I would like to know if the following statement is correct.

Statement. Let $X$ be a normal projective variety with $Pic(X)=\mathbb Z+torsion$. Let $L$ be an ample line bundle on $X$ and let $D$ be an effective $\mathbb Q$-Cartier divisor in $X$. Consider the projective cone $C$ over $X$ corresponding to $L$. Then the cone over $D$ in $C$ is a $\mathbb Q$-Cartier divisor as well. Is this true?

Note that for $X=\mathbb P^1\times \mathbb P^1$, $L=O(1)\boxtimes O(1)$, and $D$ equal to a $\mathbb P^1$ fiber the statement does not hold. So the condition $Pic(X)=\mathbb Z+torsion$ is important.


1 Answer 1


I'm going to assume that $L$ induces a projectively normal embedding (so I feel comfortable talking above divisors, or you can take Spectrum of section rings instead of cones). Then this is certainly true for the affine cone which should be all you need, since away from the cone point there is nothing to check. In other words, the index of the divisor on $X$ is the same as that on the cone (away from the cone point).

Now, your hypothesis implies that some multiple of $D$ is a multiple of $L$, say $O_X(nD) = L^j$. Set $S = \bigoplus_{i \in \mathbb{Z}} H^0(L^i)$ and $Y = \text{Spec }S$.

Claim: if $D_Y$ is the divisor on $Y$ corresponding to $D$, then $$\Gamma(Y, O_Y(D_Y)) = \bigoplus_{i \in \mathbb{Z}} H^0(L^i \otimes O_X(D)).$$

Proof of claim:

I will assume that $D = -P$ is anti-effective and irreducible, as it is straightforward to reduce to this case.

First let us explain how to construct $D_Y$. Set $U = Y \setminus${origin}. Then $\rho : U \to X$ is a $\mathbb{A}^1$-bundle over $X$, in particular it is flat. We set $D_U = \rho^* D$ (which makes sense, even for Weil divisors, since it's an $\mathbb{A}^1$-bundle). Then $D_Y$ is the unique divisor on $Y$ which agrees with $D_U$ on $U$.

Since $D$ is anti-effective, so is $D_Y$. Then $\Gamma(Y, O_Y(D_Y)) = \Gamma(Y, O_Y(-P))$ is the set of functions in $S$ that vanish along the pullback of $D_Y$. This is clearly a homogeneous prime ideal, and hence easily is seen to coincide with the ideal $\bigoplus_{i \in \mathbb{Z}} H^0(L^i \otimes O_X(D)) \subseteq S$. (Ok, I kind of waved my hands here, but hopefully this part is easy)


Similarly, the divisor corresponding to $nD$ is $(nD)_Y = nD_Y$ (this equality is just observed by pulling back to $U$ and doing the computation there). $nD_Y$ has corresponding sheaf $$\Gamma(Y, O_Y(nD_Y)) = \bigoplus_{i \in \mathbb{Z}} H^0(L^i \otimes O_X(nD)).$$ Note that because $O_X(nD) = L^j$, we see that $O_Y(nD_Y) = S[-j]$, in other words $O_Y(nD_Y)$ is just $S$ with a different grading. In particular, it is locally free so $nD_Y$ is Cartier.

  • $\begingroup$ Dear Karl, many thanks for you answer! Do I understand correctly that in the first formula for $O_Y(D_Y)$ the term on the right of the equality is the space of global sections of $O_Y(D_Y)$? $\endgroup$
    – aglearner
    Apr 23, 2013 at 7:29
  • $\begingroup$ Yup, space of global sections over all twists by $L$. $\endgroup$ Apr 23, 2013 at 18:26
  • $\begingroup$ I can provide detailed proofs of these points if you want, just let me know. $\endgroup$ Apr 23, 2013 at 18:30
  • $\begingroup$ Dear Karl, yes I would be grateful if you could give more details. $\endgroup$
    – aglearner
    Apr 23, 2013 at 18:54
  • $\begingroup$ Let me know if this clarifies things at all. $\endgroup$ Apr 24, 2013 at 17:44

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