Recent MathOverflow Questions

Markov chain proof about stationary distributaion [migrated]

Math Overflow Recent Questions - Mon, 01/08/2018 - 14:53

i have proof question that i cant seem to understand why it is true, it goes like this:

For a finite ergodic markov chain Xn with J states (1,2,..,J) and transition matrix P. show that if we find vector (X1,X2,..,XJ) such that SUM(Xi)=1 and XiPij=XjPji for every i,j then (X1,X2,..,XJ) is the stationary distribution.

now, i tried to check for matrix of size 2/3 with parameters as matrix element and to find some sort of law that for it i will be able to proof using induction (dont really think that this is the way),

what i was able to get for now: for Matrix of size n we want to solve the equation: V(P-I)=0 {V is the vector (X1,X2,..,XJ)}

so we will have n equations moreover we have n relations XiPij=XjPji and one more for SUM(Xi)=1

so we have 2n+1 equations, but i dont understand how does it helps me

anyone has some ideas how to get forward with that?

Gambler's ruin: The fair game is the longest

Math Overflow Recent Questions - Mon, 01/08/2018 - 14:30

Consider a gambler who, in every trial of a game, wins or loses a dollar with probability $p\in\left( 0,1\right) $ and $q=1-p$, respectively. Let his initial capital be $z>0$ and let him play against an adversary with the same capital $z>0$. The game continues until one of the players is ruined.

In the language of random variables, this amounts to consider a sequence $X_{1}^{\left( p\right) },X_{2}^{\left( p\right) },...\ $of random variables (on a sample space $\Omega$) taking on two values $+1$ and $-1$ with probabilities $\Pr\left[ X_{n}^{\left( p\right) }=+1\right] =p$ and $\Pr\left[ X_{n}^{\left( p\right) }=-1\right] =q$. In particular, $X_{n}^{\left( p\right) }$ describes the gambler's gain on the $n$th trial and $$ S_{n}^{\left( p\right) }\left( \omega\right) =X_{1}^{\left( p\right) }\left( \omega\right) +\cdots+X_{n}^{\left( p\right) }\left( \omega\right) ,\quad S_{0}^{\left( p\right) }\left( \omega\right) \equiv0\qquad\forall \,\omega\in\Omega $$ describes his net cumulated gain after $n$ trials. The duration of the game is the random number of trials before he is either ruined or wins the game (and his adversary wins or is ruined) $$ D^{\left( p\right) }\left( \omega\right) =\min\left\{ t\in\mathbb{N}% :\left\vert S_{t}^{\left( p\right) }\left( \omega\right) \right\vert =z\right\} \qquad\forall \,\omega\in\Omega. $$

Q. I need a reference or a simple proof of the folk result according to which, for each $p\in\left( 0,1\right) $, $$ \Pr\left[ D^{\left( p\right) }\leq s\,\right] \geq\Pr\left[ D^{\left( 1/2\right) }\leq s\,\right] \qquad\forall\, s=1,2,... $$

von Neumann's characterization of orthogonally invariant matrix norms

Math Overflow Recent Questions - Mon, 01/08/2018 - 14:10

Von Neumann has a result (rather well-known in convex analysis circles) which states that every orthogonally invariant matrix norm (meaning $\| P M Q\| = \| M \|,$ for any orthogonal $P, Q$)) is a symmetric gauge of the singular values. This is a very nice and very useful result, for which the bibtex reference is:

@Misc{zbMATH02519847, Author = {J. {von Neumann}}, Title = {{Some matrix inequalities and metrization of matric space.}}, Year = {1937}, Language = {English}, HowPublished = {{Mitt. Forschungsinst. Math. Mech. Kujbyschew-Univ. Tomsk 1, 286-300 (1937).}}, Zbl = {63.0037.03} }

The question is really one of history: how did von Neumann (who was already in Princeton at the time) decide to publish in a Tomsk University proceedings? Tomsk is very far from anywhere in Russia, and I am not aware of anything else that appeared there.

Opens maps from topological manifolds whose fibers are not generically topological foliations

Math Overflow Recent Questions - Mon, 01/08/2018 - 12:18

This is a crosspost of this MSE question. It asks about how far open maps are from being "generically foliations".

Proposition. The quotient map associated to a topological foliation (projecting to the leaf space) is open.

However the fibers of an open map from a topological manifold need not foliate its domain.

Example. Consider the multiplication map $\mathbb R^2\to \mathbb R,(x,y)\mapsto xy$. It is open, but I think its fibers do not give a topological foliation of the plane because the fiber over zero is the union of the axes, which does not look like parallel lines locally about the origin.

In the above example, the problem is at an isolated point. That leads me to wonder how far the fibers of an open map can be from "generically" foliations.

Definition. Let $X$ be a topological manifold. A partition of $X$ is generically a foliation of $X$ if it restricts to a foliation of an open dense subspace (which is also a topological manifold).

Example. I think the fibers of the multiplication maps give a generic foliation of the plane as they foliate the punctured plane.

I'm fairly certain a map (set function even) $f:X\to Y$ of topological spaces is open iff for every convergent net $y_\alpha\to y$ we have $f^{-1}(y)\subset\lim \left\{ f^{-1}(y_\alpha ) \right\}$, so the fibers of an open map are "stuck together" in this sense. I would like to understand exactly how different the latter sense is from the fibers actually being generically foliating.

Question. What are some examples of a (continuous) open map from a topological manifold $X$ whose fibers are not generically a foliation of $X$? What are some examples if we moreover assume the map is a quotient map?

Remark. I am looking for maximally "geometric" (i.e minimally "fractal") examples.

density in Wasserstein space

Math Overflow Recent Questions - Mon, 01/08/2018 - 11:57

I am wondering whether the following result is true:

Let $\mathcal W_p(\mathbb R^d)$ be the Wasserstein space of order $p$ and let $\eta$ and $\gamma$ be two probability measures in $\mathcal W_p(\mathbb R^d)$, such that supp($\eta$) $\subset$ supp($\gamma$). Then there is a sequence of probability measures $(\eta_n)$ that converges to $\eta$ in $\mathcal W_p(\mathbb R^d)$ and such that $\eta_n$ is absolutely continuous w.r.t. $\gamma$, with $\frac{d\eta_n}{d\gamma}$ being bounded.

It seems to be a well-known result but I was not able to prove it nor to find a reference. Any help and suggestion are appreciated!

Clarification of the claim 3 in the proof of the irrationality of $\pi ^2$ [on hold]

Math Overflow Recent Questions - Mon, 01/08/2018 - 11:13

I am reading Laczkovich's proof (can be found here:)

Let $f_k(x) = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \qquad (k\notin\{0,-1,-2,\ldots\})$

Claim 3: If $x\neq 0$ and if $x^2$ is rational, then $(\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(x)\neq0\text{ and }\frac{f_{k+1}(x)}{f_k(x)}\notin\mathbb{Q}.$

And there is this part that I didn't understand:

When he proves this claim he starts with.. Otherwise, there would be a number $y\neq 0$ and integers $a$ and $b$ such that $f_{k}(x)=ay$ and $f_{k+1}(x)=by$. In order to see why, take $y=f_{k+1}(x)$, $a=0$ and $b=1$ if $f_{k}(x)=0$; otherwise, choose integers $a$ and $b$ such that ${f_{k+1}(x)\over f_{k}(x)}={b\over a}$ and define $y= {f_{k}(x)\over a}={f_{k+1}(x)\over b}$. $(*)$

first of all does he assume here that $f_k(x)=0\text{ and }\frac{f_{k+1}(x)}{f_k(x)}\in\mathbb{Q}?$ When he writes this "there would be a number $y\neq 0$ and integers $a$ and $b$ such that $f_{k}(x)=ay$ and $f_{k+1}(x)=by$ "is it because he assumes what if $\frac{f_{k+1}(x)}{f_k(x)}\in\mathbb{Q}?$ but does he also assume that what if $f_k(x)=0\text ?$

I mean what he is doing here is the use of contrapositive of the claim3? I just want to know how is the process of thinking made here. Can someone explain/paraphrase the reasoning in $(*)?$ I am literally stuck there. Thanks in advance

When is the Thom spectrum of a virtual vector bundle effective?

Math Overflow Recent Questions - Mon, 01/08/2018 - 11:00

Remark: My question is valid in the classic setting of the stable homotopy category of spectra of CW-complexes. An answer on that setting will also be valid.

Denote as $SH(X)$ Voevodsky's stable homotopy category over a scheme $X$. Denote $SH(X)^{\mathrm{eff}}$ its effective variant, that is to say, the smalles triangulated subcategory of $SH(X)$ which is closed under direct sums and contains suspension spectra of spaces but not their desuspensions.

Let $V\to X$ be a vector bundle. Denote as $\mathrm{Th}(V)$ its Thom space, and denote as well its infinite supension, which belongs to $SH(X)$. More concretely, it belongs to $SH(X)^{\mathrm{eff}}$.

Let $\xi$ be a virtual vector bundle over $X$ of rank $r\in \mathbb{Z}$ and denote $\mathrm{Th} (\xi)$ its associated Thom spectrum (cf. section 4.1 of this paper). My question is:

Is it true that $$ \mathrm{Th}(\xi) \in SH(X) ^{\mathrm{eff}}\Leftrightarrow r\geq 0 \hspace{.5cm} ?$$

Low-dimensional irreducible 2-modular representations of the symmetric group

Math Overflow Recent Questions - Mon, 01/08/2018 - 10:08

I apologize if this question is a little too basic for MathOverflow, but it's somewhat outside of my background and I'm frustrated that the answer doesn't seem to be explicit in the literature even though I suspect it's easily known by the experts.

What are the lowest-dimensional irreducible representations over $\mathbb{F}_2$ of the full symmetric group $S_n$ and in particular whether there exist any of dimension $\leq n - 2$ (resp. $\leq n - 3$) if $n$ is odd (resp. even)?

I know that for each $n \geq 3$ we have the $(n - 1)$-dimensional standard representation over any field. Moreover, if $n$ is even, we get an $(n - 2$)-dimensional representation as a quotient of this over characteristic $2$ (because then the vector with all $1$'s lies in the standard representation but is fixed by all of $S_n$). It's easy to see that these representations are irreducible. What I would specifically like to know is whether or not they are the lowest-dimensional irreducible representations of $S_n$ over $\mathbb{F}_{2}$.

(I'm curious about slightly more general questions as well, like whether or not the analogous representations to the ones I described above are the lowest-dimensional over any given positive characteristic.)

Limit associated with complementary sequences

Math Overflow Recent Questions - Mon, 01/08/2018 - 08:09

Define $A=(a_n)$ and $B=(b_n)$ as follows: $a_0=1$, $a_1=2$, $b_0=3$, $b_1=4$, and $$a_n=a_0b_{n-1}+a_1b_{n-2}$$ for $n \geq 2$, where $A$ and $B$ are increasing and every positive integer occurs exactly once in $A$ or $B.$ Can someone prove that $\lim_{n \to \infty} a_n/n = 4$?

Here are first terms and some evidence regarding the limit: $$A = (1,2,10,13,16,19,22,25,29,34,38,43,47,52, \dots)$$ $$B = (3,4,5,6,7,8,9,11,12,14,15,17,18,20,21,23, \dots)$$ I've checked that $-9<a_n-4n<9$ for $n=1,\dots,10^6$.

Iterated forcing and the super tree property at $\omega_2$

Math Overflow Recent Questions - Mon, 01/08/2018 - 07:41

It is a theorem of Baumgartner and Laver that iterating Sacks forcings of weakly compact length gives rise to the tree property at $\omega_2$. Natural questions (at least for me) are: do we get stronger tree properties at $\omega_2$ if we start with larger cardinals? In particular, if the length is strongly compact, do we get the strong tree property at $\omega_2$? If the length is supercompact, do we get the super tree property at $\omega_2$? I suspect it is known so I appreciate pointers to references.

For uncountable cardinals $\kappa\leq \lambda$ with $\kappa$ being regular, we say $F\subset \bigcup_{u\in [\lambda]^{<\kappa}} 2^u$ is a $(\kappa,\lambda)$-tree if it is closed under taking restrictions, and for each $u\in [\lambda]^{<\kappa}$, $|F_u:=\{f\in F: dom(f)=u\}|<\kappa$. A function $d: \lambda\to 2$ is a cofinal branch if for all $u\in [\lambda]^{<\kappa}$, $d\restriction u\in F_u$. Given $\bar{f} = \langle f_u\in F: dom(f_u)=u\rangle$ (called a level sequence), we say $d: \lambda\to 2$ is an $\bar{f}$-ineffable branch if $\{u\in [\lambda]^{<\kappa}: d\restriction u = f_u\}$ is stationary in $[\lambda]^{<\kappa}$.

We say $\kappa$ has the strong tree property if for any $\lambda\geq \kappa$, any $(\kappa, \lambda)$-tree has a cofinal branch.

We say $\kappa$ has the super tree property if for any $\lambda\geq \kappa$, any $(\kappa, \lambda)$-tree $F$ and any level sequence $\bar{f}$, there exists an $\bar{f}$-ineffable branch.

Edit: I just realized the answer in the case where the length is strongly compact is yes: the countable support iteration of Sacks forcings of strongly compact length forces the Semistationary Reflection Principle (equivalent to the statement that all stationary preserving forcings are semiproper and in fact stronger statement, i.e. the Baire version of Rado's conjecture, holds), along with $\neg CH$, implies the strong tree property holds at $\omega_2$ by Torres-Perez and Wu ('s_Conjecture_Semi-Stationary_Reflection_the_Strong_Tree_Property_and_two-cardinal_square_principles). The question remains for the case where the length is supercompact.

Is $\mathbb{R}\cong\text{Cont}(X,Y)$ for some non-trivial spaces $X,Y$?

Math Overflow Recent Questions - Mon, 01/08/2018 - 01:09

For topological spaces $X,Y$ let $\text{Cont}(X,Y)$ be the collection of continuous functions $f:X\to Y.$ We endow $\text{Cont}(X,Y)$ with the topology inherited from the product topology on $Y^X.$

Are there spaces $X,Y$ such that $X$ has more than one point and $Y\not\cong\mathbb{R}$ such that $\mathbb{R}\cong\text{Cont}(X,Y)$?

Is every "nice" abelian category with enough projectives an additive presheaf category?

Math Overflow Recent Questions - Sun, 01/07/2018 - 08:14

A "nice" category $\mathcal{C}$ should be (for the purposes of this question) locally presentable at a minimum, and maybe a bit more. One might require $\mathcal{C}$ to be (in roughly order of increasing restrictiveness)

  • ABn for some $n$.

  • Grothendieck

  • locally finitely presentable

  • the category $\mathsf{QCoh}(X)$ of quasicoherent sheaves on a scheme $X$ (possibly with further adjectives)

  • etc.

In particular: if $\mathsf{QCoh}(X)$ has enough projectives, then is $X$ a disjoint union of affine varieties?

Clarification: Just to be clear, I'm well aware of the Freyd-Mitchell embedding theorem. This is not a question about how close abelian categories are to module categories -- it's a question about how restrictive it is for an abelian category to have enough projectives. The local presentability hypothesis rules out the duals of categories of sheaves for instance.

Motivation / Evidence: I'm thinking of things like this result: the category of sheaves on a locally connected topological space has enough projectives iff that space is an Alexandroff space -- a very restrictive condition. I suspect that the category of sheaves on an Alexandroff space is a module category additive presheaf category. Although on reflection, the category of sheaves on an Alexandroff space need not be an additive presheaf category -- in particular, it need not be locally finitely presentable. So perhaps one should assume that $\mathcal C$ is locally finitely presentable for the purposes of this question.

For another example in this direction, consider the fact that the category of quasicoherent sheaves on a smooth projective variety of dimension >0 over a field never has enough projectives. (See my CW answer below for a more general result).

Alternative formulation: If we assume that $\mathcal C$, in addition to being a "nice" abelian category with enough projectives, has a compact generator, then is $\mathcal C$ a module category? This would follow from the formulation above, since an additive presheaf category with a compact generator is a module category; but it could conceivably be easier to show. It would also settle a form of the question about categories $\mathsf{QCoh}(X)$.

Realizing $\mathcal{A}(2)//\mathcal{A}(1)$ by a finite spectrum

Math Overflow Recent Questions - Sat, 01/06/2018 - 21:39

Let $\cal A$ denote the mod 2 Steenrod algebra. Can the $\mathcal{A}(2)$-module structure on $\mathcal{A}(2)//\mathcal{A}(1)$ be enriched to an $\cal A$-module structure? If so, is there a finite spectrum $X$ such that $H^*(X) = \mathcal{A}(2)//\mathcal{A}(1)$?

Fastest deterministic factoring algorithm in subexponential space?

Math Overflow Recent Questions - Sat, 01/06/2018 - 18:13

Strassen's factoring algorithm shows that $\text{FACTORING} \in \text{DTIME}(N^{\frac{1}{4}+o(1)})$, but if I'm not mistaken in my analysis it also uses a similar amount of space. By making a trade-off I think it is possible to show $\text{FACTORING} \in \text{DTISP}(N^{k+o(1)}, N^{\frac{1}{2}-k+o(1)})$ for $\frac{1}{4} \leq k \leq \frac{1}{2}$.

On the other hand, trial division up to $\sqrt{N}$ demonstrates $\text{FACTORING} \in \text{DTISP}(N^{\frac{1}{2}+o(1)}, \log(N)^{O(1)})$. The only extra space we need is to keep a counter and perform the divisibility test.

Is there any deterministic factoring algorithm known to be in $\text{DTISP}(N^{k + o(1)}, N^{o(1)})$ for $k \lt \frac{1}{2}$?

I know that there is an $N^{\frac{1}{3}+o(1)}$-time algorithm due to Michael Rubinstein but I can't tell what the space usage would be. This would qualify as an example if the space can be made subexponential.

Otherwise, is trial division the best we can do in $N^{o(1)}$ space?

We won't be able to prove $\text{FACTORING} \notin \text{DTISP}(\log(N)^{O(1)}, O(\log(N))) = \text{L}$, since that would imply $\text{NP} \neq \text{L}$, an open problem.

Distribution of ratio between complex Gaussian and Chi-square R.V.s

Math Overflow Recent Questions - Sat, 01/06/2018 - 13:04

What would be the distribution (p.d.f.) of the following ratio?

$z = \frac{x_{1}}{|x_{1}|^2 + |x_{2}|^2 + ... + |x_{M}|^2}$

where $x_{i} \sim \mathcal{CN}(0,a), \forall i$ and $a > 1$. As can be seen, the denominator follows a Chi-square distribution with $2M$ degrees of freedom as $x_{i}$ are i.i.d. R.V.s.

Remark 1: I've run some simulations in Matlab, as shown in the pictures below for $M$ = 10, and the resulting distribution has a bell-shaped histogram. Could it be that the resulting distribution follows one of these distributions: Gaussian/Cauchy/Student's-t?

Remark 2: This is a link to the Matlab/Octave script used to plot the pictures below. Matlab/Octave simulation of the histogram of z

Growth of polynomial augmented geometric series

Math Overflow Recent Questions - Fri, 01/05/2018 - 20:17

Fix $a \geq 0$. Let $$F(n,a) = 2^{-n}\sum_{k=1}^{n-1} 2^{k} k^a.$$ Notice if $a=0$ this is just a finite geometric series. For $a>0$ my intuition and numerical approximations suggest that $$\frac{F(n,a)}{n^a} \nearrow 1.$$ How does one prove this? Taking this for granted, what is the rate of convergence in terms of $a$? More precisely, is there a function $f(n,a)$ so that $$1- \frac{F(n,a)}{n^a} \approx f(n,a).$$ I would only be interested in the leading order. Numerical approximations suggest that $f(n,a) \approx t^{-1}$ irregardless of $a$. I am somewhat distrustful of this, since it feels like the exponent should affect the convergence rate.

Bäcklund transforms of the pseudosphere: Intuitive description?

Math Overflow Recent Questions - Fri, 01/05/2018 - 17:49

Is it possible to describe what are all the Bäcklund transforms of the pseudosphere at a high, intuitive level? The shapes in the aggregate set of all the transforms? Typically one sees images of very pseudosphere-like solutions , but I was surprised by this image:          
          (Image from Tim Hoffmann: "a Bäcklund transform of the ... pseudosphere.") I admit I have not penetrated the Bobenko & Pinkall paper, which might explain this image:

"Discrete Surfaces with Constant Negative Gaussian Curvature and the Hirota Equation" (J. Diff. Geom., 1996, 43, pages 527--611). (ProjectEuclid PDF download.)

How to naturalize the complex plane, the complex numbers as something to work with geometrically? [on hold]

Math Overflow Recent Questions - Fri, 01/05/2018 - 17:39

I am interested in hearing how working research mathematicians work with the complex plane, the complex numbers as something to work with geometrically, visualize in their day-to-day life. The more fields/areas of math that are addressed the better. Of course this could be asked on MSE, but due to the research mathematicians specification I'm asking on MO. Hope that's alright.

Bar construction and the $\infty$-categorical Barr-Beck theorem

Math Overflow Recent Questions - Fri, 01/05/2018 - 17:34

I am studying the proof of the $\infty$-categorical version of the Barr-Beck theorem in Lurie's Higher Algebra, but there is a step of the proof that is puzzling me. In Lemma, a simplicial object $M_{\bullet}$ in the $\infty$-category $\mathrm{LMod}_T(\mathcal C)$ is constucted as $\operatorname{Bar}_T(T,M)_{\bullet}$.

My question is: what are the correct choices of $\mathcal C^{\otimes}$, $F_0$ and $f$ in Construction in order to obtain the $\operatorname{Bar}_T(T,M)_{\bullet}$ used in the proof of Lemma

As far as I understand, the idea is to apply the bar construction (Construction, to the coCartesian fibration of $\infty$-operads $\mathcal A^{\otimes} \to \mathcal{LM}^{\otimes}$ (here I am using $\mathcal A$ instead of $\mathcal C$ to avoid the conflict of notation with the $\mathcal C$ of Lemma expressing $\mathcal C$ as weakly enriched over $\mathrm{End}(\mathcal C)^{\otimes}$ (that is, with $\mathcal A^{\otimes}_{\mathfrak m} \simeq \mathcal C$ and $\mathcal A^{\otimes}_{\mathfrak a} \simeq \mathrm{End}(\mathcal C)^{\otimes}$). My problem is now how to define $F_0$ and $f$ (with the notations of Construction in order to get $\operatorname{Bar}_T(T,M)_{\bullet}$, and even how to make sense of the "bimodule instance" of $T$ in that expression.
To be more precise, $F_0$ should determine two bimodules in $\mathcal C$, but since $\mathcal C$ is only left tensored over $\mathrm{End}(\mathcal C)^{\otimes}$, I don't know how to make sense of the construction (in fact, I don't know how to make sense of the construction for any base $\infty$-operad other than $\mathcal O^{\otimes} = \mathcal{BM}^{\otimes}$). Moreover, even interpreting $M$ as some sort of bimodule (maybe with some kind of trivial right action, but I am just guessing), I can't find a way to interpret the second instance of $T$ in the expression $\operatorname{Bar}_T(T,M)_{\bullet}$. Or maybe I am just considering the wrong coCartesian fibration of $\infty$-operads to begin with, but then I have no clue about any other possible option.

Proof of the coherence of ${\Bbb F}_q[[X_1,\dots,X_{\infty}]]$

Math Overflow Recent Questions - Fri, 01/05/2018 - 17:05

We shall define the infinitely-many-variable formal power series ring $A = {\Bbb F}_q[[X_1,\ldots,X_{\infty}]]$ over a finite field ${\Bbb F}_q$ as the following$\colon$

$A \colon= \underset{n \geq 1}{\varprojlim}\, {\Bbb F}_q[[X_1,\ldots,X_n]]$.

For example, $\Sigma_{n = 1}^{n = \infty} X_n = X_1 + X_2 + \ldots \in A$. The ring $A$ is a non-noetherian local ring with the unique closed maximal ideal. I would like to pose the following theorem on which please let me know the correctness$\colon$

Theorem. $A$ is coherent.

Proof. Let us fix a positive integer $l \geq 1$ and for an arbitrarily chosen elements $a_1,\ldots,a_l$ consider the linear equation

$(*) \quad a_1Y_1 + \ldots + a_lY_l = 0 \quad (a_1,\ldots,a_l \in A).$

We shall define ${\mathrm M}_{\infty}$ as the set of the whole solutions of $(*)$ in the ring $A$. We shall show the finiteness of the number of generators of ${\mathrm M}_{\infty}$ as an $A$-module.

We define ${\mathrm M}_{n,m}$ as the image of ${\mathrm M}_{\infty}$ in the quotient ring $A_{n,m} \colon= A/((X_1,\ldots,X_n)^m,X_{n+1},X_{n+2},\ldots)$. We have simply $A_{n,m} = {\Bbb F}_q[[X_1,\ldots,X_n]]/(X_1,\ldots,X_n)^m$.

Lemma. For sufficiently large $n, m$, the $A_{n,m}$-module ${\mathrm M}_{n,m}$ has generators whose number is less than or equal to $l$.

Proof. Obviously, ${\mathrm M}_{n,m}$ can be viewed as the subset of the whole solutions of the linear equation $(*)$ in the quotient ring $A_{n,m}$. But the whole solutions of the linear equation $(*)$ in the quotient ring $A_{n,m}$ has its cardinality less than or equal to $|A_{n,m}|^l$. Consequently we have

$|{\mathrm M}_{n,m}| \leq |A_{n,m}|^l. $

When we view ${\mathrm M}_{n,m}$ as a $A_{n,m}$-module and set the number of the generators of ${\mathrm M}_{n,m}$ to be $\delta_{n,m}$, the following inequality holds$\colon$

$|(A_{n,m})^{*}|^{\delta_{n,m}} \leq |{\mathrm M}_{n,m}|. $

In short we have

$|(A_{n,m})^{*}|^{\delta_{n,m}} \leq |A_{n,m}|^l.$

We can see the equality $|(A_{n,m})^{*}| = \frac{(q-1)}{q}|A_{n,m}|$ because $A_{n,m}$ can be divided into $q$ disjoint cosets as $c + {\frak m}$ with $c \in {\Bbb F}_q$ and ${\frak m}$ being the maximal ideal of $A_{n,m}$. Thus we obtain

${\delta_{n,m}} {\mathrm log}_e(\frac{(q-1)}{q}|A_{n,m}|) \leq l\,{\mathrm log}_e(|A_{n,m}|)$.

So, we have

${\delta_{n,m}} \leq l \,{\mathrm log}_e(|A_{n,m}|)/{\mathrm log}_e(\frac{(q-1)}{q}|A_{n,m}|)$.

When $n,m \to \infty$, we have $|A_{n,m}| \to \infty$, which ensures that the positive integer ${\delta_{n,m}}$ must be less than or equal to $l$ for sufficiently large $n, m$. Q.E.D.

Now, by Lemma we can conclude that ${\mathrm M}_{\infty}$ has the set of generators with its cardinality less than or equal to $l$ as $A$-module because we have the equality ${\mathrm M}_{\infty} = \underset{n,m \geq 1}{\varprojlim} {\mathrm M}_{n,m}$, where all natural transition maps ${\mathrm M}_{n',m'} \to {\mathrm M}_{n,m}$ with $n' > n, m' > m$ are surjective. Q.E.D.


Subscribe to curious little things aggregator