@tb Thanks, I'll need it. I have a full day today: walk the dog, take the kitten to the vet, take my wife shopping, and take my son shopping. That will probably take all day, after which I will again walk the dog :-)

This is so absurd it's now funny to me.

@robjohn Have you bought your C. tree and stuff?

@Srivatsan we have an artificial tree. Thanks for reminding me; I need to set that up today as well.

@Srivatsan I set up the lights outside on the day after Thanksgiving.

@Srivatsan this is a joke from MO

I will never understand those people who apply de l'Hôpital in this way

Talk about delayed acceptance.

@tb Well, well. That joke was started by the same person. Why am I not surprised? =)

@robjohn talk about five months?

Er, not really. the OP did say "ask Johnson". But adding a tag is a little...

@tb Is that a duplicate of this?

@tb It probably took the OP that long to look at all those diagrams :-p

@robjohn I was thinking the same then got distracted by the abuse of l'Hospital. But I agree now that it is.

Oh, BTW rob: if you're already taking vitamin C, you might also want to add echinacea.

call for votes

I think that is my first act as a judge :-)

@robjohn Is there a difference between the purported dup and the original? One question is asking for a proof and the other why the natural (to the OP) guess of $0$ is wrong...

@JM Thanks. I usually take mega doses of C, but echinacea in moderate doses.

@Srivatsan I could be convinced. Let me think.

@rob: Oh, it's cool that you're already taking echinacea. :)

@robjohn You could be convinced by what?

@Srivatsan how much do you have on you? :-)

:=)

@JM I silently made a martingales tag.

@Srivatsan Is there a way to recant a vote to close?

@robjohn No, there's none. Now that you have cast your vote, you can relax =)

There's nothing you can do.

@Srivatsan The best I can do is comment, then.

@Srivatsan Ah, nice. Well, let's wait for it to be used more...

It's quite sad that I accidentally chose a dup question to tag =)

@robjohn Yes, that would be a good idea.

I am kind of iffy about whether it's a dup or not.

@Srivatsan Heh, my previous comment about duplicity has been upvoted.

Ello : )

@Srivatsan I have commented to the contrary now.

Hi Matt!

Hi, Matt! (I could swear I saw you jump out of chat just a minute ago...)

Man, SE 2.0 has gotten me spoiled; it now annoys me that MO can't keep track of all of my comments...

@robjohn Well, it perhaps doesn't matter either way... =)

@tb Impossible : ) I've only just come home and I left my laptop at home.

How was the Christmas orgy?

On the flip side, I'm missing the "votes" tab on MO that was removed in SE 2.0...

votes tab?

@Srivatsan I will now just sit back and watch :-)

@Srivatsan If you have an MO profile page, there's a tab that tracks stuff you've voted on.

what the hell?

@Srivatsan For you, that would be this.

@Matt It was ok, thanks for asking. One down, four to go...

Thanks JM. My voting in MO is a joke =)

@tb You'd think that one would ask the aggrieved before asking strangers... :)

Off to walk the dog. bbl

See you, robjohn!

If I may ask: Were you able to evade the hitting, Matt?

Later, robjohn

@tb That sounds like Matt is being bullied around =)

@tb Sure you may ask! I was : ) I went there late, took the seat next to my favourite guy and then left early. During the break I walked away to the back to talk to someone I know from last term.

@Srivatsan What do you mean, "like"? : D

@Srivatsan I wish I'd only be hit on by people I like. That would be much easier to deal with.

@Matt Ah, that would sure be convenient.

I'm not so sure about that...

Huh? Why not?

If the answer's not positive you'll have to hurt the feelings of someone you like...

(Heh. Looks like I was logged in all day. Must be the fact that we have multiple users on this laptop.)

A is similar to it's "companion matrix" and and she in turn is similar to B Heh =)

@Matt I told you that I saw you jump out of the chat room right before you said "Ello!"

@Srivatsan for what?

@Ilya We were thinking that tag might be useful . You don't think so?

@Srivatsan well, I think that stochastic-processes tag has quite small number of questions

@tb No. If you're in a relationship and then someone you know starts to like you a bit more then they will be clever enough not to ask you because they already know the answer. No hurt feelings there, I think.

@Matt they may not know the answer. that's one of the most intriguing sides in the start of relationships

Yes, I don't quite believe that.

I think people are pretty obvious. So yes, they will know the answer.

@Matt you are quite demanding, aren't you ;)

at least some girls are not pretty obvious (even after they say: no)

@Ilya With that I meant that I can immediately tell if someone is hitting on me.

@Matt well, am I?

@Ilya Aha... You should then discuss this with the higher members of the chat tag committee =)

@Ilya I meant IRL : )

@Srivatsan I don't participate in MSE as for now much (besides the chat)

Oolps

hell, I deleted the right post

Is there any one in Australia

@Matt well, if feelings were that simple to control, I'd agree. (Un)fortunately, life's not so simple.

Do you think you could fall for someone you've only met online?

@Matt You know, that's one thing I've always found weird about ladies. They wear all these fancy clothes and then complain when they get too much attention...

Math\

Math is better

@IloveMath Hi, IloveMath

@Matt me - yes (at least when I was younger - but you cannot be sure in my age, right?) and if you reject that possibility - how can you state that you can say immediately if someone is hitting on you?

@IloveMath I think last time I checked there were over 22 million people there...

@tb Feelings aren't but actions are : ) Or at least I'd like to think so.

Are you from Australia, @IloveMath?

No

@tb maybe it's not simple just for two of us, not for Matt :)

I try to apply Ph.D

@JM I can't argue with that.

It is frustrating.

@JM well, I've never met such ladies

@Ilya I'm not sure I understand.

Why are you disscussing relationship in Math room

@IloveMath I don't know ) just better than food topic

@IloveMath Why not?

To me , it is nothing

How old are you?

@IloveMath It's pretty run-of-the-mill. We've discussed weirder stuff before. Check the transcript if you don't believe me. :)

@JM Vanity is definitely my favorite sin...

@Matt your argument: you can immediately tell if someone is hitting on me. Then you said that you're talking about RL. So if somebody is hitting you in (ones) RL just after meeting you in Internet you cannot say that this person is hitting on you

@tb I loved that movie, too. :)

@IloveMath 24 and I'm proud of it

@Ilya Um.... yes?

@Ilya Ladies who wear fancy stuff, or ladies who moan about unwanted attention?

@Ilya this is getting a wee bit too abstract for me...

@tb I'm trying to learn category theory at the moment if you remember )

@JM the latter. well, I met only those who didn't have the sense of style and were wondering about the unwanted attention (wihtout having attention in fact)

If so, lucky you. :)

I am hoping we don't chase IloveMath away from the chatroom with this conversation. =)

@Matt finally, to you. Let us now project this situation to... RL. We deviate a bit the fact that a person just met you in the Internet (e.g. this person just met you a couple of times in Uni) and this person may fall in love with you if the person is young and romantic enough. On the other hand, this situation is so close to brief chat in the Internet that I guess you wouldn't realize if the person is hitting on you or not, would you?

@tb to finish what I was trying to say: as long as they don't ask they won't get hurt feelings.

(Just reading some Bern German transcript : ) )

@Srivatsan he is maybe checking out the transcript as JM has suggested :-p

@Ilya Yes I would. Read the body language. As for the chat: it's impossible to tell because it's asynchronous and therefore easier to hide yourself if you are trying to.

@Matt well... nevermind ) I didn't want to convince you

: )

how is your Mac?

Still in repair : (

I don't have to answer any questions anymore: I can just sit here and watch my rep go up as people keep up voting my algebra answer from yesterday.

Resting on your laurels, eh?

I wasn't actually serious. Resting on laurels is boring and not satisfying.

I know, I was kidding too. :)

Also if you sit too long, your arse becomes squishy.

: )

I am thinking about this homomorphisms thing

a field homomorphism must be injective (otherwise 1 = 0)

to prove you have a homomorphism you need to prove $1 \not = 0$, but it's not simple to do that.. I think it's better to prove the thing is injective instead.

Which question are you talking about?

This one math.stackexchange.com/questions/93577/…

I'll undelete it for a moment, but it's wrong

Ok, I'm looking at it.

I only showed that $f(\zeta^i) \not = f(\zeta^j)$ but not forall $x,y,$ $x \not = y$ implies $f(x) \not = f(y)$

Maybe only need to show that $\sum_i^n q_i \zeta^i = 0$ implies forall $i$, $q_i = 0$.

To prove that $1 \not= 0$ I have to show that every number that is equal to 1 is not equal to 0

@QED: That's easy. A ring homomorphism must preserve 0 and 1, so if the target ring has $0 \ne 1$ we're done.

@ZhenLin, but what if my homomorphism maps 0 to 0 and 1 to 1 but zeta^i and zeta^j to the same point? Then we have 0 = 1 because (f(zeta^i)-f(zeta^j))/f(zeta^i-zeta^j) can be viewed as 0 or 1 depending how you compute it

@QED The kernel of a ring homomorphism is an ideal, and a field only has two ideals.

hmm

I don't really understand

I got to go run

It's really simple. You're confusing yourself by trying to prove injectivity the hard way.

A ring homomorphism is injective if and only if the kernel is 0.

alright, so all I need to prove is that ker(f) = {0}

I'll think about what exactly that requires while I'm out - thanks!

If we have a ring homomorphism $\phi : F \to R$, and $F$ is a field, and $0 \ne 1$ in $R$, then we must have $\ker \phi = (0)$.

I need it the other way around in this case, $\ker \phi = (0)$ implies $0 \not= 1$

There's nothing to prove. Either $0 \ne 1$ or not in $R$.

but I think were some objections raised to my answer

sorry - I will be back later on

does someone here know Markov chains?

I don't, sorry.

@Ilya Apparently you're the only guy here with stochastic skills, so maybe you'll need to ask your question on main...

@JM that's pretty easy question if I able to formulate it in LA language, let me try

well, indeed, let me ask it on the main page

Just making some lasagne here. If it's a joint effort, cooking is not so bad : )

@Matt Joint efforts are good, if there aren't too many of you. "Too many cooks...", you know. :)

Hm. The question I linked to this morning has been deleted.

There was a discussion in the comments about accepting answers.

I find comments like "You might want to increase your acceptance rate." unhelpful and upsetting. Am I being too sensitive?

I can't say. One could make a case for those comments being distracting, in the same vein as "is this homework?"...

But "Is this homework?" is a neutral question while the other is aggressive.

@Matt Hmmmm. Wish I could be there... :(

Ditto.

@Matt I have nothing against nudging users to accept a few answers. Problem is that users with accept rate 0 often don't know how to do it and don't bother to figure it out, so your comment on the professor's question was much more helpful and actually effective.

@Matt I've seen people plenty annoyed with "homework police" myself. I personally don't feel strongly about those or the acceptance rate comments, but I can see why people are irritated by it.

@tb That would've been nice : )

I'm indifferent towards whether it's homework or whether they accept anything. I just don't like harsh and / or unhelpful comments.

I am back

@QED Arnold?

No, that one says "I'll be back."

haha

@Matt he said it some time ago, so...

I try to treat everything the same.

@tb do you like lasagne? or you are hungry? or both? or neither?

@DylanMoreland, I still haven't quite been able to resolve the issue you raised with my post

The one time I think it might help to make a comment is when the OP is just shotgunning questions that they would probably know how to do if they took the time to understand the first.

@Ilya Both... Matt's described the recipe on several occasions, so I'm curious..

I think it just remains to show that the kernel is (0)

@QED I still worry about whether it's defined.

Maybe I'm not seeing things clearly.

what's defined?

@tb I'll have to break through Leiden's inhabitants today to find some fastfood place for a dinner

they are already celebrating Christmas, quite messy on the streets (

@Ilya You're in Leiden! Do you see Lenstra around?

@Ilya Yeah, I'm glad the chaos will be over in a few days. I assume you will be celebrating on January 6th?

They celebrate xmas on the streets in NL? I picture a crowd of bonged out pot heads. : )

@DylanMoreland Which one? :D

@tb we don't celebrate xmas much, the New Year is the main holiday

What's on January 6th?

I have a question about this diagonalisability question. Shouldn't the OP actually check that the minimal polynomial is $x^3-1$ over the given field?

@Matt I will make a couple of photos for you. on 25-th and 1st as well

@Matt ay-ay-ay

Nice, thank you : )

@Ilya ?

What is it that might not be defined?

Oh noes. Now upstairs' kid started to practice the flute again. I think this means it's time for me to practice the guitar. BB soonish.

@Matt Oh, now it's January 7th of course. Eastern orthodox churches still use the Julian calendar for determining Christmas.

I might have defined it here math.stackexchange.com/a/93584/16697

@DylanMoreland he is the director of CWI and famous mathematician as wiki says, but I've never heard about him before

@tb I see. Although I might not remember for very long : )

@Matt I will send you a reminder ]:->

: D

@tb: could you tell me, what does the function (which is unique) $f:\emptyset\to A$ mean?

@Ilya the empty function. Usually you identify functions $X \to Y$ with their graphs in $X \times Y$. Now think about what $\emptyset \times A$ is.

@tb sure, I had to think about $f$ as a relation, i.e. in that case $f=\emptyset$ regardless of $A$, right?

yes, because $\emptyset \times A = \emptyset$.

@tb e.g. in $\mathrm{Rel}$ the initial object is again $\emptyset$. What's the terminal object then? Singletons are not terminal then

yes. and what you asked amounts to saying that $\emptyset$ is the initial object in the category of sets.

everybody has gone?

I'm here

We just got a visitor. So: BBL (probably)

@QED Sorry, got interrupted. Will take a look.

Thank you

@Matt bye, enjoy your lasagne!

@Matt probabilistic statements fits me, bye )

@tb so may I ask you what are the terminal objects for $\mathrm{Rel}$?

Bye Matt. Mmm, lasagne...

@Ilya, Why don't you prove that terminal objects are unique

@QED yes, sorry. The question should be - what is the terminal object for the category of relations

my guess is that it is $\emptyset$ again since any relation is just a subset of the product

but I'm not sure if my argument is correct

Then see if it fits the definition of terminal object

Can someone explain this answer in the cyclotomic polynomial question? How does the $x \to 2x$ substitution help?

there should be a unique relation $R:A\to\emptyset$ for any set $A$. I don't know if an empty set is a relation on $A,\emptyset$

What is a relation?

Do you have a definition for that?

@QED yes, it's a subset of $A\times\emptyset$

yes so {} is the only subset of A x {}

indeed

"I ask question, dont respond with question please." =/

Pfft...

Today hasn't been too kind to me somehow. :)

@Ilya, In fact A x {} = {}

@Srivatsan heh, qiorjlxmf has quite some chutzpah!

@Ilya I see that you've figured it out now.

It's easy for unregistered anonymous users to have chutzpah! :)

@tb well, I'm like a wine - you left me for a while and find me better than I was

Speaking of wine... where's Jonas?

@tb :) I think I will just be out for some time. See you all.

@JM Perhaps he is not in a position to speak right now... =)

@Srivatsan I see what you did there... :D

@JM dunno, I am almost the only one who still comes to our temple of knowledge these days

@tb it's done! he introduced generalized objects and now the equality of arrows becomes more clear. though your explanation helped

@JM Is the cosine integral and Euler-Mascheroni connection still eluding you?

@Srivatsan The solution I've found thus far requires a detour through the exponential integral, and is thus a bit long to set up properly. Do you have a shorter solution?

@JM Nope. I didn't think about it much. I was trying to cheat by searching for the answer for a few minutes, but I didn't land a good reference either.

I'm drawing a blank on a shorter route, so please tell me if you find anything. :)

wake me up please when Theo will return (

@JM What's your approach?

@Ilya here I am

@Ilya If I may ask: why this sudden interest in category theory?

@Srivatsan I always had such interest - and occasionally I need to read one PhD thesis which uses it a lot. So I take my chance and don't hesitate to [and so on]

OK fine. That answered a question that I didn't ask as well, thanks... :)

@Srivatsan which one? )

@Ilya Is this some course or self-study or something else?

@tb There are two posets $A = \{a\leq b\leq c\}$ and $X = \{x\leq y,x\leq z\}$ and in the book it's said that there are $5$ arrows $\mathbf{2}\to A$ and $6$ for $\mathbf 2\to X$, $\mathbf{2}$ being a Boolean algebra $\{0,1\}$. I've found only 3 for $A$

Can you list the ones you found?

@tb :D I had a Skype call - but the window with constant maps popped up ;)

well, with constant maps it will be $6$ for $A$: $0\to a,1\to a$, $0\to b,1\to b$, $0\to c,1\to c$ and $0\to a,1\to b$, $0\to a$, $1\to c$, $0\to b,1\to c$

weird, more compact: $(a,a),(b,b),(c,c),(a,b),(a,c),(b,c)$

well, that's what I got, too.

but I get only $5$ for $X$: 3 constants and (x,y) and (x,z)

@tb $X$ has $5$ since the map $(y,z)$ is not monotone, right?

I'd say so, yes :)

@Srivatsan Well, Euler-Mascheroni is easily made to pop up by manipulating the exponential integral. The cosine integral and the exponential integral have a relationship that is analogous to the cosine and the exponential function. Do a few cancellations, and out pops Euler-Mascheroni.

@JM Is the "few cancelations" a few too many? Why are you not happy with this approach?

Are we enough 3k+ users to close and reopen this one?

@Srivatsan Because I still have to explain the exponential integral. Also, I'm hoping for a solution that can stay on the real line instead of needing a complex plane detour...

@JM Um, ok.

@tb closed. how should it be reopened now and what's the reason?

@Ilya robjohn's comment?

So we have four of us here: 3k+. Do we need five?

@tb do you want to protect it from eventually closure?

Two more paramedics to revive the question

@JM 1 is left

@Ilya well, I don't like a Damokles sword dangling over a question. One might miss it when it's closed later on, that's why. It won't be long before the fifth vote is cast.

@robjohn would you make the final and very symbolic reopening of this?

@Ilya I think he's still with his dog.

@JM I guess - but at least he will read it when he will come back

@tb: after reading CT 'epic fail' becomes ambiguous

@Ilya epic fail?

doesn't make sense to me

@tb epic fail does not make sense to you at all or in the terms of CT?

both, to be honest.

(clearly, 1st implies both)

do you mean that you've never heard such phrase?

Yes - well I could guess what might be meant (and urban dictionary confirmed my suspicion). I always thought "fail" was only a verb.

@tb well, that's not the weirdest verb that happened to be a noun

tb: Have you seen the fail blog? It should be the first hit on google. [I don't quite recommend it, but it is funny at times.]

I will leave in a couple of minutes, so have a nice weekend and Merry Xmas to you guys and everybody

Same to you, Ilya, thanks! Find a decent fast food place!

@Ilya Wish you the same, Ilya.

@Srivatsan yes, I briefly looked at it, thanks.

@tb I'll certainly try :-p

That was my introduction to fail used as a noun, I think.

@tb: before I leave, he writes that if $f$ has a left inverse $g$ then $f$ is monic and $g$ is epic

then he introduces split mono being an arrow with a left inverse

@Ilya yes: If $fh = fk$ then $h = gfh = gfk =k$.

so, does it mean that each split mono is mono but not each mono is split mono?

I guess it's because if $f:A\to B$ then there is no necessary $g:B\to A$ at all and still $f$ can be mono, right?

exactly.

@Srivatsan It's not proper English, but it does get thrown around a lot on the Internet. So, "Internet English"... :)

@JM thanks for this :)

@tb brrr. ok, let's see how much of this will stay in my head after the weekend.

@Ilya well, for example, there's no field homomorphism $\mathbb{C} \to \mathbb{R}$

(but one in the other direction)

but there is one in the other direction, you mean?

sure, $\mathbb{R} \subset \mathbb{C}$, isn't it?

and field homomorphisms are always monic

I was referring to you sentence only - didn't get it

well, for the fields I know just the definition - but I guess that homomorphism have to preserve all the operations

hm... seems that there is no indeed such map

I leave that for you as a Christmas exercise.

nonono! take it back :D

you can use topology can't you?

what do you mean?

the (hypothetical) field homomorphism obviously fixes Q, and by continuity fixes R

so you are left with nowhere for C\R to go: the map can't be injective so it can't exist

thanks, seems that you have broken Theo's plan to make me busy this weekend

you'd still need to argue why your homomorphism is continuous.

bye )

bye, Ilya

I don't know how to get that it must be continuous

it's "obvious".

maybe we should forget about continuity and look at cauchy sequences instead

I think that proves R is fixed right away

I suggest to think about where $i$ can be mapped

that's even better, skips all the side tracking I was doing and gets right to the point

@JM Just for the record: the partner complained last week that I didn't dress nicely enough and should go clothes shopping.

@Matt Bah, expectations... :D

Anyway, I'm sleepy now. See you all later!

see you, JM.

@JM Good night!

Wow, you guys are a chatty bunch. Most of the SE chats are empty most of the time.

Does statistical physics and field theory contain beautiful mathematics?

@Matt Huh? Not nicely enough? What do you wear then?

What is field theory?

Quantum field theory, statistical field theory.

I know the very basics, so I wonder if it could use someone that knows harmonic analysis and PDE 8-).

@Srivatsan It is a subset of set theory.

You have a set, and some operations on that set; which are also sets.

But it is way less rudimentary.

@Srivatsan Sorry for the noise in your answer, but I had a chuckle that Jonas and me posted more or less verbatim the same comment some seconds after each other.

@JonasTeuwen But it's also rude.

@tb No problem at all, tb. =)

I met Ilijas Farah today.

@tb In fact, for some weird reason, my mind was so fixed on that upper bound that I forgot all about the lower bound :)

He gave a nice lecture about set theory and C*-algebra. Some use of infinitary combinatorics to construct counterexamples (only mentioned, though).

It was nice.

@Srivatsan I saw that. :) I added a comment using the Pythagorean theorem because that's more geometric, I think.

@tb You should write it as an answer, I think. It's certainly nicer.

I'm going for a beer. I hope you all suffer from excellent whisky and whatnot when I'm gone.

@Srivatsan well, it's in the comment, now :)

@tb But when I read the question, I assumed that the OP was speaking about the geometric method when he said something about complicated methods =)

I can't think of a complicated way to prove this, actually.

@AsafKaragila No, I had that yesterday... Today I'm behaving well, because I'm at my mom's place :) Enjoy your beer!

You don't solve integrals, you evaluate them! =)

@Srivatsan $\rm {S}{r}{i}{v}{a}{t}{s}{a}{n}{\ }{H}{a}{r}{d}{y}{?}$

@tb You must went through a lot of effort for those many braces =)

Anyway, I will be out for some time. Bye

Bye