Timeline for Properties of morphisms induced by divisors on curves
Current License: CC BY-SA 3.0
4 events
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Oct 11, 2011 at 17:38 | comment | added | rghthndsd | Ah this makes so much more sense! I was not aware of the "definition" you mentioned; if it's in Hartshorne then I missed it. So for the third, the action of a Galois element fixes the elements of $\mathbb{P}^1$, which I'm identifying with effective divisors linearly equivalent to $D$. Since these divisors also define the fibers of my map, it must be that the fibers are permuted. | |
Oct 10, 2011 at 17:54 | comment | added | Felipe Voloch | Your first question is already answered. For the second, by definition, $f(P)=D'$ where $D'$ is the unique effective divisor linearly equivalent to $D$ with $D' \ge P$. So it is clear that $f(P)=D$ if $P$ is in the support of $D$. The third question is similarly straightforward. | |
Oct 10, 2011 at 14:40 | comment | added | J.C. Ottem | By Hartshorne II.6.9 we have more generally $\deg f^∗E=\deg f\cdot \deg E$ for any $E$ on $P^1$. | |
Oct 10, 2011 at 13:07 | history | asked | rghthndsd | CC BY-SA 3.0 |