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Definition. A locally finitely presented morphism of schemes $f\colon X\to Y$ is smooth (resp. unramified, resp. étale) if for any affine scheme $T$, any closed subscheme $T_0$ defined by a square zero ideal $I$, and any morphisms $T_0\to X$ and $T\to Y$ making the following diagram commute

    g
T0 --> X
|      |
|      |f
v      v
T ---> Y

there exists (resp. exists at most one, resp. exists exactly one) morphism $T\to X$ which fills the diagram in so that it still commutes.

For checking that $f$ is unramified or étale, it doesn't matter that I required $T$ to be affine. The reason is that for an arbitrary $T$, I can cover $T$ by affines, check if there exists (a unique) morphism on each affine, and then "glue the result". If there's at most one morphism locally, then there's at most one globally. If there's a unique morphism locally, then there's a unique morphism globally (uniqueness allows you to glue on overlaps).

But for checking that $f$ is smooth, it's really important to require $T$ to be affine in the definition, because it could be that there exist morphisms $T\to X$ locally on $T$, but it's impossible to find these local morphisms in such a way that they glue to give a global morphism.

Question: What is an example of a smooth morphism $f\colon X\to Y$, a square zero nilpotent thickening $T_0\subseteq T$ and a commutative square as above so that there does not exist a morphism $T\to X$ filling in the diagram?

I'm sure I worked out such an example with somebody years ago, but I can't seem to reproduce it now (and maybe it was wrong). One thing that may be worth noting is that the set of such filling morphisms $T\to X$, if it is non-empty, is a torsor under $Hom_{\mathcal O_{T_0}}(g^*\Omega_{X/Y},I)=\Gamma(T_0,g^*\mathcal T_{X/Y}\otimes I)$, where $\mathcal T_{X/Y}$ is the relative tangent bundle. So the obstruction to finding such a lift will represent an element of $H^1(T_0,g^*\mathcal T_{X/Y}\otimes I)$ (you can see this with Čech cocycles if you want). So in any example, this group will have to be non-zero.

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Is it the same if you substitute "Artinian local ring" for "affine scheme" in the definitions? I would expect smoothness, unramifiedness, and étaleness to be local properties... –  Qfwfq Apr 21 '10 at 6:16
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unknown(google) this is trivial for formally unramified morphisms, a hard theorem for formally étale morphisms, and an open problem for formally smooth morphisms. This is why we require it for all rings in the finitely presented case, because we want étale (resp. smooth, resp. unramified) morphisms to be finitely presented and formally smooth (resp. étale, resp. unramified). You shouldn't build reductions into the definition when they only hold in special cases. –  Harry Gindi Apr 21 '10 at 20:09
    
See my question on formally étale morphisms (and the answer from Gabber that I posted). –  Harry Gindi Apr 21 '10 at 20:13
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@unknown: Yes, you get the same notions if you just use Artinian local rings; see EGA IV, section 17.4 (particularly Proposition 17.4.2). This is actually much more local than I first expected these properties to be. It wouldn't be unreasonable to expect that you can only reduce down to the case where you replace "affine scheme" by "spectrum of a local ring". –  Anton Geraschenko Apr 22 '10 at 3:31
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By the way, I was implying that yes, you can make the substitution that you noted in the finitely presented case, but not in the formal case, so my comment is consistent with Anton's comment. I was remarking that the definition is stated the way it is because it is a special case of the formal case. –  Harry Gindi Apr 22 '10 at 20:13

4 Answers 4

up vote 22 down vote accepted

Using some of BCnrd's ideas together with a different construction, I'll give a positive answer to Kevin Buzzard's stronger question; i.e., there is a counterexample for any non-etale smooth morphism.

Call a morphism $X \to Y$ wicked smooth if it is locally of finite presentation and for every (square-zero) nilpotent thickening $T_0 \subseteq T$ of $Y$-schemes, every $Y$-morphism $T_0 \to X$ lifts to a $Y$-morphism $T \to X$.

Theorem: A morphism is wicked smooth if and only if it is etale.

Proof: Anton already explained why etale implies wicked smooth.

Now suppose that $X \to Y$ is wicked smooth. In particular, $X \to Y$ is smooth, so it remains to show that the geometric fibers are $0$-dimensional. Wicked smooth morphisms are preserved by base change, so by base extending by each $y \colon \operatorname{Spec} k \to Y$ with $k$ an algebraically closed field, we reduce to the case $Y=\operatorname{Spec} k$. Moreover, we may replace $X$ by an open subscheme to assume that $X$ is etale over $\mathbb{A}^n_k$ for some $n \ge 0$.

Fix a projective variety $P$ and a surjection $\mathcal{F} \to \mathcal{G}$ of coherent sheaves on $P$ such that some $g \in \Gamma(P,\mathcal{G})$ is not in the image of $\Gamma(P,\mathcal{F})$. (For instance, take $P = \mathbb{P}^1$, let $\mathcal{F} = \mathcal{O}_P$, and let $\mathcal{G}$ be the quotient corresponding to a subscheme consisting of two $k$-points.) Make $\mathcal{O}_P \oplus \mathcal{F}$ an $\mathcal{O}_P$-algebra by declaring that $\mathcal{F} \cdot \mathcal{F} = 0$, and let $T = \operatorname{\bf Spec}(\mathcal{O}_P \oplus \mathcal{F})$. Similarly, define $T_0 = \operatorname{\bf Spec}(\mathcal{O}_P \oplus \mathcal{G})$, which is a closed subscheme of $T$ defined by a nilpotent ideal sheaf. We then may view $g = 0+g \in \Gamma(P,\mathcal{O}_P \oplus \mathcal{G}) = \Gamma(T_0,\mathcal{O}_{T_0})$.

Choose $x \in X(k)$; without loss of generality its image in $\mathbb{A}^n(k)$ is the origin. Using the infinitesimal lifting property for the etale morphism $X \to \mathbb{A}^n$ and the nilpotent thickening $P \subseteq T_0$, we lift the point $(g,g,\ldots,g) \in \mathcal{A}^n(T_0)$ mapping to $(0,0,\ldots,0) \in \mathbb{A}^n(P)$ to some $x_0 \in X(T_0)$ mapping to $x \in X(k) \subseteq X(P)$. By wicked smoothness, $x_0$ lifts to some $x_T \in X(T)$. The image of $x_T$ in $\mathbb{A}^n(T)$ lifts $(g,g,\ldots,g)$, so each coordinate of $x_T$ is a global section of $\mathcal{F}$ mapping to $g$, which is a contradiction unless $n=0$. Thus $X \to Y$ is etale.

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Wicked :-) I think that's a pretty definitive answer to Anton's question! –  Kevin Buzzard Apr 21 '10 at 19:48
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Bjorn, that's great. I'd been trying to make something like that work (to ultimately reduce to something with affine space in dim > 0), but I kept getting stuck. Then I gave up and just posted the partial idea as comment to your earlier answer. Good to see this kind of strategy works in the end, even simpler than the P^1 case. Wicked, indeed. –  BCnrd Apr 22 '10 at 3:26

Let $Y=\operatorname{Spec} k$ and let $X=\mathbb{P}^1_k$, viewed as $\operatorname{Spec} k[t]$ glued to $\operatorname{Spec} k[t^{-1}]$. Let $T = \operatorname{\bf Spec}(\mathcal{O}_X + \mathcal{O}_X(-2)\epsilon + \mathcal{O}_X(-4) \epsilon^2)$ where $\epsilon^3=0$, so $T$ is $$\operatorname{Spec}(k[t] + k[t]\epsilon + k[t]\epsilon^2)$$ glued to $$\operatorname{Spec}(k[t^{-1}] + t^{-2} k[t^{-1}]\epsilon + t^{-4} k[t^{-1}]\epsilon^2).$$ Let $I$ be the ideal sheaf of $\mathcal{O}_T$ generated by $\epsilon^2$, and let $T_0$ be the associated subscheme. Consider the $k$-morphism $T_0 \to X$ given by $$t \mapsto t + \epsilon$$ $$t^{-1} \mapsto t^{-1} - t^{-2} \epsilon.$$ (Check that this is well-defined, i.e., that $(t+\epsilon)(t^{-1} - t^{-2} \epsilon) = 1$ in $k[t,t^{-1}][\epsilon]/(\epsilon^2)$.) A lift of this to a morphism $T \to X$ has the form $$t \mapsto t + \epsilon + f(t) \epsilon^2$$ $$t^{-1} \mapsto t^{-1} - t^{-2} \epsilon + t^{-4} g(t^{-1}) \epsilon^2$$ for some polynomials $f$ and $g$, but the compatibility condition is now $$t^{-3} g(t^{-1}) - t^{-2} + t^{-1} f(t) = 0,$$ which has no solution.

Note: If we replaced $\mathcal{O}(-4)$ with $\mathcal{O}(-3)$, then there would be a lifting, given by the Taylor polynomials of $t$ and $t^{-1}$, i.e., $$t \mapsto t + \epsilon$$ $$t^{-1} \mapsto \frac{1}{t+\epsilon} = t^{-1} - t^{-2} \epsilon + t^{-3} \epsilon^2.$$

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Bjorn: your example seems to beg the question: if $X\to Y$ is any smooth but not etale morphism, does there always exist $T_0$ and $T$ as above such that the map doesn't lift?! –  Kevin Buzzard Apr 21 '10 at 9:10
    
Kevin, if we pick a point $y \in Y$ such that $X_y$ has positive dimension, we can pass to that fiber to reduce to the case $Y = {\rm{Spec}}(k)$ for a field $k$. Can work locally on $X$, so $X$ affine. The alg. closure of $k$ in $X$ is finite sepble, so can replace $k$ with that so $X$ geom. connected. By Bertini (and Bjorn's version for finite $k$), can then slice $X$ down to a geom. connected smooth affine curve. So Bjorn's case is not as far from the most general case as it may have initially seemed. –  BCnrd Apr 21 '10 at 16:08
    
I added an argument for Kevin's general version as a separate answer. –  Bjorn Poonen Apr 21 '10 at 21:23

The kind of example that comes easily to mind is where $X=L$ is a line bundle over $Y$, a smooth projective variety over, say, $\mathbb{Z}_p$. We take $T=Y$ and $T_0=Y_0$, the special fiber of $Y$. Then $L$ can have plenty of sections over $Y_0$ that refuse to lift to $Y$. Note that this implies what you want, since if the sections could be lifted repeatedly over square-zero ideals, then they could be lifted all the way to $Y$. (By formal GAGA, if you want.) That is, replace the original $Y_0$ by $Y\otimes \mathbb{Z}/p^n$ for larger $n$.

One place you can see this spelled out with $L$ the tensor powers of the canonical bundle $\omega_{Y/\mathbb{Z}_p}$ ('jump of plurigenera') is a paper by Junecue Suh:

Compos. Math. 144 (2008), no. 5, 1214–1226. (Unfortunately, I have no link that doesn't require a log-in.)

I think he constructs examples where the jump is arbitrarily large even for Shimura surfaces.

This example is probably overly pathological, and I suspect you can construct more commonplace $L$.

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Suppose that $X \to Y$ is a smooth map of varieties, and denote by $Y'$ the relative spectrum of $\mathcal O_Y[t]/(t^2)$. The liftings of $X \to Y$ to $Y'$ are parametrized by $\mathrm H^1(X, \mathrm T_{X/Y})$. Now suppose that $X' \to Y'$ is a lifting $s\colon Y \to X$ is a section; it is easy to see that the obstruction to lifting $s$ to a section $Y' \to X'$ is the image of the element of $\mathrm H^1(X, \mathrm T_X/Y)$ into $\mathrm H^1(Y, \mathrm s^*T_{X/Y})$; so to give an example where the section doesn't lift it is enough to give examples in which the map $\mathrm H^1(X, \mathrm T_{X/Y}) \to \mathrm H^1(Y, \mathrm s^*T_X/Y)$ is not 0. This is easy; for example, one can take $L$ to be a line bundle on $Y$ with $\mathrm H^1(Y,L) \neq 0$, and $f\colon X \to Y$ to be the total space of $L$. In this case what is happening is that $X' \to Y'$ is a non-trivial $L$-torsor, so the trivial section $Y \to L$ does not lift.

Another type of example is of the type suggested by Minhyong. Take $Y$ to be a projective variety over $k$ with $\mathrm H^0(Y, \mathcal O) = k$ and $\mathrm H^1(Y, \mathcal O) \neq 0$; let $Y'$ be as before. There exists a non-trivial line bundle $L'$ on $Y'$ whose restriction to $Y$ is $\mathcal O_Y$; then the only section of $\mathcal O_Y$ that lifts is the zero section.

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