A variety is $\mathbb{Q}$-factorial if every global Weil divisor is $\mathbb{Q}$-Cartier. How bad singularities are allowed so that the algebraic variety is still $\mathbb{Q}$-factorial? Is a singular curve $\mathbb{Q}$-factorial? For example is a nodal-cuspidal plane curve $\mathbb{Q}$-factorial?
[Math] When is an algebraic variety $\mathbb{Q}$-factorial
ag.algebraic-geometry
Related Solutions
Note: I added an addendum below in response to quinque's comment and the subsequent discussion of the issue (s)he raised on math.stackexchange (see the link in quinque's comment).
An easy way to produce a $\mathbb Q$-Cartier but not Cartier canonical divisor is by a quotient. For instance for the quotient $$X=\mathbb A^3/(x,y,z)\sim (-x,-y,-z)$$ $2K_X$ is Cartier, but $K_X$ is not.
I leave it for you to prove that $2K_X$ is Cartier. Here is how to see that $K_X$ is not: Clearly $X={\rm Spec}k[x^2,y^2,z^2,xy,yz,xz]$ in other words, $X$ is the affine cone over the Veronese surface $\mathbb P^2\simeq V\subset \mathbb P^5$. Blowing up the cone point gives a resolution of singularities $\pi: Y\to X$ with exceptional divisor $E\simeq V$. In fact $E^2\sim -2L$ where $L$ is the class of a line. This follows by considering the blow up as a blow up of the ambient $\mathbb A^6$ (the cone over $\mathbb P^5$) and noticing that the square of the exceptional divisor of the blow up of $\mathbb A^6$ is $-1$-times the hyperplane in $\mathbb P^5$ which restricts to a conic on $Y$. Now write $$K_Y\sim_{\mathbb Q} \pi^*K_X + aE \tag{$\star$}$$ and use the adjunction formula ($Y$ is smooth!) to get $$ (a+1)E^2\sim K_E=K_{\mathbb P^2} \sim -3L. $$ Solving for $a$ shows that $a=\dfrac 12$ which shows that $K_X$ cannot be Cartier.
Addendum (a.k.a. Intermezzo): quinque raised the interesting point that $\dfrac 12 E$ is actually $\mathbb Q$-linearly equivalent to a Cartier divisor, so (s)he was worried that then the above does not prove that $K_X$ is not Cartier. It actually does, but this points to an interesting consequence, namely that this means that $\pi^*K_X$ is actually numerically equivalent to a Cartier divisor, so using only intersection numbers one will not be able to prove that it is not Cartier.
Anyway, here is why the above implies that $K_X$ is not Cartier: Suppose it is. That means that $\omega_X$ is the trivial line bundle in a neighborhood of the singular point. Consider the pull-back of a local generator. On $Y\setminus E$ this will generate $\omega_Y$ and hence the corresponding (integral!) divisor $\pi^*K_X$ is equal (not just linearly equivalent!!) to $K_Y+bE$ for some $b\in \mathbb Z$. Then the same calculation as above implies that $b=-a=-\dfrac 12$ which is a contradiction.
The point is that if $K_X$ is Cartier, then ($\star$) holds with linear equivalence, and in fact with equality, instead of $\mathbb Q$-linear equivalence and hence $a$ has to be an integer in that case.
Also interesting to note that the same construction does not give a desired example in dimension $2$: The quotient $\mathbb A^2/(x,y)\sim(-x,-y)$ is a cone over a conic which is a surface in $\mathbb P^3$. In particular it is Gorenstein and hence $K_X$ is Cartier.
On the other hand, one gets a $2$-dimensional example by $\mathbb A^2/\mu_3$ where $\mu_3$ acts by multiplication by a primitive third root of unity. The proof is essentially the same as above. It is relatively easy to see (just as above) that this is the same as the cone over a twisted cubic.
As for the adjunction formula, it definitely works as long as $K_X+D$ is Cartier and it works up to torsion if it is $\mathbb Q$-Cartier. If it is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $\mathscr Ext$'s but this is almost Grothendieck Duality then.
Ok, the first thing you have to specify is what does it mean for to talk about a Cartier divisor (or Weil divisors). There are a number of options here. The one I like best for Cartier divisors would be invertible subsheaves of $K(X)$, the fraction field of $X$. This coincides with the one in Fulton's intersection theory if I recall correctly.
Now, what do you mean by Weil divisors? One option is formal sums of points. This is ok, but there are problems (mentioned below). Another option is $O_X$-module subsheaves of $K(X)$ in general (if you weren't dealing with curves maybe we should require these to be S2). In higher dimension sometimes people require them to be Cartier in codimension 1. But this just gives us Cartier divisors...
Why is the naive notion of Weil divisor problematic? The map from Cartier divisors to Weil divisors is not necessarily injective for non-normal varieties. Let me give an example. Consider the curve singularity $k[x,y]/(xy + x^3 + y^3)$ and consider the Cartier divisors corresponding to the principal ideals $(x)$ and $(y)$ respectively. These are different subsheaves of $K(X)$ (they are different ideals). However, they both give Weil divisors $3P$ (I assume you compute the map from Cartier to Weil divisors by computing lengths at points).
This implies that there isn't a well defined notion of $O_X(nP)$. For example, if there are two Cartier divisors giving the same $nP$, which sheaf do you choose to represent $O_X(nP)$?
The subsheaf notion is also problematic since not all subsheaves are invertible. So, I think the best behaved notion of Weil is the one that requires divisors to be Cartier in codimension 1. Unfortunately, for curves, this is not interesting.
Now, let me answer your questions.
Yes, depending on what you mean by Cartier divisor. Embed your curve in projective space. Take a hyperplane $H$ (not containing $X$) passing through $P$ and some other smooth points on $X$. $O_{\mathbb{P}^n}(H) \cdot O_X$ is certainly invertible. Let $D$ denote the effective Cartier divisor on $X$ such that $O_X(D)$ agrees with $O_{\mathbb{P}^n}(H) \cdot O_X$ away from $P$ and agrees with $O_X$ at $P$. Now simply consider $O_{\mathbb{P}^n}(H) \cdot O_X(-D)$. This agrees with $O_X$ away from $P$, and is not trivial at $P$, so I would say it is a Cartier divisor supported at $P$.
I don't think so. This doesn't work in higher dimensions either. I assume you want some properties? Like preserving linear equivalence?
Since $O_X(nP)$ is not well defined, then I don't think you can define the notion of $\mathbb{Q}$-Cartier. I guess you could say that there is some Cartier divisor that maps to $nP$ in the natural map from Cartier divisors to Weil divisors?
Best Answer
I saw this a while ago, and I assumed that someone would give you an answer, but it doesn't look like anyone will. So I'll make an attempt. As Ulrich suggests, it is better to stick to normal varieties, so your nodal/cuspidal curves are immediately eliminated. So we should move to dimension at least two.
A very simple class of examples are cones. Let $V\subset \mathbb{P}^N$ be a nontrivial smooth variety, and let $R$ be the local ring of the vertex of the cone over $V$. This is normal and $\dim R\ge 2$. $R$ (or $Spec R$) is $\mathbb{Q}$-factorial exactly when $Cl(R)\otimes\mathbb{Q}=0$ where $Cl(R)$ is the (Weil) divisor class group. Now according to Lipman "Unique factorization in complete local rings", Alg. Geom, Arcata (1974) $$Cl(R)= Cl(V)/\langle\text{hyperplane}\rangle$$
This is generally not true. For example, if $V$ is a curve, $R$ is $\mathbb{Q}$-factorial if and only if $V$ has genus $0$.
Added Notes: (1) In the last sentence, I meant that a cone over a higher genus curve $V$ cannot be $\mathbb{Q}$-factorial because $\dim Cl(V)\otimes \mathbb{Q}=\infty$ in this case. (2) As Ulrich & Francesco point out $V$ should be projectively normal. For example, it would suffice to assume the $V$ is a hypersurface. (3) I forgot to say, that I'm working over an algebraically closed ground field.