Timeline for smooth projective curve
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2010 at 8:47 | vote | accept | Yashica | ||
| Dec 10, 2010 at 22:41 | comment | added | quim | @Emerton, @roy smith: thanks for the clarification. I also imagined $h$ had to be nonconstant. | |
| Dec 10, 2010 at 19:35 | comment | added | Emerton | Dear Roy, I misread the question, and assumed that the author wanted $h$ to be non-constant as well as $g$. I agree that if one is willing to allow $h$ to map $C'$ to a point of $C$, then simpler solutions may be possible, as in your comment and your answer. Best wishes, Matt | |
| Dec 10, 2010 at 19:28 | history | edited | Emerton | CC BY-SA 2.5 |
added 337 characters in body; deleted 53 characters in body
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| Dec 10, 2010 at 19:07 | comment | added | roy smith | As noted below, the minimum degree of h is zero when all fibers of phi have positive dimension, so this answer almost never minimizes the degree, if those "trivial" solutions are allowed. | |
| Dec 10, 2010 at 18:56 | comment | added | roy smith | If we modify the question so that both g and h are non constant, then this seems a perfect answer. I.e. imagine this is jeopardy, and this is the answer in search of a question. | |
| Dec 10, 2010 at 18:29 | comment | added | mdeland | Correct - in general taking hyperplane sections won't give you the lowest possible degree. If X = P^1 x P^1 (embedded as a quadric surface), Y = C = P^1... then using the hyperplane section will give us a degree 2 map C' --> C but there is a section of the map X --> Y (which isn't realized as the intersection of X with a hyperplane). | |
| Dec 10, 2010 at 17:06 | comment | added | Emerton | I would be surprised: presumably one can find lower degree (say for some fixed choice of projective embedding of $Z$) choices of $D$ which don't happen to be very ample, if not in general then surely in some particular cases. | |
| Dec 10, 2010 at 13:04 | comment | added | quim | I guess moreover that your construction (for an adequate choice of the irreducible component of Z) minimizes the degree of h. Is this true? | |
| Dec 10, 2010 at 6:49 | history | answered | Emerton | CC BY-SA 2.5 |