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In Andrew Kobin's script on Algebraic Geometry I found on page 355 a comment I would like better understand. It states

Another way to view formal smoothness is as an abstraction of Hensel's Lemma.

Formal smoothness of a scheme $X \to S$ is characterized by the property that for every $S$-scheme $Y$ and every infinitesimal subscheme $Y_0 \subset Y$ (that is defined by a nilpotent ideal sheaf in $Y$) , the canonical morphism

$$Hom_S(Y,X) \to Hom_S(Y_0, X)$$

is surjective. Well, why can this property be regarded as abstraction of Hensels 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 that $m$ is nilpotent. In which sense can the above be regarded as an abstraction?

Another question on general properties of Henselian rings. I heard fleetingly (but forgot the concrete context) that schemes over henselian local rings have generally a rich divisor theory that on one hand somehow allows in certain way to "reduce" the analysis of divisors on $R$-scheme $X$ ($R$ local hensel with maximal ideal $m$ and residue field $\kappa=R/m$) to that on the special fiber $X \times_R \kappa$ in the sense that a "lot of information" of the theory of divisors on $X$ maybe already be extracted from the study of divisors on $X \times_R \kappa$ using formal techniques, by lifting results from $X \times_R R/m^i$ to $X \times_R R/m^{i+1}$ in much more "fruitful way" than if we work without Henselian context. On the other hand if that's true, then what I heard for example in the case of $X$'s surface there is much more "flexibility" in the constructions of divisors with desired intersection behaviour.

At least it seems that henselian assumption "guarantees" that much more information is conserved by passing to the special fiber as without this assumption. Could somebody give short insight into this correspondence principle and motivate how exactly henselianess enters fruitfully into the game? Or give a recomendable reference where these ideas are made precise and explaned in details?

Sorry if the formulation is too vague, I've only heard it once about this ideas, but unfortunately couldn't find anything about it and still quite curious what I can take from it.

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    $\begingroup$ Lifting to a complete local ring with maximal ideal $m$ is equivalent, by the definition of complete, to lifting to the ring mod $m^n$ for all $n$, and on those quotient rings, $m$ is nilpotent. $\endgroup$
    – Will Sawin
    Commented Mar 26, 2021 at 3:51
  • $\begingroup$ so here over henselian rings works a lot already in the category of schemes what in general works only for formal schemes? $\endgroup$
    – user267839
    Commented Mar 26, 2021 at 3:56
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    $\begingroup$ I'm not an expert on formal schemes. I would say that for $X \to S$ formally smooth, $S$-points of a complete local ring lift to $X$ if they lift modulo the maximal ideal, while if $X \to S$ is smooth, $S$-points of a henselian local ring lift to $X$ if they lift modulo the maximal ideal. So, at least for lifting points, henselian is as good as complete if your morphisms are really smooth and not just formally smooth. $\endgroup$
    – Will Sawin
    Commented Mar 26, 2021 at 4:06

1 Answer 1

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First about the ralation between henselian and formal smoothness property, I think a good idea is to look at what was the first version of Hensel lemma: it says that if $f(\bar{a})=0,f'(\bar{a})\, modp=0$ then there is a lift of $\bar{a}$ in $\mathbb{Z}_p$ such that $f(a)=0$. this is basically says that if you define a curve as $C=V(f)$ then $f'(a)\not = 0$ implies that $C$ has the formal smoothness property with respect to thickening of the form $F_p\to \mathbb{Z}_p/P^n$ over the point $\bar{a}$. so Hensel lemma for complete ring is a consequence of formal smoothness.

but why we define the Henselian rings and don't only work with complete rings? I think the answer is the important concept of Henselisation: if you have a local ring $(O,m)$ you can consider its completion $(\bar O,\bar{m})$ or you can consider its Henselisation $O^{h}$. there are two ways to define $O^h$ you can define it as the smallest Henselian extension of $(O,m)$ inside $\bar{O}$ or as the limit of all etale extension $O'$ of $O$ with an isomorphism $O/m\to O'/m'$. you can also consider the strict Henselisation $O^{sh}$ as the limit of all etale extensions.

there are two reasons why we are interested in $O^h,O^{sh}$ instead of $\bar{O}$. the first is that in non-Notherian setting $\bar{O}$ is not faithfully flat over $O$ but $O^h,O^{sh}$ are almost by definition always faithfully flat. the second reason is that $O_x^{sh}$ works in etale topology like $O_x$ in Zariski topology.

but how we study Henselian rings? the Henselian property itself is very powerful and you can deduce a lot of things as a consequence of that property but there is another important tool: Artin approximation property and its consequences. First, let us look at the simplest case: the Hensilsation of $\mathbb{C}[x]$ at $x$ consists of algebraic power series: power series the satisfy a polynomial equation then it is not hard to deduce from Henselian property that if you have a system of equations $f^1=0,...f^r=0$ with invertible jacobian then any solution $y$ of this system in $\mathbb{C}[[x]]$ and any constant $c$ there is a solution $y'$ in $\mathbb C[[x]]^{alg}$ such that $y=y' mod x^c$. in summary, you can approximate the solution in $\bar{O}$ with the solutions in $O^h$.

now back to your question let $A$ be a henselian ring, there is general version of Artin theorem: consider any "finitely presented" functor from the $A-algebras$ to $Sets$ for example the functor that sends $B$ to $Div(X_B)$. then Artin theorem says that for each $y\in F(\bar{A})$ and constant $c$ there is a $y'\in F(A)$ such that $y=y' mod\, m^c$. This is important because the are powefull tools in deformation theory to relate $F(\bar{A})$ to $F(\bar{A}/m)=F(A/m)$. in the case of divisors $Div(X_\bar{A})=Div(X_{A/m})$ so you get the desired relation between $Div(X)$ and $Div(X_k)$ in your language.

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  • $\begingroup$ Thank you for your answer. This one is exactly was looking for! Two questions on your notations: In your first part you wrote $ F_p \to W_n(F_p)$ where as I know $W_n$ means the Witt ring wrt $F_p$, an in this case that's $\mathbb{Z}_p$. Was the choice of Witt-notation intended to indicate in which direction the notion of formal smoothness generalizes the usual Henselian property in your example with curve $C= V(f)$ to $A \to W_n(A)$ where $A$ is is an arbitrary ring? $\endgroup$
    – user267839
    Commented Mar 29, 2021 at 19:58
  • $\begingroup$ Secondly: what do you mean by $\mathbb C[[x]]^{alg}$? I know this notation only for fields. $\endgroup$
    – user267839
    Commented Mar 29, 2021 at 19:58
  • $\begingroup$ @Rachmaninow98 about your first question no I really meant $Z_p/p^n$ I just use witt vector as the force of habit. if you have a complete local ring $A$ with residue field $k$ then of course you should use $A/m^n$ not $W_n(k)$. $\endgroup$
    – ali
    Commented Mar 29, 2021 at 19:59
  • $\begingroup$ for your second question I Just mean all the power series that satisfy a polynomial equation over $\mathbb{C}$ which is as I said the Henselisation of $(C[X],X)$ $\endgroup$
    – ali
    Commented Mar 29, 2021 at 20:01
  • $\begingroup$ so in general there is no direct connection between formal smoothness and the concept of Witt vectors? $\endgroup$
    – user267839
    Commented Mar 29, 2021 at 20:05

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