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

According to general theory, for a square zero thickening defined by an ideal I: SpecA -> SpecA', there is an obstruction of lifting a smooth scheme X over A to a smooth scheme over A' living in H^2(X,T_X \otimes I).

Can anyone give an example of being obstructed? e.g. a smooth scheme over F_p which does not lift smoothly to Z/(p^2)?

share|improve this question
    
This is almost a duplicate of my question here: mathoverflow.net/questions/423/… –  David Zureick-Brown Jan 4 '10 at 16:44
add comment

1 Answer

up vote 8 down vote accepted

Ravi Vakil's paper Murphy's Law in Algebraic Geometry... gives many references of such things: see Section 2 of

http://arxiv.org/PS_cache/math/pdf/0411/0411469v1.pdf

The first example is due to Serre:


Serre, Jean-Pierre Exemples de variétés projectives en caractéristique $p$ non relevables en caractéristique zéro. (French) Proc. Nat. Acad. Sci. U.S.A. 47 1961 108--109.

Here is the MathReview by I. Barsotti:

An example of a non-singular projective variety $X_0$, over an algebraically closed field $k$ of characteristic $p$, which is not the image, $\text{mod}\,p$, of any variety $X$ over a complete local ring of characteristic 0 with $k$ as residue field. The variety $X_0$ is obtained by selecting, in a 5-dimensional projective space $S$, and for $p>5$, a non-singular variety $Y_0$ which has no fixed point for an abelian finite subgroup $G$ with at least 5 generators of period $p$, of the group $\Pi(k)$ of projective transformations of $S$, but which is transformed into itself by $G$; then $X_0=Y_0/G$. The reason for the impossibility is that $\Pi(K)$, for a $K$ of characteristic 0, does not contain a subgroup isomorphic to $G$. {Misprint: on the last line on p. 108 one should read $s(\sigma)=\exp(h(\sigma)N)$.}


ADDENDUM:

After looking back at the question, my reference to Serre's paper is inappropriate: this is an example of a variety which does not lift to characteristic zero, whereas the poster asked for one which didn't lift (even) to $\mathbb{Z}/p^2 \mathbb{Z}$. It is still true that this sort of thing is discussed in Vakil's paper, but now the canonical primary source seems to be


Deligne, Pierre; Illusie, Luc
Relèvements modulo $p^2$ et décomposition du complexe de de Rham. (French) [Liftings modulo $p^2$ and decomposition of the de Rham complex] Invent. Math. 89 (1987), no. 2, 247--270.

MathReview by Thomas Zink:

The degeneration of the Hodge spectral sequence H^q(X,\Omega^p_{X/k}) \Rightarrow H^n_{\roman{DR}}(X/k) for a smooth and projective algebraic variety over a field of characteristic zero is a basic fact in algebraic geometry. Nevertheless, only recently has one found an algebraic proof in connection with the comparison of étale and crystalline cohomology (Faltings, Fontaine, Messing).

In this beautiful paper the authors give a short, elementary, algebraic proof for the degeneration by methods in characteristic $p>0$.

Let $k$ be a perfect field of characteristic $p>0$, and let $X$ be a smooth variety over $k$, which lifts to the Witt ring $W_2(k)$. Denote by $X'$ the variety obtained by base change via the Frobenius automorphism, and let $F\:X\to X'$ be the relative Frobenius morphism...More precisely, such splittings correspond to liftings of $X'$ to $W_2(k)$. This theorem implies the degeneration of the Hodge spectral sequence for $X$ in dimension $<p$, and, by Raynaud, a Kodaira vanishing theorem for $X$. The corresponding facts in characteristic zero may be deduced by the usual reduction process.

The authors show that their argument extends to the case where $k$ is replaced by an arbitrary base $S$ of characteristic $p$ and \Omega^{\bfcdot}_{X/S} is replaced by differentials with logarithmic poles.


share|improve this answer
    
That's a beautiful paper for many reasons, including this question. –  Charles Siegel Jan 4 '10 at 14:43
add comment

Your Answer

 
discard

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.