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He says that this result is today called Hensel's Lemma and that Hensel's standard proof applies. … It was my initial thought that the Hensel-Kurschak Lemma would follow easily from one of the more standard forms of Hensel's Lemma. …

For such $G$ Grothendieck's version of Hensel's Lemma holds (EGA IV 4: 18.5.19): $$\{\text{connected components of $G$}\}= \{\text{connected components of $G_k$}\}.$$ I'm familar with some "conventional … I'm really curious if there exist a proof that is more "geometric" proof in the sense that it uses Grothendieck's version of Hensel's Lemma to show the existence & exactness of sequence above. …

I've been reading some papers on Igusa zeta functions, and they seem to be implicitly using a "quantitative version" of Hensel's Lemma, which also asserts the number of lifts of a $\mathbb{Z}/p\mathbb{ …

I have two questions regarding Hensel's lemma. 1) Is there a known statement for Hensel's lemma for situation like this where we consider more than one polynomial? … 2) How does one deduce it for $f_1$ and $f_2$ as above from the classical Hensel's lemma? Any comments, hints, references are greatly appreciated! Thank you very much! …

The factorization of $5$ lifts to $\mathbb{Z}[x]/((x^2+1)^2)\mathbb{Z}[x]$, but it isn't a simple consequence of Hensel's lemma. … The trouble with using Hensel's lemma is that it requires a form of Bezout's identity, which does not generally hold over $\mathbb{Z}[x]$. …

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. … 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. …

I am thinking about the simplest version of Hensel's lemma. Fix a prime $p$. Let $f(x)\in \mathbf{Z}[x]$ be a polynomial. … Is there a proof of Hensel's lemma along this line? I know this is like using a big machine to solve a simple problem. …

The following form of Hensel's Lemma in Algebraic Geometry is well-documented in the literature: $\textbf{Theorem 1}$: Let $R$ be an Henselian local ring with maximal ideal $\mathfrak{m}$, and let …

This is further reduced by Hensel's Lemma to checking whether the curve has a point in some finite ring $R_v/ \mathcal{M}_v^e$ for some easily computable integer $e$. … Now, as far as I know and understand, Hensel's Lemma says that for a polynomial $f(x) \in R_v[x]$, if it has a root in $R_v/\mathcal{M}$, then it lifts to a unique root in $R_v$ and hence $K_v$. …

Let $k$ be a $p$-adic field with integer ring $\mathcal{O}_k \subseteq k$, maximal ideal $m_k \subseteq \mathcal{O}_k$ and residue field $\mathbb{F}_q = \mathcal{O}_k/m_k$. Let $X$ be a smooth, geomet …

If $X$ is smooth we conclude by Hensel lemma. If $k=A/\mathfrak{m}$ is finite then it's true because inverse limit of non-empty finite sets is non-empty. … If $A$ is a DVR then we know from a general Hensel lemma from "RATIONAL POINTS IN HENSELIAN DISCRETE VALUATION RINGS" by MARVIN J. GREENBERG. …

Now assume $X(O_K/m_K^{n+1}) \rightarrow X(O_K/m_K^n)$ is surjective with same size fibers for all $n$ and all $K$ (the strongest Hensel lemma), then is there a counterexample that $X \rightarrow \mathbb …

And especially of ones that can be applied to polynomials like $XY-(X+Y)(X^2+Y^2)$ (I found it reducible by using Hensel's lemma after a suitable change of variable, if I did not make mistake). …

It states Another way to view formal smoothness is as an abstraction of Hensel's Lemma. … The Hensel's lemma I am familar with on lifting polynomials under certain conditions from $(R/m)[X]$ to $R[X]$, where $A$ is local complete with maximal ideal $m$, that doesn't involve any assumptions …

If yes, I guess this would somehow follow from Hensel's lemma, but I'm not sure how exactly. …