# Multivariate Hensel's Lemma, but with only one polynomial

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!

• Without non-degeneracy condition, the answer is no. Consider $R=\mathbb C[t]$, $Q(X)=X^2-t$ (in one variable), and $I=tR$. There is no lifting mod $I^2$ of the zero solution mod $I$. However, there is a weaker condition (non-degeneracy when tensoring with the total ring of fractions of $R$) which insures that there is a constant $c$ such that for any $d>0$, any solution mod $I^d$ is congruent mod $I^{d-c}$ to a solution in $R$ (see Renée Elkik's paper in Ann. Sci. ENS in the 70's). – Qing Liu Oct 3 '12 at 7:28

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 $Q(X_1+a_1,\dots, X_n+a_n) \in I^{t+1}$.

• I have a quick clarification question: When you say "Suppose $Q(X_1,\ldots,X_n)$ is in $I^t$", you mean suppose there is are $\beta_1,\ldots,\beta_n\in R$ such that $Q(\beta_1,\ldots,\beta_n)\in I^t$? And then every time you write $\frac{dQ}{dX_1}$, you mean it is evaluated at the $\beta_1,\ldots,\beta_n$? If that's the case, then I interpret the mod $I^{t+1}$ solution as $\beta_1 + a_1,\ldots,\beta_n+a_n$. Thanks! – Henry Yuen Oct 8 '12 at 18:04
• yeah I was switching from using $X_i$ as formal variable to actual variables which take some specific value. Sorry for the confusion! – Will Sawin Oct 8 '12 at 18:28
• I want to stress that in several variables case the lift isn't unique anymore. – Anna Abasheva Sep 5 '18 at 13:35