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Henry Yuen
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Multivariate Hensel LiftingHensel's Lemma, but with only one polynomial

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Henry Yuen
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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 invertiblea 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!

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 invertible), 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!

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!

Source Link
Henry Yuen
  • 2k
  • 15
  • 23

Multivariate Hensel Lifting, but 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 invertible), 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!