ be a commutative ring and $g,h \in R[X]$ strongly coprime polynomials, where $g$ is monic. Then every polynomial $r \in R[X]$ may be written uniquely in the form
  $$ r = gs + ht, $$
  where $s,t \in R[X]$ with $\deg{t} < \deg{g}$.

Let's apply that lemma in the setting above.

As before, we choose $g',h' \in R[X]$ monic, lifting $g,h \in (R/I)[X]$. As $g$ and $h$ are strongly coprime, $g'$ and $h'$ are strongly coprime as well. We apply the lemma to get polynomials $s,t \in R[X]$ with $\deg(t) < \deg(g') = \deg(g)$ such that
  $$ f - g'h' = g's + h't. $$
  Reduction modulo $I$ gives
  $$ 0 = g \overline{s} + h \overline{t}. $$
  By the uniqueness part of the lemma we conclude $\overline{s} = \overline{t} = 0$, whence $s,t \in I[X]$. As before, it is easily verified that
  $$ (g'+t)(h'+s) = g'h' + g's + h't + st = f, $$
  so $g'' = g'+t$ and $h'' = h'+s$ are polynomials lifting $g,h$ whose product is $f$. But this time $g''$ is monic by construction (since $\deg(t) < \deg(g')$), and so $h''$ must be monic as well.

