the point I was trying to make is merely that if we start trying to prove that $\otimes$ and $\prod$ "commute", we're immediately stuck when trying to adapt that proof.

oh right.

yes

thanks theo for showing me all this

I think I am in very deep water now!!

well, there's a lot of stuff to digest, yes :)

no problem...

hahahahahaha

a LOT to digest

I mean I'm only in second year uni, already I'm seeing what possibly graduate level stuff!!

Don't be scared :)

@tb Common man when I first looked at AM chap 2

TENSOR PRODUCT

DIRECT LIMIT

man I was seriously freaking out man!!

here is australia we would say: that's gnarly!!!

I understand :)

And it gets even harder

I look at what chap 2 or 3 or somehting

construction of sheaves sheaves or whatever

I am like I don't know what it is but I always hear the pros talking about it

it's just words :)

hahahahahahahahahahahahaha

crap that is like AG no???

I'm not saying it is easy or trivial because of that, but never be afraid of things just because people toss the words around.

Of course, it's one major ingredient of algebraic geometry

@tb hahahahahahahahahaha I am like so smiling now, you have inspired me to tackle direct limits more!!!!

I remember my tutor trying to explain pushouts and fibred products to me :D :D :D D: D: D

@BenjaminLim Sheaves is where geometry enters the scene.

@tb Maybe one day I will become an algebraic geometer....

Maybe, maybe you'll do probability. You never know...

NONONONONONONONONO

@tb It would be cool one day to meet in person :D :D

I have met a few people on this site in person

It probably won't happen anytime soon... But it certainly would be nice.

come to australia man

get an excuse to visit amnon neeman or something

well, maybe, but I've got lots of other stuff to do right now...

research related

BUT YOU'RE ON CHAT.

I know...

I should go too

have my breakfast

10.11am here

Okay, then! See you around soon!

2:11 AM here.

yeah see you then!!

I am off to the Griffith Observatory. bbl

what an experience!!

@robjohn see you robjohn! (now I can look at your BV answer :))

(have fun with the stars planets and nebulae)

It amazes me how this keeps generating upvotes.

I will. I always to :-) good luck with the BV answer ;-)

It amazes me that 19 people bothered to (try to) answer...

That too.

What are the point of math tests? They seem useless

I failed my calc 1 class again, I just can't pass tests no matter how well I know the material

No matter what I do I can never do well on a math test

Is there any office or process to appeal a test grade in colleges?

Greatly depends on the circumstances.

I am so sick of school, it is just disgusting every time I get a test grade back

every test I get back just reminds me how worthless of a person I am, this will be my 2nd time failing calculus 1, a class 16 years olds sleep through in high school

Maybe you should consider: (1) taking fewer total hours so you have more time to devote to each class, (2) spend more time studying for tests and ask more questions in class and in office hours, (3) develop better and more effective study tools and skills, (4) hire a professional tutor

The most important thing that you can do really is: CHANGE YOUR ATTITUDE!

I study a lot, I studed over 40 hours for this test

@Jordan Umm high school calculus is not the same as college calculus.

@Jordan See (3)

well I am working on getting better, but it never seems to pay off

So we know that you don't particularly care for you math class. What classes do you like? What is your end-goal? If you don't mind my asking.

why are you doing math?

just quit, move on with your life. stop whining

and do what?

what else would I do with my life that doesnt require math

be a truck driver

I would rather kill myself then do manual labor again

There are many career paths that do not require calculus.

hahah truck driver, nice one, live in poverty my whole doing a low reward, high stress low pay, low respect job that slowly kills me

The only job I have done in my whole life that I found enjoyable was programming so I want to try that

What kind of programming?

let $a=bi=\alpha\in \mathbb{Z}[i]$ be a Gaussian integer. why is it that $N(\alpha)=a^2+b^2$ is equal to the cardinality of $\mathbb{Z}[i]/(\alpha)$? even a source would be appreciated

I just started a c++ class and I enjoy programming, I dont know much

i believe this is the case in quadratic extensions generally, when you quotient the ring of integers by an ideal the cardinality should be the norm

Well if you intend to pursue a 4 year degree in the computer sciences, most programs will require you to take calculus.

I know so that is why I want to take it

why get a degree at all. you can get programming jobs without a degree (i think)

do you genuinely like programming or do you like the idea of doing something abstract like this?

but that depends mostly on merit and ability to work on your own and produce results

both

I have never been motivated or inspired to do homework outside of neccesary amounts of it except in my programming class where I actually spent time making random programs on my own time

When you entered community college, did you take a placement exam?

yes

And they deemed that you were prepared for calculus?

no

This is like my 6th semester of college

So what math classes have you taken prior to calculus?

pre algebra, college algebra, college algebra, trig, calc 1, calc 1

No pre-calculus?

i still don't know what precalculus really is. seems totally unnecessary

no pre calc

It is often quite redundant. But it doesn't have to be.

So, do you think that you are naturally bad at math?

yes

Well you are wrong.

how do you know, David?

no use pretending, maybe he is bad

I think I am okay at math, but exceptionally bad at tests. I have test anxiety and I am dyslexic

dyslexia sucks

He may be unskilled and untrained, but that doesn't preclude him from being successful. I'm currently bad at pool, but that doesn't mean that I can't get good at if it I were to try.

some people will always be bad at sports or pool. hand eye coordination is necessary condition, for example. something that cannot really be trained

I was once just like you, I thought I was just bad at it and that there was no overcoming that.

Umm, hand eye coordination is highly trainable.

it's good to be optimistic and hopeful as a general rule

david there is a difference between a natural and someone who puts in a lot of effort. if you've ever played a sport at a high level you can see this

i will never be a world class sprinter. you have to be born with the fast twitch muscles

Well of course. And there is also a difference between putting in a lot of effort and being unable to succeed at all.

Sure, but that doesn't mean you can't run at all. (I assume that you are physically able)

sure. it could be either or for jordan. it just seems like he whines a lot and i'm just suggesting he consider his motives

some people do things because they feel they should, rather than truly wanting to

This is the point I am getting at. Only I'm trying to take a more socratic approach than just telling him to cheer up or get lost.

I completely agree.

lol, point taken

I know I like to whine, I am a bitchy person and recently I feel useless as a person. I completely lack social or atheletic skills and I have no friends so I should be good at something academic but I am not and that is frustrating to me. I can't find a purpose or use for my life

@Jordan Look, Eric's comment "it's good to be optimistic and hopeful as a general rule" is the point I am trying to make here. Your attitude is your biggest hinderance. Now, test anxiety and dyslexia are very real issues, and they can be seriously debilitating for your academic career. If you really have thes conditions, then you need to go to the appropriate department at your college and get the appropriate accomodations.

They just send you to the testing center, it is far worse than the class room I think. A large somali man yells at people in a room to finish their test and leave

If you have a "I'm just bad at it" attitude, then you will never really think that you can get better.

@Jordan If this is true, then this is a direct violation of ADA federal guidlines.

@DavidK People have been telling him that ever since he got here...

really? What are the guidelines?

@Jordan There are federal guidelines that directly address appropriate academic accommodations. Particularly for test taking accommodations. Your school is under the jurisdiction of an ADA federal liaison, and if you feel that you are unable to get the appropriate accommodations that you require you are entitled to contact them.

awesome, thanks I will look into that

You can also file complaints with the school about any staff member "yelling" at you.

Math, for most people, is just like anything else: it requires practice and training. But more importantly it requires an optimistic and confident attitude.

It sounds cheesy, but improvement really starts with you...

Now, I need to come up with an example of subsets $A$ and $B$ of $\mathbb{R}^{2}$ such that $A$ and $B$ are connected, but $A\cap B$ is not.

can't you just take an annullus?

and then partition it into two connected sets that overlap on opposite sides?

not partition*, strictly speaking

So like $A\cup B=$ annulus?

but express the annulus as the union of two horseshoes

yeah

with $A\cap B\neq\varnothing$ so they "overlap"

right, and you want them to overlap on one side of the annulus and on the other

Yep that works. Duh.

Thanks!

yw

@Jordan Hope you read the post I made for you about mean value theorem and Rolle's theorem. Please ping me if something is not clear.

(I'll be around here for sometime.)

Math classes are such a gamble, the degree between a 2.0 and a 4.0 is so small

well actually a 2.0 would be a withdrawal since no college would ever accept anyone with a 2.0 in calc1

cool, my favorite subject

hey @MarianoSuárezAlvarez

hello :)

i'm going to write up a solution to the question i asked. i think i know how i wanted to see it answered and you've helped me see how to do it

@Mariano Hi. A favour again. (Understanding about connection of $Q$ being a division algebra and it not having a non-trivial nilpotent element.)

just can't do it tonight

you can't have zero divisors

in a division ring

A division ring is a ring where every non-zero element is invertible. So, you're right.

so x^n=0 implies there are some zero divisors

Yes, a nilpotent element is a zero divisor or zero in fact,

But what about $Q$ over $\Bbb F_2$?

I think I am asking silly questions and hope it does not hurt.

what do you mean $\mathbb{Q}$ over $\mathbb{F}_2$?

@MarianoSuárezAlvarez I think I have proven the thing on elements in $M_i$ being eventually zero

I said quaternions over $\Bbb F_2$.

hi guys

i actually have to go kannap, have to skype with my girl

Hi @Ben

@EricGregor Sure, later.

hey

Thank you for the help about division algebras. @Eric

@MarianoSuárezAlvarez I think I have shown that $\bigoplus M_i/\langle x_i - \mu_{ij}(x_i) \rangle \cong \bigoplus M_i / \sim$

I confess not to remember what the M_i are :P

what is the latex code for an equivalence relation

\sim

or \equiv

Oh, I gather Mariano won't talk to me. : (

Anyway, later guys.

(it is not worth doing this here, but technically, you have to write that M_i / \mathord\sim, to get $M_i / \mathord\sim$ instead of $M_i / \sim$)

@KannappanSampath, hm?

ok

@KannappanSampath, I don't know what you were going to ask! :D

$x_i \sim x_j$ if

there is a $k \geq i,j$ such that $\mu_{ik}(x_i) = \mu_{jk}(x_j)$

@MarianoSuárezAlvarez Does it make sense to do $\bigoplus M_i /\sim$?

it would not hurt if you reminded me what the context is

@MarianoSuárezAlvarez I want to prove

that

if $\mu_i(x_i)$ is zero

where $\mu_i : M_i \to \varinjlim M_i$

then there is $j\geq i$ such that $\mu_{ij}(x_i) = 0$

the eventually zero thingy

ok

so

I think hmmmm

I can't do $\bigoplus M_i/\sim$

why not?

@MarianoSuárezAlvarez because $\sim$ is defined on individual elements $x_i$

So like in the direct sum how would I say what $x_1 + x_3 + x_4$ is equivalent to?????

well, you have to extend it additively

as in....

it is better to construct a subgroup of $\bigoplus M_i$ and then mod out by that

well but now I'm not going to be modding out by the submodule generated by the $x_i - \mu_{ij}(x_i)$ right?

consider the subgroup generated by the union of the kernels of $\mu_{i,k}-\mu_{j,k}$ for all triples $i$, $j$ and $k$ such that $i\leq k$ and $j\leq k$.

ok

yes

ahhhhhh

okokookokokokokokokok

wait

does it make sense to take the union of kernels??

is that even a submodule?

the submodule generated by the union

oh sorry I did not see "generated"

Call this submodule $D$

I want to show that $\bigoplus M_i/D \cong \varinjlim M_i$

and what do you mean by $\varinjlim M_i$?

I define it here

to be $\bigoplus M_i/C$

where $C$ is the submodule generated by elements $x_i - \mu_{ij}(x_i)$

heh

show that C=D

pfffff

so as you can see I am trying to invoke some universal property, things like that

@MarianoSuárezAlvarez To show the two submodules are equal would not be so easy right

well, show that $\bigoplus M_i/D$ has the same universal property of the limit

have you tried? :)

Show a generator in $C$ is in $D$

and vice versa?

so Mariano

I think the universal property bit is easier for me

by the way what did you mean before by extending linearly?

how do I know after I extend linearly it is still an equivalence relation

wait let me have a go at this

@MarianoSuárezAlvarez I don't even need to extend linearly

forget about that

forget about

explaining three different things at the same time is a bad idea

Mariano I got it already

I got it

The only thing I need to know now

is how to extend the equivalence relation $\sim$ by linearity

a relation on an abelian group A is a subset of A\times A

So suppose we have $\sum_{i \in I} x_i$ and $\sum_{j \in I} x_j$

the linear extension of the relation is the the subgroup generated by that subset

ok

can you give an example

it is the minimal relation which contains the original one and which is compatible with addition

hmmmmm

let $\sim$ be the relation on $Z^2$ such that $(x,0)\sim (y,0)$ and $(0,x)\sim(0,y)$ for all $x$, $y\in Z$ and no more pairs are related

ok

then the linear extension is the trivial relation in which all pairs are related

so in the case of defining when $\sum_{i \in I}x_i $ is equivalent to $\sum_{j \in J} x_j$

$\sum_{i \in I}x_i $ is equivalent to $\sum_{j \in J} x_j$ if there exist $a_1,\dots,a_k,b_1,\dots,b_k$ such that $a_i\sim b_i$ and $\sum_{i \in I}x_i - \sum_{j \in J} x_j = \sum_{q=1}^k (a_i-b_i)$.

hmmmmmm

you know what

it is just easier to say quotiening out by the submodule you defined earlier generated by union of kernels

show that that quotient has the required univ. property

yes

So let $\pi_i : M_i \rightarrow \bigoplus M_i/D$

I'll be back after dinner :)

where $D$ is the submodule generated by the union of kernels of $\mu_{ij} - \mu_{kj}$ for any triple $i,j,k$ such that $j \geq i$, $j \geq k$

@MarianoSuárezAlvarez ok

hi everybody

I have an algebra question

@MarianoSuárezAlvarez I just realised one thing

How do we even talk about the kernel of $\mu_{ij} - \mu_{kj}$?

The domain of each of those terms is not the same!!!

Unless that is defined by $(\mu_{ij} - \mu_{kj}) ( \text{element in $M_i$} + \text{ element in $M_k$}) = \mu_{ij} (\text{element in $M_i$}) + \mu_{kj}(\text{element in $M_k$})$?

@MarianoSuárezAlvarez Is this correct?

Given a group $G$, let $Z(G)=\{h\in G: \forall g\in G(hg=gh)\}$ the center of $G$. Let $GL(n,F)$ the set of $n\times n$ invertible matrix over the field $F$. I want to proof that $Z(GL(n,F))=\{\lambda I_n:\lambda\in F^*\}$.

@BenjaminLim, extend each $\mu_{i,j}$ in the unique obvious way to a map $\bigoplus M_i$ to itself

what do you mean?

Is what I said above correct?

@MarianoSuárezAlvarez

you have a sign wrong

oh ok

anyhow, there is a unique obvious way to extend the map $\mu_{i,j}$ to a map $\bigoplus M_i\to\bigoplus M_i$

that is what I meant

right

I try to write exactly what I mean :D

@MarianoSuárezAlvarez Ok in that case here is my go:

I am still in the midst of dinner

:)

ok

anyway I think I can do it now

what I defined above is correct

I am not the only one here, though

perhaps I should post on the main site

@BenjaminLim, can you help me on this?

@leo My group theory is quite poor .......

I see. Thanks anyway

@leo This is just an algebra bash.

oh wait

you want to find the center of $GL_n$?

@BenjaminLim yep

just use elementary matrices

that is easy enough

you see this comes from the fact that

any invertible matrix

can be row reduced to the identity

row reduction is the same thing as multiplying on the left or right by an elementary matrix

@leo Do you get what I mean?

@BenjaminLim elementary matrix I think you do mean

@BenjaminLim yes

@leo Furthermore there are only three types of elementary matrices

it is enough to show that the only matrices conmuting with all the elementary matrices are the set of diagonal matrices

the center must be the set of constant diagonal matrices, by the way

@BenjaminLim what are them?

@leo What are the row operations you can do?

adding one row to another

scaling

and swapping rows

see?

yep

@BenjaminLim I was thinking in other approach. Let me show you and let me know if it is too complicated.

@leo Hold on, I am in the middle of posting a question on the main site!

@leo you will see immediately

by using elementary matrices

that it must be a diagonal matrix immediately

this is because if you some $A$ in the center

and $L$ some elementary matrix

then $LA$ has an effect on the columns

$AL$ and effect on the rows

since $A$ is in the center

or something like that

the point is you use an argument like that

@BenjaminLim Thanks

These have to be equal

see?

@BenjaminLim Thanks. Indeed this approach is simpler

hope they help you with your post :-)

thanks man

@Ben skype--are you available now?

ok

@KannappanSampath $T$

$T^2 + 3T + 1$

$\Bbb{C}[t]$

suppose $\chi(t)$

$T$

this means that $\chi(T)=0$

Suppose characteristic polynomial of $T$

$f(T)$

$f(T)= 0$

$f(t) = \chi(t)q(t) + r(t)$

$t = T$

$\chi(T)q(T) = -r(T)$

what is the degree of $r(t)$

$ 0 \leq \deg r < \deg \chi$

$r(T) = 0$

implies that $r = 0$

implies that $\chi(t) | f(t)$

Suppose $T$ is a linear operator on $V$, such that any space of dimension $\dim V-1$ is $T$ invariant, prove that $T=\lambda I$.

(for some $\lambda \in$ the ground field)

saying that $\ker(T - \lambda I) \neq V$

is the same thing as saying that there is $v \in V$ such that

$(T-\lambda I) v \neq 0$

if they are linearly dependent

then you can write $c_1T(v) = c_2\lambda v$

$T(v) = (c_2)(c_1) \lambda v$

(This just means that $v$ and $T(v)$ are linearly independent. )

$(v,T(v), w_3,w_4, \cdots, w_n)$ is a bases for $V$.

subspace of dim $V$ -1

$v, w_1, \ldots _{n-2}$

$a_1v+ a_2w_1 + \ldots a_{n-1}w_{n-2}$

No, it might be better to just ask what $T(v)$ is?

$a_1T(v) + \ldots a_{n-1}T(w_{n-2})$

a_1 = 1

T(v) = linear combination of guys

Signing off for today.

Bye @Ben

bye!!

@BenjaminLim Consider this problem: Let $T$ an operator over a finite dimensional vector space such that for every operator $S$ over $V$, $TS=ST$. Show that there is a $\lambda$ in the field so that $\ker(T-\lambda I)=V$

@leo I am busy now

hold on

@BenjaminLim no problem

I ask just for curiosity

i am so confused

why

with your problem

?

yeah

pfffffffffffffff

@BenjaminLim what is the plural of index (index like $i$ in $x_i$)

dunno

?

@leo indices

@robjohn thanks

I was not sure

you could probably use indexes as well, but I think indices is better.

so confused right now

fffffffu

(Plural "indices" or "indexes")

@BenjaminLim re wut?

Morning folks.

See you later.