Mathematics - 2015-11-28 (page 1 of 2)

a polynomial f(x) is irreducible if it is irreducible as an element of R[x]

yeah

for instance 6 is not irreducible in Z[x] because it is 2x3, even though it cannot be written as a product of polynomials of lesser degree

but ofc 6 is a unit in Q[x]

well can't 2 for example be considered as polynomial of degree zero

?

12:23 AM

2 and 3 do not have degree less than the degree of 6

oh I see.

oh I guess that would be the case when it is a field.

what I am saying

i.e what I said above applies to fields

Q: Proof that the solutions are algebraic functions

Could someone of you take a look at my question:

0

I am looking at the following: $$$$ $$$$ I haven't really understood the proof... Why do we consider the differential equation $y'=P(x)y$ ? Why does the sentence: "If $(3)_{\mathfrak{p}}$ has a solution in $\overline{K}_{\mathfrak{p}}(x)$, then $(3)_{\mathfrak{p}}$ h...

?

I guess in a unique factorization domain that is true aswell

user174558

12:46 AM

Why does transferring a large file to a flashdrive slow down to very slow? In contrast, transferring it to a harddrive does not.

hard disks are probably faster writers

I also find it takes longer to transfer a lot of small files than it does one large one

@anon if we have the ring R[x] the units are the constant polynomials ?

the units are the units of R

constant polynomials aren't necessarily units tho

yeah

why are the units of R[x] the units of R?

1:01 AM

exercise

ok I see it now

because we have that the identity of R[x] is just the 0 polynomial

you can easily then prove that any non-constant polynomials will never be units,and then easily deduce after that the units of R[x] will be the units of R

yeah

I guess we must add the fact that the ring R is an integral domain or we would be in trouble.

also the identity here is 1 not 0 I was getting confused for a sec.

for completeness here is the proof:

a unit in R will be a unit in R[x] (so that is trivial). Now for the other direction consider $\phi(x)$ being a unit, where $\phi \in R[x]$, so that means there exists function $g \in F[x]$ such that $\phi(x)g(x) = 1$, but this means $deg(1) = 0 = deg(\phi*g) = deg(\phi(x)) + deg(g(x))$ , so we must have $\phi(x)$ and $g(x)$ both having degree zero, so that is precisely the units in R.

@anon that will not be true in a general ring right ?

that isn't an integral domain.

I guess what I should say not generally true in a general ring.

1:18 AM

right, for instance 2x+1 is its own inverse in (Z/4Z)[x]

I see

cool I am clearing up my head on those irreducible stuff.

AlpArslan

Hey guys, I just have a small question to do with simplicial homologies.

If K is a simplicial complex and \sigma a simplex in K

Then, let L be K\{\sigma}

This is a simplicial complex.

If \sigma has dimension n, then, d_n(\sigma) is an n-1 cycle and consists of simplices in L, and so, represents a class in H_{n-1}(L).

This defines a homomorphism \phi: \mathbb{Z} \to H_{n-1}(L) by sending 1 to [d_n(\sigma)]

I need to find a homomorphism \psi : H_n(K) \to \mathbb{Z} such that the sequence

0 \to H_n(L) \to H_n(K) \to \mathbb{Z} \to H_{n-1}(L) \to H_{n-1}(K) \to 0 is exact.

I understand that, any class, [\tau] in H_n(K), needs to go to 0 for \tau not \sigma,

But, I'm not sure where to send \sigma.

I understand that it's image must be an integer m, say, such that, m [d_n(\sigma)] lies inside the image of d_n: C_n(L) \to C_{n-1}(L)

By the way, \sigma must be such that it is not a proper face of any other simplex in K! I'm sorry I forgot to mention that.

Which means that there needs to be an n-simplex in L , say \tau, with boundary, d_n(\tau) such that [d_n(\tau)] = m [d_n(\sigma)].

But how is this possible unless m = 0?

1:55 AM

you're supposed to put dollar signs around $\LaTeX$

doesn't work otherwise

enthdegree

2:13 AM

Hello all. What is a tool that will generate a random binary matrix where each row sums to R and each column sums to C?

I've got matlab

2:35 AM

@anon gauss lemma proof is very nice

atleast for the part when d isn't a unit

3:45 AM

Hi @PVAL.

I'm mostly just an idiot

5:42 AM

How do I show that $\text{rad}({\frak i}) = \text{rad}({\frak{g}})\cap {\frak i}$

I can see that the RHS is an ideal(since both things in the union are ideals)

When $\mathfrak{i}$ is solvable, $\text{rad}({\frak{i}})={\frak i}$

And since $\text{rad}({\frak g})$ is the maximal solvable ideal of $\mathfrak{g}$, if it is equal to $\mathfrak{i}$, we get equality, else the RHS will shrink in cardinality

I'm mostly just an idiot

6:00 AM

Question: Does the popularity of tags on math overflows give a good representation of the popularity of the fields in research mathematics?

I.e. is Algebraic geometry the most active area of mathematics at the moment?

Q: arXiv vs MathOverflow - popularity of disciplines

No.

42

Inspired by the comparison of programming languages by GitHub and Stack Overflow activity (e.g. this one for 2015) I decided to look at the popularity of mathematical disciplines by using data from both arXiv and MathOverflow (see also my motivation for getting a dump of arXiv metadata). Here it ...

discussion tags

2

One particular thing of note is that applied mathematics is universally underrepresented.

So are various fields of analysis, and algebraic fields are more popular on MathOverflow than in publication frequency.

(Algebraic geometry is indeed quite active, but not in the proportions MO posts would indicate.)

I'm mostly just an idiot

Second most active in Arxiv preprints, but yeah, thanks for that, I see that mathematical physics is very underrepresented

Sure, but I mean, MO posts would indicate that algebraic geometry is about 8x as popular as PDEs, whereas it's more like... 30-40% more popular.

One other method would be to see proportions of people in the AMS member list subscribed to various fields.

Robert Cardona

9:20 AM

If I have two Noetherian subrings, $S, T$ of a ring $R$, is their product $ST$ also Noetherian? I'm trying to show every ring is the direct limit of it's Noetherian subrings, but I'm first trying to show the collection of Noetherian subrings is a directed set. Is this the case?

9:34 AM

Could someone explain how achille replaces $n$ with $x\frac{d}{dx}$ here? Is this obvious somehow? math.stackexchange.com/a/1547255/109879

@ChantryCargill (i) what does d/dx do to x^n? (ii) how can you go from (-x)^n to (n+1)(-x)^n using d/dx, multiplying by x, etc.?

applying D and (xD) to powers of x^n is something you want to understand the consequences of, because then you can reverse the process and write poly(n)x^n as poly(D)x^n (not the same kind of polynomial but w/e)

9:56 AM

@anon Thanks. It's immediately obvious after you tell me that.

10:48 AM

Hello @anon !! Are you familiar with math.stackexchange.com/questions/1548963/… ?

user174558

11:14 AM

@ChantryCargill Yes, I was hungry that day, lol.

12:03 PM

@JasperLoy Haha, I figured. I know the feeling though. No food in the fridge or the cupboards, ordered pizza too many times to justify doing it again, too hungry to make it to the grocery store. Just hoping someone will come to the rescue, otherwise giving in to your demise. Looking back, there's only one reason I made it this far in life.

12:43 PM

@anon But then why isn't the distance attained?

Hello @robjohn @TobiasKildetoft !! Could you take a look at my question: math.stackexchange.com/questions/1548963/… ?

1:45 PM

@anon We have to show that there is no y such that $||x-y||=0$, right? How can we do this?

@DanielFischer Do you maybe have an idea?

hello

@Brennan.Tobias Hello there.

how are you

Not bad. Just having a few beers and chillin.

What are you up to?

nice :) I'm good, been learning some number theory

1:59 PM

Hmm, is it normal for the Community user to delete downvoted meta questions (with upvoted answers) with no explanation?

Henning Makholm, excuse me if I am mixing things up but I think I recognize your name: Did you write a C partial evaluator?

@Brennan.Tobias Yes, partially (!) -- back in the late 1990s I and some other students were paid to reimplement C-Mix at the University of Copenhagen.

@HenningMakholm It's normal when the account is deleted or destroyed. Which happened here.

great :D

I'm really fascinated by partial evaluation, I started to read a book on it (Partial Evaluation and Automatic Program Generation)

@DanielFischer Does that hold for the main site too? The meta discussion probably had no lasting value, but it worries me if someone who asks a homework question, gets downvoted but does get a complete solution can get Community to cover his tracks simply by deleting the account he asked from.

@Brennan.Tobias Yes, that book was the Olden Knowledge even when I entered the field. One of the authors (Jones) was my Ph.D. supervisor.

2:08 PM

I'd love to hear your input about it: Can we really make compilers automatically out of interpreters using the futamura technique?

@HenningMakholm Yes, that also applies to main. Doesn't happen too often, it's more common that they delete the question before the answer got an upvote.

(If you notice such occurrences: please flag.)

@Brennan.Tobias With 20 year of hindsight: No, I don't really think something we could recognize as a production-ready compiler for a real language can be produced that way -- it worked in toy settings, but I think the more lasting value of the research that was done then was to inspire improved optimizations in hand-written compilers. Much of what optimizing compilers do nowadays can be described as a sort of partial evaluation, and we did help push the limits of that.

ah that's very interesting to know!

I have a dream to write a compiler one day :)

@Brennan.Tobias It's an experience :-)

user177666

Hello everybody! This is my first comment here to test my chatting system.

2:24 PM

hi Amir

user177666

@Brennan.Tobias What do you guys do here normally? I mean, should the content of chatting be mathematical necessarily or chit chatting is also allowed?

I'm not sure, but there are a lot of really nice people here

I asked about various topics and people were very kind to share their knowledge with me

Paradox 101

2:58 PM

If we need to solve the Legendre equation about points $x=1$ and $x=-1$ we assume that for the power series solution $x_o=1$ and $x_o=-1$?

Tim

hello. could someone help to look up a book in their university library?

Carrier-Grade VoIP, Third Edition
Richard Swale

McGraw-Hill Professional, 2014.

@DanielFischer Could I ask you something?
I am looking at the proof of the claim: If Y is a subspace of X then $\overline{Y}$ is a subspace of X.

Let $x,y \in X$ and $\lambda, \mu \in \mathbb{R}$. It remains to show that $\lambda x + \mu y \in \overline{Y}$.

Why do we have to show this? So does this mean tha if we have a space and want to show that a set is a subspace we have to show that each linear combination of two elements of the space belongs to the set?

@evinda Yes, a subset of a vector space is a (linear) subspace if and only if it is nonempty and closed under linear combinations. Since we're taking the closure of a subspace, the nonemptiness is clear, so only closure under linear combinations remains to be shown.

3:16 PM

@evinda: Shouldn't $x,y \in \overline{Y}$?

In my notes it says $x,y \in X$. So is it wrong? @Huy

You tell me.

@Huy In order to show that a set is closed under linear combinations we take two elements of the set and show that their linear combination belongs also to the set. So it should be $\overline{Y}$. Right?

@evinda Oh, didn't notice that. That's wrong then, it's as Huy said.

Morning.

3:22 PM

Exactly, @evinda.

Morning, @MikeMiller.

Morning, @Mike.

@Huy @DanielFischer Ok...

Then according to my notes:

There are sequences $(x_n) , (y_n)$ in Y such that

$x_n \to x$ and $y_n \to y$.
So we have $\lambda x_n \to \lambda x$ and $\mu y_n \to \mu y$
$\Rightarrow \lambda x_n+ \mu y_n \to \lambda x+ \mu y \Rightarrow \lambda x+ \mu y_n \in \overline{Y}$.

@Huy @DanielFischer Couldn't we say the following ?

There are sequences $(x_n) , (y_n)$ in $\overline{Y}$ such that

$x_n \to x$ and $y_n \to y$.

3:39 PM

@evinda That wouldn't work, we need to use the properties of $Y$ to reach the desired conclusion, and so we must use elements of $Y$, not of $\overline{Y}$. It's not a good idea to use sequences, in my opinion, since that doesn't work for general topological vector spaces. The argument using the continuity of addition and scalar multiplication is much cleaner.

@DanielFischer This proof holds only for Hilbert spaces, right?

We know that $Y \subset \overline{Y}$, right?

But previously we have supposed that $x,y \in \overline{Y}$ so we don't know if $x,y \in Y$. Or am I wrong?

If so, then did they mabe take because of this $x,y \in X$ ?

@evinda More general than just Hilbert spaces, it works for all metrisable vector spaces (in particular normed spaces). Yes, we know $Y\subset \overline{Y}$. That's what immediately gives us $\overline{Y}\neq \varnothing$. Next, one takes arbitrary $x,y\in \overline{Y}$. Then one takes sequences $(x_n),\,(y_n)$ in $Y$ that converge to $x$ resp. to $y$, and by means of those concludes that $\lambda x + \mu y \in \overline{Y}$.

4:01 PM

@DanielFischer: What do I do in situations like this?

The question itself is not so interesting that it should be reverted and answered.

@MikeMiller Check whether the user is registered. If they are, comment and tell them about the "delete" link. If not, make a moderator aware of it so the post can be deleted.

@DanielFischer We have that $x \in \overline{Y}$ .
We know that there is a sequence $(x_n)$ in Y such that $x_n \to \text{an element of Y}$. But this element will also belong to $\overline{Y}$.
So we are sure that for all elements of $\overline{Y}$ there will be a sequence in Y that will converge to them. Right?

Thanks.

@evinda I can't make sense of your second line there.

If $a$ is an element of $Y$ there will be a sequence $(x_n)$ in Y such that $x_n \to a$.
But since $Y \subset \overline{Y} \Rightarrow a \in \overline{Y}$ . @DanielFischer

4:10 PM

@evinda But that is completely uninteresting. What you need is that for every $a \in \overline{Y}$ there is a sequence $(x_n)$ in $Y$ with $x_n \to a$.

@DanielFischer Could you explain me why this holds?
Does it always hold if $B \subset D$ that for every $x \in D$ there is a sequence $(x_n)$ in B with $x_n \to x$ ?

@evinda Of course not, consider $\{0\} \subset \mathbb{R}$ (or even $\varnothing \subset X$). But we're looking at $\overline{Y}$. You need to understand what the closure of a set is.

4:44 PM

Hey everyone! Does anyone know how to configure tex maker or is there a better offline latex editor because I am in a slump now and really need to get to work

@Mike @Daniel @evinda @Alec @robjohn @anon

@JulianRachman There is an online program and you don't have to install anything

@JulianRachman sharelatex.com

@JulianRachman I use kile (or plain kwrite for short things).

@DanielFischer Y is a set with some restrictions and $\overline{Y}$ is the same set without the restrictions, or not?

@evinda I have been using sharelatex but I want to do offline work

Ah ok... @JulianRachman

What do you want to write in latex? @JulianRachman

user174558

4:54 PM

@JulianRachman The best editor is TeXworks which comes with a full TeXLive install on Windows.

Aha @JasperLoy

I am writing a paper @evinda

@evinda No. $\overline{Y}$ is the closure of $Y$. You need to understand what that means. Without some basic understanding of topology, you'll be hopelessly lost in functional analysis.

user174558

@evinda Hi, are you graduating soon?

user174558

@JulianRachman So far, I have only written on toilet paper.

4:55 PM

@JasperLoy whag is TeXLive

Really? With 14? Bravo!!! @JulianRachman

@JasperLoy Yes... Did I tell you about it?

user174558

@JulianRachman It is a TeX distribution you can install on your computer.

user174558

@evinda No, you did not.

@evinda thank you! I am like super excited! :)

Hi

user174558

4:56 PM

@JulianRachman You must install a TeX distribution on your computer to use TeX, and then you need an editor to help you type text and compile the source code.

@JulianRachman How long did you do research ?

user174558

@JeSuis Hello, are you Jesus?

@JasperLoy I have written on napkins but never before toilet paper bc my pen bleeds

@JasperLoy Then , how do you know it?

user174558

@evinda I am guessing.

4:57 PM

Why? :D @JasperLoy

I am going into my 3rd month once Dec 17th hits @evinda

user174558

@JulianRachman TeXLive and TeXworks are good because they work on Windows, Linux and Mac.

I need to prove that the equation $ux^2+vy^2=w$ for $u\ne 0,v\ne 0,w \in \Bbb{Z}/p\Bbb{Z}$

user174558

@JulianRachman 3rd month of pregnancy?

@JulianRachman If you finish your paper , send it to us!

@JulianRachman Have you finished earlier with school?

4:59 PM

Alright. I am going to try it out right now. @JasperLoy lol I am making history with my pregnancy. :/

@JasperLoy no, unfortunately.. because otherwise I would not need help.

@evinda ya! I will make sure of it! And no. I am going to do the full 4 years because I enjoy being a teen. :)

@DanielFischer Did you see my question ? I can prove that there is $(p+1)/2$ square in Z/pZ but I don't see how it's related.

user174558

@JulianRachman TeXmaker has a lot of commands that I don't like since I type them in full myself. I prefer the simplicity of TeXworks.

@DanielFischer Ok, I will read about it, because I haven't had topology... Then we deduce that $\lambda x_n+ \mu y_n \to \lambda x+ \mu y$ and since $\lambda x_n+ \mu y_n \in Y$, do we deduce that $\lambda x+ \mu y \in Y$ ?

@JulianRachman A ok.. What do the teachers tell you about your high IQ?

5:02 PM

@JasperLoy Alright. Tex maker just wasn't working for me

@evinda we really don't do IQ in our area

user174558

@JulianRachman TeXmaker is also cross platform. Both editors are fine for me. I spent hours figuring out how to install TeX and use it!

And I have never taken an IQ test

user174558

@JulianRachman IQ tests are stupid.

@JasperLoy ok I will let you know how it goes.

I meant that you occupy with higher mathematics... @JulianRachman

user174558

5:04 PM

They should not even have been invented. Different people have different intelligences.

user174558

EQ is as stupid as IQ.

@evinda they say I am capable and all I need to do is prove it

user174558

If you can play piano, you are good at piano, but it is not part of an IQ test. See how silly it is?

user174558

Just another stupid statistic to label someone with.

Everyone has the same intelligence. Some know how to use it, some do not.

PerplexedGuest

5:05 PM

Hello all!

user174558

I thought Professor Balarka has ignored me.

@anon Can you help ?

No, why would I ignore you?

(I am no professor)

user174558

Because I am an asshole.

Nah.

user174558

5:07 PM

I emailed Jonas twice but no reply. I hope he is alive.

Or could we only deduce it if we had uniform convergence? @DanielFischer

user174558

Jonas is a real bro to me.

Alright, I need to run.

PerplexedGuest

Every time I think I make progress on this problem it just reveals another layer of nastiness to work through. @____@

@DanielFischer: I want to define a Laplacian on $H_0^1(M)$ of some Riemannian manifold. One way would be to go to $C_c^\infty(M)$ and define some differentiation operator $D$. Call its closure $T$. Then, $T^* T$ would be (maybe up to sign) the Laplacian. My prof asked me to do this for the upper-half plane and then verify that $T^*T f = - \Delta_{hyp} f$ for all $f \in C_c^\infty$. What could he mean by the Laplacian on the RHS?

@DanielFischer: I've looked at some textbooks on this topic and they all seem to define the Laplacian through divergence and gradient, but to define the gradient they'll need a map $d: C^\infty(M) \to \Lambda^1 T^* M$. Isn't that essentially the same thing evantually?

Old John

5:18 PM

@JasperLoy He was certainly alive and well a couple of weeks ago - I had a beer with him in Amsterdam :)

Is someone of you familiar with Grothendieck's problem?

@evinda are you still talking about Y={finite sequences} and x=(1,1/2,1/3,1/4,...)? The distance between elements of l^2(N) is 0 iff the two things are the same right?

@JeSuis I am not the person to ask to get tex to work

Right. So do we say the following? @anon

We assume that there is a $ y \in Y $ such that $||x-y||=d(x,Y)=0$.
$ ||x-y||=0 \Leftrightarrow x=y $, contradiction since it doesn't hold for all $ x \in \ell^2(\mathbb{N}) $ that there is a $ n \in \mathbb{N} $ such that $ \forall j> n: x_j=0 $.

that last part I don't understand

just say contradiction since $x\not\in Y$

oh, I see what you mean in the last part

@Huy: I'm a little confused. Your $d$ is just the exterior derivative.

There are a few notions of Laplacian I know. There's just the notion of $\tr \nabla^2$ which is used in Riemannian geometry which I know nothing about. I think this is the same as the divergence of the gradient. There's the Hodge laplacian, given by $dd^*+d^*d$. And then one can define this for $d_A$ for any connection $A$ on a bundle $E$.

I would guess he's asking you to show the first and the second are the same.

5:35 PM

@anon So the following is a complete answer, right?

Let $x=(x_1, x_2, \dots, x_n, \dots) \in \ell^2(\mathbb{N})$. So $\sum_{k=1}^{\infty} x_k^2<+\infty$.

We fix a $ n \in \mathbb{N} $ und we have the following.

$\sum_{k=n+1}^{\infty} x_k^2= \sum_{k=1}^{\infty} x_k^2- \sum_{k=1}^n x_k^2$

We let $n \to +\infty$ and get:

$\lim_{n \to +\infty}\sum_{k=n+1}^{\infty} x_k^2= \sum_{k=1}^{\infty} x_k^2- \sum_{k=1}^{+\infty} x_k^2=0$

Thus the element of $ Y $ , $ (x_1, x_2, \dots, x_n, 0, \dots) $ converges to $ x $ while $ n \to +\infty $.

TeXLive is taking FOREVER to finish downloading!

Note that these legitimately seem like different operators: the Hodge laplacian never uses the divergence operator and never leaves the land of forms.

@MikeMiller: Sorry for the confusion but I'm also confused. I just looked up the "different Laplacians" in DG: https://en.wikipedia.org/wiki/Laplace_operators_in_differential_geometry

Don't you think he rather means the Bochner Laplacian on the LHS (and the connection Laplacian on the RHS)?

The $\delta$ is the adjoint of $d$.

yes, I just noticed

5:39 PM

The Bochner laplacian is just a generalization of Hodge, because Hodge is just a hell of a laplacian.

oh, really?

how can that be? according to that wiki page, the Hodge one has a more complicated relationship to the connection, whereas the Bochner just differs by a sign

$E = \Bbb R$ with trivial connection (covariant derivative just $d$ the exterior derivative)

Maybe I'm confused... but I feel what I said should be correct

@Huy: I bet the complicated relationship degenerates when you're working with 0-forms (which the Bochner laplacian does)

But I'm a little uncomfortable with it.

that would be quite a surprising behaviour though, in general, no?

@anon Or could I improve something?

Which?

user46225

5:46 PM

hi all. Consider $k[x]\otimes_k k[x]$ as a right $k[x]$-module with the following struture: ($p\otimes q)\cdot x=px\otimes q- p\otimes qx$. Is it projective?

some complicated relationship becoming much much simpler by generalizing

oh or do you mean it only degenerates for 0-forms?

@evinda it's fine. it's a bit much to ask for a check-over of your entire argument every time you do little tweaks. try and to the small things yourself.

Ok... Thank you very much!!! @anon

@Huy: Right, it should be complicated for higher forms. If you generalise the Bochner laplacian to an operator on $\Omega^p(E)$ I bet it would have a complicated relationship to the standard laplacian.

ok. thanks for mentioning the Hodge Laplacian, that's probably the right direction. I'll look into it and ask my prof if that's what he meant.

5:54 PM

hi @anon @BalarkaSen

hi

hey

@anon Could I also ask you something else?

How do we show that $\overline{Y}^{||\cdot||_{\infty}}=C([0,1])$ where $Y=\{ p|_{[0,1]}: \text{ p is a polynomial }\}$ ?

@anon what do you mean?

@evinda even tho it's a basic fact, I actually don't know that one

@JeSuis I mean I am not experienced and knowledgeable about getting LaTeX-document-writing-programs working and all that jazz

6:06 PM

I never asked about Latex..

you weren't the one who asked about getting TeX-Maker to work?

ah, that was Julian

you just asked "can you help," which is very cryptic and nondescript

I have no idea if I can help if I don't know what it's with!

@anon Ok.. I have to notice the difference with that what we were talking about previously.
Difference: The space here is not a Hilbert space.

Do you have an idea what we could notice?

Nop, i was asking about the solution of $ux*2+vy^2=w$ on Z/pZ, this is because I post one minute before the question, and that you are connected. My bad

which are the variables and which the constants?

hmm

hmm suppose that R is U.F.D and let p(x) in R[x] such that the gcd of the coeffients of p(x) is 1. why is it that the assumptions of the coefficients having gcd 1 forces that if p(x) = a(x)b(x) then neither a(x) nor b(x) are constant poly?

I just woke up so I don't see it right away

6:12 PM

I get a good answer on main:), $x$ and $y$ are the variables, i did not think that we have two sets of p+1/2 elemtents...

well we have 1 coeffient if it was constant

ohh I see

because we can see the polynomial as gcd(0,0,0,0,...,c), where c is the constant and so this has gcd c

ok ok I see it

@anon At the example we saw th distance d(x,Y) is never attained, but in this case it is attained if f=p.

Has it do with the fact that now we do not have a Hilbert space?

right ?

@anon?

I assume p(x) is nonconstant?

yeah

but wait the assumpton of gcd will give that it is non-constant to begin with

6:17 PM

if one of a or b is a nonunit constant then multiplying the coefficients of the other (b or a resp.) makes their gcd a times whatever it was befor

@L33ter 1*1=1

I suppose we don't need to assume p(x) is nonconstant, just not a unit constant

yeah

because we have the assumption that the common divisor of the coeffients of p(x) = 1, so I guess what it means that $p(x) = a_nx^n + ... + a_0$, where we have $gcd(a_n,...,a_0) = 1$, but then in your example it would be $gcd(0,0,0,....,1) = 1$, so yeah we must have it to be non-unit.

yeah I guess here the book already captures that detail b y saying if p(x) were reducible

so it captures that detail

@evinda No, we deduce that $\lambda x + \mu y \in \overline{Y}$, which is what we need. It will in general not be in $Y$.

@Huy Sorry, not my department. What little I once knew about that stuff I've long forgotten. Mike should know.

@DanielFischer Could you explain me how we deduce that $\lambda x + \mu y \in \overline{Y}$ , knowing that $\lambda x_n + \mu y_n \to \lambda x+ \mu y$ and $\lambda x_n+ \mu y_n \in Y$?

@anon I guess it can be even weaker right because not all units would work for example $gcd(0,0,....,7) = 7$, so it must be the constant can be values other than the identity right?

gcd(0,7)=7

6:25 PM

I am sorry I meant to write 7

@evinda One characterisation of $\overline{A}$ in metric spaces is that there is a sequence in $A$ converging to the point. In any case, if there is a sequence $(z_n)$ in $A$ with $z_n \to z$, then $z \in \overline{A}$.

but anyway p(x) being reducible will capture that information anyway.

Ok... Thank you!!! @DanielFischer

Do you maybe have an idea how we show that $\overline{Y}^{||\cdot||_{\infty}}=C([0,1])$ where $Y=\{ p|_{[0,1]}: \text{ p is a polynomial }\}$ ?

@JeSuis Pigeon hole. Consider the sets $\{ ux^2 : x \in \mathbb{Z}/p\mathbb{Z}\}$ and $\{ w - vy^2 : y \in \mathbb{Z}/p\mathbb{Z}\}$.

@evinda Weierstraß approximation theorem. If you can just use it, there's nothing left to do. If proving it is part of the exercise, there should be an outline how to prove it, it's not totally obvious how to do it.

Paradox 101

If we need to solve the Legendre equation about points $x=1$ and $x=-1$ should we assume that for the power series solution $x_o=1$ and $x_o=-1$?

Can anyone please confirm this?

6:34 PM

@DanielFischer I think that I can just prove it. It is stated previously. How do we use it?

@evinda Did you accidentally swap "use" and "prove" there?

@DanielFischer Oh yes, I am sorry...

Anthony

If the characteristic of a field is p, and a polynomial, $f(x)$ is irreducible and shares a root with its derivative, why is it that we can write it as some other polynomial $g(x^p)$?

think i found it: math.stackexchange.com/questions/396232/…

enthdegree

ugh, yikes. last night i submitted a paper to a conference and had a nightmare that the pdf had an unresolved citation

6:55 PM

@evinda I would suggest you look up the proof in a textbook, or maybe on wikipedia. It would not fit well in chat.

7:13 PM

Ok.. I have to notice this equality with the exercise: Find in $\ell^2 (\mathbb{N})$ a subspace $Y$ and a $x \in \ell^2(\mathbb{N})$ such that d(x,Y) is not attained.

Difference: The space here is not a Hilbert space.

Do you have an idea what we could notice?

At the exercise we saw that the distance d(x,Y) is never attained, but in this case it is attained if f=p.

Has it do with the fact that now we do not have a Hilbert space? @DanielFischer

@evinda ? It's unclear. Note that $\ell^2(\mathbb{N})$ is a Hilbert space.

Frank Science

@MikeMiller Did you learn Verdier's duality?

Does it maybe hold that if we have a space that is not a Hilbert space we can always find a x such that d(x,Y) is attained? @DanielFischer

How do I configure TeXLive or MikTeX to TeXWorks??

7:28 PM

hi @FrankScience

Ah, I didn't know there was a sheaf theoretic Poincare duality.

Good stuff.

@evinda No. In a Hilbert space, if $Y$ is a closed convex set (in particular if it is a closed linear subspace), then the distance $d(x,Y)$ is attained for every $x$. In any metric space, if $A$ is a non-closed subset, there is always some $x$ such that the distance $d(x,A)$ is not attained (take an $x \in \overline{A}\setminus A$).

Frank Science

amazon.com/Cohomology-Sheaves-Universitext-Birger-Iversen/dp/…

@DanielFischer I see.. And what can we say for the case that we have space, that is not a Hilbert space?

@evinda If the subset is not closed, there's always a point such that the distance is not attained. If the subset is closed, there may or may not be one.

A ok... Thank you!!! @DanielFischer

7:42 PM

@robjohn Thanks!! :D

Has there been any study on the homotopy type of $Emb(S^1, S^3)$? That's some sort of classification space of all knots, right? Each path component tells me about isotopy classes of a knot. I don't think the usual compact-open topology should be the right one here though - maybe neighborhoods should consist of knots which can be taken to one another using "small" ambient isotopies, whatever that might mean.

@DanielFischer I couldn't figure it out (not until @robjohn answered in a comment) .. I was wondering (atm) if we could boil it down to something like $H(F(z),F(\sigma(z))) = 0$ where $H$ is something nice that might lead us to a functional equation of $F$ and $\sigma$ is a transformation form $\mathbb{C}\setminus [0,\infty) \to \mathbb{C}\setminus [0,\infty)$ ..

@r9m Ah, so you don't know if it even has an interesting functional equation? That makes things difficult.

7:59 PM

@FrankScience Nope.

@BalarkaSen Yep.

Thanks @MikeMiller. Any result/reference you have in mind?

Google.

Hey @anon their is something I don't understand let us consider $\phi : \mathbb{Z} \rightarrow \mathbb{Q}$ where $n \mapsto n * 1_Q$ their is something I don't intuitively get we know in this case we will have $Z \equiv \phi(Z)$

what about it

isomorphic I guess

not that

and so how come that will be Q ?

8:01 PM

how come what will be Q?

how come that will be isomorphic to Q

how come what will be isomorphic to Q?

ah, seems like there're some papers by Ryan Budney. Thank you @MikeMiller.

He would be the one.

well we know that the prime subfield will be either isomorphic to Q or $F_p$ depending on whether ch(F) = 0 or ch(F) = p

so in that case ch(F) = 0

8:03 PM

prime subfield of what?

prime subfield of the field F, so in the example above it is Q

the prime subfield of Q is Q, sure

is that all you're asking?

The space of (long) knots of some type should be $K(G,1)$s if I remember. Links are harder and don't work as well.

but using the map above I don't understand why that is even surjective ?

like it seems that will generate Z again in Q

What's a long knot?

8:04 PM

Z->Q is not surjective. obviously. what does Z->Q have to do with prime subfields?

Oh, R \hookrightarrow R^3

Just saw that in the paper.

Not quite, but your googling will find the correct statement.

well isn't the prime subfield of F is the subfield of F generated by the multiplicative identity

yes. and what does that have to do with Z->Q?

Ah, path components of $Emb(\Bbb R, \Bbb R^3)$ are all $K(G, 1)$. Coolio.

8:07 PM

just to make things clear what does it mean generated by the multiplicative identity ?

means the smallest subfield containing the identity

is that the same as if we get all elements of F and consider all elements of the form $p \in F$, where $p.1_f = p + p + p ......(p\ times)$?

wut

@DanielF: I'm getting annoyed. This is the third time this question has been asked. All previous ones have been closed. See here and here, and who knows how many copies of this I just haven't seen.

@DanielFischer but that approach (it seems) might not work .. since it turns out $F$ is $\operatorname{Li}_3$ .. I could make that approach work for $G(z) = \int_0^1 \frac{\log (1-x)}{z - x}\,dx$ and was wondering what would happen if we try and raise the power of $\log$ :) .. (sorry .. got disconnected there)

8:10 PM

This question is slightly better than the previous ones. It's still garbage.

suppose we denote the prime subfield of F as M, so is $M = \{p \in F | p.1_f = p + p + p .... (p \ times)\}$?

in order for "p times" to even make sense it needs to be an integer. are you assuming F has characteristic zero? and even then you're only specifying the prime subring, not the prime subfield.

I see what if we pick p as an integer will that make sense ?

oh I see ok I get it now

what if F doesn't have any integers in it? even if it does have integers, are you saying the only elements of the subfield are integers?

no, I mean to say $p \in Z$

8:13 PM

@r9m Well. I don't know much about polylogarithms. Do any of them have nice functional equations?

so if I define $M = \{p \in Z | p.1_f = 1_f + 1_f + 1_f (p \ times)\}$

I should have defined it as follows

I think

Consider $\Bbb Q$. You've just defined the prime subfield of $\Bbb Q$ to be $\Bbb Z$. and you haven't given any definition of the prime subfield in positive characteristic.

the prime subring is the set of all sums and differences of $1_F$ with itself. there is a unique unital ring homomorphism $\Bbb Z\to R$ and the image is the prime subring. it is the smallest unital subring containing $1_F$. the prime subfield is the smallest subfield containing the prime subring, or equivalently the subfield of all fractions $a/b$ where $a,b$ are in the prime subring.

oh I see

ok I see now ok I think everything is clear now, so what the book is trying to say is that depending on whether or not ch(F) = 0 or ch(F) = p. We will get either the field of fractions of Z or the field of fractions of Z/pZ

If char(R)=p then the prime subring of R will be F_p. If R is a field with char(R)=p, then its prime subring and prime subfield are the same, F_p. If char(R)=0 then the prime subring is Z, and if R is also a field then the prime subfield is Q.

yes, I understand now.

thank you

8:17 PM

@DanielFischer yes they do! :) You can check the wolfram pages and Lewin's book! They satisfy some nice functional equations!

thanks alot anon

mmhmm

@DanielFischer: Thanks, I appreciate it.

I noticed that ring theory and field theory has alot more information than group theory

user174558

8:39 PM

Group theory is a large field too.

user174558

@L33ter A lot

yeah @JasperLoy

@DanielFischer Let $(X, \rho)$ be a metric space and $x \in X, A \subset X (A \neq \varnothing)$.

I want to prove that $ x \in \overline{A}$ if and only if $d(x,A)=0$.

@DanielFischer I have tried the following:

Let $x \in \overline{A}$.

Then there is a sequence $(x_n)$ in $A$ such that $x_n \to x$ or equivalently $\rho(x_n, x ) \to 0$.

Thus $d(x,A)= \inf \{ \rho(x,y): y \in A \}=0$.

Let $d(x,A)=0$.

This means that $\inf \{ \rho(x,y): y \in A \}=0$.

How could we continue?

Is the first part right? Could I improve something?

8:59 PM

@evinda If $d(x,A) = 0$, and you take a small ball around $x$, what can you say about that ball?

9:28 PM

@DanielFischer So is the first part right?
The ball is defined as follows:

$B_{\rho}(x, \epsilon)=\{ y \in X: \rho(x,y)< \epsilon \}$

Right? But what can we say about it?

@evinda Yes, the first part is right. So, you want to show that $x \in \overline{A}$. The first step is that you know what that means. If you know what that means (in any of a number of equivalent characterisations of $\overline{A}$), you should roughly see how to continue.

I found this in wikipedia: x is a point of closure of S if the distance d(x, S) := inf{d(x, s) : s in S} = 0

But I think that we cannot use it without proving it, can we? @DanielFischer