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?
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asked Sep 26, 2018 at 22:07
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2$\begingroup$ Duplicate of math.stackexchange.com/questions/60591/… , I think. $\endgroup$– David E SpeyerSep 26, 2018 at 22:10
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3$\begingroup$ 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 $Y$. $\endgroup$– Jason StarrSep 26, 2018 at 22:16
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4$\begingroup$ @JasonStarr Isn't the pullback of $N$ in this case the strict transform plus $\frac 1 2 E$, where $E$ is the exceptional curve? $\endgroup$– StefanoSep 26, 2018 at 22:22
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2$\begingroup$ @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). $\endgroup$– Jason StarrSep 26, 2018 at 22:43
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3$\begingroup$ 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. $\endgroup$– HaconSep 26, 2018 at 22:59
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1 Answer
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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 $A=\widetilde{N}+E$.