Okay, the day is gone. You may resume upvoting my answers.

I am going to sleep now.

If I die in my sleep please make sure I get cremated and my ashed spread in Ø. "Set theorists don't die, they become elements of the empty set."

Hi

@DylanMoreland Are you around?

I have a small problem with AM exercise 1.26 (iii)

I don't understand how $U_f = \{x \in X : f(x) \neq 0 \}$ is a basis for the topology on $X$....

@BenjaminLim Am now.

Okay, found the book. I think you need to use Urysohn again.

Like, take a point $x \in X$ and an open neighborhood $U$ of $x$. There's a continuous function $f\colon X \to \mathbf R$ which is $0$ on $X - U$ and $1$ on $\{x\}$.

Then $x \in U_f \subset U$. So that does it.

@DylanMoreland Ok nice thanks :D

Just interested how did you figure out to use Urysohn again? Just a hunch?

@BenjaminLim I needed a continuous function. I don't really know of other ways of making these in a random (but still normal) topological space.

hahah

nice

That and Tietze (which is stronger).

I get the $x \in U_f$

I have seen two interesting applications of the Urysohn lemma :D

@DylanMoreland Trivial Question: How does one deduce $U_f \subset U$?

Because I arranged for $f$ to be zero on $X - U$.

Ahh that's it.

Yeap. Because if $x \in U_f$, we must necessarily have that $f(x) \neq 0 $, so that $x$ cannot be in $X - U$.

I think you can use Urysohn's lemma to prove his metrization theorem, which is significant. I think you can prove some facts about embedding topological manifolds in $\mathbf R^n$ as well.

@DylanMoreland I only know that you can use it to prove the Tietze Extension Theorem

This is probably all in Munkres' book somewhere.

Actually in that AM exercise I still needed to prove that compact Hausdorff implies normality

But that was not too hard

@DylanMoreland Thanks for your insight, I should get back to finishing the rest of the exercise :D

Perfect Saturday morning. An empty chat room only proves that you guys enjoy Saturday. Good luck!

Does it?

@DylanMoreland Hey can I ask some things on AM 1.27? I've not encountered a problem in commutative algebra like that before and I'm quite confused.

I don't really see why when we do $k[t_1, \ldots, t_n]/I(X)$ (in the notation of AM) we get a polynomial ring whose elements are functions of $X$.

I get how two elements are equivalent in the quotient but don't really see how the equivalence classes of polynomials "become" functions of $X$. I thought this meant that a function became constant on $X$ but this is wrong...

What confuses me even more is how the equivalence class of $t_i$ in the quotient becomes a coordinate function that picks out the $i-th$ coordinate of a point $x \in X$....

@AsafKaragila Johnstein? :-p

@Ilya I hope you are having a good Saturday. It is now afternoon there, isn't it?

@Benjamin Lim : Essentially each class is giving a "genuinely different" polynomial function on X. Each element of the polynomial ring k[t_1, ..., t_n] can be considered as a function on X (by restriction of what it does to the underlying space that X is embedded in)...but it is clear that altering such a polynomial by adding/subtracting something that is 0 on X doesn't really change the function at all (it will just be doing the same but adding 0 on to it).

@robjohn Yes. That is your new last name.

@AsafKaragila Have I become Jewish?

@robjohn Please lie down on the table and I'll use this ax to surgically remove your turtleneck. :-)

@AsafKaragila Good... I thought you were going to remove my forehead. :-D

@robjohn Well.. fore something.

@fretty I don't get what you mean....

@Asaf: the accepted answer for this question is also downvoted. Brutal. I hope they are kinder with my answer :-)

Oops, i didn't notice the second part... back to work :-)

@Benjamin Lim : Ok so let's have an example. Suppose we are studying the solutions in R^2 of the equation y = x^2. Now this is a simple thing to solve but let's look at what happens in the algebra. We are in two dimensions so we have to start with R[x,y]. Ok so in here we have the following (different) functions on R^2: 3x, 3x + (y-x^2), 3x - 6(y-x^2). But when we look at what they do when we restrict to points on the curve y = x^2 what happens?

@fretty they are all the same?

because $y - x^2$ goes to zero.

@fretty you should get mathjax

@robjohn can help with that :D

Yes, because all I have done is taken 3x and added/subtracted bits on the end that are zero on that curve. The bits I added on weren't zero on the whole of R^2 but just on the bit of R^2 we were looking at, the curve.

Ok, so this example was to show that...

@robjohn I capped out yesterday with 90 points that went to Arturo but today I got nothing so far.

So when we do the quotient R[x,y]/<y-x^2> we are essentially getting rid of this behaviour

because all of those polynomials will now be equivalent

Ok. I get that is essentially what quotiening out means.

But why the polynomials in the quotient being now functions of $X$?

the different polynomials that gave the same function on the curve have now been made into the same class

i.e. each class is giving a function on the curve

huh? Sorry I don't get this :(

It takes a while to understand, this is essentially where Algebraic Geometry begins

really i am now at a roadblock ....

ok so in our example the class [3x] in this polynomial ring stands for all functions that are "the same" as the function 3x on the curve

right

@AsafKaragila how do post-cap points go to Arturo? I am confused.

that is easy enough

so like we said all functions like 3x + (y-x^2), 3x - 6(y-x^2) ...

yes

@robjohn Well... how would he get to 100k otherwise? Hard work?? I think not! :-)

they all do the same thing when you look at points on the curve, so this is represented by looking at the class [3x] in the quotient

ok

then how does $[3x]$ become a function on $X$?

modding out is just stopping things giving the same function

yeah I know that

well you can now take ANY polynomial in that class and it will define a function on the curve

and it wont matter which you take, they all do the same thing

because they only matter upto something that is zero on the curve

yeah I get that part, only thing is why now the quotient becomes a polynomial ring in terms of $X$..

all that stuff about modding out, what goes to zero, etc I get...

@AsafKaragila He must have written the Answer-Bot. It scans all other answers and puts together an answer from related answers ;-)

well there is slightly more than just a modding out being done, this is capturing the geometry

ok, suppose I now take this on faith

I get slightly why the quotient becomes a function of $X$

but anyway, suppose I get this

I(X) is just the "polynomials that are zero on X", it is capturing the geometry of X in an algebraic way

the modding out by this is just giving you the "different" functions on X

I don't get why if we look at the equivalence class of $[t_i]$

say $\xi_i$ as given in AM

Then the $\xi_i$ acts as the function that picks out the $i-th$ coordinate of a point $x \in X$...

well what are these as functions on the bigger space R^n

No, suppose we have $t_i$ in the ring $k[t_1, \ldots t_n]$

for example in R^2 what does the function f(x,y) = x do?

in the quotient?

well this is just a function on R^2, it picks out the x coord

but for any curve inside R^2 this function will still pick out the x coord

No, that's not really what I am confused about...

it is just the restriction of the function

the t_i's are the same thing here

I don't get how if we look at the image of $t_i$ in the quotient say call it $\xi_i$.

Then $\xi_i (x) $ for $x \in X$ picks out the $i-th$ coordinate of $x$....

well t_i is in its own class and this picks out the i'th coord

and as we just saw functions in the same class give the same function on X

i don't get this...

so the class [t_i] is in some sense giving the function that picks out the ith coord

why for only $x \in X$?

because we are only looking at functions on X

Oh wait is it something like

not on the bigger space

Yeah so in ordinary terms, we have that the function $x$ picks out the $x$ coordinate of a point in $R^2$ say (this is really redundant)

so when we look at the quotient now

the function $[t_i]$ picks out the $i-th$ coordinate of a point $x \in X$, am I right in saying this?

yes, but you see why

why...??

t_i itself as a function on X picks out the ith coord

so all functions equivalent to t_i must do the same

well sort of yeah

I get how it does it in the big space before taking the quotient

so for example in R[x,y] we may take the function x

this gives the x-coord of a point

ok

right

now lets look at a curve in R^2

say y-x^2 = 0

for example y = x^2 again

yes, now the functions x, x + (y - x^2), x + 67y(y - x^2), x - 8(y - x^2)

they all do the same thing when you look at only the points on the curve

ok

yeah

they all do exactly the same as the function x

so in the quotient these will all belong to the class [x]

so they just pick out the ordinary $x$ coordinate?

of any point on the curve?

yes, so even though lots of functions in [x] dont look like the ordinary "pick out the x coord" function

they all do this "on the curve"

yeap

bingo

And that is what is being captured here

@fretty Thanks man, much clearer to me now :D

If this kinda thing interests you then you might want to study Algebraic geometry...this ring is basically one of the important constructions to measure the geometry in terms of the algebra

well I'm just starting with AM

Yeah, bear in mind that some of the exercises (like this one) are from related branches. Stuff like this is usually investigated with more detail in actual books on algebraic geometry.

I like how in mathematics a lot of stuff that seems really complicated

is sometimes so simple underneath...

The first time I saw this it took me a while to understand the meaning behind it. I like to have an intuition...some people would rather just know that it is "something useful" and then derive things from that.

@tb My god where have you been?????????

@fretty This makes a lot more sense now I get a little why the coordinate functions generate $P(X)$ as a $k-$ algebra ...

It is a very elegant thing to construct...if you carry on with AG you will find that there are nice relationships between when curves are equivalent and when the coordinate rings are isomorphic.

@MattN: sorry for that incredibly tasteless youtube link. It was supposed to be tasteless, but not that tasteless. /me shudders

@BenjaminLim I'm pretty busy these days...

Research?

I wish...

I am really glad I did some topology stuff over the holidays

Really useful today in tackling AM 1.26 :D

I should go guys, it's 12.08 here in australia (am)

Well, no surprise, topology is useful in general :)

See you, Ben!

But the psychological thought of having done this stuff before

and feeling familiar with it

I like that feeling man

I understand :)

@fretty I have no idea how to go about proving surjectivity of the map in 1.27....

@tb And it was all self - taught too (with some help of Math. SE) :D

@BenjaminLim even better news!

@tb I really like for AM 1.26

What is AM 1.26?

(I mean, what's the question in there)

Oh it's like

If you have a topological space $X$ that is compact Hausdorff

and $C(X)$ the ring of all real valued continuous functions on $X$

define $\tilde{X}$ to be the subspace of Spec$(C(X))$

Then the problem is to show that $X$ is homeomorphic to $\tilde{X}$

I like how with the Urysohn Lemma you can "construct" open sets for $X$

Even though god knows what topology is on it :D

Now that was impressive man :D

Prepation came in handy in proving compact hausdorff $\implies$ normality

@BenjaminLim what is the subspace $\tilde{X}$? The image of the the map that sends $x$ to the (maximal) ideal of functions vanishing at $x$?

Oh sorry

forgot to tell you that $\tilde{X}$ is the subspace of all the maximal ideals :D

Sorry and the map between the two spaces is $\mu$ that sends a point $x \in X$ to $m_x$

where $m_x = \{f \in C(X) : f(x) = 0\}$

Sorry

A LaTeX thing: instead of writing Spec$(C(X))$, write $\operatorname{Spec}(C(X))$ That looks a bit better...

Like so: $\operatorname{Spec}(C(X))$

I know that, but a bit lazy...

@BenjaminLim no problem, I made the right educated guess :) Yes. That's a very nice exercise :)

Yeah

It's pretty cool like I said how god knows what topology is on X

but then you know $\tilde{X}$ and the topology there

then somehow you get a collection of sets that becomes a basis for $X$

and now you can actually get open sets in $X$

But what was really cool was how

this afternoon I was struggling to even prove how some collection was a basis for a topology on $X$

And then Dylan was like use Urysohn like this and then kaboom...

I see :) Urysohn and Tietze are pretty strong.

@tb: good afternoon!

Hi, robjohn!

Hi, Rob

@tb could they beat Superman? (Übermensch)

@robjohn by abducting lois, I guess :)

@KannappanSampath Everything ok with the prime ideals stuff?

@tb or using Kryptonite :-)

@BenjaminLim Yeah. I got it right. There was an easier version in the exam for $3$ marks!

I think if you start with a maximal ideal of P(X) then you can use the correspondence about ideals of quotients to get a maximal ideal of K[t_1, ..., t_n]

@fretty Right I will work it out tomorrow, right now I'm really tired....

(That asked me to do only for two prime ideals and asked if the assumption of primeness is required)! :-)

@fretty Will you be around tomorrow?

And all maximal ideals of this look like <(t_1 - a_1), ..., (t_n - a_1)> so you would have found an x corresponding to it

@KannappanSampath Well for two it is not required...

errm, maybe

wait that's true because $k$ is algebraically closed?

yes

@BenjaminLim Yes. I gave from the first a proof that did not use the primeness and I said since I could give a proof without primeness, it is not required!

wait are you saying that is the kernel of the evaluation map from $k[t_1, \ldots , t_n]$ to k by evaluating at the point $(a_1 \ldots a_n)$?

I used R quite foolishly earlier lol but it was really to demonstrate the point of functions being equivalent

@BenjaminLim there's a nice follow-up to that exercise: take a locally compact space $X$ and look at the ring $R = C_b(X)$ of bounded and continuous functions. Show that $X$ embeds as an open subset of the space $\beta X = \operatorname{Spec}{R}$. The space $\beta X$ turns out to be the Stone–Čech compactification of $X$

yes, essentially

yeah, I know but that did not matter really because the point was to understand how stuff worked

@tb wahhhhhh?????

@robjohn Oh, I didn't know that Superman had an Achilles Heel like that...

@tb Do I need any high level stuff (higher than before) to tackle that?

Not at all.

Same level of difficulty.

Ok.

@tb You know everytime I talk with you guys on chat my adrenaline gets pumping :D

I forgot to say that $X$ is open and dense in $\beta X$.

@BenjaminLim that's nice to hear :)

Right.

@tb I assume, from your comment, that you read the Wikipedia article. :-)

@BenjaminLim Hmmm. I don't see why?

Just like the proof for subgroups: If $I \subset J_1 \cup J_2$ then $I$ must be contained in either one of them

@robjohn I admit that I only read the first paragraph for lack of interest.

@tb understandable...

@tb Though admittedly I am not so familiar with locally compact spaces

@BenjaminLim I see why this. But, not that adrenaline comment. I linked the message to that comment, you may have not noticed! :-)

@KannappanSampath You see here where I am there are not many people to talk about mathematics

even if there are my coursemates they are mainly testosterone high...

: )

@KannappanSampath I tell you what: Two people that I met here on math.SE I have met in real life

pretty cool

@tb However, when I was much younger, I used to scour my comics for information on the different colors of Kryptonite. Now, all that information is collected in that aricle.

random people

Like this guy once

I met him for the first time

talked for 4 hours in a café

@robjohn Geek :)

@tb I am :-)

strange how two strangers that have never met before can just sit down and talk like that :D

@BenjaminLim Which guy? Fretty?

No Ragib Zaman

@BenjaminLim Oh the same that is active here?

yeah

Ah, you just answered my question :)

@tb What have you been busy with?

But isn't Sydney quite far from Canberra?

(like 1000 miles or so?)

No

it's only 100 miles

280 km approx.

i go up there during the holidays :D

Oh, I should look at a map :)

@BenjaminLim I'd rather not elaborate.

Ok, I understand. This is a public forum anyway

@BenjaminLim In India, to travel 280KM, takes 6 hours!

@KannappanSampath Don't get my started on how long it took me to go from Dehli to Agra

Even worse in Nepal from KTM to Pokhra...

Oh, I confused Canberra and Adelaide. Sorry about that.

My friend wanted to go from Tiruchirappalli to Chennai

@BenjaminLim 10 hours!

about that yeah

I am resident of Tiruchirappalli :) @BenjaminLim

@tb Canberra is in the ACT which is within NSW

Adelaide is like South Australia

@KannappanSampath He was so frustrated...... many times the van broke down and stuff

Nice I have not been there

Have you been on a elaborate non-Mathematical visit to India?

yes

that time when I was in Dehli

(non-Math---No conference or blah)

yes. Many people were so confused how a non-indian guy could sing bole chudiyan ....

@BenjaminLim I see. That's interesting!

I have a nice collection of hindi songs

some from movies like Krazzy 4, Dostana

although dostana did not seem indian really the whole movie was practically in miami

You can listen to Carnatic Music :-)

@tb So now the business is over ??

@KannappanSampath I don't get the bollywood, kollywood, tollywood thing....

Is Kallai Mattum Kandal carnatic?

@BenjaminLim Unfortunately not. But the worst part of it is, I think.

@tb That is nice to know. I guess we all have our troubles in life

In my first year of uni I flunked two mid semester exams and got really frustrated....

@tb Hey are the functions in $C_b(X)$ real valued?

That's NOT bad. Here in my institute, people are asked to leave if they fail a course!

@KannappanSampath Terrible. For me I have to maintain a High Distinction average for science every semester otherwise i get kicked out of my degree...

@BenjaminLim Sure it is!

@KannappanSampath I could be wrong but I believe that is one of the reasons students just go for the HDs rather than take risks

@BenjaminLim doesn't really matter. Usually people take $\mathbb{C}$.

@tb Ok.

But with the reals it works just as well.

@BenjaminLim HDs?

@KannappanSampath Like they would rather not take courses that are deeper and more risky

High distinctions

Here the framework is pretty rigid.

Hi Matt

@tb Hey. Didn't expect to see you today. (It's ok, it was not that bad. The thought counts.)

Hi Kannappan.

Ok guys I have procrastinated too much should go now

bye

really late here

bye :D

@KannappanSampath Congrats! For selection in SRFP!

Bye Ben, Good night!

@SidharthIyer How did you know! :-)

From this: web-japps.ias.ac.in:8080/fellowship2012/…

@MattN no it was way worse than expected :(

@SidharthIyer Ah, I see!

@KannappanSampath :-)

@tb How can you know the song while not knowing the song?

Glad you pinged me, btw.

@SidharthIyer Did you install the Chat Jax

(MathJaX for Chat = ChatJaX)

@MattN well, just before I left, I remembered that I heard the song about 20 years ago in the radio. Then I listened to 5 seconds of it before posting the link...

Ok.

@tb But like I said: the thought counts.

Okay, I'm relieved.

@Matt Did you see that theorem 2.6 (2.5 is a lemma we corrected! :-))

@KannappanSampath Sorry, not yet.

@KannappanSampath Yep.