Masse

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visits member for 1 year, 9 months
seen Jan 11 '13 at 19:33

Jan
9
comment Do regular noetherian schemes of dimension one only have finitely many etale covers of bounded degree
@pranavk. It seems that I underestimated the difficulty of this question. When you said "a zoo", it became clear to me that this question is more difficult than I had expected. As you can read in the question, I was motivated by the arithmetic and the geometric case. I was too optimistic and hoped for an easy "classification" of Dedekind schemes which would lead to a comprehensible answer. But it seems that this is too difficult.
Jan
8
asked Does the Riemann hypothesis for liftable varieties over a finite field imply the Riemann hypothesis for all varieties over a finite field
Jan
8
awarded  Commentator
Jan
8
comment Corresponding notion of unramified for motives (or de Rham cohomology)
@Piotr. That's very useful! I didn't realize that.
Jan
8
comment Corresponding notion of unramified for motives (or de Rham cohomology)
@Mikhail. Thank you for your comment.
Jan
7
comment Do regular noetherian schemes of dimension one only have finitely many etale covers of bounded degree
Your counterexample (an open affine subscheme of Spec $R$) is two-dimensional, no? Also, I'm not sure how to exclude such examples. It might be possible to do a case-by-case analysis.
Jan
7
asked Do regular noetherian schemes of dimension one only have finitely many etale covers of bounded degree
Jan
5
asked Corresponding notion of unramified for motives (or de Rham cohomology)
Jan
4
comment On the m-th power of the Hodge bundle and Arakelov's theorem
Also, I find the condition $f_\ast omega^{\otimes n}$ non-zero in Arakelov's inequality a bit strange. Isn't the degree of the trivial line bundle zero? Thus, isn't Arakelov's inequality trivial in this case? (See for example Theorem 1.1 in the survey of Viehweg on Arakelov inequalities.)
Jan
4
comment On the m-th power of the Hodge bundle and Arakelov's theorem
Thanks for your answer and sorry for the late reply! I think I'm not understanding something completely; if $f_\ast \omega^{\otimes n} =0$, it can happen that $f_\ast \omega^{\otimes (n+1)}$ is non-zero? I find this very strange...The question reuces to whether $f_\ast \omega^{\otimes n}$ is non-zero for ALL $n>>0$. (And here I really mean for ALL $n$ big enough. Thus, without exceptions.)
Dec
23
awarded  Editor
Dec
23
revised On the m-th power of the Hodge bundle and Arakelov's theorem
deleted 11 characters in body
Dec
23
asked Does the Albanese map satisfy Torelli's theorem
Dec
23
awarded  Scholar
Dec
23
comment Is the moduli space of genus three smooth quartics affine?
@Jason. That's what I was looking for indeed. Any nonempty effective divisor on $\mathbf P^{14}$ is ample.
Dec
23
comment Is the moduli space of genus three smooth quartics affine?
@Olivier. You're right. I didn't mean to say that. Rather, more generally, the moduli space of smooth hypersurfaces of degree $d$ in $N$-projective space is affine if $d>N+1$.
Dec
23
accepted If rational points are like entire curves, then what do algebraic points correspond to
Dec
23
comment If rational points are like entire curves, then what do algebraic points correspond to
That seems reasonable. Just to convince myself a bit more. Let $X$ be curve of genus at least $two$ over a number field $K$ of big gonality. Then the set of quadratic points on $X$ is finite by Faltings-Frey. In particular, this should correspond to the fact that there are only finitely many hyperelliptic curves dominating $X$. And that of course follows from the big gonality of $X$. (Actually there are no hyp ell curves dominating $X$.) So ok, I'm convinced that this "works".
Dec
23
asked Is the moduli space of genus three smooth quartics affine?
Dec
23
asked If rational points are like entire curves, then what do algebraic points correspond to