One version of Hensel's Lemma is the following statement: 


Let $R$ be a commutative ring with a unit. Given a polynomial $Q\in R[X]$ and a root $\alpha$ of $Q$ modulo some ideal $I$ (i.e. $Q(\alpha) \in I$), assuming some non-degeneracy conditions (e.g. $Q$ is square-free), then for every $t > 1$, there exists $\beta\_t \in R$ such that $\beta\_t = \alpha \mod I$, and $Q(\beta\_t) \in I^t$, and furthermore, $\beta\_t$ is unique. 


The multidimensional generalization of Hensel's Lemma is often presented as:


Given $f\_1,\ldots,f\_n$ in $R[X\_1,\ldots,X\_n]$ and a simultaneous root $\alpha \in R^n$ modulo an ideal $I \subset R$ (i.e. $f\_i(\alpha) \in I$ for all $i$), assuming some non-degeneracy conditions (e.g. $\det J(\alpha)$ is a unit), there exists $\beta\_t \in R^n$ such that $\beta\_{t,j} = \alpha\_j \mod I$ for all $j$, and $f\_i(\beta\_t) \in I^t$ for all $i$.


Here, $J(\alpha)$ denotes the evaluation of the Jacobian of $f\_1,\ldots,f\_n$ on $\alpha$. 


My question is: is there an intermediate generalization in between the univariate case and the multivariate case above where we only consider one polynomial $Q\in R[X\_1,\ldots,X\_n]$, and we simply want to lift roots of $Q$ modulo an ideal $I$ to roots of $Q$ modulo $I^t$? It seems intuitively like an easier thing to do (we don't require simultaneous solutions to a system of polynomial equations). Does this intermediate generalization exist, and if so, what non-degeneracy conditions would we require?


Thank you!



Multivariate Hensel's Lemma, but with only one polynomial
Yes. If the ideal generated by $(I,dQ/dX\_1,..,dQ/dX\_n)$ is the unit ideal at some mod-$I$ solution of $Q$, one can lift to a mod $I^t$ solution of $Q$.


Proof: It is clear that we merely need to check the induction step, that if we have a mod $I^t$ solution we can get a mod $I^{t+1}$ solution. Suppose $Q(X\_1,..,X\_n)$ is in $I^t$. Then, for $a\_1,...,a\_n \in I^t$


$Q(X\_1+a\_1,...,X\_n+a\_n)=Q(X\_1,...,X\_n)+a\_1 \frac{dQ}{dX\_1} + .... + a\_n \frac{dQ}{dX\_n} + \epsilon$


where $\epsilon \in I^{t+1}$.


Choose $a\_1,\dots, a\_n$ such that


$a\_1 \frac{dQ}{dX\_1}+ \dots + a\_n \frac{d Q}{d X\_n}=- Q(X\_1,...,X\_n)$ mod $I^{t+1}$.


This is always possible since we can choose the sum to equal any element in:


$I^t \frac{dQ}{dx\_1} + \dots + I^t \frac{dQ}{dx\_n} + I^{t+1} = I^t\left(\frac{dQ}{dx\_1} + \dots + \frac{dQ}{dx\_n} + I\right)=I^t (1)=I^t$.


Then $Q(X\_1+a\_1,\dots, X\_n+a\_n) \in I^{t+1}$.


