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40 votes
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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 ...
roy smith's user avatar
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27 votes
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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.
Will Sawin's user avatar
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18 votes
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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 ...
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,...
David E Speyer's user avatar
12 votes
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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 ...
R. van Dobben de Bruyn's user avatar
11 votes
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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 ...
Mohan Ramachandran's user avatar
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 ...
K.K.'s user avatar
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9 votes
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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 ...
Chen Jiang's user avatar
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9 votes
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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, ...
Sasha's user avatar
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9 votes
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A question on "Ample subvarieties of algebraic varieties"

I suspect that you are supposed to view the projective variety $X$ as being given with a chosen projective embedding $X\subset \mathbb P^n$, and therefore a distinguished ample divisor $\mathcal{O}_X(...
Tom Ducat's user avatar
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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 ...
so-called friend Don's user avatar
8 votes
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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. ...
abx's user avatar
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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{...
Donu Arapura's user avatar
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8 votes
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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 ...
Francesco Polizzi's user avatar
8 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 ...
R. van Dobben de Bruyn's user avatar
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 ...
Alexandre Eremenko's user avatar
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.
R.P.'s user avatar
  • 4,745
7 votes
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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 ...
Sándor Kovács's user avatar
7 votes
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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 $L = p_1^*(\mathcal{O}_{\mathbb{P}^1}(1))$, ...
Chiles's user avatar
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7 votes
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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 ...
R. van Dobben de Bruyn's user avatar
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: $$ ...
Sasha's user avatar
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7 votes
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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^...
Sasha's user avatar
  • 37.6k
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 $\...
Francesco Polizzi's user avatar
7 votes
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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 ...
Sasha's user avatar
  • 37.6k
6 votes
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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&...
Mohan's user avatar
  • 6,137
6 votes

existence of birational morphism and divisors

I don't know where to find a proof written, but it is not hard to give one here. One direction is easy. If $S \rightarrow \mathbf P^2$ is a birational morphism, then let $D$ be the pullback of a ...
Pop's user avatar
  • 867
6 votes

Global sections of multiples of a divisor

If $c = 2$, let $D',D'' \in H^0(X,kD)$ be different divisors. Then $2D',D'+D'',2D''$ are three different divisors in $|2kD|$. Similarly, the case $c > 2$ is impossible. So, assume $c = 1$. Let $D' \...
Sasha's user avatar
  • 37.6k
6 votes
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Cohomology of divisors on Hirzebruch surfaces

Sure. To compute $H^0$ one can first pushforward to the base $\mathbb{P}^1$. If $a \ge 0$ one obtains $$ p_*\mathcal{O}(a\Gamma + bF) \cong p_*\mathcal{O}(a\Gamma) \otimes \mathcal{O}(b) \cong S^a(\...
Sasha's user avatar
  • 37.6k
6 votes
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Two conditions on divisors on surfaces

Neither condition implies the other. $(i)\, \not\!\Rightarrow\, (ii)$: Take for $X$ a surface with $K$ ample, but $h^1(K)=h^1(\mathscr{O}_X)>0$ (e.g. a product of 2 curves of genus $>1$). Then ...
abx's user avatar
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