Mathematics - 2012-06-16 (page 1 of 3)

Jordan

00:00

I don't follow why the e^c turns into 15

robjohn

@Jordan The $e^C$ can be anything for the differential equation, but it needs to be 15 to match the original information ($x=15$ at $t=0$)

Jordan

oh because it is times 1

I didn't see the algebra going on

M.B.

00:20

That happens to me a lot too, Jordan.

Peter Tamaroff

M.B.

Indeed. My brain is seemingly unable to remember basic properties of certain algebraic objects over a significant amount of time :(

Steven Li

01:59

hello

?

Eugene

02:17

@PeterTamaroff

Dylan Moreland

@PeterTamaroff Not you too.

Eugene

@DylanMoreland do you know about valued fields?

Peter Tamaroff

@Eugene Hey!

@DylanMoreland Hahaha, what? I was kidding!

Eugene

@DylanMoreland also can you see anything wrong with this proof?

@PeterTamaroff hey

Peter Tamaroff

02:33

@Eugene I've been studying quite a lot on congruences both from Apostol and Landau. Do you have any assignment on them I can check?

Eugene

@PeterTamaroff meaning?

Peter Tamaroff

@Eugene I mean, do you have any excersises that are worth doing?

Eugene

@PeterTamaroff sure. prove that $\zeta(5)$ is irrational. =)

Peter Tamaroff

@Eugene Hahahaha lol. With congruences? How'd that be?

Eugene

@PeterTamaroff it's an open problem

Peter Tamaroff

02:36

BTW this question got a lot of attention. Innocent questions!

@Eugene But why would you relate it's irrationality to congruences?

Eugene

@PeterTamaroff you wouldn't. i don't know how to prove it. if i did it would be proven.

Peter Tamaroff

@Eugene Oh! OK.

Eugene

i still don't understand why my proof keeps on getting downvoted.

but i guess you don't know algebraic geometry.

Peter Tamaroff

@Eugene I definitely don't. "Flex" and "Hessian" sound awesome, amongst other words there.

Eugene

well they're not really crucial concepts. just useful for this proof.

Peter Tamaroff

02:44

@Eugene LOL; look at this proof in Landau's book:

He proved first the Möbius inversion formula.

Thus, since

$$n = \sum_{d \mid n} \varphi(d)$$

one has

$$\varphi(n) = \sum_{d \mid n}\mu(d) \frac n d$$

Now he states

$$\varphi(n) = n \prod_{ n=1}^r \left(1-\frac{1}{p_r}\right)$$

for $$n= p_1^{r_1}\cdots p_r^{q_r}$$

He proves it by saying

> $$\varphi(n) = \sum_{d \mid p_1 \cdots p_r}\mu(d) \frac n d =n \prod_{ n=1}^r \left(1-\frac{1}{p_r}\right)$$ as seen by expanding the $2^r$ terms in the product.

Eugene

seems like a roundabout way to prove something.

sorry i'm just not a fan of landau's =P

Peter Tamaroff

@Eugene Roundabout = circular?

Eugene

long way

Peter Tamaroff

@Eugene Oh! Well, it doesn't take that long, really.

However, I think you'll like his alternative proof.

Eugene

haha

Peter Tamaroff

02:57

@Eugene THEOREM 74: Let $a>0$, $b>0$, and $(a , b)=1$. Let $x$ and $y$ range over reduced sets of residues $\mod b$ and $\mod a$, respectively. Then $ax+by$ ranges over a reduced set of residues $\mod ab$.

Eugene

yah

Peter Tamaroff

@Eugene Like that better?

Eugene

@PeterTamaroff ?

Peter Tamaroff

@Eugene Hahaha I mean, if you like that approach better than the roundabout one. =D

Eugene

hahaa

my favorite one is the pigeonhole one

Peter Tamaroff

02:59

@Eugene Tell me moreeee!

Eugene

@PeterTamaroff it's essentially the theorem 74 you just mentioned.

Peter Tamaroff

@Eugene Landau talks about the pidgeonhole principle.

Eugene

@PeterTamaroff yup

Peter Tamaroff

When talking about residue classes if I remember correctly.

Eugene

@PeterTamaroff yup

Peter Tamaroff

03:02

Here!

If $m$ objects are palced into $m$ drawers and each drawer contains ar least one then each drawer contains exactly one.

Alternatively, if $m$ objects are put into $m$ drawers and each contains at most one, then each drawer contains exactly one.

The first one exemplifies "it suffices that one of the $m$ numbers falls into each class", the other "it suffices that each of the $m$ numers be incongruent".

Eugene

@PeterTamaroff yup

Peter Tamaroff

@Eugene So, what are you doing?

Seems congruences don't interest you much =P

Eugene

@PeterTamaroff watching drive. it looks like a dumb movie.

Peter Tamaroff

@Eugene The one of the TAXI?

Eugene

@PeterTamaroff and they don't! at a certain point they're just a tool

@PeterTamaroff this one

Peter Tamaroff

03:06

@Eugene I knoooooooooooooow!!!!!!!!!!!

@Eugene I reccommend watching Polanski's last film, Carnage.

However, it is not suitable for all audiences, some might be put off by it, since it seems boring and stuff, but I found it very good.

Eugene

i watched the pianist and didn't like it though

Peter Tamaroff

@Eugene I wasn't amused by it also.

I enjoyed the music and the scenery.

Eugene

@PeterTamaroff it was kind of a dumb film overall

wow. 9034 rep. nice

Peter Tamaroff

@Eugene When I get to 10k, I'll mindlessly close your questions XD

Eugene

@PeterTamaroff you can already do that at 3k

must be you who's downvoting me then!

Peter Tamaroff

03:11

@Eugene Oh, no no, at 10k I think you have the power to close with one vote, but I think now that only mods can do that!

@Eugene Hahaha no way, I usually upvote!

Eugene

@PeterTamaroff only mods can do that.

hey upvote this

i need one vote so this question will considered as answered. looks like the troller is really going after me.

Peter Tamaroff

@Eugene As we told you, you can get that guy. But if it is OK for you, NVM.

Eugene

@PeterTamaroff ok.

Peter Tamaroff

@Eugene F**K! It's past midnight. Got to go, physics tomorrow!

Eugene

@PeterTamaroff bye. =)

Peter Tamaroff

03:17

@Eugene Bye!

Benjamin Lim

03:29

@Eugene Only two more votes on answers for a bronze badge in commutative algebra :D :D

Eugene

@BenjaminLim haha. good on you

@BenjaminLim there you go

Benjamin Lim

@Eugene hahahahahahaha

Eugene

03:50

@BenjaminLim ?

Eugene

04:32

@BenjaminLim did you get the badge?

Benjamin Lim

no

Eugene

@BenjaminLim huh. that's weird

anon

it takes some time until the part of the system purposed with handing out badges realizes you have a badge pending

David Wheeler

ye shall have it eventually ben

anon

also, it's possible only upvotes on answers count toward tag badges, I can't remember

Benjamin Lim

05:06

@anon I have currently 98 upvotes for answers of mine for commutative algebra

leo

05:17

@Eugene yo

Eugene

yo

leo

wassuop!

Eugene

@anon that is indeed correct. that's why i upvoted two answers of ben on CA

@leo nothing much. trying to strengthen my algebraic geometry.

and you?

leo

tired now. About to sleep

Eugene

ah. well goodnight then!

leo

05:20

just here to see whats going on

nothing much as you say

good night all

Eugene

nothing seems to be going on. it seems like a quiet day today

bye

awllower

06:05

I have a question, but think it fits not to a question form, so I ask here.
[Here](http://books.google.com.tw/books?id=RT5R_29X69wC&pg=PA35&dq=The+prime+ideals+of+R+which+ramify+in+R+are+those+containing+the+discriminant.%2BJanusz&hl=zh-TW&sa=X&ei=wiHcT6HRDJCWiQfTg52VCg&ved=0CDUQ6AEwAA#v=onepage&q&f=false) it is asserted that $R'$ is a free $R$ module because it is finitely generated and torsion-free over a PID. Might I ask for a reference or an explaination as to the reasonings behind this argument? thanks.

anon

hmm

Yes, that should work.

awllower

May I ask why? Thanks.

anon

If $X$ is a generating set then $R\,'\cong\bigoplus_X R$ right?

awllower

Yes

Then?

I mean: what does this have to do with that it is torsion-free, and how oe discovered that?

anon

well, hmm, I'm not sure how to rule out nontrivial linear relations between the elements of $X$..

Eugene

06:13

wtf just happened??

anon

the moon crashed and majora won

Eugene

i saw a really weird answer posted.

it's deleted now so i don't have a link to it

anon

what question was it on?

Eugene

this one

and i mean a seriously weird answer...

anon

"I Like to shake my penis at people on the street."

Perhaps he's on the wrong forum.

Eugene

06:16

@anon indeed. seems like he was looking for chatroulette.

let's hope you don't get suspended again for quoting it though!

anon

Hey, I put quotation marks on it.

Eugene

hahaha. it's so weird how random suspension is here.

anon

I would describe my incident as a one-off thing. Although Gigili sure gets caught up in a lot of misunderstandings.

awllower

Might I then ask a seemingly even more stupid question?
Why is the integral closure of a Dedekind domain in a finite extension of its quotient field torsion free over the Dedekind domain?

Eugene

@anon does she get suspended a lot?

anon

06:20

at least once, plus the recent thing with Ragib, plus the incident with Asaf, plus with Kannappan, plus ... there was another person I can't remember off the top of my head.

Eugene

@anon what happened with rajib and asaf?

anon

The recent thing with Ragib is that Gigili gave some incorrect and downright strange advice to Jordan on a diff equ question, and Ragib interpreted this as her deliberately trying to mislead Jordan, and he then accused her of this a few times (one time is starred thrice on the right, right now).

Do you know Asaf?

Eugene

@anon i've had a tangle with him once from which the tagging chatroom was born.

anon

lol

Dylan Moreland

@awllower If you Google around for proofs of the "structure theorem for finitely generated modules over a PID" you should get something.

awllower

06:25

Oh
Thanks.

Dylan Moreland

And there's certainly a proof in Lang's Algebra. In most algebra books, I'd imagine.

awllower

Albeit I cannot chack it now, but I will, and thank again.

Dylan Moreland

If you just want to isolate the proof that finitely generated + torsion-free implies free, that's probably less work.

anon

@Eugene Asaf made some comment about a guy not having good English because he was Moroccan (sp?) or something, then Gigili made a remark about Asaf's English (which is generally between fair and perfect imo), and it culminated in a sexist remark by Asaf getting flagged and a couple of chat members (me and tb) responding to him. (After the suspension he decided not to come back.) Sort of an old story now.

Eugene

@anon wow. that i didn't know. no wonder he refused to come to chat.

anon

06:28

There was an organized campaign to try to pursuade him back though. Many a star was spent on that effort.

Eugene

@anon and he refused because?

anon

He never explained why in chat, but one of our chat regulars (Benjamin? can't remember) talked to him on Skype, and it seems he wanted to devote more time to real life stuff.

Eugene

@awllower i think this proof might be in stewart and tall's algebraic number theory. i proved this in my class last semester but to go over my notes would be too long.

@anon ouch. i'm feeling really unproductive again. i haven't done any serious research since my senior year in undergrad...

my professors say that's good since i should be learning more at this time. but that's easy for him to say since he published 18 papers in 3 years...

Dylan Moreland

@awllower I guess I was commenting on your earlier message. This part is easier. How can an element of a field be torsion over a subring of the field?

anon

heh, I though there was something funny about "torsion" and extension of a quotient field

awllower

06:34

@Eugene: Thanks. I was suspecting that this has something to do with the general properties of rings, while unable to confirm this thought. So the question reduces to saying that an integral ring over a Dedekind ring is torsion free.

Eugene

you mean to thank dylan.

awllower

@DylanMoreland: maybe the characteristic is not zero?

Dylan Moreland

Doesn't matter!

awllower

oh

Because it is an integral domain?

Dylan Moreland

Yep.

anon

06:36

indeed

Dylan Moreland

@Eugene I have friends who put out something good every month or so. It's pretty depressing.

Eugene

i know. this is the guy telling me not to worry.

easy for him to say seeing as he has two transactions...

Dylan Moreland

Also depressing is talking to any student of Richard Taylor's.

Eugene

@awllower i was referring to the theorem that if $A$ is an integrally closed ring of characteristic $0$ with field of fractions $K$, and $L$ is a finite extension of $K$, then if $B$ is the integral closure of $A$ in $L$ then $B$ is a submodule of a free $A$-module of rank $[L:K]$.

@DylanMoreland luckily i will never have that privilege. =)

ken ono's students are pretty depressing to talk to though.

Dylan Moreland

Right, the finitely generated part is the interesting bit.

Eugene

06:42

matt wage (one of his REU students) published in transactions in high school

Dylan Moreland

His REU works out really well for some people.

awllower

Ah!
I see why principal ideal domain is required! It is to find elements which generate some isomorphic image of the module!!

Dylan Moreland

He's a promoter.

Eugene

@DylanMoreland that's what i keep hearing. is that really true?

they say he's the donald trump of math.

i took 2 classes under him and did one research project under him. man is a machine. got us to publish in under 3 months.

Dylan Moreland

Can't argue with that.

Eugene

06:45

that he's the donald trump of math?

anon

@Eugene Is that supposed to be good or bad?

Benjamin Lim

Hey guys

Can I ask if my understanding of the implicit function theorem is correct

Eugene

@anon bad. it means he talks a big game.

anon

heh

Eugene

the man's got like 3-4 annals papers though. so no doubt he's a really good mathematician

anon

06:46

@BenjaminLim go ahead, my multivar calc is rusty with the technicalities though

Benjamin Lim

Ok I will state the implicit function theorem

Eugene

that's not counting his inventiones

Benjamin Lim

Let $A$ be open in $\Bbb{R}^{k+n}$ ; $f : A \rightarrow \Bbb{R}^n$ be of class $C^r$. Write $f$ in the form $f(\Bbb{x},\Bbb{y})$, for $\Bbb{x} \in \Bbb{R}^k$, $\Bbb{y} \in \Bbb{R}^n$. Suppose that $(a,b)$ is a point of $A$ such that $f(a,b) = 0$ and det $\frac{\partial f }{\partial \Bbb{y}} (a,b) \neq 0.$

Then there is a neighbourhood $B$ of $a$ in $\Bbb{R}^k$ and a unique continuous function $g : B \rightarrow \Bbb{R}^n$ such that $g(a) = b$ and $f(\Bbb{x},g(\Bbb{x})) = 0$ for all $\Bbb{x} \in B$

Now @anon

What I don't understand is the bit

where some people say

the inverse function theorem allows you to solve for some of the variables in terms of the others

is the bit about solving for variables to do with the $g(\Bbb{x})$?

Eugene

the key is the determinant bit

Benjamin Lim

@Eugene Ok

But what I don't understand is this

If I have say $u(x,y,z) = 0$, $v(x,y,z) = 0$ and $w(x,y,z) = 0$

Eugene

06:51

as i've learned the hard way (in an exam) a wrong selection can really screw it up.

anon

(For example)

derp

Benjamin Lim

@Eugene If I set $g = (u,v,w) $ that goes from R^3 to R^3

Eugene

what does derp mean?

Benjamin Lim

I calculate the derivative of $g$

anon

means I'm being foolish

Eugene

06:52

you've used it a few times but i don't understand it

ah i see

Benjamin Lim

and find that at some point say the derivative is non-zero

@eugene @anon how does that mean I can solve for $x,y,z$ in terms of $u,v,w$ by the implicit function theorem?

Eugene

@BenjaminLim this is really answered in rudin =)

pg 224 i think

Benjamin Lim

@Eugene No not really

What it states there I already get

I understand the "can be solved for" in the single variable case

I'm just saying

Eugene

@anon i should take a leaf from asaf's book...

@BenjaminLim in rudin doesn't it have an example on page 227?

anon

Writing $f(U,V,W;x,y,z)=\big(U-u(x,y,z),V-v(x,y,z),W-w(x,y,z)\big)$ allows a function $\vec{g}(U,V,W)$ such that $f(U,V,W,g_1,g_2,g_3)=0$, i.e. an inverse $\vec{g}$ of the vector function $(u,v,w)$ of $(x,y,z)$.

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