Let $X$ be a projective Calabi-Yau threefold with a single ordinary double point at $x \in X$, and smooth elsewhere. Is $X$ necessarily factorial?

I suspect that the answer is "yes", for the following reason. Some neighbourhood of $x$ is analytically isomorphic to the 'conifold' geometry $$ xy - wz = 0 ~~\mathrm{in}~~\mathbb{C}^4~. $$ This is not factorial; $x = w = 0$ is a non-Cartier divisor, and blowing up along this gives a small projective resolution of the conifold. However, analytic isomorphisms do not preserve the local class group, and $\mathcal{O}_{X,x}$ may nevertheless be a UFD. I suspect that this is necessarily true, since if not, we could blow up along the non-Cartier divisor, but one-nodal projective Calabi-Yau threefolds do not admit projective resolutions. I would like to know whether this argument can be made rigorous, or replaced by a simpler one.

Edit: It may be important to assume that $X$ is smoothable, and this is the case which matters to me.


2 Answers 2


Let $X$ be a Calabi-Yau three-fold with only ordinary double points. One can always find a small resolution $Y\rightarrow X$ where $Y$ is a (not necessarily Kaehler) complex manifold. Let $C_1,\ldots,C_n$ be the exceptional curves. Friedman proved in his paper "Simultaneous Resolution of Threefold Double Points" that $X$ has an infinitesimal smoothing if and only if there is a linear dependence relation $\sum_i a_i[C_i]=0$ in $H_2(Y,{\mathbb R})$ with all $a_i$ non-zero. Furthermore, by unobstructedness of the deformation theory of Calabi-Yau varieties with isolated ordinary double points (proved by Ran and Tian), having an infinitesimal smoothing is equivalent to having a smoothing.

In your case, $X$ just has a single ODP, so there is one exceptional curve $C$. So there is a smoothing if and only if $[C]=0$ in $H_2(Y,{\mathbb R})$. This is in turn equivalent to $X$ being factorial (there is no algebraic small resolution).

So what you want is true: if $X$ is smoothable, it is factorial. However, there are examples of non-smoothable $X$ with one ODP, and these are not factorial.

  • $\begingroup$ Thanks Mark, that's exactly what I was after. I was actually reading Friedman's paper, but with my poor grasp of deformation theory, I hadn't managed to extract what I needed! $\endgroup$ Aug 7, 2013 at 15:25
  • $\begingroup$ Blowing-up the ODP one obtains as exceptional divisor a smooth quadric in $\mathbb{P}^3$. If there is no algebraic small resolution then the two rulings of this quadric must be numerically equivalent in the blown-up threefold, right? $\endgroup$ Aug 7, 2013 at 15:25
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    $\begingroup$ Francesco, that's right. $\endgroup$
    – Mark Gross
    Aug 7, 2013 at 15:36

At the moment, I do not have in mind the answer for the general case. However, it is yes when the threefold is a complete intersection in some $\mathbb{P}^n$. In fact, in this situation a more general result holds.

Let us call a threefold singularity an ordinary $m$-ple point if the corresponding tangent cone is a cone over a smooth surface in $\mathbb{P}^3$. When $m=2$ we obtain precisely the ordinary double points. Then we have the following

Proposition. Let $Y \subset \mathbb{P}^n$ be a smooth, complete intersection fourfold and $X \subset Y$ be a reduced, irreducible threefold, which is complete intersection of $Y$ with a hypersurface of degree $d$. Assume that the singular locus of $X$ consists of $k$ ordinary multiple points $p_1, \ldots, p_k$ of multiplicity $m_1, \ldots, m_k$. If \begin{equation} \sum_{i=1}^k m_i < d \end{equation} then $X$ is factorial.

In particular, since a complete intersection Calabi-Yau threefold has always degree at least $5$, you get your result.

A proof of this proposition and related references can be find in my recent preprint with A. Rapagnetta and P. Sabatino On factoriality of threefolds with isolated singularities, arXiv:1305.4371.

  • $\begingroup$ Thank-you for the answer, Francesco. I will have a look at your preprint. However, the case of a complete intersection in $\mathbb{P}^n$ won't actually help me directly. $\endgroup$ Aug 7, 2013 at 13:48

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