reference request – D-modules as ind-coherent sheaves over positive characteristics?

There is an interpretation of D-modules over “sufficiently nice” prestacks $X$ (read: various finiteness conditions apply, perhaps even smoothness) by Gaitsgory and Rozenbylum (see chapter I.4 here and this paper), in which one views D-modules as ind-coherent sheaves (i.e. filtered colimits of coherent sheaves) over so-called de Rham spaces $X_{dR}$ attached to the previously mentioned “nice” prestacks $X$. As I understand it, this is essentially using the fact that each crystal in quasi-coherent sheaves comes canonically equipped with a flat connection, and thus can be seen as a D-module; the approach by Gaitsgory-Rozenblyum is therefore a version of infinitesimal cohomology (in the sense of Grothendieck-Ogus) wherein establishing the six functors is somewhat easier, as now the six functors for D-modules can be deduced from the general theory of ind-coherent sheaves. There is, however, a caveat: our prestack $X$ has to be an object in characteristic $0$, and preferably over a field of characteristic $0$, as smooth schemes over fields are automatically reduced.

Now I am aware of the fact that there is also a theory of “arithmetic” D-modules, developed by Berthelot, wherein one replaces infinitesimal sites and all the businesses involving de Rham spaces with crystalline sites, whose objects are pd-immersions and whose coverage is the usual Zariski coverage. Given the somewhat ad hoc definition of pd-structures, is it also possible (at least in principle) to reformulate the theory of arithmetic D-modules in the style of Gaitsgory and Rozenblyum ? Have there been any attempts at this, and if this is not possible, why so ?

rootfinding – How many positive roots can $sum_{i}frac{a_i}{x+b_i}$ have where $b_i$’s are all positive?

What is the maximum number of positive roots $sum_{i}^Nfrac{a_i}{x+b_i}$ can have where $b_i$‘s are all positive? (everything here is a real number. To provide context, I encountered this problem while doing theoretical neuroscience research where I am modeling a biological neuronal network as an artificial neural network.)

$x$ is our variable, $a_i$ is a constant (can be either positive or negative), and $b_i$ is always a positive constant. $a_i$ and $b_i$ have unique values at each $i$.

In other words, how many $x>0$ can satisfy $sum_{i}^Nfrac{a_i}{x+b_i}=0$?

If $a_i$ happens to be all positive or negative, I see that there are no roots at positive $x$. For example, the following shows $y=frac{1}{x+1}+frac{1}{x+2}+frac{1}{x+3}$. You can see that there are poles at -1, -2, and -3 (which are $-b_i$‘s), and the roots exist between the poles. Since $b_i$‘s are all positive, the roots between the poles need to be all negative.

enter image description here

However, if $a_i$‘s are a mix of positives and negatives (and $b_i$‘s are still all positive), there can be root(s) outside the poles, making it possible to have a root when $x>0$. For example, $y=frac{1}{x+1}+frac{1}{x+2}-frac{3}{x+3}$ is $0$ at $x>0$ as shown below:

enter image description here

If I zoom in to the positive $x$ part, we see the following:

enter image description here

It overshoots below 0, and then asymptotically approaches 0.

So far, no matter how large my $N$ is, a randomly generated function $sum_{i}^Nfrac{a_i}{x+b_i}$ seemed to have only one root at positive $x$, if there was any, when I swept through $x$ on my computer. However, I still believe the number of positive roots should be dependent on $N$. Any thoughts?

equation solving – Why isn’t this expression returning true to being positive when it is clearly positive?


Expre10 = (B^2/C + 2 B + B^2/D + (B C)/D + (B D)/C) + 
   1 (C + D) - (A (Beta) (Sigma))/(B C D) (C + D);
Assuming({A > 0, B > 0, C > 0, 
  D > 0, (Beta) > 0, (Sigma) > 0, (A (Beta) (Sigma))/(B C D) <=  
   1}, Simplify(Expre10 > 0))

Returns the inequality:

B (B + C) (B + D) > A (Beta) (Sigma)

But we clearly see if (A (Beta) (Sigma))/(B C D) <= 1 holds then our inequality will always be true, so why am I getting the wrong output?

real analysis – Mid-point convex measurable subset of $mathbb{R}$ with positive Lebesgue measure is an interval

The question is right as the title:

Let $E$ be a measurable subset of $mathbb{R}$ w.r.t. Lebesgue measure, and has positive measure. For any $x,yin E$, $frac{x+y}{2}in E$. Prove that $E$ is an interval(like $(a,b),(a,b),(a,b)$ etc., possibly infinity endpoint).

The hint is to find some function on it, and I tried looking at its characteristic function, but I have no clue.

factorial – If n is a positive integer that is four digits long and is relatively prime to 100!, why must n be prime?

Suppose there is some positive integer n that is four digits long and is relatively prime to 100! (meaning n and 100! have no common factors other than 1). n must be prime, but why?

100! is a composite number, but composite numbers can be relatively prime to other composite numbers, so that can’t be the reason n is prime. n being four digits long and 100! having factors that are all less than four digits must have something to do with it, but I can’t wrap my head around the exact reason.

So, why does n have to be prime in this situation?

linear algebra – Inner product matrix have positive determinant

Suppose $|v_1rangle, |v_2rangle ,cdots |v_krangle in S_N$ and suppose a matrix $G$ form by the inner product of these vectors
$$G_{ij}=langle v_i|v_jrangle $$
I’m trying to prove that $text{det} (G)geq 0$ where equality hold for if vectors are linearly dependent.

$$langle v_i|v_jrangle =langle v_j|v_irangle ^*rightarrow G_{ij}=G_{ji}^*rightarrow G=G^dagger$$
this means that eigenvalues will be positive. In the diagonal form,
$$G=text{diag}(g_1,g_2,cdots ,g_k)$$
The determinant can be written as
To prove this is positive, We can prove that all the eigenvalues are positive but that is much more restrictive. I’m not sure, How do I proceed? Please help me with this.

algebraic number theory – Extension of morphism in local fields of positive characterisic

Consider $theta:mathbb F_q(T)mapstomathbb F_q(T)$ defined by $theta(Q)=Q(T^q)$. It is a morphism of fields. Let $P$ be an irreducible polynomial of $mathbb F_q(T)$. Then, $theta$ can be uniquely extended to $mathbb F_q(T)_P$ by continuity (denote it $theta$ again), where $mathbb F_q(T)_P$ is the completion of $mathbb F_q(T)$ for the topology induced by the $P$-valuation . It is a morphism of fields yet and for every $xin K_P$, one has $theta(x)=x^q$.
My question: can $theta$ be extended continuously to $Omega_P$, the topological closure of an algebraic closure $overline{K_P}$ of $K_p$ such that $theta$ is a morphism of fields in $overline{K_P}$ and $theta(x)=x^q$ for any element $xinoverline{K_P}$?