Mathematics - 2012-03-13 (page 3 of 5)

Asaf Karagila

16:00

Probably.

I am going to do the dishes.

anon

"In fact, three clouds cover the plane if and only if CH is true." Cool.

Ilya

@anon CH? Continuum Hypothesis?

Skullpatrol

$\Huge \text{Ilya is back!}$

2

anon

naturally

Ilya

@anon naturlich?

anon

16:04

au naturel?

Ilya

@anon cosmetique?

phew, finished grading exams. There are moment I hate this signal analysis course

anon

can someone do some numerical experiments to verify what the op says in this question?

code: Integrate[Integrate[(ax-by) Exp[-(x+y)] (a^2 x+ b^2 y+ c xy)^(-3/2),{x,0,Infinity}],{y,0,Infinity}]

Asaf Karagila

@anon Please wear pants when chatting here.

robjohn

@anon A single Integrate should be more efficient, I would think.

anon

make me

robjohn

16:17

@anon just don't post any pics.

anon

I'm disappointed I can't find any pantsless mathematician porn on google. Where art thou, Rule 34?

Skullpatrol

@anon What is Rule 34?

t.b.

@Skullpatrol what is Google?

anon

awkward moment to return to chat...

Skullpatrol

@tb A rip off from the googolplex.

Googolplex

A googolplex is the number 10googol, i.e. 10^{(10^{100})}. In pure mathematics, the magnitude of a googolplex could be related to other forms of large number notation such as tetration, Knuth's up-arrow notation, Steinhaus-Moser notation, or Conway chained arrow notation. History In 1938, Edward Kasner's nine-year-old nephew, Milton Sirotta, coined the term googol, then proposed the further term googolplex to be "one, followed by writing zeroes until you get tired". Kasner decided to adopt a more formal definition "because different people get tired at different times and it would neve...

t.b.

16:27

@anon bleeh, I probably shouldn't be here if I'm in a stinky mood...

I'll try to lighten up before I come back.

Matt N.

Oh. So I'm not the only one who's in a stinky mood. I see. But it still means I said something that made him leave : /

@anon Up for another quick algebra question? : )

Kannappan Sampath

The most beautiful answer I have ever come across.

@MattN I am ready, but not sure if I'll be of some help though.

anon

why would you think I'm an authority on algebra? yeash.

Matt N.

Well I have two questions actually.

You've helped me before : )

Kannappan Sampath

And, in that problem, that's actually something like: Since the definition of rings says, $A$ under $+$ is an abelian group. So, closure means supposed to inherit the structure of $+$ and hence, the inverses will be inherited as well.

Matt N.

16:36

Anyway, so here it goes: on page 2 in Atiyah we have:
"Prop. 1.1. There is a 1-1 order-preserving correspondence between the ideals $I$ of $R$ which contain $J$ and the ideals $\bar{I}$ of $R/J$, given by $I = \phi^{-1}(\bar{I})$".

Kannappan Sampath

Yes.

So, here order preserving means for any ideals $I \subseteq J \subseteq K$ there corresponds ideals $\bar I \subseteq \bar J \subseteq \bar K$.

Matt N.

So I want to write down that bijection. In my own words/notation: Let the set $S$ denote the set of all ideals of $R$ containing $I$. Then I have a bijection $\phi : S \to \bar{S}$, $J \mapsto J/I$. Right?

Or is that bad? I mean for some reason they write $\phi^{-1}$ in Prop.1.1.

anon

$S\to R/I$ um...

Kannappan Sampath

Well, you are looking for bijections between what?

Matt N.

Oh messed up, sorry.

$J$ is mapped to the coset $(J \setminus I) + I$.

Is that an ok thing to write? I'm asking because it feels kind of clumsy.

Kannappan Sampath

16:41

No, an ideal $J$ that contains $I$ is mapped to $\phi(J)=J+I$

Matt N.

@KannappanSampath Between the ideals $J$ containing $I$ and ideals in $R/I$ that are of the form $J/I$.

@KannappanSampath Fixed. : ) Thank you.

(all $i$ in $I$ are mapped to zero)

^I see what it looks like, I just can't write it down : )

Kannappan Sampath

@MattN You really are looking for a bijection for ideals in $R$ that contains $I$ and all ideals in $R/I$.

So, here is how I'd write it down: $R \supset J \mapsto \phi(J)=J+I=\{j+I \mid j \in J\}$ where $I \subseteq J$

So, if $j \in I$ then as you said, then that would go the coset element $\Bbb 0$ in $R/I$.

Now, claim that $J+I$ is actually an ideal.

Matt N.

That's clear to me. I'm thinking about why I get all ideals in $R/I$. : )

Aha.

Kannappan Sampath

So, to see why you get all ideals in $R/I$ define an inverse map.

I think I have been speaking obvious things so long. I regret it. :/

Matt N.

Let $\bar{J} $ be an ideal of $R/I$. Then $\phi^{-1} \bar{J} = J \cup I = J$ since $I \subset J$

@KannappanSampath What do you mean?

Kannappan Sampath

16:50

Removed.

Matt N.

: )

Kannappan Sampath

Cup and cap look confusing if not for the bookmarklet.

Matt N.

I'm not even sure one is allowed to use set theory ($\cup$) to operate on groups...

Kannappan Sampath

@MattN What does this mean?

@MattN I mean, I have been telling all of that s**t as if you were beginning Algebra! : (

Matt N.

Well if you have two groups $G$ and $G^\prime$ then $G \cup G^\prime$ isn't necessarily a group again.

@KannappanSampath What?! You were helpful!

Kannappan Sampath

16:53

Yes. But if $I \subseteq J$ are ideals, then $I \cup J$ is an ideal as you observe.

Matt N.

Yes. True dat.

Kannappan Sampath

(The same goes for groups.)

Matt N.

Also true.

Ok. I'm off again. See you in a bit and thank you!

Kannappan Sampath

@MattN No second question?

Jonas Teuwen

@robjohn We just had a nice talk about the Carleson-Hunt theorem in UMD Banach spaces :-).

robjohn

16:56

@JonasTeuwen UMD?

Ah, Martingales

Jonas Teuwen

I think it only holds if and only if it is a UMD space.

Ilya

@Jonas: do you remember which lecture notes by your supervisor have you advised me? I can't fund them

Daniil

Hey.

Jonas Teuwen

@Ilya On SPDEs?

Kannappan Sampath

Is it allowed to ask someone to accept your answer because there are no other answers, it seems quite rude! :-)

Ilya

17:06

@JonasTeuwen yes

Jonas Teuwen

@Ilya fa.its.tudelft.nl/~neerven/publications/papers/ISEM.pdf

Kannappan Sampath

@Jonas Watched and heard your lecture. Order of examples impressive. If I were in that class, I'd give you a full 5 stars. :-)

Jonas Teuwen

@KannappanSampath There were very little examples! :-).

Kannappan Sampath

@JonasTeuwen But, the few were in right order. :-)

Jonas Teuwen

Oh, nice :-).

Kannappan Sampath

17:29

A group theory answer after a long time in my life.

Hey @Dylan How do you do?

We recently had a fiasco about subrings of Atiyah!

Dylan Moreland

Hi.

I'm at a conference so won't be very talkative. Did you find another disruptive typo?

Also I thought this guy would know better.

Kannappan Sampath

@DylanMoreland Meaning Grapth knows about these things better?

Dylan Moreland

I just meant that he seemed like a smart fellow.

But the question is meaningless and will at its worst be pretty offensive.

Kannappan Sampath

Well, the point is with his definition of a subring.

He says they are subsets closed under addition, multiplication and contain $1$.

Dylan Moreland

I can't imagine that it's important. Subrings should share $1$ unless you're doing something that requires you to relax that condition, like working with semisimple algebras/representations and all that.

Kannappan Sampath

17:35

No, it's not about $1$, but about $0$. Why will $0$ reside here?

Dylan Moreland

They should have just said additive subgroup.

Kannappan Sampath

This is what I could come up with:

And, in that problem, that's actually something like: Since the definition of rings says, $A$ under $+$ is an abelian group. So, closure means supposed to inherit the structure of $+$ and hence, the inverses will be inherited as well.

Dylan Moreland

$0$ is in there; don't spend another moment on this.

What's wrong with tb

Ah :(

Kannappan Sampath

I am sorry, I know you're irritated with that silly question, but why $0$ there?

Dylan Moreland

I'm not irritated! It's advice that's meant to help.

Kannappan Sampath

17:39

Well, should I assume Atiyah meant that there and move on? (I never even read this defn until Matt brought it up here.)

Dylan Moreland

I mean, for example, if $A$ is a subring of $B$ then you want the inclusion $A \to B$ to be a homomorphism of rings.

Kannappan Sampath

Yes.

Dylan Moreland

So the zeros had better be the same.

Kannappan Sampath

Well OK. So, we all agree Atiyah was being imprecise there!?!

(My sloppiness has increased manyfolds that I am failing to distinguish the right from the wrong. ) T_T

Anyway, time to sleep. :=)

Jonas Teuwen

18:07

@KannappanSampath Night night.

Matt N.

I have another question: A principal ideal in a commutative ring $R$ is defined to be an ideal $I$ generated by one element, $\langle i \rangle$.
I think this only works if $R$ has a $1$ because I need $i + i \in \langle i \rangle $ and that only holds if I can do $i + i = (1 + 1) i$ because then $(1 + 1) \in R$ and hence $(1 + 1)i = i + i \in \langle i \rangle$. Can someone confirm this?

anon

18:31

Um, so why do you need a 1 for $i+i\in \langle i\rangle$ again?

I thought you generally just worked with rings with a 1 anywho..

Matt N.

@anon Only because I don't see how to prove $i + i \in \langle i \rangle$ without a $1$. How do you prove it without $1$?

@anon True but I'm thinking about the general case.

anon

hm. never thought about it.

Consider R=3Z as a commutative ring without unity and the ideal I=3R. I don't see a 3+3 in there, so I don't think it works without a 1.

Anyway your argument seems to boil down to "it holds if I have a 1" but you don't really demonstrate the only part (am I not getting something?). A counterexample does that swiftly though.

t.b.

18:48

@MattN you need to be specific what you mean by "generated". The intersection of ideals is an ideal and the ideal generated by an element is the intersection of all ideals containing that element. You lose the explicit description of a principal ideal as $Ri$ if you don't have a unit (note that not even $i$ needs to be included in the set $Ri$ if you don't have a unit).

anon

the $\cap$ of ideals containing the element. that's a novel way to define it. why haven't I read that?

eh, it's on planetmath. for some reason I have outdated understanding.

So anyway if $i$ is in all ideals containing it, then $i+i$ is in all those same ideals (by definition) and thus is in their intersection. cake.

why do I say "anyway" so much? hrm.

Matt N.

Sorry was afk (cooking)

brb

He left again :,(

anon

I told you. Apology letter. Single-spaced, ten pages, with tear stains.

Matt N.

You're funny. But it's no laughing matter : (

Matt N.

19:12

@tb Ah right. Thank you! By generated by one element $i$ I meant $\langle i \rangle := \{ ri \mid r \in R \}$.

@anon I was asking myself the same the other day.

anon

not sure which question you're referring to

Matt N.

Mouse over my reply, then you'll see.

anon

oh

Um, not by your definition of $\langle i\rangle = Ri$ it isn't.

But $\{ri |r\in R\}$ isn't necessarily a subgroup if $1\not\in R$!

Matt N.

@anon True! Because we don't have to have $i \in \langle i \rangle$.

Yep. Getting there. Thanks!

anon

Well dicks. I was editing a rather tedious question in the reviewer tools and my browser froze. Refreshed and poof it's gone, don't even know where the question is.

t.b.

19:19

It's quite a general principle: the subthing generated by a set $S$ is the smallest subthing containing the set. What you need is that the subthings are closed under arbitrary intersections

In many cases you can give more or less explicit descriptions of the subthing generated by the set, but this is by no means always so.

@MattN I'm sorry I really didn't mean it the way you seem to interpret it. I'm not having the best of days, that's all. (I found the typo thing a bit exaggerated, but I should simply have ignored it)

anon

ah, found it. random spaces and line breaks and commas and capital letters are always fun to deal with.

Edit Summary: "everything"

t.b.

@anon Oh, yes, this person's masterful in producing stuff that needs a lot of editing. Google translate couldn't do worse, even if it tried...

Matt N.

19:37

Yay! You're back!

I think you're right about the thing being exaggerated.

In fact, I think it was quite a dim thing to write.

I was also in an irritable mood so I needed to write something dumb to make me less irritated.

t.b.

To get back to your "generation" question: did you see the above? If you think about the topology or a $\sigma$-algebra generated by a set: you have the top-down definition I gave (intersection of subthings) and in good situations you have the bottom up definition (explicit description from the generating set). But if you think about it you have to work harder (and make additional assumptions) for the latter. That's exactly the situation with ideals.

Matt N.

Yes I saw. I actually knew that about the intersection thing but I didn't think of it because the book defined it the way I posted above.

Thank you.

t.b.

On a more algebraic level: think about what happens if you don't require your ring to be commutative: a set can then generate a left ideal, a right ideal and a two-sided ideal. They can all be distinct.

Matt N.

Yes.

Asaf Karagila

@anon: I posted this meta thread, but I feel that the title is a bit off.

anon

19:51

little to none = noun, little to no = adjective

dogpile, I love it

Matt N.

: D

anon

wait a minute, was asaf... baiting us?

Matt N.

Probably.

Are you back to normal? Or do you still have to be handled with care? (Just asking, I don't mind either way.)

t.b.

Ask again tomorrow :)

Matt N.

Ok : )

anon

19:58

Buy him cupcakes. Those always work.

Matt N.

But cupcakes aren't nice.

anon

they're sweet.

Asaf Karagila

Buy him cocaine and hookers?

Matt N.

I was going to buy him some chalk.

anon

I hate chalk.

Asaf Karagila

20:00

$\textbf{Theorem:}$ Blackboard > whiteboard.
$\textbf{Corollary:}$ Chalk > markers.

anon

after an order-reversing morphism, maybe

Matt N.

If I was a lecturer I'd totally have long fingernails (at least on one hand) just to scratch across the board when the lecture starts and it's not quiet enough.

anon

oh, you decapitalized them too. very thorough.

I am this close to closing this chat tab.

Asaf Karagila

Why? D:

anon

I jest.

Asaf Karagila

20:02

You gestate.

anon

I do.

t.b.

Oh, I'm not the only one not getting this answer. Now BJ asked exactly the same question I had...

Matt N.

If you read a book, maths not fiction, do you skip over bits you don't understand or do you think about something until you figure it out? Or a mix between the two?
I'm asking because I've been quite slow reading books because when I get stuck on a tiny detail I try to figure it out.

So I've been thinking about trying the other: just ignore something I don't understand and carry on reading. (as long as it's not a major thing)

Jeff

@MattN i do the same thing. i can't move on until I get the detail.

t.b.

A bit of both. Sometimes it helps to take something for granted and see how it's used later on in order to figure out what I am confused about.

Hi Jeff

Matt N.

20:14

@Jeff That's quite unproductive.

Jeff

hi @tb. i've been here all along, just had nothing valuable to say.

@MattN you're not kidding!

@MattN in fact, i thought i was the only one who doesn't understand the stuff in textbooks! :D

Matt N.

@Jeff Well I do. Sometimes it just takes too long for me to figure out a detail.

Jeff

@matt right! when i have a class, or time constraints, i will skip over it and try to get back to it when i can.

Matt N.

I'm too OCD for that. : ) But I think I'll change my strategy.

My cat tore a page in my favourite book today. : /

Jeff

@matt damn cat. kick him (or her) out! :D

(what's the book?)

Matt N.

20:19

Gallian's Contemporary Abstract Algebra.

I taped it.

Jeff

is it readable? that's all that counts for textbooks (I tend to abuse mine - lots of writing, folded pages, etc.)

t.b.

That's a very Möbius-sy transformation

Jeff

my "ask a question" tab hung :(

Matt N.

@Jeff Very readable. I abuse mine too, although not as badly as you: I write into them (pencil only though!).

anon

well, it is a mobius transformation

Jeff

20:23

@matt yeah, i only write in pencil. except i put my name in pen inside the cover

@matt and then there are times I write a question down and go to the prof to ask it and I'm like "that's a stupid question! why would i ask that". :D

Matt N.

: )

t.b.

@anon d'oh what was I thinking. :/

Jeff

any number theory people in here who have "Elementary Number Theory" by Underwood Dudley handy?

anon

Jeff

my "ask a question" tab froze and I don't want to re-type the whole question.

@anon where is that?

anon

20:33

@Jeff: (a) Did it say it saved a draft of your question? Might want to open a new tab just to check (without closing the frozen one). Otherwise screencap and retype. (b) Dunno, it's just water under a bridge.

Matt N.

@anon : )

t.b.

It's depicting the current contents of my head...

Matt N.

Never mind : )

Jeff

@anon i don't see a notification of saving draft. i will check now. the image is scrolled out of view too much for a screen cap to be useful.

@anon where would i find a draft? (I opened a new tab)

Matt N.

@tb Maybe you need a bit of massaging.

(I'm not saying by me!)

anon

20:35

just re-click "Ask a Question" - if the draft was saved it will come back up

Jeff

@anon yup. it's there. thx

Matt N.

There aren't many questions that I can and want to answer lately.

@tb And a drink maybe?

t.b.

Both, I guess.

Oh, Harry's back...

Jeff

@matt and then there are the times, like right now, where I figure out a question while (and sometimes because) I'm typing it up.

Matt N.

@Jeff Yes. Does that count as rubber ducking?

Jeff

20:44

@MattN hahaha :D.

i'd never heard that term b4

Matt N.

: )

Jeff

@MattN and the answer is: yes

Matt N.

@tb Do you have drink at home?

Asaf Karagila

Can you please stop using SMS shortcuts?

Jeff

@AsafKaragila who? me?

Asaf Karagila

20:47

Yes.

t.b.

@MattN I do have one beer left. And I'm not in the mood for anything stronger...

Jeff

OK. Sorry. I didn't even know I was doing it.

Matt N.

@tb phew Although one seems like not a lot.

Asaf Karagila

Any of you guys have plans for tomorrow?

Jeff

Picking my dad up at the airport. Does that count?

Asaf Karagila

20:50

No.

Dennis Hayden

it's Pi day so eat pie and meet with NSA recruiters.

Matt N.

@AsafKaragila Yes. Do some CA and some Additive Combinatorics. That's during the day. Then before I'm off to a class towards the evening I'll pop into chat to say hi and after class I'm going out for dinner with a friend of mine.

Asaf Karagila

Will you be going to a pie eating contest?

Matt N.

No.

Asaf Karagila

I see.

t.b.

20:52

I wouldn't even have noticed. I tend to go by the reasonable day/month/year convention and fortunately there's no $\pi$-numerology in there.

Dennis Hayden

we need a 14th month

Asaf Karagila

We're gonna bake a pie with the letter pi on top, then watch the movie Pi whilst eating the Pie with the letter pi.

Dennis Hayden

Movie named pi?

t.b.

@MattN no it's not a lot, but my liver will thank me...

Asaf Karagila

@DennisHayden Yes.

t.b.

20:56

@DennisHayden there

Matt N.

@tb Your liver and your brain and who knows what else : ) But I don't feel bad for suggesting it : )

Asaf Karagila

@tb You should be killing your liver, not making him thank you!

Matt N.

@AsafKaragila Him? Mr. Liver?

Benjamin Lim

Hey

8am now

Matt N.

Hi Benjamin.

Benjamin Lim

20:57

what's the time and temperature now at your place?

Matt N.

10 pm, cool.

Benjamin Lim

cool means...?

Matt N.

Around 10 degrees.

Benjamin Lim

ah ok

I just checked the temperature for Zürich

Matt N.

I was going to make a joke but I don't dare.

t.b.

20:59

Go for it!

Benjamin Lim

@tb Have you been here the whole night?

ahahhhhh ...

Matt N.

: )

Transcript for

$Mathematics$ Mathematics