12:25 AM
@BalarkaSen I AM DONE EXAMS.

12:47 AM
I have to Tex this assignment for tomorrow but I also want to sleep...

1:45 AM
You're not going to believe what I found
A book

2:03 AM
Israeli Elections 3: For Real This Time is officially scheduled for March 2, 2020
(This is after two elections in 2019, both of which failed to produce a government)

hey, Akiva
is it like two parties both having almost enough to win the elections, or is it a "we can't really have a coalition of $n$ parties because $n$ is too large" thing

3:07 AM
What function is less than y=x but the closest function to y=x?
in computer science one might call such a function a sub-linear function
it must be continuous

3 hours later…
6:13 AM
@Ultradark There is no such function. y=x^a grows slower than y=x for a<1 and a can be made as close to 1 as you want
Unrelated: I'm pretty sure I saw a comment on MathOverflow about how there are probability distributions on the integers where the probability of choosing any integer is zero, but (IIRC) the proof is nonconstructive. Am I imagining things or is this real?

\o @D.ZackGarza Welcome :-)

6:49 AM
@anakhro Woo nice
Up for Eliashberg-Mishachev?

@D.ZackGarza i could swear we know each other from... somewhere ;)

@EsolangingFruit I don't follow. If probability of choosing every integer is zero, by countable additivity probability of $\Bbb N$ is zero.
Maybe you only want finite-additivity? Then it's not as clear to me anymore
It should be possible for sure
Choose some ultrafilter on $\Bbb N$ and assign $A$ to be of probability $1$ if upto finitely many elements $A$ is in the ultrafilter, something like that. Every finite set has probability zero because $\emptyset$ is not in the ultrafilter.
by "upto finitely any elements" I mean all the sets obtained from $A$ by adding/removing finitely any elements is contained in the ultrafilter

8:07 AM
Hi @ÍgjøgnumMeg

hiya @BalarkaSen

Recommend good black metal dawg
I have run out of albums

(God Seed - Live at Wacken)
rofl

this is their best album though probably:

8:10 AM
Diotima is great

very descriptive track titles

@ÍgjøgnumMeg lmao

Dornenreich - Flammenmensch is good

Let me check this out. @Soham I'll have a look at those albums as well

ypm > diotima in my estimation, just by a little bit

8:11 AM
Wolves in the Throne Room - Black Cascade is a BANGER

Oh love that album

do you like tech-death
like, say, obscura

havent heard of em

ex-necrophagist members

8:13 AM
and death

Behexen - Rituale Satanum is cool af
makes me wanna also be on my way to kill Euronymous

Hahahah

lmao

Huh I didn't know Death was tech-death

idk about the tech- tbh but then again my musical diet was different from the people who listened to Death albums when they came out so we have different standards for "tech"
this is the good shit
(not really)

8:16 AM
better than pop music

now now
no pop shaming pls

can't agree
with the pop shaming
i spent all morning singing along to lana del rey and the national

youre not my friend anymore
2

lol

if you like krallice i recommend ulcerate too

8:17 AM
@ÍgjøgnumMeg That was good!

Behexen?

Dornenreich

Ah yeah! Flammenmensch is a cool af song

tbh i should just go through the catalogue of all prophecy label associates

truth
Helrunar are cool as well
Til Jarðar is a good song

8:21 AM
on it
i like trv kvlt names

hahaha sem
Right, I better get up and chug a coffee and a cigarette and then do some maths

same

no smonk pls

smonk gud

o u have asthma gg

8:22 AM
I'm still nervously awaiting a reply from that prof rofl

on the topic of kvlt throwback to when i tried to tell someone to listen to a track from Bergtatt but couldn't actually pronounce the name and just said "google ulver capitel 3"

lol
yeah its just a fairytale in five chapters so i dont have to remember track names
thinksmart

not hard

cu-rispyyy

8:26 AM
ah this helrunar song is good
i like the vocals

yeboi

Baldr Ok Íss

Farewell by Hermóðr is good
(I will pipe black metal songs into this chat for the rest of time if you want)

do you have anything a bit more atmospheric and less "raspy"

Imma listen to Baldr Ok Íss for now

8:29 AM
oh this last one looks like it

have you listened to drudkh

if ya want something blackgazey then Lantlôs are incredible :P

deafhaven's new albums have gone down the slope :(
Roads To Judah, which came way before Sunbather, is a surprisingly good black metal album
no blackgaze tho

boi
Epitaph of Penitence by Mist of Misery #DSBM

OH fuck 2nd track from Baldr Ok Iss is A+
@ÍgjøgnumMeg hahah no DSBM
the second track has some orchestral stuff which reminds me of this Finnish band called Circle
those guys are crazy

8:35 AM
i recall looking up the lyrics to some (i was unaware that it was dsbm) album as a teenager and it was just a very detailed description of how the singer wanted to off himself
hmm this hermodr album, me like

@ÍgjøgnumMeg Check this out
its very weird metal btw

Will do hehe

doesn't strike me as weird but i like it a lot
am reminded of some of the ihsahn solo work

i mean idk wtf it is lol

it's like leprous
i wouldn't call it metal
actually yeah strike the ihsahn comparison this is very leprous indeed

8:43 AM
math Hilbert ramification theory is cool af

im not familiar with leprous

the featuring part is completely coincidental

Circle is definitely a metal band though
also i think the lyrics are nonsense
they sing in some made up language

sigur ros for people who blew their throats out screaming

aw yeah sigur ros is A+

8:45 AM
lantlos sounds sick

@Soham preach

@SohamChowdhury not sure if i find the similarity but this is good

leprous?

they have stuff in the same album (iirc, been a while) that borders on modern jazz
and weird choir parts

8:49 AM
i see the jazz elements yeah
o shit Drudhk has an album with Winterfylleth

lantlos track 2 on neon reminds me of ne obliviscaris
actually maybe this is why i like this band

That's probably one of my top 3 lantlos sogs
songs*

is that pulse/surreal

These nights were ours

8:53 AM
ah

I think Pulse/Surreal is my fave, then Eribo I collect the stars, and then these nights were ours
rofl
fangirling

didn't expect the autocomplete
pleasantly surprised

wuts this compiler bro

text editor
emacs

strange

8:59 AM
@Balarka during your ANT studies, have you come across decomposition of primes in terms of Galois theory yet?
(or maybe not in terms of Galois theory but .. with some Galois theory rofl)

pulse/surreal is an absolute banger
2
ok i have to go now tho

@Soham TRUTH

i know very little ANT, you mean Gal acting on primes lying above a certain prime?

metal? with bass?
it's more likely than you think

@Balarka yeah, as in, one studies the group $Z_\mathfrak{P} := \lbrace \sigma \in \operatorname{Gal}(L/K) : \sigma(\mathfrak{P}) = \mathfrak{P}\rbrace$ (i.e. the stabiliser of the prime under the action of the Galois group)
and this is obv a subgroup of the Galois group (as a stabiliser) so there's a fixed field, and it turns out ALL of the splitting of a prime occurs between $L$ and this fixed field, and nothing happens between that fixed field and $K$
There's also a surjection $Z_\mathfrak{P} \to \operatorname{Gal}(\Bbb F_\mathfrak{P}/\Bbb F_\mathfrak{p})$ (where the automorphisms of $\Bbb F_\mathfrak{P}$ are induced by those of $L/K$)
some cool theory lol

9:05 AM
pretty dope. I wonder if there's a geometric way to understand this action

Probably
the action is transitive so you get all of the primes over $\mathfrak{p}$ by hitting a big prime with a sigma
which leads to the fundamental equation collapsing to ramificaiton index times inertial degree times splitting number = degree of extension

If $L/K$ is an extension then we have a corresponding map $\text{Spec} O_L \to \text{Spec} O_K$ and fiber over the prime $\mathfrak{p}$ is $\text{Spec}(O_L \otimes \kappa(\mathfrak{p}))$ where $\kappa(\mathfrak{p})$ is the residue field at $\mathfrak{p}$
$\text{Gal}(L/K)$ acts on $\text{Spec}(L \otimes_K K^{alg})$, the geometric fiber of $\text{Spec} K \to \text{Spec} L$, so maybe there's a way to push this forward to an action on $\text{Spec}(O_L \otimes_{O_K} \kappa(\mathfrak{p}))$
But maybe this is not helpful

idk I haven't done anything relating to the spectra of these rings lol
(well obv I have but I mean..)

yeah fair

Only at individual primes rather than taking the whole spectrum and doing shit rofl
Consider a number field $K$ with infinite $2$-class field tower and $m=2$. — franz lemmermeyer 16 hours ago
wat
2

Let $K$ be a number field and let $\mathfrak{a}$ be a non-zero ideal of $\mathcal{O}_K$ such that $\mathfrak{a}^m = (a)$ for some $a \in \mathcal{O}_K$. Does there exist a finite extension $L/K$ such that $\mathfrak{a}\mathcal{O}_L$ is principal? The answer is yes; one can take $L = K(\sqr... this is what I meant to link lol 9:11 AM franz is one of the few big shots who hang around in MSE often Yep, Lubin is often around too. I remember typing up an answer for an ANT question and then he pops up with an answer and I just deleted everything ah yeah Lubin as well 9:26 AM I defined a new function I wonder if it’s a useful one @ÍgjøgnumMeg I have never really understood this story to be honest. I understand that if$q \subset O_L$is a prime lying above$p \subset O_K$then since$pO_L$is a product of a bunch of stuff including$q$, applying$\sigma \in \text{Gal}(L/K)$we get$pO_L = \sigma(pO_L)$is a product of a bunch of stuff including$\sigma(q)$and by unique factorization of ideals you get$\sigma(q)$also lies above$p$, so$\text{Gal}(L/K)$acts naturally on the primes lying above$p$. Yep, what that says is that the ramification indices of primes above$p$in your extension are all the same in the Galois case and all of the inertial degrees Makes sense. Which is pretty cool because your fundamental equation becomes much easier to apply in the Galois case lol I am having some difficulty, but having difficulty pinpointing what that difficulty is 9:39 AM fair, we're actually just reaching this point in our lectures atm, I've seen it before but this is my first proper treatment of it so if I can help I will but if not then I'll have learned smth too hahaha No I definitely think you can help, but let me see if I can formulate a question first Look at the residue field$\kappa(p)$of$p$in$O_K$and take a prime lying above, say$q$, and look at the corresponding residue field$\kappa(q)$in$O_L$. What can we say about the extension$\kappa(q)/\kappa(p)$? How does$\text{Gal}(L/K)$relate to$\text{Gal}(\kappa(q)/\kappa(p))$? well the groups$Z_q$that I described above are subgroups of$\operatorname{Gal}(L/K)$and there's a surjection$Z_q \to \operatorname{Gal}(\kappa(q)/\kappa(p))$(note that this extension of residue fields is defo Galois in the number field case because your residue class field extensions are just finite field extensions) The question you can ask is what happens at the extremes? I.e. when$Z_q$is the whole Galois group? Oh ok, that surjection is very weird @ÍgjøgnumMeg Yeah that seems like an interesting question If$Z_q$is the entire Galois group then$q$is the only prime lying above$p$Let me keep an example in mind maybe.$\Bbb Z[i]$over$\Bbb Z$with$(2)$below and$(1 + i)$above Oh ok then I picked the right example 9:45 AM lol yeah Huh. So$\kappa(q)/\kappa(p)$is$\Bbb F_2/\Bbb F_2$? Right, of course Wait, but Gal(Q(i)/Q) always fixes the ideal (1 + i) Well by the fundamental equation (and the fact that quadratic extensions are Galois) you get$efg = 2$and$(2)$is ramified so$2\cdot f \cdot g = 2$so$fg = 1f$is the degree of$\kappa(q)/\kappa(p)$If Gal(Q(i)/Q) fixes (1 + i) completely doesn't that mean Z_(1+i) = Gal(Q(i)/Q) = Z_2 Whereas Gal(k(q)/k(p)) = Gal(F_2/F_2) is trivial which says Z_q is not the entire Gal(k(q)/k(p)) in this case, so something is off, because you said that would be the case Hmmm Oh maybe not vice versa 9:50 AM @BalarkaSen check out the live at rockefeller hall You said if Z_q is entire Gal(k(q)/k(p)) then q is the only prime lying above p Ohhh So maybe ramification is what prevents the converse to hold? Z_q is a subgroup of Gal(Q(i)/Q), not of Gal(k(q)/k(p)) Oh you when said "entire Galois group" you meant Z_q is Gal(L/K) entirely? 9:52 AM yeah lol i.e. your prime above p is fixed by everything Oh ok I thought we were thinking about when the surjection Z_q -> Gal(k(q)/k(p)) is going to be an isomorphism Got it That of course makes sense If your prime is unramified then you actually get an isomorphism Z_q -> Gal(k(q)/k(p)) Aha The kernel of that map is called the inertia group of q, that guy also has a fixed field that sits between the fixed field of the decomposition group and the extension$L$so you've got like$L/L^{T_\mathfrak{P}}/L^{Z_\mathfrak{P}}/K$all of the splitting of$\mathfrak{p}$comes from$L^{Z_\mathfrak{P}}/K$, all of the inertial degree comes from$L^{T_\mathfrak{P}}/L^{Z_\mathfrak{P}}$, and all of the ramification comes from$L/L^{T_\mathfrak{P}}$Weird stuff. How do you get this surjection map in the first place?$Z_q$is the set of Galois automorphisms of$L$which fixes$q$, so I suppose it naturally acts on the local ring$(\mathcal{O}_L)_q$, hence on the residue field 9:58 AM$T_\mathfrak{P} := \ker(Z_\mathfrak{P} \to \operatorname{Gal}(\kappa(\mathfrak{P})/\kappa(\mathfrak{p}))$btw Which gives the map$Z_q \to \text{Gal}(k(q)/k(p))$It's non-obvious why it's even surjective Yeah that's very strange to me It's like saying every germ of an automorphism at$q$can be extended to a global automorphism You start by reducing to the case where$q$is the only prime above$p$(currently reading from my bachelor's dissertation rofl) You actually get an isomorphism$\mathcal{O}_K/q \cong \mathcal{O}_K^{Z_q}/q^\prime$where$q^\prime = q \cap \mathcal{O}_K^{Z_q}$rofl which is why you can do this reduction$O_L/q$you meant? 10:02 AM yeah sorry No sorry I mean O_K/q my notation is all screwed up because I'm reading out of my dissertation lol Yeah no worries.$L/K$is my extension,$p$is in$O_K$and$q$is in$O_L$lying above$p$for me bcur.org/journals/index.php/TPSS/article/view/671/593 I refer you to theorem 5.1 in here Thanks :D shameless self-plug Proof by reference, you're becoming a real mathematician 10:06 AM hahaha Note that I've made some stupid side notes in that proof (for instance that Aut(K^Z_p/K) is not Galois, which isn't true) also the font etc. suck because the journal editor said I had to make it look like the others which is lame, but I fear something that occurs often I like the summary of the proof before it starts. Quite clearly written Thanks man! Hm, but I am annoyed, because it feels what I would do is try to "analytically continue" the automorphism of$\kappa(q)$to all of$O_L$. Maybe this is what the proof does, but hidden in some retarded algebraic language Which means I have to struggle through the algebra to see it hahaha no offense meant to you of course, i am cursing dedekind or whoever came up with this shit has to be dedekind right 10:13 AM Ye none taken Probably dedickined No wait This is Hilbert theory lol Of course The man who claimed$\Bbb{RP}^2$does not immerse in$\Bbb R^3$heh There's some really cool magic facts to do with Hilbert class fields linking to modular forms in some surprising way which I regularly get a boner over tmi, man, tmi 10:14 AM hahaha is this cft or regular algebraic nt -_- regular algebraic nt, but on the cusp of cft I guess pun tsk tsk you're in great form today pun pun 10:16 AM heh can't think of another these puns are a class apart christ hecke* don't swear hahaha, this is vile @ÍgjøgnumMeg do you have any experience with Murty's problem books? 10:23 AM I have a copy of problems in modular forms lol what are the prerequisites for that? a first course in complex analysis I believe looks like just complex analysis ah Indeed i really like the look of all three tbh, i feel like i should pick one and work through it i've had enough of convincing myself i understand things i don't probably the alg nt 10:24 AM yeah it's very cool, some of the deeper theorems are proven but the majority of stuff is left as exercises to the reader, with solutions in the back paper copies are outrageously expensive here sadly so i don't get the full ramanujan immersion experience sad lol i'm enjoying working through amann-escher's first analysis volume atm OK. If$q$is the unique prime in$O_K$lying above$p$in$O_L$then$Z_q = \text{Gal}(K/L)$. Choose a primitive element$\alpha$for$\kappa(q)$over$\kappa(p)$, and essentially observe$\text{Gal}(L/K)$acts transitively on the roots of the minimal polynomial for$\alpha$over$K$. yis 10:28 AM Which gives all the automorphisms for$\text{Gal}(\kappa(q)/\kappa(p))$as modulo$q$that minimal polynomial is the minimal polynomial for$\alpha$over$\kappa(q)$Alright I have a dumb computational question now:$(p, \frac{1}{2}\sqrt{p})$is prime in$\mathcal{O} := \Bbb Z[\frac{1 + \sqrt{p}}{2}]$because $$\mathcal{O}/(p, \frac{1}{2}\sqrt{p}) \cong \Bbb Z[X]/(X^2 - X + \frac{1 - p}{4}, p, \frac{1}{2}X) \cong \Bbb F_p[X]/(X^2 - X + \frac{1}{4}, \frac{1}{2}X) \cong \Bbb F_p$$ I think$X^2 - X + \frac{1}{4} \equiv (X-\frac{1}{2})^2 \bmod p$which gives the last isomorphism I think.. lol I just ordered a copy of Steen Seebach in the end, I found one for 12.60€! Shouldn't$(p, \sqrt{p}/2)$be$(p, X - 1/2)$Possibly Well$(p, \omega - c)$is the ideal above$p$where$c$is the constant term in the factorsation and$\omega$is a generator for the ring of integers lol But the computation is fine I think 10:36 AM anyway, that ideal is correct I think because $$(p, \frac{1}{2}\sqrt{p})^2 = (p)(p, \frac{1}{2}\sqrt{p}, \frac{1}{4})$$ and$(p, \frac{1}{4}) = 1$(where by$1/4$I mean a lift of$4^{-1} \bmod p$to$\Bbb Z$) Yeah it does lie above$(p)$kewl okay well that's the hard part, the exercise is to factor$p$over$\Bbb Q(\sqrt{p})$and$\Bbb Q(\sqrt{-p})$lol Ah Blergh why am I doing algebraic number theory now Why isn't anyone talking about topology because topology is for nerds time to TeX my exercises and shove the responsibility for handing them in to my colleagues because I'm too lazy to go to uni isnt it winter break 10:43 AM Nah that starts on the 20th I think @BalarkaSen What kind of topology do you want to talk about? @ÍgjøgnumMeg I see @Alessandro Not too crazy like your usual shtick Oh I had a question for you @ÍgjøgnumMeg officially we have classes on the 23rd as well But all of the professors I'm taking classes from decided they won't do one on the 23rd lol Do you think this works? @BalarkaSen Should I be worried? 10:44 AM Nah @Alessandro fair, we finish on the 20th lol and I have no idea when we return There's a correspondence between ultrafilters on$X$and finitely additive two valued measures on$X$That's just by setting 0 or 1 depending on if your set is in the ultrafilter or not right 10:47 AM I don't think so, @Martin. The question was if you can find a finitely additive measure on N which takes 0 on singletons To get a real ($\sigma$-additive) measure you need what's called an$\omega$-closed filter, but it's a theorem of$\mathsf{ZFC}$that there is no$\omega$-closed filter on$\Bbb N$and it's independent of$\mathsf{ZFC}$whether there is an$\omega$-closed filter on any set (If there is one then there's a measurable cardinal) @BalarkaSen Well, I think this is exactly the same thing - to get zero on singletons we take a free ultrafilter. My idea is to define$\mu(A) = 1$if you take away or throw in finitely many points on$A$it always stays in the ultrafilter @MartinSleziak This would be an ultrafilter not containing finite subsets? Free ultrafilters are precisely the ultrafilters which contain all cofinite sets. So for any finite set$A\in\mathcal U\Leftrightarrow A\setminus F\mathcal U$. To get measure zero on the singletons you just need an ultrafilter containing no singletons (hence no finite sets either), a nonprincipal ultrafilter 10:49 AM Why does it exist Zorn's lemma applied to filters extending the cofinite filter Definition of free ultrafilter that I am used to is that$\bigcap\mathcal U=\emptyset$. Hm, I see. From Zorn's lemma you know that every system of sets with finite intersection property is contained in an ultrafilter. Existence of free ultrafilters is strictly in between ZF and ZFC, but it can't be proved in ZF alone 10:50 AM It is equivalent to BPI and some other frequent results, right? If you take an ultrafilter and throw away all the subsets$A \subset \mathcal{U}$such that there is some finite subset$P$and$Q$for which$(A \setminus P) \cup Q \in \mathcal{U}$, does that give you an ultrafilter? I think so. It's called the ultrafilter lemma sometimes Yes, ultrafilter lemma is the name used on WP: en.wikipedia.org/wiki/… You can't remove elements from an ultrafilter and still get an ultrafilter, you lose maximality 10:52 AM Oh fair enough. @BalarkaSen It seems to me that what you describe here is that$B\triangle A$is finite. Where$B=(A\setminus P)\cup Q$. Yeah This is often denoted as$A=^*B$. And if you work with free ultrafilter,$A\in\mathcal U$,$A=^*B\RightarrowB\in\mathcal U$. Changing only "small" set cannot influence whether or not the set belong to the ultrafilter. So I am wondering what exactly goes wrong in the construction. Choose an ultrafilter$\mathcal{U}$, define$\mu(A) = 1$if$A \in \mathcal{U}$and whenever$B =^* A$,$B \in \mathcal{U}$as well, using your notation. And you can say exactly the same thing if you take the dual ideal instead of finite sets. (I.e., the same is true if you say$X\setminus P,X\setminus Q\in\mathcal U$instead of saying that$P$,$Q$are finite. Ok, I'm listening. (Sorry for the digression.) We have$\mu(A)=1$for the sets from ultrafilter. What next? 10:57 AM Ah no what I defined is$\mu(A) = 1$if$B \in \mathcal{U}$whenever$B =^* A$. Isn't this equivalent to simply saying that$\mu(A)=1$for$A\in\mathcal U$? (Still assuming that we work with a free ultrafilter.) But that's the same as$\mu(A)=1$iff$A\in\mathcal{U}\$ by what Martin was saying, isn't it?