Mathematics - 2013-02-07 (page 2 of 4)

First I should say that I am aware of the existence of this question here and this question here. My question is a little different from these two because I am asking about a certain detail in the proof and besides $V$ and $W$ are now just algebraic sets and not affine varieties. Now $V \times W...

Sanchez

I think it's being answered?

BenjaLim

@Sanchez Yeah it has been but I just wanted to check if I'm understanding thigns correctly :

Firstly, if $V,W$ are algebraic sets then $V \times W$ is yea

Eric

Hello everyone

BenjaLim

5:17 AM

If $V$ is defined by the vanishing of $f_1,\ldots,f_n$, $W$ of $g_{n+1}, g_{n+m}$

skullpatrol

@Eric Hi

BenjaLim

then $V \times W$ is defined as the vanishing of $f_1,\ldots, g_{n+m}$

Sanchez

Yes

BenjaLim

each of which is now considered as a polynomial in $k[x_1,\ldots,x_{n+m}]$ @Sanchez yea?

awllower

I have to go now. Bye everyone!

Sanchez

5:17 AM

Hi @Eric

BenjaLim

@awllower bye.

Sanchez

It's right

BenjaLim

@Sanchez Also, why is it that

skullpatrol

@awllower L8er

Eric

Hello :8005777 and :8005770

BenjaLim

5:18 AM

@skullpatrol $I(V \times W) = I(V) + I(W)$

Eric

well that worked good.

Hello @skull

BenjaLim

where $I(V)$ is now considered as an ideal of $k[x_1,\ldots,x_{n+m}]$

similarly for $I(W)$

I can sort of see why @Sanchez

Sanchez

Well, $I(V) + I(W) \subset I(V \times W)$ should be obvious

BenjaLim

yea.

Sanchez

Well, let's see if this works @BenjaLim

Anything in $I(V \times W)$ can be written as $\sum f_ig_i$, where $f_i \in k[x_1,\cdots,x_n]$, $g_i \in k[x_{n+1},\cdots,x_{n+m}]$

BenjaLim

5:25 AM

@Sanchez Hmmm why?

Sanchez

Well, it's true for any polynomial in $x_1,\cdots,x_{n+m}$

BenjaLim

ah yes of course.

Sanchez

Hm, ok

Now, fix $b \in W$, and consider $\sum g_i(b) f_i$

This is a polynomial in $k[x_1,\cdots,x_n]$, that vanishes on $V$

BenjaLim

yea.

so is in $I(V)$.

and also $\sum g_if_i(v)$ for $v \in V$ is a polynomial that vanishes on $W$

Sanchez

If all of the $g_i$ vanishes at $b$, so be it.

If not, we obtain a linear dependence relation among the $f_i$'s, modulo something in $I(V)$

BenjaLim

5:29 AM

ok.

@Sanchez wait

Sanchez

So let's assume that all the $f_i$ are linearly independent in $k[x_1,\cdots,x_{n+m}]/I(V)+I(W)$

BenjaLim

why must $\sum g_i(b)f_i$ vanish on $V$?

Sanchez

Because $b \in W$

for any $v \in V$, $(v,b) \in V \times W$

BenjaLim

hmmm

Sanchez

Since $\sum g_if_i \in I(V \times W)$, it vanishes on these points

BenjaLim

5:31 AM

yes that is correct.

Sanchez

Anyway, let's continue

BenjaLim

ok I think I get it

Sanchez

What I'm trying to say is that, you can "shorten" an expression $\sum f_ig_i$, modulo things you don't care about (i.e. things in $I(V) + I(W)$, if $g_i$ doesn't all vanish on $W$, and that $f_i$ are not linearly independent in $k[x_1,\cdots,x_{n+m}]/I(V)+I(W)$

BenjaLim

i think I'm confused....

Sanchez

hm, let me see if there's a better way to put it

BenjaLim

5:34 AM

@Sanchez I have to go now. My mum's calling me. I will think about it and get to you later. At least we know it's true.

Sanchez

Sure. It is true.

Oh yea. That is what I'm saying. Start with an expression $\sum f_ig_i$. If $g_i$ doesn't all vanish on $W$, take one such point $b$, and consider $\sum g_i(b) f_i$. This would be a nonzero linear dependence relation of $f_i$, modulo an expression in $I(V)+I(W)$. WLOG say $g_1(b) \neq 0$, then you can write $f_1 = -\frac{g_2(b)}{g_1(b)}f_2 - \cdots - \frac{g_n(b)}{g_1(b)}f_n + $sth in $I(V)$. Substituting in $\sum f_ig_i$, you get a shorter expression, modulo $I(V)+I(W)$

So for a function in $I(V\times W)$, start with a shortest expression $\sum f_ig_i$ in $\frac{k[x_1,\cdots,x_{n+m}]}{I(V)+I(W)}$. Let's say it's nonzero. If $g_i$ doesn't all vanish on $W$, we can use what I said just now to shorten the expression. Same if $f_i$ doesn't all vanish on $V$. So the shortest expression must be 0, i.e. $I(V \times W) = I(V) + I(W)$ @BenjaLim

skullpatrol

24 mins ago, by BenjaLim

@skullpatrol $I(V \times W) = I(V) + I(W)$

Sanchez

lol

skullpatrol

:-)

BenjaLim

6:47 AM

@Sanchez

Sanchez

yo

BenjaLim

Ok I get that $\sum g_i(b) f_i \in I(V)$

@Sanchez so that $\sum g_i(b)f_i $ is a non-zero linear dependence relation of $f_i$ modulo something in $I(V)$.

Sanchez

@BenjaLim, okay.

BenjaLim

ah ok.

I think I get it

Sanchez

Should I redo it or you get it?

BenjaLim

6:51 AM

Let me explain it to you @Sanchez

Sanchez

I think everything is clearer if I work in $k[x_1,\cdots,x_{n+m}]/I(V)+I(W)$ in the outset

BenjaLim

If $g_i$ all vanish at $b$

then we are done.

Sanchez

er, not exactly

BenjaLim

because each $g_i \in I(W)$

@Sanchez why not because then each $g_i \in I(W)$ since $b$ was arbitrary.

Sanchez

Hm, okay, you would need to use symmetry for $f_i$ too, but yeah.

BenjaLim

6:53 AM

@Sanchez Ok for the moment suppose that there is $b \in W$ so that not all $g_i(b) = 0$

wlog it is $g_1$

then by what I said above, we already know that $\sum g_i(b)f_i$ must vanish on all of $V$

so that we can write $\sum g_i(b)f_i = p$ where $p \in I(V)$

then $$f_1 =- \frac{g_2(b)}{g_1(b)}f_2 - \ldots - \frac{g_n(b)}{g_1(b)}f_n + \frac{1}{g_1(b)}p$$ @Sanchez

Sanchez

Yes

there is a $\frac{1}{g_1(b)}$ in front of $p$, but doesn't matter

@BenjaLim, no

BenjaLim

wait I saw the wrong thing

Sanchez

oh okay

brb

quick shower

BenjaLim

ok

@Sanchez I'm not sure what you mean by "we get a shorter expression by substituting in $\sum f_ig_i$

Sanchez

7:11 AM

@BenjaLim, consider the expression $\sum f_ig_i$ in $k[x_1,\cdots,x_n]/I(V)+I(W)$

BenjaLim

@Sanchez ok.

Sanchez

then once you get the expression for $f_1$

BenjaLim

you mean $k[x_1,\ldots,x_{n+m}]$

Sanchez

Yes, sorry.

You substitute, then you get $\sum f_ig_i = f_1g_1 +f_2g_2 + ... = f_2 g_2' + \cdots f_n g_n'$

where $g_n' = g_n - g_n(b)/g_1(b) g_1$

the $p$ term is gone if we consider $k[x_1,\cdots,x_{n+m}]/I(V) + I(W)$

then you see that the expression originally has length $n$, now has length $n-1$

BenjaLim

ah ok.

so the length was shortened

after we take mod $I(V) + I(W)$. That was where I was confused.

Sanchez

7:13 AM

the point is that this shortening process can always happen, unless your $f_i \in I(V)$ and $g_i \in I(W)$ initially

i.e. your initial expression is already 0 mod $I(V)+I(W)$

So if you want to write this down rigorously, you start with a shortest nonzero expression (modulo $I(V)+I(W)$) of a function in $I(V \times W)$

and say that shortening can occur unless $f_i \in I(V)$, $g_i \in I(W)$ blah blah blah

BenjaLim

for every $i$

Sanchez

yes

Sorry for not being able to say these coherently. I guess I always have this problem when I first need to figure something out

BenjaLim

ok but it seems to me we only need say that $f_i \in I(V)$ for every $i$.

because then $\sum f_ig_i \in I(V) \subseteq I(V) + I(W)$

Sanchez

oh you are definitely right

BenjaLim

Right so this proof can be simplified. We start off with $b \in W$. If $g_i(b) = 0$ for every $i$ then our entire sum is in $I(W)$ and hence in $I(V) + I(W)$

Otherwise if $\exists b : g_1(b) = 0$ say

Then $\sum f_i g_i$ modulo $I(V)$ is something shorter

continuing this process $\sum f_ig_i \equiv 0 \mod I(V)$

and so is $0$ mod $I(V) + I(W)$

done.

@Sanchez

@Sanchez Hmm but in this answer here: math.stackexchange.com/a/295271/38268

Sanchez

7:19 AM

if exists $g_1(b) \neq 0$

BenjaLim

go on.

Sanchez

I mean you had a typo

BenjaLim

ah ok.

@Sanchez Hmm in the answer in the link

Sanchez

yes, that's my idea

BenjaLim

they claim that $I(V \times W) = \sqrt{I(V) + I(W)}$....

Sanchez

7:24 AM

is the sum of two radical ideals radical?

BenjaLim

hmmm

@Sanchez Now I'm curious..

but I think our proof was correct. Or maybe the objects in question are different? Now we are talking of algebraic sets and not varieties...

Sanchez

I find it weird, I'm not sure what goes wrong when $k$ is not algebraically closed above.

BenjaLim

yeah. It seems to me that they are invoking the Nullstellensatz or something like that.

@Sanchez I will post a question on main.

Sanchez

sure

Sanchez

7:49 AM

gotta go, see you @BenjaLim

BenjaLim

@Sanchez c ya. I was just typing a question on main.

skullpatrol

@Manishearth Wazzup mu?

Manishearth

Nothin much :)

skullpatrol

What's nu?

Manishearth

$\nu$

^^him

not me

:P

skullpatrol

7:56 AM

:-D

Manishearth

@skullpatrol Doing a lot of stuff in college -- programming an RPi right now

skullpatrol

@Manishearth What does "RPi" stand for?

Manishearth

Raspberry Pi

The Raspberry Pi is a credit-card-sized single-board computer developed in the UK by the Raspberry Pi Foundation with the intention of promoting the teaching of basic computer science in schools. The Raspberry Pi is manufactured through licensed manufacturing deals with Element 14/Premier Farnell and RS Components. Both of these companies sell the Raspberry Pi online. The Raspberry Pi has a Broadcom BCM2835 system on a chip (SoC), VideoCore IV GPU, and originally shipped with 256 megabytes of RAM, later upgraded to 512MB. It does not include a built-in hard disk or solid-state drive,...

skullpatrol

Cool.

Orange Harvester

@Manishearth Is it possible to strip it of the linux kernel and install your own OS? If yes, how good is the BIOS documentation? Especially, how good is the video support for basic text-based console?

Manishearth

8:07 AM

@OrangeHarvester yep

@OrangeHarvester umm, there is full video support

Raspbian has a GUI

skullpatrol

How does this promote "the teaching of basic computer science in schools?"

Orange Harvester

@skullpatrol current computers are too complex to explore even.

It is astoundingly difficult for someone to learn stuff about I/O systems and similar hardware related issues on current PCs.

Manishearth

@skullpatrol Anyway, I'm programming this not to learn, but to use it for something

(namely an electronic noticeboard)

skullpatrol

icic

Orange Harvester

@Manishearth Are you writing code on top of Raspbian?

Manishearth

8:17 AM

python

yep

Orange Harvester

@Manishearth Okay. Actually, I wanted to remove the Raspbian too. I have written some basic OS code and I wanted to push it on some kind of raw hardware. I have not yet written any kind of drivers, except for very basic line i/o kind of thing. Also, I am not interested in writing a complex driver. So, my question is, does Raspberry Pi have some kind of BIOS? If no, would it be easy enough to write a line I/O driver for it?

I am currently working on a qemu emulator.

Manishearth

@OrangeHarvester not sure

Orange Harvester

@Manishearth okay. np.

Jonas Teuwen

9:30 AM

Morning.

@OrangeHarvester Argh, I made drivers (and hardware) in the past. It is a pain in the little star.

skullpatrol

Hi.

Jonas Teuwen

Girlfriends are very deteriorate for good maths. Girl needs to be very good as compensation.

skullpatrol

Is that the opinion that got you the suspension?

Jonas Teuwen

No.

Opinions are like assholes: I have many of them.

skullpatrol

We all have one.

@Novice Yo palio, wazzup?

user19161

10:18 AM

@JonasTeuwen Haha.

Jonas Teuwen

11:12 AM

Hah.

skullpatrol

Ho.

Orange Harvester

@JonasTeuwen Yes. hug for that. Every one trolls the driver makers. The people who manufacture the hardware, the userspace people etc etc.

skullpatrol

11:34 AM

Even the trolls troll the trolls who troll the driver makers :P

Jonas Teuwen

12:03 PM

@OrangeHarvester I made a spectrum analyser, and a driver...

Orange Harvester

@JonasTeuwen Cool. Spectrum range?

Jonas Teuwen

I also tried figuring out if one could make a framework for Linux drivers in other *nix-like things, but that gave me a quadratischen Glatzkopf.

@OrangeHarvester Only up to 200MHz then, the samplers are kinda expensive.

Gam Erix

12:28 PM

Maybe this sounds easy for you guys: math.stackexchange.com/questions/297135/… heh. I kinda get a headache trying to figure it out :$

Jonas Teuwen

@OrangeHarvester Yes.

@GamErix Take some paracetamol and try harder!

Gam Erix

it's that point with 4 indexes which makes it so hard for me :x

skullpatrol

12:45 PM

skullpatrol

1:03 PM

He's gone... another successfully answered question by the Math.SE chat room :-D

user58512

@JasonBourne can you help me with group theory pls

user19161

Testing $\left(\frac{a}{b}\right)$

user58512

llol you are still on the fractions thing

user19161

@user58512 I am leaving the chat now, sorry, doing some installation

skullpatrol

$\left(\frac{Bye}{Jason}\right)$

peoplepower

1:07 PM

@user58512 What's your question?

user58512

@peoplepower, what is a suborbit?

If I have a group G acting on a set X, then I take the stabilizer G_x of some x

then a suborbit is "the orbits of G_x" but I can't understand that

peoplepower

I have not seen the term before, and that definition is indeed ambiguous.

user58512

I showed any transitive action is equivalent to a coset action, and apparently a corollary of that is that suborbits correspond to double cosets G_x g G_x

any idea how to make this definition meaningful?

peoplepower

I found a reference in Isaacs' Finite Group Theory (Gorenstein uses it implicitly in his Finite groups book).

user58512

ok thanks

peoplepower

1:14 PM

Yeah, there is an induced action on $X$ by the stabilizer of $x$.

You are just taking orbits of that action.

user58512

G_x acts on X?

oh right, G_x is just a subset of X

and it's a group

ok thanks a lot!

peoplepower

If there is just one suborbit in $X-\{x\}$ the action is called doubly transitive. That is about all I know.

I also recognize the form G_x g G_x, so I can look up Gorenstein's proof of that for doubly transitive actions and see if it helps here.

user58512

you already completely solved my problem! :)

are elements of G_x in bijection with cosets g G_x?

that wouldn't make sense

maybe xG with cosets g G_x

peoplepower

Not usually.

Tobias

@user58512 cosets of G_x are in bijection with elements of the orbit containing x

peoplepower

1:29 PM

I probably should have linked to the statement I was talking about (the second one).

skullpatrol

user58512

I can't get the double cosets thing from this at all

Tobias

@user58512 so you have a transitive action, and you want to look at the orbits of one of the stabilizer subgroups?

user58512

yeah

Tobias

so any element in the set you are acting on corresponds to a coset gG_x (from now on, x is fixed)

and the orbit then corresponds to the coset G_x(gG_x)

(since the orbit of y under the action of G_x is G_x y)

user58512

1:43 PM

I see! thanks a lot!

Tobias

(note that one ought to assume the action transitive to call something a suborbit, as otherwise the notion depends on the choice of an orbit to start with)

user58512

2:22 PM

wait im confused

I write the orbit of y under G_x as y G_x

so I get (g G_x) G_x

but thats not a double coset

Tobias

@user58512 then start with a right coset instead

(or don't write actions from the right)

awllower

I have a little question:
In the treatment of Class-field theory by Jürgen Neukirch, it appears that the whole theory is group-theoretic. So could we say that CFT is a part of group-theory?

Tobias

I think a lot of people would disagree with that

awllower

Really!

But why?

Tobias

my knowledge of CFT is not really good enough that I have an opinion on it myself

awllower

2:30 PM

I am also trying to form an opinion.

In view of what A.Weil has said: CFT is as basic as Galois-theory.(It might not be what he said exactly.)

@Tobias By the way, is there really some problem with my reasonings in my answer?

user58512

@awllower, I have a question about Dirichlet characters, is it ok ?

awllower

!!
I will try at least.

My pleasure to hear some interesting problem of course.

user58512

@awllower, by counting - the real dirichlet characters are exactly the kronecker symbols - but what about higher reciprocity?

Tobias

@awllower no, it looks good to me (I upvoted it when I finally got the argument)

user58512

so we can't use the same technique of L functions for cubic fields?

awllower

2:38 PM

@Tobias Thanks then.

I think the cubic fields are also solved?

I mean: we already have the cubic reciprocity?

Or are you asking could we apply the techniques of L-function?

user58512

yeah, but can you put it into into an L function and use analytic techniques to find its class number?

awllower

I see...

I am not so sure either. Per chance you could post a question?

Sorry I cannot help you.

user58512

it's just a curiosity

awllower

Oh.
It might still work in my view.

Charlie

A new prime number was found!

4

awllower

2:54 PM

Really. When?

Charlie

Oh my i'm so excited! cbc.ca/news/technology/story/2013/02/06/…

3

It's a mersenne prime

2

this is so nice !

awllower

Indeed!

Charlie

:-D

@awllower Darn...just you and I to celebrate ths moment...

awllower

3:11 PM

I think others are here as well, but having nothing to say.

Charlie

They only appear to say bullshit

awllower

@Charlie I have a little question: In your user-field, you quoted Hamlet. And I would like to know where the upper half comes from.

Thanks in advance.

I mean :"There's a lot things about me.Things you wouldn't understand, things you couldn't understand, things you shouldn't understand. I'm a loner, dude, a rebel."

Or does this come directly from you?

Charlie

@awllower thanks for asking! It's Peewee Herman

awllower

I see. Thanks for explaining.

Charlie

:)

ahenderson

3:22 PM

Hi all, i posted a question yesterday that is not attracting much attention. math.stackexchange.com/questions/296630/….

user58512

@ahenderson, just multiply by the time interval

@ahenderson, think of it like areas of rectangles

ahenderson

@user58512 hmm, that was a very fast and obvious answer. Thank you.

@user58512 If you care to place that as an answer i will accept it.

user58512

ok

awllower

I have another little question now:
http://math.stackexchange.com/questions/297214/proof-that-a-finite-separable-extension-has-only-finite-many-intermediate-fields

I am considering:
Suppose the degree is $n$. Then there is a basis consisting of $n$ elements ${x_i}$. Hence all intermidiate fields must have a set of basis consistingof $m< n$ elements, each of which is a linear combination of the $x_i$. Then notice that it suffices to prove that there is a finite number of intermidiate fields of a fixed degree. So suppose that $m$ is fixed. By what we have shown, the intermediate fields correspond to $m*n$ matrices of rank $m$, up to $m*m$ invertible matrices.

Does that work?

While I still fail to show that equivalence relations of m-by-n matrices modulo m*m invertible matrices are finite in number...

Any opinion on how to solve the problem?

Well, unfortunately I have to sleep now. I shall swell upon it for more time thus.

Charlie

3:46 PM

Good night!