Mathematics - 2012-12-20 (page 2 of 3)

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3:00 AM

Please try and stick to English in the chat, there are many people from many parts of the globe and we all like to be able to read what is being said. English, of course, means actual English and not l33t spik.

Pero por ejemplo yo si creo Latinoamérica debería hacer un poco mas del 2% de la producción científica mundial. Que en las universidades la gente que estudia ciencias sociales/políticas debería ser menos y deberían ser más los que estudian ciencias puras o áreas técnicas

etc

@PeterTamaroff y que comen en sus casas en Argentina? :-)

@leo Si, para m tambien. Muchos subestiman la importancia de la ciencia

Y de la matematica en particular

@leo Hmmm, tradicionalmente, empanadas, milanesas, pure, y claro, asado.

@PeterTamaroff Si es un problema eso. La matemática está directamente relacionada al desarrollo.

@PeterTamaroff Pero como en el almuerzo de todos los días, qué es lo común?

eso mismo?

que bueno el asado.

@leo Hmmm, no siempre

@leo Digamos que las milanesas y las emp en mi casa si.

Pero despues carne, fideos, pizza.

3:16 AM

@PeterTamaroff las milanesas son comunes. En programas argentinos siempre las mencionan

@PeterTamaroff aquí parecido

pero la base de la comida es arroz y frijoles. En México es el maíz.

1. is exactly the same thing that say $f(x)\neq g(x)$ for all $x$ in $B$ in here

4:03 AM

no

@leo

hey

@BenjaLim hey there!

Merry Christmas!

@leo Hmmm

someone wants to talk of fusion categories

but doesn't even know why $\Bbb{Q}[x]/(x^2 - 2) \otimes_{\Bbb{Q}} \Bbb{R} \cong \Bbb{R}[x]/(x^2 -2)$.

@BenjaLim me neither. @mariano knows that kind of things

@leo I can tell you why that is so

@leo Consider the ses

4:15 AM

I'm leaving to get some sleep

$0 \to (x^2 - 2) \to \Bbb{Q}[x] \to \Bbb{Q}[x]/(x^2 - 2) \to 0$

@BenjaLim how the ses?

short exact sequence

Then tensoring with $\Bbb{R}$ which is exact gives

oh

$ 0 \to (x^2 -2) \otimes \Bbb{R} \to \Bbb{Q}[x] \otimes \Bbb{R} \to \Bbb{Q}[x]/(x^2 - 2) \otimes \Bbb{R} \to 0$

@leo tensor product is over $\Bbb{Q}$.

4:16 AM

@BenjaLim I see.

But now $\Bbb{Q}[x] \otimes \Bbb{R} \cong \Bbb{R}[x]$

@PeterTamaroff buenas noches!

good night!@Ben

and the image of $(x^2 -2) \otimes \Bbb{R}$

ok.

heh

@anon hey

4:17 AM

yo

@anon just having a rant above.

I wonder if blah has a more field-theoretic solution.

Hello, I was wondering if there is a Physics chat room here? I tried searching but couldn't find it and I am new, so I might be doing something wrong.

@MarianoSuárez-Alvarez Hey

I was ranting above about tensor products :D

Anyone?

4:21 AM

@DemCodeLines go to physics. se

I put Physics in the search bar and there were not any active rooms or ones for Physics

Mariano Suárez-Alvarez

Hello, Ben

@MarianoSuárez-Alvarez hey

@MarianoSuárez-Alvarez I figured out that problem in algebraic number theory the other day.

Mariano Suárez-Alvarez

good :-)

@MarianoSuárez-Alvarez oh do you have a reference for this fact?

Mariano Suárez-Alvarez

4:27 AM

which one?

@MarianoSuárez-Alvarez

www.jmilne.org/math/CourseNotes/ANT.pdf

remark 3.39

page 61

@MarianoSuárez-Alvarez I don't get the proof of Serre because I haven't studied completions.

Mariano Suárez-Alvarez

/me looks

Ah.

I've seen that proved in a couple of places, using the exact same argument as in Serre

@MarianoSuárez-Alvarez He uses completions and stuff

Mariano Suárez-Alvarez

yup

he does that in order to be able to do it locally

is there a simpler more "elementary proof of this fact"?

Mariano Suárez-Alvarez

4:32 AM

well, completions are quite elementary! :-)

trying to eliminate them will only complicate things

hmmm.....

Mariano Suárez-Alvarez

as they are introduced in order to be able to aproximate things modulo high powers of ideals easily

it is just language, really

@MarianoSuárez-Alvarez Because

Mariano Suárez-Alvarez

but an incredibly convenient one

@MarianoSuárez-Alvarez wait does serre introduce completions in his book?

Mariano Suárez-Alvarez

4:33 AM

you could ask on the main site for a proof without completions

I'm sure matt or david S can point you to one if there is one available

probably not: completions are «well-known»

see A-M

or, if you have more time, Eisenbud's book on comm. alg.

@MarianoSuárez-Alvarez Actually

I think Marcus problem 20 on pg 87 is what I need

Mariano Suárez-Alvarez

that part of Serre's book, which is the more elementary ppart of the book, is really beautifully written

I just found it :D

Now I just need to do it :D

@MarianoSuárez-Alvarez Actually

I am learning algebraic number theory now from Marcus

@MarianoSuárez-Alvarez I should finish chapter 5 in two months or so.

@MarianoSuárez-Alvarez Do you think I can go on to Serre's local fields after that?

Mariano Suárez-Alvarez

why would you want to do it?

at this point you want to aim at something, probably :-)

@MarianoSuárez-Alvarez Because I have thought of a career in number theory.

4:43 AM

hey @benja, what is $$\sum_\lambda \mathrm{tr}(\mathbf{S}_\lambda A)x_1^{\lambda_1}x_2^{\lambda_2}\cdots?$$

what is the notation? @anon

you want to know the trace?

Mariano Suárez-Alvarez

serre's book aims at local class field theory

I am pretty sure any ANTist should know that

@MarianoSuárez-Alvarez ok.

Mariano Suárez-Alvarez

but if you have a friiendly ANTist at hand whom you trust, you can talk to him/her

$\mathbf{S}_\lambda$ is the schur functor, $A$ a linear operator, $\lambda$ runs over all integer partitions. I know the $\mathrm{tr}(\mathbf{S}_\lambda A)$ is the schur polynomial $s_\lambda$ in the eigenvalues of $A$.

Mariano Suárez-Alvarez

4:45 AM

I read that book because I wanted to unerstand stuff related too Brauer groups, for example

@MarianoSuárez-Alvarez I don't know what this guy does:

@anon ok assume that $A$ is diagonal first.

@anon in your question above

Mariano Suárez-Alvarez

yoy should ask him! :-D

are you asking what polynomial that thing is?

Mariano Suárez-Alvarez

what do you mean, howcome?

@MarianoSuárez-Alvarez he did suggest we can look class field theory after this course.

@anon that is correct what you said about the schur polynomial.

@anon I don't understand what you would like to know in the problem above.

4:49 AM

anon

wana see a lambert series for the vonmangoldt function

@BenjaLim I am wondering if it has a nice closedish form. It is a generating function, obviously. It is supposed to generalize formulas like $$\sum_k\mathrm{tr}(\mathrm{Sym}^k(A))q^k=\frac{1}{\det(1-qA)}$$

ok.

@anon You may want to check out appendix in fulton and harris.

appendix A.

@anon I can give you an expression in the opposite way.

That is I can tell you what $\textrm{Tr} S_\lambda A$ is when $A$ is diagonal.

This will be an expression in polynomials.

I thought that trace only depended on $A$'s eigenvalues (in the algebraic closure of whatever the base field is), and thus no restriction on being diagonal was necessary?

but there is a nice interpretation from lie theory

because if $A$ is diagonal

I can write it in terms of exponential of some kind of diagonal matrix

that may lie in the Cartan subalgebra of some Lie algebra.

For example if $A$ is invertible

Say it is a complex matrix

then we know that $\exp : \mathfrak{gl}_n \to \textrm{GL}_n$ is surjective.

@anon that formula above

I have not seen it.

Where did you get it?

Procesi pg263

4:57 AM

ah the proof is nice.

Reminds of a proof of Pieri's formula.

@anon Hmm unfortunately I don't know of the top of my head

if there is an expression for the generating function that you wrote above.

5:20 AM

hello

is anyone in this chat

n;ah

2 hours later…

7:30 AM

Hey guys. I've got a crypto-question (RSA) I hope you can help me with:
I know p=467, q=479, n=pq, e=73443 and d=70167. I have to decrypt "037095105399056356" with ASCII-values.
Really hope some of you can lead me on the right track. :)

7:40 AM

rsa

what "ASCII-values"?

are*

Have you used the rsa algorithm before?

Are you familear with eulers phi function etc?

I'm familiar with the basics of RSA, phi-function etc. I guess I have to split the "string" up and decrypt it and convert the resulting numbers to their corresponding ASCII-value.

Is this homework or somthing?

those are small values for p and q

lol

What are you asking just how to do it?

Or are you asking for the decryption?

It is some kind of homework, yes. I've fought with it for a while and would love to know how to decrypt it properly.

sec lol, I dont remember the algorithm totally, let me just review

calculate

why is there a zero at the front of your digit string

it should be an integer?

It is a zero, and it have annoyed me too.

7:49 AM

Are you using some other numeral system?

Not as far as I'm aware of.

is the cipher text that string of digits?

ok listen who ever encoded that, did so with a corresponding system

when you 'decrypt' that

you will get another string of digits

but these should match up to some system already defined

such as 11=a,12=b,13=c

so you can take the decrypted digit string

and convert it into letters

but that system should already be known by the people using your system to encode messages

did the instructor give you such a system?

I don't think that is the right cipher text

I'm pretty sure that that "system" would be the ASCII-table.

im not very tech savvy

lol

let me look that up

well whatever, lets just say I get you the resulting string of digits

you can figure that part out

the problem is your cipher text isn't an integer

therefore you cant compute [ciphertext]^d mod n

As [ciphertext] should be less than n, I have tried to split it up in groups of 2, 3 and 6 digits but that doesn't seem to work either. :/

7:57 AM

oh mabe its string of digits your right

I havn't done this in a while

How would you even break it up?

is the sum of the digits divisible by 2,3,6?

i mean the number of decimal places

Floor[log[10,n]]+1

My first thought would be to break it up in groups and decrypt them:
037095105399056356 -> 037 095 105 399 056 356
or 037095 105399 056356 etc. But all of those numbers decrypt to a number in 10^6 which I think I can't use.

it should be less then n

obv

lol

Is this an assignment or somthing?

user19161

Hey @ethan, nice new pic!

lol

user19161

You like to lol here don't you? Just like me.

8:05 AM

wana see a lambert series for the vonmangoldt function

here

user19161

Nah, I don't know these things.

oh

lol

user19161

But I can just admire its looks.

I thought you new about number theory

the vonmangoldt function $\Lambda(n)$ is zero everywhere unless n is a prime power say $n=p^j$ in which case $\Lambda(n)=ln(p)$

it should be a capital lambda

oops

user19161

@BenjaLim The way you ask me about CW complexes that day sounded arrogant to me, which is why I did not want to answer you. I was only having a conversation with you, not trying to tell you whether I know about CW complexes or not. And note that I always call myself a banana, and say that I only know 1+1=2.

8:08 AM

whats a CW

lol

oh

the topological space

nvm

Jasper

what are you good in?

user19161

@Ethan Nothing. I am only a banana.

lol

do you have a specialtie

user19161

No, my specialty on this site is...

is...

user19161

...algebra-precalculus, LOL.

8:11 AM

jasper

user19161

See, I even got a badge for it!

what do you do for a living

user19161

Me? I am not doing anything now...

user19161

Just taking a long break.

Are you a student at a university?

lol

user19161

8:13 AM

Well, I finished my undergrad long ago, worked for a while and now am on a long break like I said.

oh

what did you study?

user19161

I studied math. But I only know very little.

do you have any of your own side work or research you work on?

user19161

Hmm, no, I don't think I can be said to have done any research level stuff. I wrote something in my undergrad days though. I can share with you if you email me. First name dot last name at gmail dot com if you want.

what field is it?

lol you should do your own work, studying is very boreing

I think learning anything is boreing

but applying what youve learned

thats fun

I mean not many people like to pay for a nice car, but most enjoy riding it

user19161

8:16 AM

It was about measure and integration.

do you self study?

lol

user19161

Well, now and then. I may apply to graduate school a few years from now, and hopefully I will get a place and can start a new career in math...

whats a "career in math"

?

user19161

Well, I mean teaching and research.

lol i hate teaching

ive never really taught

but i hate tutoring

user19161

8:19 AM

Well, it is my opinion that great researchers are also great teachers.

tho

I guess I dont hate teaching

you can be good at somthing and not enjoy it

tutoring

or whatever

user19161

@Ethan Yes, I know what gcd and lcm are, LOL.

?????

user19161

You can click on the left arrow to see which line I am replying to.

user19161

You can click on the right arrow to reply to a line.

user19161

8:22 AM

Move your mouse over the lines and you will see the arrows.

user19161

Many chat newbies don't know about these arrows, so I have to tell them.

oh

lol

8:48 AM

@JasperLoy I did not mean it in that way.

Hi @BenjaLim

@OldJohn Hey

@OldJohn I am going nuts

@BenjaLim How are things?

Why?

@OldJohn too many algebraic number theory problems to finish.

the problem is

40 of them

to solve one you have to use results from other problems

@BenjaLim Sheesh!

darn :(

8:50 AM

@Ethan If you want to know about CW complexes you can ask me

@OldJohn that's why I'm like farrkkkkkkkkkk

but if you get stuck on one, I guess you can just assume the result for the next one?

yeah I guess so

@OldJohn Oh

I can ask you something on the CRT?

I don't understand a statement in my book

Sure - if I know

Marcus?

yes

@OldJohn how did you guess?

pg 87

problem 20

@BenjaLim I think you told me - the other day :)

8:52 AM

you have the book?

Hang on - my copy is upstairs ...

ok

The first 6 digits (037095) is decrypted to $037095^d mod n = 037095^70167 mod 223693 = 70105$. This is eqv. to "F" and "i" in the ASCII-table.
Thanks for the help. :)

@BenjaLim I am afraid I am not going to be much help here - I haven't got as far as "primes lying over $P$ ... I am not as far through ANT as you are :(

@OldJohn Ok, no worries

@OldJohn but here is a question

that doesn't require one to know about lying over theorems and stuff

@OldJohn Suppose I have a product of prime ideals $Q_1^{e_1} \ldots Q_r^{e_r}$.

8:56 AM

OK

Then in Marcus, they say that

"For each $i = 1,\ldots, r$ and for each $j = 1,\ldots, e_i$ we can choose an element $\alpha_{ij} \in (Q_i^{j-1} - Q_i^j) \cap (\cap_{h \neq j} Q_h^{e_h})$.

@OldJohn he says this is possible by the chinese remainder theorem.

I am not seeing how it comes from the CRT.

From what I understand the CRT says that if $I_1,\ldots, I_k$ are ideals that are mutually coprime;

then the product of all of them is equal to their intersection

and then $R/(I_1\ldots I_k) \cong R/I_1 \times \ldots R/I_k$

Yep - that is how I understand it too

Can't see how CRT ensures $a_{ij}$ exists - looks more like Hensel's lemma, in a way

pfffffffffffffffffffffffffffff

@OldJohn how about this:

By unique factorisation, given any $i,j$ we must necessarily have that $Q_i^{j-1} - Q_i^j$ be non - empty.

So we can choose an element $x_{ij} \in Q_i^{j-1} - Q_i^j$.

Now I think this is where we need to apply the CRT to get our $\alpha_{ij}$.

ah! - that might be the way

@OldJohn ok we are now closeR to the end......

9:04 AM

yep :)

@BenjaLim bugger - have to go deal with a guy coming to sort out the kitchen - back later, sorry

@OldJohn no problem.

Good luck!

@OldJohn I may have a solution by then :D

9:17 AM

@OldJohn I have a solution :D

Ok remember I said above that let us fix $i$ and $j$. Then by unique factorisation we can choose $x_{ij} \in Q_i^{j-1} - Q_i^j$.

Since $Q_i^j$ will be coprime to $Q_h^{e_h}$ for any $h \neq i$

We can find an $\alpha_{ij}$ that satisfies the following congruences:

$\alpha_{ij} \equiv x_{ij} \pmod{Q_i^j}$

$\alpha_{ij} \equiv 0 \pmod{Q_h^{e_h}}$ as $h$ ranges over all $h \neq i$.

the last line above says that $\alpha_{ij} \in \bigcap_{h \neq i} Q_h^{e_h}$

So we need to check that the first line tells us that

$\alpha_{ij}$ actually lands in $Q_i^{j-1} - Q_i^j$.

Firstly $\alpha_{ij} \notin Q_i^j$

For if it were actually in $Q_i^j$ then this would contradict $x_{ij} \notin Q_i^j$

and also $\alpha_{ij} \in Q_i^{j-1}$

because $\alpha_{ij} - x_{ij} \in Q_i^j \subseteq Q_i^{j-1}$ and $x_{ij} \in Q_i^{j-1}$.

@OldJohn boom

@OldJohn I'm listening to this now: youtube.com/watch?v=xvX_5ym_ajI

9:49 AM

So many good maths jokes on here. I'm tempted to collect them! xD

user19161

Finally answered something today.

Hi

My Go/No-Go became a Go.

user19161

@Chris'ssister You changed your email!

user19161

@JonasTeuwen Well done bro, well done.

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