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12:00 AM
@IanMateus You're thinking too hard, I suppose. $\Bbb C$ is complete, so Cauchy sequences are convergent.
I hate how in English any noun can be made into a verb.
 
@Karl What do you mean by that? I'd assumed you already knew the finite stuff
 
@PedroTamaroff ah, verbing?
 
@PedroTamaroff I am not familiar with such concepts, I'll read them. Convergence is quite "obvious" geometrically, but I couldn't prove it.
 
@seaturtles I guess that is the name?
 
twas a pun
or irony, or whatever
 
12:03 AM
@seaturtles OH, YOU!
 
@Karl I guess I've just gone and lost all respect for you ;(
 
oh, me
 
@Mike It's the infinite case that I'm learning about
 
Ah, cool
respect regained
 
@PedroTamaroff thank you for the keywords, I found some proofs online :-)
 
12:06 AM
@IanMateus CAWL.
 
@Ian You need to know about completeness for Cauchy sequences... because such series need not converge if your (wherever you're doing math) isn't complete!
 
Some really cool shit there, @Mike, though I haven't done much more than the Galois Descent theorem so far.
 
@Mike You don't like Galois theory? :)
 
@AlexYoucis I peeked a little about $I$-adic topology for a ring.
@seaturtles I think anon is the meaner side of your personality?
 
@AlexYoucis No, I was just giving him crap for not already knowing the basics. But he does. So I rescinded the crap.
 
12:09 AM
Like Charlie and Hank
@seaturtles
 
shrugs
 
@Karl I don't know any of it. I probably should.
 
@PedroTamaroff You peeked at it? Is it like taboo or something in South America?
 
@seaturtles I can never engage you in conversation.
 
come at me bro
 
12:10 AM
@AlexYoucis Just stuff I might learn furhter on.
 
whats up?
 
@seaturtles i cannot decide what to read about
 
@PedroTamaroff Haha, yeah, it's important stuff. For sure.
 
@Pedro anon is his anime side
 
@seaturtles It turns out that the class group has another, more advanced meaning (re: your old question).
 
12:11 AM
@AlexYoucis The idea that $\mathfrak m^n$ are nbhds of $0$ became clearer when looking say at $k[[X]]$ and $(x)$.
 
But, it's kind of obnoxiously advanced, if not neat. For, if you change that definition into the context of ellpitic curves you get the Tate-Shafarevich group.
 
> more advanced
e gadd
 
@PedroTamaroff Sure. Whenever $\mathfrak{m}$ is the maximal ideal of a valuation ring, you can just understand it as the valuation theoretic completion
@seaturtles Lol, what does that mean?
 
@Alex Write about it.
 
12:13 AM
@Mike You write about it.
 
No knowledge, no time. :(
 
@seaturtles I know what egads means, I was wondering what that means haha
@Mike Lies, and misinformation.
 
well, my first exposure to the word was professor elvin
 
Or rather: no knowledge.
 
@seaturtles Melvin!
 
12:14 AM
elvin
 
@seaturtles Just googled that. You're trying to swap nostalgic video game lingo with the wrong hombre.
@seaturtles Never played it.
 
heh
 
he's a great professor
 
@seaturtles I know.
I can read!
 
@AlexYoucis so what are your hobbies
do you read fine literature
 
12:16 AM
@Mike Math, music, friends. The big three.
 
You should read fine literature too.
 
@Mike Fine literature eh? Talk to me about fine literature when you're done reading R.L. Stine son.
 
Rotman should count.
And Lang.
And Hungerford.
 
That may be the most offensive thing someone's ever said to me.
 
@Mike Word up.
 
12:20 AM
I thing I'm obligated to murder you to regain my honor.
 
WAT
R
U
TKING
BOUT
SON
 
@Pedro If Alex ends up dead, it was me
 
@Mike By all previous accounts, it seems as though I'd be able to incapacitate you with a B-level insult. I have no fear.
 
If liking R.L. Stine is a B-level vice, then
well uh
we have to kill pretty much everyone except people on their A game
 
@Mike I think I can see your university from all the way up here.
 
12:25 AM
@Alex Anyway, you should read Gaddis.
That message is intended for every Alex that gets pinged.
 
@AlexYoucis Dude.
This might probably be the slickest proof of existence of the cyclic decomposition of finite abelian groups.
@seaturtles @KarlKronenfeld @Mike It is due to Eugene Schenkman.
 
@Pedro You should read too.
Borges, say.
 
@Mike I read everyday.
 
12:41 AM
Do you read Borges every day?
 
@Mike No, I read maths.
 
Borges is great
 
Hi @ all. Here's a cool question for you. Give me a sequence $a_n\in\Bbb Z$ so that $\phi(a_n)/a_n \to 0$.
I wonder if our teenage number theorists know this off the top of their heads ...
 
Hi Ted
 
Hi @user127001
 
12:49 AM
no longer teenage :-(
 
never was a teenage number theorist :-(
 
Aw, poor antique anon! :)
 
@TedShifrin Well, $$\frac{\varphi(n)}n=\prod_{p\mid n}(1-p^{-1})$$
 
Good, @Karl :)
 
@Pedro you're not a teenager either, quiet
 
12:51 AM
@seaturtles snif
 
Yup @Pedro.
 
Why?
 
That could refer to half a million things, @Mike
 
Why what? @Mike
 
@Ted Your function is still multiplicative, so use primordials. and the question wa why should I find such a sequence?
 
12:55 AM
Primorials.
 
Just get out your primordial soup, everyone!
 
Cuz I asked my high school Spivak student the question.
 
@TedShifrin What did he answer?
 
He guessed $n!$.
 
@Pedro Phones.
 
12:56 AM
smacks @Mike harder
What are primorials?
Oh, I guessed.
 
All hail Euler.
Cause $1/\zeta(1)=0$ son.
 
That's still true, @Ted
Just ain't efficient.
 
Edmund Landau's thesis was 14 pages long.
I dream of such brevity.
 
I didn't see an easy proof for the fa torisl, but I didn't try hard.
 
@TedShifrin Now make him find $$\sum_{k\geqslant 1}\frac{\mu(k)}k$$
 
1:00 AM
No, don't.
 
But $$\prod_{p\;\;\rm prime}\left(1-\frac 1 p\right)$$
=)
@Mike
 
Nah. Our course is done and he's got his hands full for the final.
 
It's those nice stuff in ANT.
 
>nice stuff
>ANT
pick one
 
That's not nice.
 
1:02 AM
No number theory for me, @Fernando
 
@FernandoMartin implying there's only one ANT
 
@Ted did you see my link?
 
>Fernando
>Not bitching about what he's not interested in.
ANT is really lovely.
 
>implying it wasn't obvious
 
@PedroTamaroff: you got me
 
1:03 AM
Yeah, @Mike, people who do the PDE side of geometry care for sure.
 
Atiyah-McDonald is a damn good book
 
That's a lovely example.
 
@Ted Jeez. To me it just looks like a bunch of disgusting formulae.
 
Not my style of geometry...
 
1:05 AM
@Pedro: "4. Out of the impossibility of perpetual motion of second kind comes the
proof of the second law of thermodynamics."
???
(that's from the paper you linked)
 
wtf is a null space
 
I guess Landau was a peculiar lad.
@Mike Kernel.
 
You just don't appreciate the power of ANT @FernandoMartin
 
I sure as hell don't @Karl
 
Seriously @Mike?
 
1:07 AM
@KarlKronenfeld They can lift 10 times their own weight.
 
@Mike: even I know that one
 
LOL @Fernando
 
@Ted "Suppose that A is a C*-subalgebra of B(H) with trivial null space."
 
OK, not the null space I had in mind
 
B(H) is the algebra of bounded operators on a Hilbert space.
 
1:08 AM
Yeah, think it is.
So no common kernel.
 
Oh, I see.
 
@Karl: I don't appreciate the main problems in ANT to be honest. I don't know why RH is relevant, for instance.
I'm sure it's a very important problem, but I don't know why
 
@FernandoMartin I actually said something along the same lines when talking to @Pedro once.
 
@Fernando For its consequences, I think.
 
Yeah I know @Mike, I meant to say I don't know what those consequences are
I know that if it holds, then primes are "well-distributed" in some sense
or they behave as we would want to
 
1:12 AM
For example, if RH is true, the K-theory of the category of subsemigroups of the naturals is trivial
 
I know some of those words
 
I don't, I made that up
 
$K$-theory. The theory of $K$'s
 
I think it's more along the lines of PNT. PNT gives us a good asymptotic. But knowing RH would prove a far tighter one
 
@FernandoMartin That's not ANT.
 
1:17 AM
then what is a major problem in ANT?
 
drawing $\Bbb Z$
artistic number tbeory
 
@FernandoMartin It depends on how you define algebraic number theory. ABC is one, whether or not that is conjectural anymore is to be seen. Mordell's conjecture was a big one. The sections conjecture could loosely be defined to be number theory.
 
I was referring to analytic NT @AlexYoucis
 
@FernandoMartin Oh, haha. ANT is, in my very small world, almost universally defined to be algebraic number theory.
 
16 mins ago, by Karl Kronenfeld
@FernandoMartin implying there's only one ANT
 
1:19 AM
goddamnit
 
@KarlKronenfeld Scrolling up is for the feeble-minded.
:)
 
I know, I thought it was clear from context I was talking about analytic NT
 
context people
It was, Nando
 
@FernandoMartin Oh well haha.
 
What's another big problem in analytic NT besides RH?
 
1:20 AM
Waring's problem
 
@FernandoMartin Once again, I guess that depends on how loosely you define analytic number theory. If you mean things that involve analysis, how about BSD?
 
I heard once that number theory isn't really a field
It's just a collection of hard problems and people who work on them
and some of those problems and people are noninvertible
 
@Mike: ba dum tshhhh
 
@Mike That's reasonable. .
@Mike Is that because they are singularly genius?
 
@AlexYoucis: Do you work in algebraic NT?
 
1:24 AM
Does ${\rm Spec}(A)$ contains $A$?
 
no @EnjoysMath
 
@FernandoMartin I'm interested in arithmetic geometry, which is like algebraic geometry+algebraic number theory-numbers.
 
Alex works wherever he wants
 
@EnjoysMath I'm not even sure I know what that means haha.
 
@EnjoysMath that doesn't make sense
 
1:24 AM
Isn't $A$ a prime ideal in itself?
 
@EnjoysMath Is 1 a prime?
 
Read the definition - prime ideals are proper
 
By convention, no.
 
1:25 AM
oh ok now I see what you mean
 
Oh darn! That's wy $V(A) = \varnothing$.
 
@EnjoysMath Mhmm.
 
It's worth noting that by the same token, $0$ is neither a field nor an integral domain.
 
I thought you were asking if A was a subset
rather than an element
 
A ring that is a prime-transitive set @Mike. Hot topic, bro.
 
1:27 AM
@KarlKronenfeld What does that mean?
 
It doesn't.
 
@AlexYoucis making terms up--based on the real term transitive set.
 
Let $V(\mathfrak{a})$ be all elements of ${\rm Spec}(A)$ containing $\mathfrak{a}$. How does an arbitrary intersection of $V(\mathfrak{a}_i)$'s equal another $V(\mathfrak{a}')$ ?
 
@KarlKronenfeld You're going far kid--that's half the battle.
@EnjoysMath It's $V$ of the sum.
 
oh thx
 
1:29 AM
@EnjoysMath: exercise 1.15.iii.
 
wtf, how did you do that?
:|
 
Magic.
 
I'm studying J. Dieudonne and it referred to commutative algebra stuff, so was only glancing at Atiyah-McD's
 
@EnjoysMath What are you reading by Dieudonne?
 
TREATISE ON ANALYSIS II
 
1:32 AM
what of Dieudonne?
 
one of the dark green books
that come in a series of math encyclopedia
 
@EnjoysMath Interesting. I know the series, never heard of the book.
 
So $\mathfrak{p} \supset \mathfrak{a,b} \implies \mathfrak{p} \supset \mathfrak{a+b}$?
$\mathfrak{p}$ prime
 
is it not analysis?
 
@EnjoysMath that's true even on the level of additive groups
 
1:34 AM
yes, you don't need it to be prime for that to hold
 
@EnjoysMath Yes, the sum of ideals is supremum in the lattice of ideals of a ring.
 
It's a problem about topology on the set ${\rm Spec}(A)$ of a comm. ring $A$
quick proof?
 
$\frak p$ is closed under addition
 
@EnjoysMath Have you tried it?
 
1:34 AM
done
 
Done.
 
oh wow :D
!!!
 
See? Punctuation.
 
1:35 AM
c-c-c-done
 
lols
It seemed hard at first, but was actually a no-brainer
 
I'm a little slower, but I'm done too
 
@EnjoysMath what's the problem?
 
I haven't started.
 
It's a bunch of questions
 
1:36 AM
@KarlKronenfeld You a junior or a senior?
 
@AlexYoucis I may have told you once in the past. I am a self-learner.
 
I think all of us here are
 
@KarlKronenfeld Does that mean it's not your major, or you don't attend school at all?
 
I'm not
 
It takes so much studying outside of 'class' that we're all self-learners
 
1:38 AM
@AlexYoucis I don't attend school.
 
oh here we go
 
@KarlKronenfeld Then, why study math?
 
grabs popcorn
 
@EnjoysMath ?
 
puts cayenne on popcorn
 
1:39 AM
Karl is actually the leader of a shadowy spy group. This is his hobby
 
@AlexYoucis Does school in some way motivate you to study math?
 
Karl is secretly Bourbaki. All of them
 
@Ted delicious
 
@KarlKronenfeld It motivates me in the sense that I get feedback from those more knowledgeable than me, I get to talk to peers at length including doing study groups with them, and I have time to do it, in the sense that I am not also working another job.
 
cayenne can stop a heart attack
 
1:40 AM
Now you tell me, @EnjoysMath
 
@Ted Why, Were you trying to induce one?
 
carne asada tacos with green salsa and jalepenos
 
@KarlKronenfeld It's not that there is no enjoyment in the idea of studying math not in school, I would also probably do it, it's just a much more difficult endeavor. Both in terms of motivational factors and logistically.
 
No, I've had two robotic heart surgeries.
 
O_O
robots... operated on u...?
 
1:43 AM
surgeries of the robotic heart?
 
Yup :D
 
@TedShifrin Da Vinci?
 
Look up daVinci surgery
Yes, @AlexY
A human did my cancer surgery, which was a hell of a lot scarier
 
@AlexYoucis I am actually going to school in the fall for precisely the first two reasons you gave, so I am not ardently against school. That said, I highly suspect my appreciation for math and my motivation to do math will remain unchanged even in that environment.
 
@KarlKronenfeld Yeah, it's not about local concern, it's easy to do math if you know you'll eventually be around others to chat with, it's just hard globally. Math is the best, of course, but it really is, in my humble opinion, a team sport.
 
1:57 AM
@Karl You intend to remain anonymous?
 
@Mike For the time being.
@AlexYoucis Agreed. It took a while, but yeah I recently realized how true that is.
 
Cool. (I'm not trying to pressure either way. I'm just curious.)
 

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