Lifting to Characteristic 0 not over W

I thought of this several months ago and forgot about it. Now I rethought of it again and I just can't find it anywhere in the literature, so I'll ask here.

Is it known whether or not there exists a (smooth, proper, ...) variety over a field $k$ (perfect? alg. closed?) of positive characteristic that lifts to characteristic $0$ over some ramified extension of $W(k)$ and also lifts to $W_2(k)$, but does not lift over $W(k)$ itself?

In other words, if it lifts to characteristic $0$ (in a way related to $W(k)$) and it lifts to $W_2(k)$ must it lift via $W(k)$ itself?

I have looked at examples that lift to $W_2(k)$ but not to $W_3(k)$ and hence not over $W(k)$, but they don't seem to lift to char $0$ at all. I've also looked examples that lift over a ramified extension but not $W$, but these can easily be shown to not lift to $W_2$ as well.

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Does it even lift? –  meij Dec 30 '12 at 17:41
I couldn't bring myself to give an actual upvote (to meij's comment, when it was an answer), but a moral +1 is definitely well deserved! –  Jacob Bell Dec 30 '12 at 18:06
Am I missing some joke (based on Jacob's comment)...? –  Matt Dec 30 '12 at 20:45
Ah, thanks. That's pretty good I guess. –  Matt Dec 31 '12 at 15:04

Apply this theorem to the "germ" of $\text{Spec}(\mathbb{Z}[x]/\langle p(x^2-p) \rangle)$ over $\text{Spec}(\mathbb{Z})$.
Sorry. I think I'm being dumb. This paper has always confused me (I looked at it when first trying to find this example to find examples that lift to $W_2$ but not $W_3$). Are you saying the example is some component of the Hilbert scheme of nonsingular curves? Or are you saying that we create it from starting over $\text{Spec}(\mathbb{Z})$? Anything that starts over $\text{Spec}(\mathbb{Z})$ shouldn't work (but I think I'm not understanding) because base change to $W$ should give you the lift to $W$. Four people upvoted this immediately, so it is probably me missing the obvious. –  Matt Dec 30 '12 at 21:33
By "lift", I assume you want a proper, flat morphism to your base scheme. If you do not, then obviously every scheme over $\text{Spec}\mathbb{Z}/p\mathbb{Z}$ "lifts" to $\text{Spec}\mathbb{Z}_p$ by just composing the projection with the closed immersion $\text{Spec}(\mathbb{Z}/p\mathbb{Z}) \hookrightarrow \text{Spec}(\mathbb{Z}_p)$. Vakil's theorem proves that for some explicit smooth curve in $\mathbb{P}^3$, the stalk of the Hilbert scheme is an essentially smooth algebra over $\mathbb{Z}[x]/p(x^2-p)$. contd. –  Jason Starr Dec 31 '12 at 11:21
contd. Thus there is a morphism from this algebra to $\mathbb{Z}/p^2$, i.e., set $x=0$. Also there is a morphism from this algebra to the (complete) ring $\mathbb{Z}_p[\sqrt{p}]$. However, there is no morphism from this algebra to $\mathbb{Z}/p^3$. Now consider the blowing ups of $\mathbb{P}^3$ along these curves. By Theorem 5.4 (which is older), it follows that there is a flat, proper deformation of this blowing up over $\mathbb{Z}/p^2$ and over $\mathbb{Z}_p[\sqrt{p}]$, yet not over $\mathbb{Z}/p^3$. –  Jason Starr Dec 31 '12 at 11:26