Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

An important component of algorithms for factoring multivariate polynomials over a commutative ring $R$ is Hensel lifting. Here's a brief, concrete example to set the stage for my question:

Let $f \in F[X,Y]$, where $F$ is an algebraically closed field. Suppose that for some $b\in F$, $f(b,Y)$ is square-free; call such points base points. Let $\alpha^{(b)}_1,\ldots,\alpha^{(b)}_n$ be roots of the univariate polynomial $f(b,Y)$, or equivalently roots of $f(X,Y)$ modulo the ideal $(X - b)$.

Hensel lifting gives an algorithmic way to lift the roots $\alpha^{(b)}_i$ of $f$ modulo $(X - b)$ to roots $g^{(b)}_i$ modulo $(X-b)^t$ for any $t$, where we have the property that $g^{(b)}_i = \alpha^{(b)}_i \mod (X-b)$. Call these roots Hensel roots.

(In a polynomial factorization application, there would be a way to take the $g^{(b)}_i$'s and convert them to actual factors of $f$).

Here's my question: given two base points $b$ and $b'$ of $f$, we've performed the Hensel lifting procedure to obtain the Hensel factors $\{ g^{(b)}_i \}$ and $\{ g^{(b')}_i \}$ (lifted to the same level). Since $F$ is algebraically closed, there are the same numbers of Hensel factors over $b$ as there are over $b'$.

Is there any relation between $g^{(b)}_i$ and $g^{(b')}_i$ that's meaningful? Can we say that $g^{(b)}_i(X - b) = g^{(b')}_i(X - b')$? That is, are Hensel roots preserved across change of base points?

Thank you!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Hmm, let's try an example: $f(X,Y) = X - 2 Y + Y^2$ over $\mathbb C$. For any $b \ne 1$ the roots of $f(b,Y)$ are $\alpha_i^{(b)} = 1 \pm \sqrt{1-b}$. These lift one level to $g_i^{(b)} = -1 \pm \sqrt{1-b} \mp \dfrac{X-b}{2 \sqrt{1-b}}$. They certainly depend on $b$, so it's not true that $g_i^{(b)}(X-b) = g_i^{(b')}(X-b')$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.