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?

Consider:

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.


Since
$$langle v_i|v_jrangle =langle v_j|v_irangle ^*rightarrow G_{ij}=G_{ji}^*rightarrow G=G^dagger$$
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
$$text{det}(G)=prod_ig_i$$
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 $xinoverline{K_P}$?