bio | website | math.univ-toulouse.fr/… |
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location | Toulouse | |
age | 44 | |
visits | member for | 3 years, 4 months |
seen | 3 hours ago | |
stats | profile views | 2,146 |
Chercheur au CNRS, working in Toulouse.
I don't usually comment or answer questions from anonymous users.
Dec 14 |
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Quotient of a (non-linear) algebraic group by a closed subgroup
Here is another reference: Raynaud, 'Passage au quotient par une relation d'équivalence plate', Proceedings of a conference on local fields (Driebergen, 1967), p. 82 Ex. a) (i). |
Nov 29 |
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Fermat's last theorem over larger fields
Sorry, your are right, Mazur's conjecture does not quite cover your question (but it gives a partial answer). The conjecture in Zarhin-Parshin corresponds precisely to what follows from Mordell-Lang + Mazur's conjecture. |
Nov 28 |
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Fermat's last theorem over larger fields
See also Zarhin-Parshin, "Finiteness theorems...", p. 92 for a related conjecture (also made by Mazur). |
Nov 26 |
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How to see that this pairing of line bundles is multiplicative?
Do you want a canonical isomorphism (inv. under base-change) ? Or would you be happy with any isomorphism ? |
Nov 19 |
accepted | On a proposition in Hartshorne's paper “Ample vector bundles on curves” |
Nov 16 |
awarded | Nice Question |
Nov 16 |
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On a proposition in Hartshorne's paper “Ample vector bundles on curves”
Thank you very much for taking the time to write this answer. If I understand your answer right, your construction is in the same line as what ulrich described in his comment. I would still like to see a proof or a counterexample to Prop. 4.1, though but if nobody provides one in the next few days I will accept your answer. |
Nov 15 |
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On a proposition in Hartshorne's paper “Ample vector bundles on curves”
@ulrich. Thank you very much for this interesting comment. That settles it then, as far as (X) is concerned. Here is a possible simplification of your argument: as you write, $C\to f(C)$ must be generically separable because $H^0(A,\Omega_A)\to H^0(C,\Omega_C)$ does not vanish. Now since $C\to f_*(C)$ is also generically inseparable, it must be birational. Note that $f_*(C)$ might be singular, so its geometric genus might not be defined (or is there a way to show that it must be non-singular ?). |
Nov 14 |
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On a proposition in Hartshorne's paper “Ample vector bundles on curves”
Thank you for this reference. I had a look at Fulton's paper but I cannot find an argument in there that shows that (X) (in my post) or Prop. 4.1 is wrong in a specific situation. The counterexample of Serre that Fulton mentions is apparently a construction that appears in the very paper by Hartshorne that I am concerned with (in par. 3). It is an example of a curve and a vector bundle on it, which is not ample and yet has only positive quotients. How does all this connect to (X) or Prop. 4.1 ? |
Nov 14 |
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On a proposition in Hartshorne's paper “Ample vector bundles on curves”
@Donu Arapura: $\phi$ is birational onto its image (I edited the question 30 minutes ago - you might have read the post just before that - sorry). |
Nov 14 |
asked | On a proposition in Hartshorne's paper “Ample vector bundles on curves” |
Aug 25 |
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Rational points techniques on curves not using their Jacobian
The method of Kim is very interesting but he can only prove Siegel's theorem for the projective line, as far I can see. In that case, (much) more elementary proofs can be made quite effective (see Silverman's book on elliptic curves) but even this does not give a method to find all the rational points on a curve. |
Aug 24 |
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Rational points techniques on curves not using their Jacobian
The very conjectural "effective Mordell conjecture" (see Astérisque 183) would provide such a method. The analog of this conjecture is proven over function fields (proofs by Arakelov and Szpiro) and such a method is thus available over function fields. Over number fields, I don't know any method that does not use the Jacobian. |
Aug 22 |
awarded | Yearling |
Jul 2 |
awarded | Curious |
Jun 20 |
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splitting of the Hodge filtration
I think the OP means: "When is the canonical Hodge splitting already defined over $k$ ?". This happens rarely and the failure to split over $k$ is basically encoded in the Mumford-Tate group, about which there are many deep conjectures. For instance if $X$ is a CM abelian variety and $k$ contains the Galois closure of the CM field, then the sequence splits (and the Mumford-Tate group is a torus). |
May 14 |
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Ampleness of Hodge bundles over complex curves
@Jason Starr. Thank you for your remark. I would expect something like that but I cannot find any coherent bibliographical reference for this kind of thing. |
May 14 |
asked | Ampleness of Hodge bundles over complex curves |
Apr 1 |
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Are there perverse sheaves on abelian varieties with small Euler characteristic?
Isn't $d=g$ in your last statement ? |
Apr 1 |
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Is it possible to prove Mordell's conjecture geometrically?
This is a good question. It has always struck me that the only known proofs of the Mordell conjecture over number fields, as well as of the Manin-Mumford conjecture require arithmetic models. The only thing I know in the "purely complex" (or "geometrical") direction is Zannier-Pila's approach to Manin-Mumford (but not Mordell...), which still requires arithmetics, but less so than Raynaud's (or Ullmo-Szpiro-Zhang, or Pink-R.'s). |