44
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 ...
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.
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 ...
Community wiki
15
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,...
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 ...
11
votes
Accepted
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 ...
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 ...
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, ...
9
votes
Accepted
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(...
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 ...
8
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 ...
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. ...
8
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 ...
8
votes
Accepted
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
$$2=h^0(L)=h^0(p_*L)=h^0(\mathcal{O}(e_1))+h^0(\mathcal{...
8
votes
Accepted
Volume of a divisor on a smooth projective surface
At least for effective divisors, the answer is strongly related to Zariski decomposition.
If $D$ is an effective divisor on a smooth surface $X$, Zariski proved in [Z62] that there exists a unique ...
8
votes
A constructive proof of the theorem of the cube
This is not really an answer, but a rephrasing together with some comments on why this is difficult. I end with one example where you can actually compute something (purely algebraically) on $E \times ...
7
votes
Accepted
The kernel of a nef line bundle
Consider $V=\mathbb{P}^1\times \mathbb{P}^2$ with projections $p_1\colon V \rightarrow \mathbb{P}^1 \text{ and } p_2\colon V \rightarrow\mathbb{P}^2.$ Let $L = p_1^*(\mathcal{O}_{\mathbb{P}^1}(1))$, ...
7
votes
Elementary proof of Riemann-Roch for compact Riemann surfaces
The proof given in Otto Forster, Lectures on Riemann Surfaces (Graduate Texts in Mathematics 81), chapter 16, seems very much suited to your list of prerequisites.
7
votes
Accepted
Pull-back of the canonical divisor via a rational map
There are several issues with this question.
One issue is that if $f$ is a rational map and not a morphism, then you have to say what you mean by $f^*$. Another issue is that if $K_Y$ is not at least ...
7
votes
Picard group of symplectic group modulo orthogonal group
With the suggested choice of the symplectic and orthogonal form, there is a direct sum decomposition of $\mathbb{C}^{2n}$ into the sum of two Lagrangian (with respect to the both forms) subspaces:
$$
...
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
$$
\tilde{X} = \mathbb{P}_{\mathbb{P}^2}(S^...
7
votes
Divisors whose restriction is big
Take $Y=\mathbb{P}^1$ and let $X=\mathbb{F}_n=\mathbb{P}(\mathcal{O} \oplus \mathcal{O}(-n))$ be the Hirzebruch surface with a section $C_0$ such that $C_0^2=-n$.
Let $H$ be an ample divisor on $\...
7
votes
Accepted
Cohomology of divisors on Hirzebruch surfaces
Sure. To compute $H^0$ one can first pushforward to the base $\mathbb{P}^1$. If $a \ge 0$ one obtains
$$
p_*\mathcal{O}(a\Gamma + bF) \cong
p_*\mathcal{O}(a\Gamma) \otimes \mathcal{O}(b) \cong
S^a(\...
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 ...
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 ...
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' \...
6
votes
Accepted
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 ...
6
votes
Accepted
Relation between canonical bundles under étale maps
The usual definition of $\omega_X^\bullet$ is $\pi_X^!(k[0])$, where $\pi_X \colon X \to \operatorname{Spec} k$ is the structure morphism. If $f \colon X \to Y$ is a map of $k$-schemes, we therefore ...
5
votes
Elementary proof of Riemann-Roch for compact Riemann surfaces
I wrote up notes for the 4 lectures I did going through a completely algebraic proof at the end of a Shavarevich based algebraic geometry course. I think this is a nice approach in that it introduces ...
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