Mathematics - 2012-03-30 (page 5 of 5)

ymar

22:00

I meant det != 0...

David Wheeler

well i was looking at the n = 2 case.... where we have: $q_1(a,b) + q_2(a',b') = (0,0)$

if we had LD, we can assume wlog $q_2 \neq 0$

ymar

$q_i\in\mathbb R$?

David Wheeler

yes

so $(q_1/q_2)a + a' = 0$, for example

Matt N.

test: $\Bbb R$ $\mathbb R$

Benjamin Lim

Hey guys

ymar

22:06

@DavidWheeler Yes, but what now?

Matt N.

Hello Benjamin : )

ymar

Hi, Ben.

Matt N.

Did you solve that algebra question?

David Wheeler

so $q_1/q_2$ is rational, right?

Benjamin Lim

no......

Matt N.

22:07

@BenjaminLim What are the remaining questions you need to have answered?

Sorry for not being of much help.

ymar

Yes, right.

Benjamin Lim

2

Q: Exercise on separable polynomials over fields of prime characteristic

Having learned about separable polynomials today in class, I tried to do the following exercise concerning separable polynomials, namely: Suppose $f$ is the minimal polynomial of $a$ over a field $F$ of prime characteristic. Let $K = F[a]$. Then $f$ is separable iff $F[a^p] = K$. Now one di...

field-theory

@MattN

Matt N.

Oh. I was talking about $\Bbb Z_4 [x] / (x^1 + 1)$

David Wheeler

so we multiply through by $1/q_2$, and we've reduced it to LI over Q

Benjamin Lim

Anyone please help

@MattN That is not a field. $(x^2 + 1) \subset ( 2,x+1)$ that is not the whole ring

David Wheeler

22:09

now, i could be wrong, but i think we can use induction on n, now

ymar

@DavidWheeler But this we can do anyway. What we want to do is reduce all the way to LI over $\mathbb Z$

Matt N.

@BenjaminLim Awesome. Great that you solved it : )

David Wheeler

but we already showed LI over Z = LI over Q

Benjamin Lim

actually zhen lin was the one that provided that answer

ymar

@DavidWheeler Yeah, but my question is how to avoid going through Q.

With Q, we can do what anon said.

Benjamin Lim

22:11

@ymar Can you help with the question I posted above?

David Wheeler

well, i think you're going to run into Q anyways...because the integer coefficients might not have any common factors

t.b.

Can't you use Smith Normal Form ?

ymar

@BenjaminLim Not immediately. I'll have to think about it.

Benjamin Lim

thanks

@ymar It's not attracting a lot of views too

David Wheeler

to go a bit further, if $q_1/q_2$ is rational, then for some INTEGER k, $kq_1/q_2$ is an integer as well.

ymar

22:17

@tb How can I use it? I think the problem comes down to proving that if the Grammian is 0, then the vectors are linearly dependent.

@BenjaminLim I'll see what I can do. It's a nice question.

Benjamin Lim

zhen lin was telling me yesterday about showing that the if $g$ is the minimal polynomial of $a^p$ over $F$.

David Wheeler

so instead of multiplying through by $1/q_2$, we can multiply through by $k/q_2$ obtaining LI in Z.

Benjamin Lim

Then $g(x^p) = f(x)^{n}$ where $n$ is some positive integer greater than 1. $f$ is the minimal polynomial of $a$ over $F$.

t.b.

@ymar you can assume right from the start that your vectors are multiples of the standard generators for which both the implications seem obvious. If they are linearly (in)dependent over $\mathbb{Z}$ then they are linearly (in)dependent over $\mathbb{Q}$.

ymar

@DavidWheeler Well, yes but you're still mentioning Q,

@tb Standard generators being vectors with coprime coordinates?

t.b.

22:23

$(0,\ldots,0,1,0,\ldots,0)$.

Benjamin Lim

@tb You guys are talking about modules over PIDs?

ymar

Yup, free, finite-dimensional modules over $\mathbb Z$ to be exact.

Benjamin Lim

Sigh, I hope I can join in.....

David Wheeler

i don't see the problem with invoking Q. it's like factoring over Z[x], by passing to Q[x].

Benjamin Lim

but unfortunately now I have to revise set theory

I have an analysis mid term on tuesday

ymar

22:26

@DavidWheeler It's not a big problem. I would just like to know if it can be done. I think it would be nice.

@tb I don't understand why I can assume that $v_i$ are multiples of those!

t.b.

@DavidWheeler I agree. You can resolve many questions on integral domains by passing to the field of fractions.

David Wheeler

because they generate $\Bbb{Z}^n$

Benjamin Lim

bye

t.b.

@ymar That's what the Smith normal form tells you: let $A$ be the matrix having the vectors you're interested in as entries. Smith's algorithm gives you invertible $S$ and $T$ such that $SAT$ is diagonal.

Benjamin Lim

bye guys

t.b.

22:28

bye ben

ymar

Bye!

David Wheeler

(a,b,c...) = a(1,0,0,...) + b(0,1,0,...) + c(0,0,1,...) and so forth

ymar

@tb And multplying a matrix by an invertible matrix doesn't change linear (in)dependence of the columns, yes?

David Wheeler

$\Bbb{Z}^n$ is "nice" it has a basis

t.b.

@ymar exactly. That's what invertibility of $S$ and $T$ gives you.

ymar

22:30

@DavidWheeler Yes, but this is not enough. t.b. wasn't talking about linear combinations but multiples.

@tb Yes, I can see it now.

Thank you very much!

David Wheeler

i understand that, but we can "re-arrange things" until we get multiples

ymar

@DavidWheeler But the existence of the basis is far from enough for this...

David Wheeler

to go back to my toy example: suppose we started with (a,b) and (a',b') so a typical linear combination is r(a,b) + s(a',b')

we'll just assume Z-linear for now

that's the same as saying (ra+sa')(1,0) + (rb+sb')(0,1) = (0,0)

ra+sa' and rb+sb' are still integers, we've just "shuffled things around"

so i don't see how we lose anything by just considering the standard generators

ymar

Yes, they are but I don't get the point.

Matt N.

Heh. You guys are very serious.

ymar

22:39

Of course every vector in $\mathbb Z^n$ is a combination of $e_i$

Matt N.

It's 1 am and you're still talking about maths : )

You've definitely not had enough to drink.

ymar

@MattN I haven't. :(

Matt N.

@ymar Then you know what to do : )

David Wheeler

i'm saying a linear combination of arbitrary vectors can be considered as a (different) linear combination of the $e_i$ so we may as well consider those from the get-go

anon

Prove: If $\rm X=SAT$ with $\rm S,T$ invertible, then $\rm \exists v: Xv=0 \iff \exists w:Aw=0$. Use smith normal form and invoke this equivalence - this is the approach t.b. suggested. What's not to like about this?

ymar

22:42

@anon I like it very much.

anon

This is why you can go from arbitrary vectors to basis vectors.

David Wheeler

A is diagonal right?

anon

It doesn't have to be for the immediate equivalence to be proven, but it does have to be in order to finish the proof at hand.

David Wheeler

in smith normal form, though

Jonas Teuwen

Back from the slow coffee workshop :-).

anon

22:44

Yes.

Matt N.

@JonasTeuwen Heh! So you've been making coffee until now?

ymar

@DavidWheeler Yes, and this is what t.b. said. He didn't say linear combinations of $e_i$ but multiples.

Jonas Teuwen

Well, no until 22h30 :-).

Just took me a while to get home.

David Wheeler

since S,T are invertible, they can't take non-zero vectors to a zero one...seems clear to me

Matt N.

@JonasTeuwen What have you learned? : )

ymar

22:47

Great, so $\mathbb Q$ is indeed unnecessary here.

Jonas Teuwen

Nothing. Was just fun :-).

Gigili

@DavidWheeler Seems he got an answer to his question already, why are you still trying to help? I'd thank you for all the lines you typed in LaTeX.

David Wheeler

@Gigili...we're just talking. i daresay he's helping me as much as the other way 'round.

ymar

Right, that wasn't very courteous of me not to thank everybody,

Thanks, anon, David and t.b. :)

Gigili

@DavidWheeler Aha, sorry then.

David Wheeler

22:51

@ymar @tb @anon no, thank YOU. it's a pleasure to see how you people think.

ymar

Gigili, see what you've done? Now everybody's acting weird.

:)

t.b.

let's cool it :) The room had enough drama for a week...

Jonas Teuwen

My salary will probably be "frozen" 8-(.

Matt N.

What? Why?

robjohn

@JonasTeuwen that could be good.

Matt N.

22:53

Did you not produce enough theorems this month? : D

Gigili

@ymar Umm, I was wondering what he's doing when your problem's solved.

Matt N.

@robjohn Yeah, it might prevent him from buying Port Ellen.

robjohn

@MattN could keep it from going down

Jonas Teuwen

@robjohn I would get a 12% increase!

ymar

@tb Well, I think there was a misunderstanding somewhere.

David Wheeler

22:54

there was drama and i missed it? fastnabberglab!

Jonas Teuwen

@MattN No, budget cuts.

robjohn

@tb yeah, for once we're mod free ;-)

ymar

@DavidWheeler I seem to have missed it too!

Matt N.

@JonasTeuwen Oh noes. I didn't realise we're still in a recession.

robjohn

@JonasTeuwen oh, then that is not good.

Gigili

22:55

@DavidWheeler Drama as in they kicked someone out.

David Wheeler

@Gigili who? what? where? when?

anon

Err, was Rob kicked out while I wasn't paying attention?

robjohn

@anon he was suspended for an hour if I understand

Gigili

@DavidWheeler Follow the starred message over there ----->

David Wheeler

the artist formerly known as Skullpatrol?

robjohn

22:58

@DavidWheeler that'd be he

ymar

I think he has two accounts as Rob, doesn't he?

I seem to remember two avatars in a short period of time.

robjohn

@ymar he changed the other to Skullpatrol then deleted it

He still has the chat Skullpatrol, but the main site Skullpatrol is gone

I may be wrong about the chat Skullpatrol

anon

It's Rob on his chat profile, last time I checked today

Gigili

I can't believe someone saying I would like him not being around about someone, they really need to feel confident which I do not.

Matt N.

This topic has been talked to death /me thinks.

Gigili

23:01

Actually, I don't want the one who said it around here but I never said it.

@MattN I don't care what you think, I needed to say it.

David Wheeler

i am bored with it already. let's talk more about me.

Jonas Teuwen

@MattN Here a picture: yfrog.com/z/oc51811027j

@DavidWheeler Okay. Let's talk about your sexual preferences.

Or rather, let's not.

Matt N.

@JonasTeuwen Looks like fun! : )

David Wheeler

i was "this close" to complying, too

anon

I call for a caption contest.

Matt N.

23:04

I think I should log out. Drink makes me too honest.

Jonas Teuwen

@MattN No problem!

David Wheeler

really, you wouldn't want to hear about my sexual preferences. i'm boring.

Matt N.

I'm already too honest without drink.

ymar

@DavidWheeler Yet, you are still talking about it. :-)

@anon What's a caption contest?

Matt N.

@ymar Thanks for the topic change.

David Wheeler

23:05

@ymar ouch. sir, you slay me.

anon

@ymar: People try to caption a picture the funniest way possible.

t.b.

@anon don't think so. He jumped out of the room at the exact moment I entered a few hours ago.

Matt N.

The pictures in my head are too vivid. I wouldn't want to see any of it in most cases.

robjohn

okay, so it is the same account but a new account on the main site.

ymar

@DavidWheeler Actually, I would like to talk about that, but I guess it can't be done here.

Matt N.

23:06

@tb Looks like he hates you : )

Jonas Teuwen

Hate is not good. Let's have a beer together.

Matt N.

I've already had a bottle of fizzy wine on my own.

I think that's it for me today.

ymar

@anon Then I'll be happy to watch the contest.

Matt N.

: )

Jonas Teuwen

@MattN As long as you can still make coherent sounding sentences, there is no problem.

Matt N.

23:08

Seeing as I want to have another productive day tomorrow.

@JonasTeuwen I'm impressed with myself in fact. : )

Jonas Teuwen

You could try coffee and whisky and see which one wins.

Well, maybe make that beer...

t.b.

Since when can't you order the starred list chronologically? What is the order now anyway? Is it the last time a star was added?

Jonas Teuwen

I didn't know you could.

robjohn

I can't believe that I am ahead of Robert Israel on votes for a question.

wonders never cease

Matt N.

@JonasTeuwen No. I think it's just right for me : )

robjohn

23:12

@MattN why do you want to have one of those?

Jonas Teuwen

Ping! Food!

Matt N.

@robjohn Because I like them.

t.b.

@robjohn His condensed posts are often quite hard to parse... Not in this case, where I hold the picture responsible :)

Matt N.

And they're scarce.

robjohn

@tb I think pictures really help to draw in the votes. They definitely draw attention

ymar

23:13

@JonasTeuwen Is there a problem beyond that point?

robjohn

@tb and hopefully they make the point clearer

Jonas Teuwen

I never have any problems.

Matt N.

I am not that keen on pictures at all. I like well written text. But I did give you two plus ones today for two of your answers : )

Jonas Teuwen

You don't like my picture? 8-(.

Matt N.

*) in answers to my questions

: )

t.b.

23:16

@robjohn a picture is worth two R.I. lines?

Matt N.

Oh. It's late.

ymar

I got up twelve ours ago. No chance for a sleep now. I'll go get myself a beer -- see you!

t.b.

See you ymar

Matt N.

@ymar Enjoy!

robjohn

@tb could be :-)

@ymar later

Matt N.

23:19

@robjohn What's R.I.?

robjohn

@MattN Robert Israel

@MattN Or perhaps it was a reference to a couple of Rhode Island rail lines...

t.b.

back to trains?

Matt N.

@robjohn Probably the latter...

robjohn

@MattN no.

t.b.

no, he only builds railroad tracks all around the globe. Preferably on Sunday nights, as far as I know.

robjohn

23:21

@MattN I used to count train cars with my son when he was younger.

@tb Yes, monopolizing the world through crayons!

Matt N.

well well.

Before I end up typing into the wrong window: night folks. : )

t.b.

good night

t.b.

23:41

Aaah, mystery resolved

Dylan Moreland

@tb Hah.

t.b.

But it makes me wonder what masochistic TA asks his students to hand in homework in TeX. I always asked the students to please, please not do it.

anon

Why not?

Err, are you talking about raw latex markup? Do people actually hand that in?

t.b.

Because they were even more incapable of typesetting than of getting an argument straight. Focusing on the LaTeX bit distracted attention from the more important part. Both in the write-up as well as in the grading.

anon

I see, I hadn't thought of that.

A novice was trying to fix a broken Lisp machine by turning the power off and on. Knight, seeing what the student was doing, spoke sternly: "You cannot fix a machine by just power-cycling it with no understanding of what is going wrong." Knight turned the machine off and on. The machine worked.

eh, it was funnier here

t.b.

23:57

Re: your comment on the Kleiman reference request: I think the reason simply is that G.'s articles rely on the others, so they had to include them to make the articles comprehensible.

Zhen Lin

I see someone asked about the same article I did...

t.b.

yeah and went through the same disappointment...

Zhen Lin

I found a different article by Kleiman which had what I wanted though, so that was OK.

robjohn

Is it wrong to answer a L'Hopital question without using L'Hopital? :-)

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$Mathematics$ Mathematics