38
votes
Accepted
Elementary Proof of Riemann-Roch for Compact Riemann Surfaces
RRT
There is a big difference in difficulty between the compact Riemann surface case and the projective curve case, for reasons already mentioned. Namely a projective curve comes equipped with a ...
- 11.7k
27
votes
Accepted
Is the divisibility graph of the proper divisors of n more often planar than not?
No, because almost all numbers have at least $4$ distinct prime factors, making the divisibility graph contain a hypercube and thus be nonplanar.
- 126k
18
votes
Accepted
Are there topological versions of the idea of divisor?
Disclaimer. I am no expert at all in algebraic geometry. Therefore much of the following will be oversimplified or maybe even simply wrong. You are still invited to improve it.
EDIT. There is a paper ...
16
votes
Accepted
How do i show that:$\prod\frac{p^2+1}{p^2-1}=\frac{5}{2}$ without using properties of Riemann zeta function?
This is a well-known problem, attributed to Sam Wagstaff in Richard Guy's Unsolved Problems in Number Theory. Section B48 "Products taken over primes" includes a paragraph
Wagstaff asked for an ...
- 75.2k
13
votes
Elementary Proof of Riemann-Roch for Compact Riemann Surfaces
Joe Harris, as recorded in his course notes here, gives the following slick proof when both $D$ and $K-D$ are effective; it has the advantage of never mentioning $H^1$. See lecture 1 for this argument,...
- 145k
13
votes
On Q-Cartier Divisors
Maybe I can say something useful here. The main confusion seems to be how to find the sheaves/ideals/modules associated to multiples of divisors. As Martin Bright points out, symbolic power of a ...
- 19.9k
12
votes
Accepted
Bertini's theorem over non-algebraically closed field
This is true both over finite and infinite fields.
For infinite fields, see [Jou, Cor. I.6.11(2)]. It works for a general section of any very ample line bundle $\mathscr L$, using that over an ...
- 24.4k
11
votes
Accepted
Is there a divisor in $\mathbb P^2$ such that all analytic maps into its complement algebraize?
On page 73 of Kobayashi's book Hyperbolic Complex spaces he shows that if D is a certain configuration of 6 lines in the plane then its complement is complete hyperbolic and hyperbolically embedded ...
- 4,041
11
votes
Accepted
Generalization of the rigidity lemma in birational geometry
EDIT: I've just realized that this holds under somewhat weaker assumptions. It is not necessary that the fibers of $g$ are connected.
EDIT#2: Apparently, in my previous edit I weakened the conditions ...
- 41.6k
11
votes
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Fundamental groups of complements of divisors in $\mathbb P^2$
I'd leave this as a comment, but I don't have enough reputation. Consider the long exact sequence in homology of the pair $(\mathbb{P}^2, \mathbb{P}^2-D)$. Since $H_1(\mathbb{P}^2,\mathbb{Z}) = 0$ and ...
- 666
9
votes
Accepted
Anti-canonical divisor of a Fano variety
If you want to consider smooth (weak) Fano variety, then Fukuda has effective estimation of the birationality of anti-canonical systems for any dimension (but not optimal), see [S. FUKUDA, A note on ...
- 1,054
9
votes
Accepted
Is the reduced scheme associated to a Cartier divisor always Cartier?
No. Consider, for instance, the quadratic cone
$$
X = \{xz - y^2 = 0\} \subset \mathbb{A}^3
$$
and the double line
$$
D = X \cap \{x = 0\} = \{x = y^2 = 0\}
$$
on $X$. Then $D$ is a Cartier divisor, ...
- 34.7k
8
votes
Is it possible to show that :for $n \geq 1:\sigma(n!-1) $ never be prime and why $\sigma(n!-1)\bmod 10 $ at most is $0$?
If $n\ge 4$, then $24 \mid n!$. It is an easy exercise to show that if $24 \mid N$, then $24 \mid \sigma(N-1)$. (Pair each factor of $N-1$ with its cofactor, and use that every unit modulo $24$ is its ...
- 5,894
8
votes
Accepted
Push-forward of nef divisors via finite morphisms
I suppose you want $f$ to be surjective, otherwise $f_*D$ is not defined. Then $f_*D$ is nef: for any curve $C\subset Y$, $\ (f_*D\cdot C)=(D\cdot f^*C)\geq 0$. But it might very well be ample. ...
- 35.7k
8
votes
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Pushforward of a very ample line bundle on a curve to $\mathbb{P}^1$
No, not in general. Take $C=\mathbb{P}^1$, $L=\mathcal{O}(1)$, $p$ to be map $x\mapsto x^2$ in affine coordinates. Then $p_*L$ has rank $2$, but
$$2=h^0(L)=h^0(p_*L)=h^0(\mathcal{O}(e_1))+h^0(\mathcal{...
- 32.7k
8
votes
Accepted
Volume of a divisor on a smooth projective surface
At least for effective divisors, the answer is strongly related to Zariski decomposition.
If $D$ is an effective divisor on a smooth surface $X$, Zariski proved in [Z62] that there exists a unique ...
- 63.9k
7
votes
Elementary Proof of Riemann-Roch for Compact Riemann Surfaces
There are (at least) 2 kinds of proofs: analytic ones (which use the existence of Abelian differentials with certain properties) and algebraic ones.
The proofs of the first kind use powerful analytic ...
7
votes
Elementary Proof of Riemann-Roch for Compact Riemann Surfaces
The proof given in Otto Forster, Lectures on Riemann Surfaces (Graduate Texts in Mathematics 81), chapter 16, seems very much suited to your list of prerequisites.
- 4,590
7
votes
Accepted
Pull-back of the canonical divisor via a rational map
There are several issues with this question.
One issue is that if $f$ is a rational map and not a morphism, then you have to say what you mean by $f^*$. Another issue is that if $K_Y$ is not at least ...
- 41.6k
7
votes
Accepted
The kernel of a nef line bundle
Consider $V=\mathbb{P}^1\times \mathbb{P}^2$ with projections $p_1\colon V \rightarrow \mathbb{P}^1 \text{ and } p_2\colon V \rightarrow\mathbb{P}^2.$ Let $L = p_1^*(\mathcal{O}_{\mathbb{P}^1}(1))$, ...
- 86
7
votes
Accepted
Infinitely small intersections with nef $\mathbb R$-Cartier divisors
This is possible. Basically, if $N$ is a point on the boundary of the nef cone that is not in the span of the rational points on the boundary, then we can approximate $N$ arbitrarily closely by ...
- 24.4k
7
votes
Picard group of symplectic group modulo orthogonal group
With the suggested choice of the symplectic and orthogonal form, there is a direct sum decomposition of $\mathbb{C}^{2n}$ into the sum of two Lagrangian (with respect to the both forms) subspaces:
$$
...
- 34.7k
7
votes
Accepted
Picard group of a cubic hypersurface
It is cyclic, generated by $\mathscr{O}(1)$. Indeed this is true for $X$ by the Lefschetz theorem (SGA2, Exp. XII, Cor. 3.7), and the restriction map $\operatorname{Pic}(X)\rightarrow \operatorname{...
7
votes
Picard group of a cubic hypersurface
Another way to find $\mathrm{Pic}(X)$ is the following. Note that the cubic $X$ is the symmetric determinantal cubic and it has a resolution of singularities
$$
\tilde{X} = \mathbb{P}_{\mathbb{P}^2}(S^...
- 34.7k
7
votes
Divisors whose restriction is big
Take $Y=\mathbb{P}^1$ and let $X=\mathbb{F}_n=\mathbb{P}(\mathcal{O} \oplus \mathcal{O}(-n))$ be the Hirzebruch surface with a section $C_0$ such that $C_0^2=-n$.
Let $H$ be an ample divisor on $\...
- 63.9k
7
votes
Accepted
Square root of a line bundle up to a finite surjective morphism
Assume $\mathcal{L}$ is associated with an effective Cartier divisor $D$. Let $D'$ be another Cartier divisor such that $D + D'$ is divisible by 2 in $\mathrm{Pic}(X)$. Let
$$
g \colon X' \to X
$$
be ...
- 34.7k
7
votes
A constructive proof of the theorem of the cube
This is not really an answer, but a rephrasing together with some comments on why this is difficult. I end with one example where you can actually compute something (purely algebraically) on $E \times ...
- 24.4k
6
votes
Accepted
Divisor on variety determined by its restriction to curves
If you assume projective and smooth, this is not hard. Induct on dimension, $\dim X=1$ being the hypothesis. If $\dim X=n\geq 2$ and result proved for smaller dimensions, if $Y\in \mathcal{O}(mH)$, $m&...
- 5,852
6
votes
Fibrations of projective varieties
Another proof: The assumptions imply that $X\times_YX$ is irreducible and that the two composite maps $X\times_YX\rightrightarrows X\xrightarrow{g}Z$ coincide over the generic point of $Y$. By density ...
- 19.1k
Only top scored, non community-wiki answers of a minimum length are eligible