## ag.algebraic geometry – Distinguishing ample divisors by minimally intersecting curves on a projective simplicial toric variety

My question has an easily formulated generalization, which I will state first. Let $$sigma subseteq mathbf{R}^n$$ be a strongly convex polyhedral cone. For each minimally generating lattice point $$m in sigma^o cap mathbf{Z}^n$$ of the interior cone $$sigma^o subseteq sigma$$, let $$S(m) subseteq sigma^{vee} cap mathbf{Z}^n$$ denote the set of lattice points $$u$$ with $$langle u,m rangle = 1$$. My question is:

Does $$S(m) = S(m’)$$ imply that $$m = m’$$?

As a special case, assume that $$sigma$$ is the nef cone of a simplicial projective toric variety $$X_{Sigma}$$. Then my question seems to amount to the following:

If $$D_1$$ and $$D_2$$ are two ample divisors minimally generating in the ample cone, then does $$D_1 cdot C = 1 Leftrightarrow D_2 cdot C = 1$$ for all effective curves $$C$$ imply that $$D_1 = D_2$$?

This is the case I am most interested in.

## ag.algebraic geometry – Yoga on coherent flat sheaves \$mathcal{F}\$ over projective space \$mathbb{P}^n\$

I’m reading Mumfords’s Lectures on Curves on an Algebraic Surface (jstor-link: https://www.jstor.org/stable/j.ctt1b9x2g3)
and I found in Lecture 7 (RESUME OF THE COHOMOLOGY OF COHERENT SHEAVES ON
$$mathbb{P}^n$$; p 47) dealing with yoga on coherent
sheaves $$F$$ over pojective space $$mathbb{P}^n$$ I found on page 52
a proof I not understand:

Corollary 3: Given a coherent sheaf $$mathcal{F}$$ on $$mathbb{P}^n times S$$,
$$mathcal{F}$$ is flat over $$S$$
if and only if there exists an $$m_0$$ such that if $$m ge m_0$$,
$$p_* mathcal{F}(m)$$ is
locally free. Hence, in this case, the Hilbert polynomial of on
$$mathbb{P}^n _s$$ is locally constant.

Proof: If $$F$$ is flat over $$S$$, then let $$m_0$$ be large enough so
that derived image $$R^i p_*(mathcal{F}(m))=(0)$$, if $$i>0, m ge m_0$$.
Using Corollary $$1$$ and $$1 frac{1}{2}$$ one deduces that
$$p_*(mathcal{F}(m)) otimes k(s)$$ maps onto $$H^0(mathbb{P}^n _s, mathcal{F}_s(m))$$
for all $$sin S, m ge m_0$$. Then by (iii), $$p_* mathcal{F}(m)$$ is locally free.
As for converse (…)

In original:

Problem: The “…Then by (iii), $$p_* mathcal{F}(m)$$ is locally free…” part I not understand.
(iii) (on page 51) states:

By above we know $$p_*(mathcal{F}(m)) otimes k(s) to H^0(mathbb{P}^n _s, mathcal{F}_s(m))$$
is surjective, that is we can apply (iii) to $$i=1$$ and deduce
$$R^1p_*(mathcal{F}(m))$$ is locally free sheaf. But Mumford claims this for
$$p_* mathcal{F}(m)= R^0 p_* mathcal{F}(m)$$.

Is this an error in the proof or do I miss something?

## algebraic geometry – Genus of union of elliptic curve and projective line

Let $$Esubset mathbb P^2$$ be a smooth irreducible complex cubic curve and $$Lsubset mathbb P^2$$ a complex projective line in general position. Let $$C:=Ecup L$$ be their union.
The assumption that $$L$$ is of general position means that there are three intersection points, each of multiplicity one, and they are all nodes, i.e. $$C$$ is a nodal curve. Is $$C$$ a stable curve?

To check stability, one considers the normalisation $$widetilde C to C$$. The normalisation map is bijective on smooth points, whereas nodes have two preimages. The smooth curve $$widetilde C$$ consists of three components, each of them containing two preimages of the nodes. Then $$C$$ is stable iff all components of $$widetilde C$$ are stable. An irreducible smooth curve is stable iff it has genus at least 2, or genus 1 with at least one marked point, or genus 0 with three marked points. (In our case, the marked points are just preimages of nodes.)

My question is: what are the genera of irreducible components of $$widetilde C$$? It is clear that they should all have the same genera, and my intuition tells me that they should all have genus one.

## ag.algebraic geometry – Holomorphic map from a Riemann surface to a smooth projective variety with Zariski-dense image

Let $$V$$ be a connected smooth projective complex variety. Does there exist a holomorphic map $$Sto V$$ with Zariski-dense image where $$S$$ is a Riemann surface without boundary (possibly of infinite genus)?

If $$V$$ is unirational we can compose the dominant rational map $$mathbb{A}^{mathrm{dim}: X}to V$$ with the map $$mathbb{A}^1to mathbb{A}^{mathrm{dim}:X}$$ given by $$zto big(e^z, e^{e^z}, dotsbig)$$ after possibly throwing out finitely points from $$mathbb{A}^1$$. If $$V$$ is an abelian variety the same idea works if we consider the surjective holomorphic (non-algebraic) map $$mathbb{A}^{mathrm{dim}:X}to V$$. It is not clear to me what happens for K3 surfaces.

## ag.algebraic geometry – Kan liftings and projective varieties

Regard the following two bicategories:

• $$operatorname{dg-mathcal{B}imod}$$, with objects dg categories, and morphisms categories from $$C$$ to $$D$$ being the categories of $$C$$$$D$$-bimodules. Composition is given by the dg tensor product. Note that this might be considered as the bicategory of Chain-complex-enriched profunctors/ relators (see Coend Calculus), so the machinery developed in this paper should (as chain complexes form a Bénabou-cosmos) show that every 1-morphism in this category has both a left and a right adjoint, and Left Kan Liftings exist here
• $$mathcal{V}ar$$, with objects smooth projective varieties and morphism categories from $$X$$ to $$Y$$ the derived categories $$D^b (X times Y)$$, as considered here. The 1-morphisms here encode kernels for Fourier-Mukai-Transformations. Using the Serre kernel, it is shown in that paper that every 1-morphism in this bicategory has a left and a right adjoint.

These bicategories are very closely related; it was shown e.g. by Orlov that if one takes the dg enhancements of the derived categories $$D^b_{dg}(X)$$, $$D^b_{dg}(Y)$$, then the dg functors between those are in one-to-one correspondence with the elements of $$D^b(X times Y)$$, to name just one similarity. Therefore, I would suspect that also the second category possesses Left Kan Liftings, but I am not sure how to construct them (as already the proof that the Serre kernel lets us construct adjoints there is very nontrivial).

Is this indeed true? How are they constructed? And most importantly (as from the fact that $$mathcal{V}ar$$ has adjoints, a lot can be deduced about Grothendieck duality etc.) do they also have interesting uses in the study of smooth projective varieties, or can their construction even be extended to more general schemes?

## Rational sections of projective bundle

Why does a projective bundle $$mathbb{P}(E) to X$$, with $$E$$ a vector bundle on $$X$$, always have rational sections? How do you construct them?

## algebraic geometry – Smooth projective surface whose hyperplane sections are elliptic curves is ruled

This is a problem from chapter 6 in Beauville’s book Complex Algebraic Surfaces. I have a smooth projective surface $$S$$ whose smooth hyperplane sections $$H$$ are elliptic curves. I want to show that $$S$$ is either a del Pezzo surface or is an elliptic ruled surface.

I want to proceed by the hint in the book. I have managed to show that $$q=h^1(S,mathcal O_S)le 1$$ by looking at two holomorphic 1-forms, restricting them to the hyperplane sections, doing some stuff and getting that they have to be proportional everywhere. Now I want to take care of the cases $$q=0$$ and $$q=1$$ separately. When $$q=0$$, Beauville suggests that I show that $$Kequiv -H$$, i.e. that the canonical class is the negative of $$H$$ in $$mathrm{Pic}(S)$$. This would solve this part of the problem because I have a result from an exercise a while ago saying if $$Kequiv -H$$ then $$S$$ is a del Pezzo surface. The problem is that I’m not sure how to go about showing this.

The best I can think of is to show that $$H=-K$$ in cohomology, which would suffice because the Chern class map $$mathrm{Pic}(S)to H^2(S,mathbb Z)$$ is an injection since $$q=0$$. However I’m not sure how this helps. To be frank, I don’t know how to go abut showing two divisor classes are the same if I don’t have some explicit description for them.

For the case $$q=1$$, I’m hoping that the previous part will show me that $$H.K<0$$ in general or something, as this would imply that $$S$$ is ruled. I haven’t thought too much about this part to be honest.

Any help is appreciated. I’m more looking for a hint or nudge that will get me thinking in the right direction, but even just ideas which are not fleshed out at all could be helpful.

## ag.algebraic geometry – On relating \$l(A), l(B)\$ and \$l(A+B)\$ for Weil divisors on a smooth projective curve where one of the divisors is effective

Let $$X$$ be a smooth projective curve over an Algebraically closed field $$k$$. Let $$k(X)$$ denote its function field.

If $$A, B$$ are Weil divisors on $$X$$ such that $$A$$ is effective (i.e. $$Age 0$$) , then is there any (in)equality between $$l(A+B)$$ and $$l(B), l(A)$$ ?

Here, for a Weil divisor $$D$$ on $$X$$, by $$l(D)$$ we denote the $$k$$-vector space dimension of the Riemann-Roch space $$L(D):={fin k(X)^*: D+ div(f)ge 0}cup {0}$$.

For a divisor $$D$$ on $$X$$, the complete linear system $$|D|$$ be the collection of all effective divisors which are linearly equivalent with $$D$$. $$|D|$$ can be given the structure of a projective space by identifying it with $$( L(D)setminus {0})/k^*$$ and by that structure, $$dim |D|=l(D)-1$$. Now it is known (Hartshorne, Chapter IV, Lemma 5.5) that if $$D,E$$ are both effective divisors, then $$dim |D|+dim |E|le dim |D+E|$$ i.e. $$l(D)+l(E)le l(D+E)+1$$ . What I’m basically asking is that if something similar holds if we assume only one of the divisors is effective…

## ac.commutative algebra – Second summand to make projective module free

Suppose there’s a projective $$R$$-module $$P$$ (non-free). We know that there is another $$R$$-module $$M$$ such that $$Poplus M$$ is free over $$R$$. Is there a way to write down such an $$M$$ in terms of $$P$$?

If this is not always tractable, is it possible in certain specialized circumstances? The setting that comes to mind is where $$P$$ is a non-principal ideal of the ring of integers of a number field (as Wikipedia says this is an instance of $$P$$ being projective but not free).

Thanks!

(This was originally posted on math.stackexchange but I realized that here might be the better place to ask, apologies if not.)

## ac.commutative algebra – Example of a projective module with non-superfluous radical

Let $$R$$ be a ring with unit. A submodule $$N$$ of an $$R$$-module $$M$$ is called superfluous if the only sumbodule $$T$$ of $$M$$ for which $$N+T = M$$ is $$M$$ itself.

It is shown, for example, in

(1) F. W._Anderson, K. R. Fuller “Rings and Categories of Modules” (1974)

that if every submodule of $$M$$ is contained in a maximal submodule, then the radical of $$M$$ is superfluous (Proposition 9.18). This, in particular, implies that for every finitely generated module $$M$$ its radical is superfluous. In exercise 9.2. it is explained that divisible abelian groups coincide with their radicals, and therefore their radicals are not superfluous. Divisible abelian groups are not projective objects.

I was curious if it is possible to construct a projective module with non-superfluous radical.

Question: is there an example of a ring $$R$$ and a projective $$R$$-module
$$P$$ such that the radical $$JP$$ of $$P$$ is not superfluous?

The existence of such modules (or, at least, that its non-existence is non-obvious) is somehow hinted by the formulation of Corollary 17.12 in (1):

Let $$J = J(R)$$. If $$P$$ is a projective left $$R$$-module such that $$JP$$
is superfluous in $$P$$ (e.g., if $${}_RP$$ is finitely generated), then
$$J(End({}_RP)) = Hom_R(P,JP)$$ and $$End({}_RP)/J(End_RP) cong End({}_RP/JP)$$.