# Pull-back divisor being Cartier

Let $$\pi \colon X \rightarrow Y$$ be a projective morphism with connected fibers between normal quasi-projective varieties. Let $$N$$ be a $$\mathbb{Q}$$-Cartier divisor on $$Y$$ so that $$\pi^*(N)$$ is Cartier. Does it follows that $$N$$ is itself Cartier?

• Duplicate of math.stackexchange.com/questions/60591/… , I think. Sep 26, 2018 at 22:10
• Let $Y\subset \mathbb{P}^3$ be a projective cone over a smooth plane conic, i.e., $Y$ is a quadric hypersurface of rank $3$. Let $\pi$ be the minimal desingularization. Let $N$ be one of the lines of ruling on $Y$. Sep 26, 2018 at 22:16
• @JasonStarr Isn't the pullback of $N$ in this case the strict transform plus $\frac 1 2 E$, where $E$ is the exceptional curve? Sep 26, 2018 at 22:22
• @Stefano. You are correct. I did not read the previous comment about the definition of pullback in this context (I assumed the pullback was inverse image ideal sheaf). Sep 26, 2018 at 22:43
• What if $Y$ is a smooth curve and $\pi$ is a stable family that has a double fiber over $y\in Y$. Then isn't $\pi ^*((1/2)y)$ Cartier? If we assume $N$ is integral, then we have an inclusion $\pi _* O_X(\pi ^* N)\to O_Y(N)$ where the LHS is torsion free and the RHS is reflexive. This is an isomorphism over the smooth locus of $Y$ (by the projection formula). But it is also surjective since if $f\in k(Y)$ satisfies $(f)+N\geq 0$, then $f\circ \pi \in k(X)$ satisfies $(f\circ \pi)+f^*N\geq 0$ which shows the inclusion is a surjection. Sep 26, 2018 at 22:59

## 1 Answer

Following the clarification in the comments, I am interpreting the question as follows.

Question. For an effective Weil divisor $$N$$ on $$Y$$, for an effective Cartier divisor $$A$$ on $$X$$, for a positive integer $$\ell$$ such that the effective Weil divisor $$\ell N$$ is Cartier and such that the pullback effective Cartier divisor $$\pi^*(\ell N)$$ equals $$\ell A$$ as effective Cartier divisors, is $$N$$ Cartier?

The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.

The minimal resolution of a cone over a smooth plane cubic is a ruled surface. Let $$C\subset \mathbb{P}^2$$ be a smooth, plane cubic (a genus $$1$$ curve). Let $$Y$$ be the projective cone in $$\mathbb{P}^3$$ over $$C$$. The minimal desingularization of $$Y$$, $$\pi:X\to Y,$$ is the closure in $$Y\times C$$ of the graph of the linear projection from $$Y$$ to $$C$$. Denote by $$\rho$$ the projection, $$\rho:X\to C.$$ This morphism is a $$\mathbb{P}^1$$-bundle. The morphism $$\rho$$ maps the exceptional locus $$E$$ of $$\pi$$ isomorphically to $$C$$, and $$E$$ is a relative hyperplane class for $$\rho$$. The normal sheaf $$\mathcal{O}_X(\underline{E})|_E$$ is isomorphic to $$\mathcal{O}_{\mathbb{P}^2}(-1)|_C$$.

Effective Weil divisors "torsion-equivalent" to a conical hyperplane class. Let $$H\subset C$$ be the restriction to $$C$$ of a general hyperplane in $$\mathbb{P}^2$$. Let $$D\subset C$$ be a degree $$3$$ effective divisor such that the divisor $$H-D$$ has finite order $$\ell>1$$ in the Picard group of $$C$$, i.e., $$\ell H - \ell D$$ is the divisor of a rational function $$f$$ on $$C$$.

Let $$M\subset Y$$, resp. $$N \subset Y$$, be the cone over $$H$$, resp. $$D$$. Note that $$\ell N$$ and $$\ell M$$ are linearly equivalent Cartier divisors on $$Y$$ in the linear equivalence class of $$\mathcal{O}_{\mathbb{P}^3}(\ell)|_Y$$. In fact, $$\ell$$ is the smallest positive integer such that $$\ell N$$ is a Cartier divisor on $$Y$$.

Denote by $$\widetilde{M}\subset X$$, resp. by $$\widetilde{N}\subset X$$, the strict transform under $$\pi$$ of $$M$$, resp. of $$N$$. Note that $$\ell \widetilde{M}-\ell\widetilde{N}$$ equals the Cartier divisor of $$f\circ \rho$$, as does the total pullback under $$\pi$$ of the Cartier divisor $$\ell N-\ell M$$. Thus, the coefficient of $$E$$ in the pullback of $$\ell M$$ equals the coefficient of $$E$$ in the pullback of $$\ell N$$.

Already $$M$$ is an effective Cartier divisor on $$Y$$, and the pullback of $$M$$ is the strict transform $$\widetilde{M}$$ plus the exceptional divisor $$E$$. One way to see this is to deform $$M$$ to a hyperplane section of $$Y$$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $$M$$ with $$E$$ equals $$0$$, it follows that the coefficient of $$E$$ equals $$1$$. Thus, the total pullback of $$\ell M$$ equals $$\ell \widetilde{M} + \ell E$$.

Therefore also $$\ell N$$ equals t$$\ell \widetilde{N} +\ell E$$. Although $$N$$ is not Cartier on $$Y$$, the pullback of $$\ell N$$ equals $$\ell A$$ on $$X$$ for the effective Cartier divisor $$A=\widetilde{N}+E$$.