## 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.

## ac.commutative algebra – Non-zero divisors on an $I$-completely flat module

Let $$A$$ be a commutative ring (not necessarily Noetherian), $$I=(f_1, f_2, dots, f_n) subseteq A$$ a finitely generated ideal that is weakly proregular, or better yet, generated by a regular sequence. Let $$M$$ be a derived $$I$$-complete $$A$$-module (which means $$mathrm{Ext}_A^{i}(A_{f_j}, M)=0$$ for all $$igeq 0$$), not necessarily finitely generated. Assume further that $$A$$ is (derived or classically) $$I$$-adically complete, and suppose further that $$M$$ is $$I$$-completely flat (sometimes also called $$I$$-adically flat), meaning that $$mathrm{Tor}_{>0}^{A}(A/I, M)=0$$ and $$M/IM$$ is a flat $$A/I$$-module.

My question is

If $$x in A$$ is a non-zero divisor of $$A$$, is $$x$$ necessarily a non-zero divisor on $$M$$?

This seems true at least under some further assumptions on $$x$$, such as $$xA cap I^k=xI^k$$ for all $$k$$, but I am actually interested in the somewhat perpendicular case when $$x in I$$, or better yet, the general question whether such $$I$$-completely flat modules are torsion-free.

## ag.algebraic geometry – Holomorphic tubular neighborhood of divisors at infinity

For the discussion of holomorphic tubular neighborhoods and some criteria for their existence see this question.

Let $$X$$ be a smooth quasi-projective variety over $$mathbb{C}$$. Hironaka tells us that there exists a smooth projective compactification $$X subset bar{X}$$, such that $$D =bar{X}backslash X$$ is a strictly normal crossing divisor. Let $$D=cup_{i}^N D_i$$ be its decomposition into irreducible smooth divisors. What is the obstruction to finding such $$bar{X}$$, such that each $$D_i$$ has a holomorphic tubular neighborhood in $$bar{X}$$?

Example 1: Let $$S$$ be a smooth projective variety, $$E$$ a vector bundle on $$S$$. The space $$X=text{Tot}(E)$$ is compactified to $$bar{X}=mathbb{P}_S(Eoplus mathcal{O}_S)$$ with the divisor at infinity $$D =mathbb{P}_S(E)subsetbar{X}$$. Its normal bundle should be $$mathcal{O}_{mathbb{P}(E)}(1)$$ and one should be able to find a holomorphic tubular neighborhood by constructing one in each fiber and gluing together.

Example 2: Using the answer by Joey, I have convinced myself that if the divisor $$D_i$$ is $$mathbb{P}^n$$ for some $$ngeq 2$$ and $$n+1=text{dim}(X)$$, then it also has such a neighborhood. If $$n=1$$, then one needs additionally that $$(D_i)^2<0$$.

The above examples make me believe that it could be always achievable, as each blow up at a smooth point introduces a new $$mathbb{P}^{n}$$ with normal bundle $$mathcal{O}(-1)$$.

## 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…

## abstract algebra – Group with exactly n elements of order n, then n has at most two prime divisors

This question is in a introductory abstract algebra book.
Let $$n > 1$$, and G a group with exactly $$n$$ elements of order $$n$$, show that at most two distinct primes divide $$n$$.

My attempt:
Let $$k$$ be the number of cyclic groups of order $$n$$, we can generate these subgroups from the $$n$$ elements of order $$n$$, if that’s the case then every one of those groups has exactly $$phi(n)$$ elements of order $$n$$ that can’t be shared with other of the $$k$$ cyclic subgroups.

Then as we have exactly $$n$$ element’s of order $$n$$ then it must happen that $$n = k cdot phi(n)$$ and that’s where I’m stuck. I tried expressing $$phi(n)$$ as $$prod_{p|n} (1 – frac{1}{p})$$ but i don’t see how I can get the “at most two primes” from there. I just arrived to $$k cdot prod_{p|n} (1 – frac{1}{p}) = 1$$

## algebraic curves – A question about principal divisors and poles

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## ag.algebraic geometry – Pullback of boundary divisors under forgetful maps

Let $$overline{mathbf{M}}_{0,n}$$ be the moduli space of stable $$n-$$pointed smooth rational curve of genus zero and $$overline{mathbf{U}}_{0,n}$$ the universal family described by $$pi_n:overline{mathbf{U}}_{0,n}longrightarrowoverline{mathbf{M}}_{0,n}$$ with the n disjoints sections $$sigma_i: overline{mathbf{M}}_{0,n}longrightarrow overline{mathbf{U}}_{0,n}.$$ Joachim Kock in “An invitatation to quantum cohomology” Example 1.5.11 gave the relationship between a boundary $$mathbf{F}_n$$ of $$overline{mathbf{M}}_{0,n}$$ with the boundary $$mathbf{F}_{n+1}$$ of $$overline{mathbf{M}}_{0,n+1}$$ by the following formula
$$mathbf{F}_{n+1} = varepsilon^*mathbf{F}_n + sum_{i}sigma_i$$
where $$varepsilon: overline{mathbf{M}}_{0,n+1}longrightarrow overline{mathbf{M}}_{0,n}$$ is the forgetful maps.
Can someone explain me more how to get the formula?

## List of divisors for a Python number

Hello, I am making a code that gives a number p, I find out what its positive divisors are and insert them in a list. My problem is when I try to add only the integers to the list, because the program returns an empty list to me. Follow my code below using number 10 as an example:

p = 10
d = ()
c = ()
for i in range(1,11):
a = p/i
d.append(a)
for i in range(len(d)):
if d(i) == int:
c.append(d(i))
print(c)


the result I would like would be for the program to return the list c = (10,5,2,1)

## What are all the units, zero divisors, and nilpotent elements of Z / < 24 >.

I'm having trouble finding nilpotent elements of Z /<24>.
There are none?
Units are 1 + Z /<24> y -1 + Z /<24>.
The divisors zero are 6 + Z /<24>, 4 + Z /<24>, 3 + Z /<24>, 8 + Z /<24>, 12 + Z /<24>, 2 + Z /<24>. I'm right?

## python – Euler 12: counting divisors of a triangular number

This is the question:

The sequence of triangular numbers is generated by adding the natural numbers. So the seventh triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …

Let's list the factors of the first seven triangular numbers:

 1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28


We can see that 28 is the first triangle number that has more than five divisors.

What is the value of the first triangle number that has more than five hundred divisors?

import math
def triangulated(num):
x = 0
for num in range(1, num + 1):
x = x + num
return x

l = ()

def factors(g):
for n in range(1, triangulated(g) + 1):
if triangulated(g) % n == 0:
l.append(n)
if len(l) > 500:
print(triangulated(g))
print(l)
l.clear()

for k in range(1, 10000000000):
factors(k)
print(k)


It helps to optimize this problem.