Timeline for Pull-back divisor being Cartier
Current License: CC BY-SA 4.0
11 events
when toggle format | what | by | license | comment | |
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Sep 27, 2018 at 1:14 | vote | accept | Joaquín Moraga | ||
Sep 26, 2018 at 22:59 | comment | added | Hacon | 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 22:57 | answer | added | Jason Starr | timeline score: 5 | |
Sep 26, 2018 at 22:43 | comment | added | Jason Starr | @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 22:22 | comment | added | Stefano | @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 22:16 | comment | added | Jason Starr | 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 22:14 | comment | added | Joaquín Moraga | $\pi^*(N):=\frac{\pi^*(mN)}{m}$, for $m$ divisible enough so that $mN$ is Cartier, where $\pi^*(nM)$ is the pull-back of Cartier divisors. | |
Sep 26, 2018 at 22:12 | comment | added | David E Speyer | Thanks! While we're at it, can you clarify: Does pulling back $N$ mean taking the sum of the components of $\pi^{-1}(N)$ with appropriate multiplicity, or do you mean that $\pi^{\ast} \mathcal{O}(N)$ is locally principal? | |
Sep 26, 2018 at 22:11 | comment | added | Joaquín Moraga | I think your counter-example does not have connected fibers. | |
Sep 26, 2018 at 22:10 | comment | added | David E Speyer | Duplicate of math.stackexchange.com/questions/60591/… , I think. | |
Sep 26, 2018 at 22:07 | history | asked | Joaquín Moraga | CC BY-SA 4.0 |