A comment in this post seems to suggest the statement above, but I do not see this.
One direction is trivial.
How can we show that if $r_1 m_1 + cdots + r_n m_n in PM$ then each $r_i in P$.
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A comment in this post seems to suggest the statement above, but I do not see this.
One direction is trivial.
How can we show that if $r_1 m_1 + cdots + r_n m_n in PM$ then each $r_i in P$.
which of the following statement are necessarily true ?
$1.$ There exist a finitely generated group containing some element of infinite order
$2.$There exist an infinite group which is not finitely generated but all whose element have finite order
$3.$ There exist a finitely generated infinite group no element of which have infinite order
My atempt : For $1$ , I could not able to find te counterexample
I think $ 2,3$ are true
For $2$. take $mathbb{Q}$
For $3$. take $mathbb{Z}$
Show if $S subset mathbb{Q}(sqrt{d})$ (as a subring) and $(S,+)$ finitely generated then $S subset R_d$
This is an exersis which was supposed to help me with more complicated problems in $R_d$. Here is my attempt;
Suppose $S,+)=<s_1…s_n>=<a_1+b_1sqrt{d}…a_n+b_nsqrt{d}>$, $a_i,b_i in mathbb{Q}$
Let $(A_1,+)=<z_1…z_n>,(A_2,+)=<xi_1…xi_n>$ be subgroups of $(mathbb{Z},+)$ generated by $n$ integers So that $A_1 times A_2 leq mathbb{Z} times mathbb{Z}$ (please read ”times” as ”circle plus” as in direct product, I cannot seem to do the symbol here).
Define $phi:S rightarrow A_1 times A_2$ by $phi(s_i)=phi(a_i+b_isqrt{d})=(z_i,xi_i)$ (This defines the homomorphism on the group right? since these are generators?)
This and an isomorphism (I did check this but I will omit details unless requested). So I have the following relationships;
$R_d leq mathbb{Q}(sqrt{d})$
$(A_1 times A_2,+) leq (mathbb{Z} times mathbb{Z},+) leq (mathbb{Q} times mathbb{Q},+)$
$S simeq (A_1 times A_2,+), R_d simeq mathbb{Z} times mathbb{Z}$
So by corresponce theorm; I can say $S leq R_d$? Or do I need to write more. It already feels like a long proof for what was supposed to be a ”simple exercise”. Also I never use the structure of the ring. I wrote out some examples and it seemed to me that is where we need $a_i, b_i$ to be integers. IE. if $S=<s_1,s_2>$ then it seemed $s_1s_2$ could not have been generated by $c_1s_1+c_2s_2$ (I wanted to aviod this kind of proof since it seemed entirely mechanical and I wanted to get a better understand of structure).
Given a smooth affine variety $X$ defined over $mathbb{Q}$, its singular cohomology is isomorphic to the algebraic de Rham cohomology, which is the cohomology of the complex $Omega_X^0toOmega_X^1todots$ of differential forms on $X$ with polynomial coefficients. If $X$ is given by equations $f_1,ldots,f_minmathbb{Q}(x_1,ldots,x_n)$ then the algebra of differential forms has the following explicit presentation:
$$
Omega_X^* = mathbb{Q}(x_1,ldots,x_n,d x_1,ldots,d x_n)/(f_1,ldots,f_m,d f_1,ldots, d f_m),
$$
and the differential $d$ is defined in the obvious way. So the problem of computing the cohomology of $X$ is reduced to the problem of computing the cohomology of the explicitly presented (super-)commutative dg algebra above.
It is not clear if there is a general algorithm for computing the cohomology of a finitely presented commutative dg algebra. I suspect there is none. On the other hand, the de Rham complex can be interpreted as the complex representing the derived push-forward $R^* pi_* O_X$, where $pi:mathbb{Q}^nto mathrm{point}$ is the projection. There is an algorithm to compute this kind of push-forwards, it basically relies on the notion of a complex of D-modules with holonomic cohomology: the D-module $O_X$ is holonomic, and the push-forward of such a complex is again such a complex. The push-forward is computed by successively applying the push-forwards via
$$
mathbb{Q}^ntomathbb{Q}^{n-1}tomathbb{Q}^{n-2}todotsto mathrm{point}.
$$
For details, see this paper and other papers by these authors and references therein:
Oaku, Toshinori; Takayama, Nobuki, An algorithm for de Rham cohomology groups of the complement of an affine variety via (D)-module computation, J. Pure Appl. Algebra 139, No. 1-3, 201-233 (1999). ZBL0960.14008.
The question is: what makes the dg algebra $Omega^*_X$ so special that its cohomology can be computed? Is there a notion of a ‘nice’ dg algebra so that 1) all dg-algebras of the form $Omega^*_X$ are nice and 2) there is an algorithm for computing the cohomology of a nice dg-algebra?
I am looking for natural groups with undecidable conjugacy problem. By natural, I mean that the word problem should be decidable, and the group should be given by some natural action. I know that $mathbb{Z}^d rtimes F_m$ (with a suitable action of $F_m$) has undecidable conjugacy problem. That’s very nice, but I’d like to know other examples. I do not care about finite presentation, and I’m also fine with the group being a f.g. subgroup of something natural and geometric, which maybe simplifies things. A concrete case I was not able to resolve is whether all f.g. subgroups of right-angled Artin groups have decidable word problem.
Šunić, Zoran; Ventura, Enric, The conjugacy problem in automaton groups is not solvable., J. Algebra 364, 148-154 (2012). ZBL1261.20034.
Fix a positive integer $n$. Let $X = {X_i}_{i in mathbb{N}}$ be a discrete time stochastic process such that each $X_i$ is a ${0,dots,n-1}$-valued random variable. Suppose that the joint probability distributions of any finite sequence of $X_i$‘s only depends on the order of their indices, or to be more precise suppose that $X$ satisfies the following:
Call such a stochastic process ‘strongly homogeneous.’ I’m trying to understand what the set of strongly homogeneous stochastic processes looks like. This is my approach so far:
The set of ${0,dots,n-1}$-valued discrete time stochastic processes can be understood as the set of Borel probability measures on the (compact) space $A = {0,dots,n-1}^{mathbb{N}}$. This is a subset of Banach space of (regular Borel) signed measures on $A$. Let $S$ be the set of such measures corresponding to a strongly homogeneous stochastic process. It’s not hard to check that $S$ is a convex, weak* closed set, and therefore that the Krein-Milman theorem applies to it. This gives us that every element of $S$ is in the weak* closure of the convex hull of the set of extreme points of $S$ (where a point is extreme if it is not the convex combination of any other elements of $S$). This leads to my precise question.
Question: What are the extreme points of $S$?
Note that the extreme points of the set of all probability measures on $A$ is precisely the set of Dirac measures on $A$, but a similar statement here is not sufficient. For instance, if $n=2$, then the only strongly homogeneous Dirac measures are those concentrated on the constant $0$ sequence or the constant $1$ sequence, but convex combinations of these do not have the measure corresponding to a sequence of i.i.d. fair coin flips in their weak* closure.
My suspicion is that the measures corresponding to i.i.d. sequences are the extreme points, but I haven’t proved either that they are all extreme points or that all extreme points are of that form. (Proving that they are all extreme points should be easy, however.)
Let $A,B$ be compact subsets of $mathbb R$. Let $a=mathrm{dim}_F(A)$, $b=mathrm{dim}_F(B)$ be their Fourier dimensions, respectively. My questions are as follows:
Is it true that $mathrm{dim}_F(Acup B)=max{a,b}$?
If $A$ and $B$ are positively separated, is it true that $mathrm{dim}_F(Acup B)=max{a,b}$?
In addition to 2, if $Asubseteq (0,1)$ and $Bsubseteq (2,3)$, is it true that $mathrm{dim}_F(Acup B)=max{a,b}$?
In addition to 3, if $a=0$, is it true that $mathrm{dim}_F(Acup B)=b$?
I have searched online but I cannot find any references for that.
$A,B$ are $k$-algebras and finitely generated as $k$-modules. Let $X$ be a simple $A otimes_k B$ module. Since $A otimes_k B$ is finitely generated as a $k$-module, $X$ is finitely generated as a $k$-module, hence also as a module over $A$ via the canonical map $A to A otimes_k B$.
I am not sure how to see that $X$ is finitely generated as a module over $A$. Any help would be appreciated!
Let $X$ and $Y$ be topological spaces. Assume $X$ no dense finite subset and $Y$ is path-connected. Given $n$ pairs of points $(x_i, y_i)$ where $x_iin X$ and $y_iin Y$ for $1leq ileq n$ and a continuous map $f:Xto Y$ can we find a continuous map $f’:Xto Y$ homotopic to $f$ such that $f(x_i)=y_i$?
I asked this question one month Ago in MSE but no answer for existence of argument which show if such counter example exists we would have infinity of them or just finitely many examples
Lehmer’s totient problem asks whether there is any composite number $n$ such that Euler’s totient function $φ(n)$ divides $n − 1$. which it is unsolved problem or we may reformulate that question as : if $φ(n)$ divides $n − 1$ then $n$ must be a prime , Now my question here is :if a such counter example exists for Lehmer’s totient problem could we prove that there are infinity of them or just finitely of them ?