37 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 ...
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  • 11.5k
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.
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  • 114k
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 ...
16 votes
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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 ...
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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 ...
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  • 19.1k
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,...
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11 votes
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Bertini's Theorem

As pointed out by Alex in his comment, this is in general not true. For instance, consider the case $N=d=n=m=2$. Then $|L|$ is the linear system of plane curves of degree $2$ passing through $p_1$ ...
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11 votes
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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 ...
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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 ...
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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 ...
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  • 666
11 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 ...
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9 votes
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Cone over the Veronese surface

Let me start being a little nitpicking with the formulation of the question. The fact that $X$ is $\mathbb Q$-factorial does not in itself imply that such $a$ and $b$ exists. One also needs the fact ...
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9 votes

Cone over the Veronese surface

The answers are the following. (1) It is well known that the singularity at the vertex of the cone over the Veronese surface is isomorphic to a quotient singularity of type $\frac{1}{2}(1, \, 1, \,1)$...
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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 ...
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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, ...
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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 ...
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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{...
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7 votes
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Simple normal crossing divisors

You gave the definition of normal crossing divisor. The definition of simple normal crossing is the following. A Weil divisor $D = \sum_{i}D_i \subset X$ on a smooth variety $X$ of dimension $n$ is ...
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  • 6,653
7 votes
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Top self-intersection of exceptional divisors

So you want to compute the intersection numbers $(H^p\cdot E^q)$, $p+q=n$. Let me start with some notation. Let $b: X\rightarrow \mathbb{P}^n$ be the blowing up, $i:E \hookrightarrow X$ the ...
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7 votes
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Blowing-up a point in the singular locus

Note that without any assumption on the singularities the answer to your question is yes. For instance, consider the hypersurface $Y = \{x_0^2+x_1^3+x_2^4\}\subset\mathbb{P}^3$. Then $Sing(Y) = [0:0:0:...
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  • 6,653
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 ...
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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.
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  • 4,320
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 ...
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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 $L = p_1^*(\mathcal{O}_{\mathbb{P}^1}(1))$, ...
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7 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. ...
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  • 34.3k
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 ...
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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: $$ ...
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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
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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 ...
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