@BrianMScott I guess when I try to recall the proof that a compact subset of a metric space is closed

I always remember the result that it holds in a hausdorff space

the hausdorffness lets me recall the way to the proof :D :D

Well, you need Hausdorff...

Exactly.

@BrianMScott Now that was slick!

@tb if you say metric space I don't remember the gist of the proof

but throw in hausdorff and I do :D

Then you should learn about nets, I suppose ;-)

Yep. Extraneous hypotheses can sometimes really get in the way.

@tb 2) above was a real handicap. So was forgetting about boundedness!!

I have to run; the grocery store won’t stay open forever.

@BrianMScott ok bye!!!

Take care, see you later

See you later!

@tb maybe when I have the time I will look at nets at the end of chap 4 munkres

Ben, note that the proof Brian gave goes through when you throw in closed. So a nested sequence of closed compact sets in an arbitrary topological space has non-empty intersection.

@tb So ok it is not true in a general topological space that compact implies closed

so we need to throw in that condition (the space must be at least T2)?

I have not seen an example of a compact non-closed set in a general topological space

Take a set with at least two points and the trivial topology.

Then every subset is compact.

yes

But only the empty set and the set itself are closed.

oh nice!!!!!!!!!!!!!!!!!!!!!!!!!!

I'm starring that!!!

Some call it nice, others call it contrived...

haha I love these things when use like the trivial topology or something :D :D :D

@tb Actually I wanna tell you that I love topology

when I started reading munkres

nothing had ever seemed so clear before

maybe because of the book

@tb I have had visions of becoming a topologist !!!

Topology is really nice, I like it :)

And it's so powerful too!!

stuff all these metric spaces man!!!

Oh, those are very powerful, too!

well they don't generalise

Metric spaces have some pretty nice properties that general topological spaces don't quite have :P

@Rofler Can you use metric spaces to tackle say the zariski topology?

Of course not.

wait is it even metrisable?

It's only $T_1$.

The Zariski topology is pretty terrible

@tb Actually it's only $T_0$

yeah so there's the answer

Depends in what generality. I'm pretty happy with varieties and there it is T_1

@tb math.stackexchange.com/questions/108503/…

I know that the spectrum of a ring is only T_0 in general. I was talking about the Zariski topology of a variety over an algebraically closed field.

Oh I don't know what happens there though. Finite point sets become closed? @tb And I thought you did not like AC now you are talking about varieties???????

I've done my share of that stuff. I had some great courses on algebraic groups.

hahahahahahahahahahahahahahahahahahahahahahahaha

That doesn't force me to like Noetherian rings and modules.

hahahahahahahahahhahahahahahahahahahahaha

man this modules stuff is killing me man

What about modules is killing you?

the zariski topology is terrible only if you try to do things with it which it is not supposed to help you with

@Rofler math.stackexchange.com/a/55067/5783

Ah, hi @MarianoSuárezAlvarez

I don't understand the proof at the end

@MarianoSuárezAlvarez @Rofler Finally the algebraists have arrived!!

technically, I am cooking

@MarianoSuárezAlvarez In Argentina do you have machaca??

probably not, because I don't know what that is :D

maybe not it's a mexican thing

en.wikipedia.org/wiki/Machaca ?

yes

must be good :D

hey since the minimal polynomial divides the characteristic polynomial of a matrix

the minimal polynomial must annihilate the matrix too right?

it is defined like that, in fact!

it is the minimal polynomial among those that annihilate the matrix

(minimal with respect to the degree, or the divisibility relation)

@MarianoSuárezAlvarez Can you have a look at this link? math.stackexchange.com/a/55067/5783

@tb I have replied to Joriki on meta.

(one normalizes it to be monic, for fun)

I don't understand the sentence at the end

the last sentence of Pierre's answer?

@MarianoSuárezAlvarez yeah

when you apply $bf(b)$ to $e_n$

why is there a contradiction?

@robjohn Thanks! That's probably the best one can say.

@tb The 79 characters is highly suggestive.

you have that $be_n=0$

how?

That is the part that has been driving me around in circles!!!!!

then your problem is not the last sentence...

@BenjaminLim a flat tire can do that, too.

Ok we know by the Cayley Hamilton Theorem that the characteristic polynomial of $b$ must not have zero constant term

@MarianoSuárezAlvarez I looked at the MO thread and they say the same thing too

but how ?????

how what?

what don't you understand?

how is it that $be_n = 0$?

because it is defined that way :P

are you talking about the surjective part, or the injective part?

huhuh?

injective

the surjective part you just do the tensoring business

@MarianoSuárezAlvarez fffffffuuuuuuuuuuuuuuuuuu

you are getting confused by the fact that two things are being done at the same time, methinks

ok

so we know that $bf(b)$ is the zero endomorphism

right

therefore there is an a in A which kills e_n

but multiplication by a is not

that's absurd

huhh?

wait two guys talking at once is confusing me

multiplication by a non-zero element of A cannot annihilate a basis element of A^n

ok

@MarianoSuárezAlvarez that $b$ there is not an element of $A$

@tb Nope. I typed it without the weird characters and they showed back up. :-(

It is in the comment parser or something.

$1,2,2,9,2,46,2,250,37,254,2,31052,2,1480,896,306174,2,2097506,2,6025516,6638,59‌​930,2$

I think Benjamin's problem is with how we're adjoining $b$ to the ring $A$

@Rofler I get that bit.

Wiggy... it shows up here, too. I know that the characters were not in the pasted string

@robjohn I'm telling you that there's something terribly wrong... :)

@BenjaminLim What specific part of the proof is the problem then?

b in an endomorphism of A^n, but the equality a=bf(b) means that multiplication by a is the same as applying bf(b) on every element of A^n

$1,2,2,9,2,46,2,250,37,254,2,31052,2,1480,896,306174,2,2097506,2,6025516,6638,59‌​930,2$

strictly, it should be written a Id = b f(b)

but that's overly pedantic

@MarianoSuárezAlvarez Ok I get that

ok so if $b(e_n) = 0$

then the right hand side is zero

whereas $a$ is not

so we get a contradiction

It is not in the MathJax. I have it turned off (refreshing the window and not running the bookmark), paste the clean string in, and copy the string to check it, and the weird spaces are there.

1,2,2,9,2,46,2,250,37,254,2,31052,2,1480,896,306174,2,2097506,2,6025516,6638,599‌​30,2

@MarianoSuárezAlvarez why is $b(e_n) = 0$?

@Rofler that is the problem now

Even without the $...$ they get added.

it is in the text input method that they are getting messed up.

so confused now

2,2,9,2,46,2,250,37,254,2,31052,2,1480,896,306174,2,2097506,2,6025516,6638,59930‌​,2

@robjohn and it's always those 200c and 200d's

@MarianoSuárezAlvarez why is $b(e_n) = 0$?

If I leave out the first two characters, the added spaces move up two characters. It is always 79 characters in

@MarianoSuárezAlvarez Sorry that is correct $(b f(b))(e_n) = 0$ because $(b f(b))$ is the zero endomorphism

@robjohn yes, I was just about to say that.

Are you asking about the proof of (a) or of (b) ?

i am asking about (b) in pierre's answer

injectivity part

Like I said the usual tensoring business proves (a)

well, then, in the proof of (b) he is not saying that b(e_n)=0...

ok yeah

ok let's forget that bit then

@robjohn same thing in the comment boxes on main. The garbage is added after 79 characters there with or without dollars

@MarianoSuárezAlvarez I got it now

I got it now

I just need to understand why $(b f(b) ) (e_n) = 0$

it doesn't

but that is not true nor claimed to be true in the proof of (b)!

it lies in $A^{n-1}$

while $ae_n$ does not

that's where the contradiction lies

@Rofler Oh!!!!!!!!!!!!!!!!!!!!!!!

so wait

$b(e_n)$ lies in $A^{n-1}$

> Assume (b) is false. Then there is an $n$ and a injective endomorphism $b$ of $A^n$ satisfying $b(e_n)\in A^{n-1}$, where $e_n$ is the last vector of the canonical basis. Then $b$ is integral over $A$ by Cayley-Hamilton, and we get a contradiction by using the lemma (with $B:=A[b]$) and applying $a=bf(b)$ to $e_n$.

@tb yep.

that is the argument given for (b)

yes

so Mariano

like @Rofler has said

I get it now

Huzzah!

I only need to understand why $(b(f(b))(e_n)$ lies in $A^{n-1}$

:D

I know that $b(e_n) \in A^{n-1}$

because $b$ has image lying inside $A^{n-1}$

so in particular, $b^k$ has image lying in $A^{n-1}$

yes!!!!

so any sum of powers of $b$ has image lying in $A^{n-1}$

each $b^k$ is an endomorphism of $A^{n-1}$

But why the need to write it as $b(f(b)$?

um

why not just some polynomial?

$b$ is an endomorphism of $A^n$ which has image lying inside $A^{n-1}$

yes

So we've shown that $b$ is integral by Cayley Hamilton, right?

yes

And it's not a zero-divisor since it's injective

Hence we have by the lemma that there exists some $a$ in $R$ such that $a=bf(b)$ for some monic $f$

right

$bf(b)$ is a polynomial in $b$, and each power of $b$ has image inside $A^{n-1}$, so the polynomial itself has image inside $A^{n-1}$

@Rofler Let me put it in my own words now

And put this in the grave once and for all

let him finish

($bf(b)$ is an element of $A[b]$ for the record)

By the Cayley Hamilton theorem, we have that $b^n + a_1b^{n-1} + \ldots a_n=0$ for $a_i \in A$

Now $a_n$ cannot be zero for then this would contradict the kernel of $b$ being trivial.

So assume $a_n$ is not zero

then we have that $b^n + \ldots a_{n-1}b = -a_n$

Apply $e_n$ to both sides

the left side lies in $A^{n-1}$, the right side in $A^n$

contradiction.

no

huh??

the right hand side does not lie in A^{n-1}

man!!!!!!

what Ben has said is essentially right

the sentence «the left side lies in $A^{n-1}$, the right side in $A^n$» is not a contradiction

he means that the right side does not lie in $A^{n-1}$

if I had a penny for everything that a student meant but did not say

well on the left we have that $(b^n + \ldots a_{n-1}b)(e_n) \in A^{n-1}$

I would be much richer now

On the right $-a_n(e_n) \in A^n$

@MarianoSuárezAlvarez That is what I meant to say

@BenjaminLim, yes. but what you *need to say

huhuh?

FINALLY MY GOD!!!!!!!!!!11

in order to reach a contradiction is that a_n e_n is not in A^{n-1}

Also, the justification for $a_n$ not being $0$ is that we can MAKE it non-zero by cancelling b's from the left (using the fact that $b$ is injective)

Ok that is right

because, as said, the sentence «the left side lies in $A^{n-1}$, the right side in $A^n$» is not a contradiction

@MarianoSuárezAlvarez Now you're picking on my words!!!!!

ffffffuuuuuuuuuuu

well, we are talking about your proof

on what should we be picking on?!

Are you twelve? Who tells someone who's helping them "fffffffuuuuuuuuuuuu"?

@Rofler The justification is that if $a_n$ is zero, then we can write the characteristic polynomial as $b(\text{some stuff})$ = 0

@Rofler Y u no look at internet memes?

anyway

@MarianoSuárezAlvarez that was just a joke :D

The justification for $a_n$ is non-zero is that we can make it non-zero (the standard proof of cayley-hamilton might give $f$ with zero-constant term, you have to cancel the $b$'s)

@Rofler If we have $b(\text{some stuff}) = 0

That's the key point where we actually used $b$ was injective

then you apply the left hand side to any non-zero vector and you have something in the kernel of $b$

otherwise, we would have never used the hypothesis!

@MarianoSuárezAlvarez What I said above is correct?

indeed, if b f(b)=0, then the image of f(b) is contained in the kernel of b

since b is injective, this implies that f(b) = 0

yes

then we just take $f$ now to be the characteristic polynomial

and pull out factors of x as much as you can

yeah so in beginning if the constant term is zero

we look at the highest power of $b$ that divides every term in the characteristic polynomial and pull that out

then that $f$ there will have non-zero constant term

yes

right

pheww!!!!!

finally

I was going round in circles

thanks @Rofler @MarianoSuárezAlvarez

I guess I was looking for a wrong contradiction

yes :)

ffffffffuuuuuuuuuuu @benjamin lim

alright

I think I better go study for my theory of driving test

I don't like the proof given for the surjective case though in that link

:P

@Rofler the usual tensoring business works

because $1 \otimes \phi$ is surjective on elementary tensors

$b$ might be a zero-divisor; I know of alternative proofs xD

extending by linearity shows that $1 \otimes \phi$ is surjective on all tensors

@Rofler oh crap yes......

@Rofler how is that giving us a problem though?

The lemma needs $b$ a non-zero divisor

pierre's lemma?

yup

But in my proof above I did not use his lemma

I know; but the proof given in the link uses it

@Rofler ok

@Rofler Also in my proof above, how is it that $-a_n(e_n)$ is not zero, viz. that the constant does not annihilate $e_n$?

we need to check that

no, we don't need to check that

$e_n$ is a basis element in a free basis

because we want to say that $-a_n(e_n) \in A^{n}\setminus A^{n-1}$

right

so why will an element of the ring not annihilate it?

because $ae_n$ is the element $(0,0,...,0,a)$ which is not zero

since there is a non-zero coordinate

Anyway, @MarianoSuárezAlvarez, I think I understand $\kappa(P) \otimes M$ (we're just extending the ring of functions by inverting some convenient functions), but I still don't get $I\otimes M$

@Rofler I get it, what I said above does not make sense

anyway thanks for your help

@MarianoSuárezAlvarez you too

@BenjaminLim no problem :)

I thought that $\kappa(P) \otimes M$ would shed some light on $I \otimes M$, but they the problems plaguing the latter are surprisingly separate from the former, since there are no nasty non-elementary tensors to deal with in the former

bye guys!!

how to do power -3

like x^-3

it doesn't work

Are you asking how to type it for other people, how to type it for a computer to evaluate, or how to compute it yourself?

x^(-3) is the same as (x^3)^(-1), or (x^-1)^3. Do you see how to compute it?

I am assuming you want to evaluate x^(-3), given x.

hi anon

yo

aww..no. I'm asking to ( write the question in here)

@tb It is the same problem we had with the smileys. I added a space before the offending number, and that reset the 79 character count. I added a comment.

@Srivatsan good day!

@robjohn Hi

I assume you are still out of the US.

@SbSangpi You can just write x^(-3). If you want to LaTeX it on the mainsite, you want to write $x^{-3}$. You can latex in here but you need to use the special bookmarklet.

I mean, when I write the question like x^-3 i doesn't show x power -3, it show x power - and 3 separate

@robjohn Yes..

@SbSangpi x^{-3}. The braces tell the TeX parser that -3 is a single expression.

x^{-3} it work thanks @anon

@SbSangpi you need to collect the exponent in {...}

all gud thx all

Is Kannappan not around these days?

@Kannappan: Happy New Year, by the way... :)

@Srivatsan He says that he is not coming back

@robjohn Time drain?

It was a disagreement with others here.

I still talk to him on Skype.

Hm.. One second.

and he is posting on the main site and other chats.

@robjohn That's quite unfortunate. Not sure what is happening with chat, but we seem to be losing quite a few regulars.

I saw him two days ago in here. Did something transpire since then while I was sleeping or something?

@Srivatsan I told him I would miss him, but he said that he would be on skype. That is not quite the same.

@anon That was about when he left.

@Srivatsan I used to talk with him. What happened?

@anon He got into something with Gigili and David Wallace, and others.

Ah. I remember Gigili and two Davids talking about Kanna and Jordan, but I went to bed before I saw Kanna come back.

I wish that people would lay off J. I feel sorry for him.

@BrianMScott Do you mean the starring of his comments? Or was there something else?

@robjohn You mean a quarrel¿

@Srivatsan And using his name as a general term for blanking on something.

@BrianMScott Quite mean indeed.

@Srivatsan That was Jordan who was being starred. I was not here, or I could have removed the stars.

It's one of the few room owner powers.

Way to make one curious: chat.stackexchange.com/transcript/36?m=4188115#4188115. I wonder what Kannappan was talking about...

Probably this.

I was trying to track the conversation further into the past, but I get lost.

Anyway.

I think these things are based on several separate threads. They can be hard to follow.

@robjohn Hm, a feature request for SE :).

I was kidding, of course.

Ok, got to go.

Bye!

@Srivatsan See you later!

wooooo thunderstorm

@TheChaz Hi.

thunderstorms are great fun

my favorite part of summer, honestly. although, it's hardly spring :p

you must live somewhere much warmer than I do

howdy rob

Howdy Tyler.

how's it going?

alright and you?

sleepy :)

and have an exam tomorrow. Friday exams!

Friday the 13th, no less. who does that? :)

@anon What did you think of the answer to the twins riddle?

Your avatar made me think of it.

meh. depends on your definition of twin.

Are two members of triplets called twins?

Provided they are all still alive of course.

Just some food for thought.

@BrianMScott Hi, I think I found the source of confusion for the label "dummy" variable, it is better to use the adjective "index" variable, it helps to inform the student that this is a book keeping device.

woudn't it be better to inform them of the meaning of the term 'dummy variable'?

no one uses 'index variable' so it will be more productive for them to learn what 'dummy variable' means!

No one?

have you seen that term used for the variable with respect to which an integral is taken, for example?