03:00 - 18:0018:00 - 23:00

3:11 AM
My god. This happens again. I spent much time typing answer to this question math.stackexchange.com/questions/3470455/… and just before finish, it is deleted...

@WhatsUp when you try your best but you don't succeed

Ja... quite depressing.

3:33 AM
@WhatsUp just get in the habit of opening notepad or whatever it is for you and pasting as u go
I swear if my neighbours don't stfu im going to get a nose bleed

3:59 AM
Doesn't help if the person deletes the question. It always infuriates me when someone deletes after I've answered.

Hi @Ted

hi @BalarkaSen

Hi @Leaky

@BalarkaSen you can now directly challenge others on chess.com puzzle battle
also:

cool

4:09 AM
do you wanna do a puzzle battle?

not now i think
need to read math, have been sleeping for too long

4:26 AM
Let $K = \Bbb Q(\alpha)$ be an algebraic number field of degree $n$ where $\alpha$ is an algebraic integer, and suppose the minimal polynomial for $\alpha$ satisfies the Eisenstein's criterion with a prime $p$. Then if for some $f(x) \in \Bbb Z[x]$ of degree $n-1$, $f(\alpha) \equiv 0 \pmod{p\mathcal{O}_K}$ then $f(x) \in p\Bbb Z[x]$.

is that a question?

No, I know the proof. Trying to interpret it.

whats the proof?

Eisenstein basically says $\alpha^n \equiv 0 \pmod{p\mathcal{O}_K}$, so consider $\alpha^{n-1}f(\alpha) \equiv 0 \pmod{p\mathcal{O}_K}$. This forces $\alpha^{n-1}f_0 \equiv 0 \pmod{p\mathcal{O}_K}$, where $f_0$ is the initial coefficient of $f$.
So $\alpha^{n-1} f_0 = p \beta$ for some $\beta \in \mathcal{O}_K$. Take norm to get $N_K(\alpha)^{n-1} f_0^n = p^n N_K(\beta)$.
We also know by Eisenstein that $N_K(\alpha)$ is divisible by $p$ exactly once, so $p$ has to divide $f_0$.
Continue similarly
Essentially it's saying if a polynomial is zero on $\mathcal{O}_K/p\mathcal{O}_K$ then it's zero as a polynomial over $\Bbb Z/p\Bbb Z$, I suppose.
Which is not true in general; consider simply $x^p - x$ as a polynomial which is zero on $\Bbb Z/p\Bbb Z$ but is not zero as a polynomial. The Eisenstein criterion is saying when this is true?

4:45 AM
interesting

@LeakyNun Apparently this says, in the same setup, $p$ doesn't divide $|\mathcal{O}_K : \Bbb Z[\alpha]|$. This is because if $\beta$ is some element of $p$-torsion in $\mathcal{O}_K/\Bbb Z[\alpha]$, then $p\beta \in \Bbb Z[\alpha]$. Write $\beta = r_0 + r_1\alpha + \cdots + r_{n-1}\alpha^{n-1}$ for $r_i \in \Bbb Q$. One of the $r_i$'s is not an integer as $\beta \notin \Bbb Z[\alpha]$, yet $pr_i \in \Bbb Z$.

How long does it take to receive a reply after sending a message through "Contact Support"?

So $p$ divides the denominator of each $r_i$. Write $r_0 + r_1\alpha + \cdots + r_{n-1} \alpha^{n-1}$ as a single fraction, $(f_0 + f_1 \alpha + \cdots + f_{n-1} \alpha^{n-1})/c$ where $p$ divides $c$. This belongs to $\mathcal{O}_K$, so $f_0 + \cdots + f_{n-1}\alpha^{n-1} \in p\mathcal{O}_K$
But this forces each $f_i$ to be divisible by $p$ by earlier, which is contradiction, because we normalized the fraction.

and I think from this you can derive the condition earlier?

Which test is hard comparatively?

1. Top 10% with a score of 7/10 when total students participated in test were 1000
or

2. Top 10% with a score of 9/10 when total students participated in test were 100 ?

4:52 AM
We can use this to prove the ring of integers of $K = \Bbb Q(\sqrt[3]{2})$ is $A = \Bbb Z[\sqrt[3]{2}]$. Because $2^2 3^3 = 108 = \text{disc}(A) = |\mathcal{O}_K : A| \text{disc}(A)$, and neither $2$ nor $3$ divides $|\mathcal{O}_K : A|$ as the minimal polynomial here is $x^3 - 2$ which is Eisenstein at $2$ and a linear change $(x - 1)^3 - 2$ is Eisenstein at $3$.
So $|\mathcal{O}_K : A| = 1$

cool!

Weird stuff
@LeakyNun Yeah sounds right, let me think about that
Whoops I meant $\text{disc}(A) = |\mathcal{O}_K : A|^2 \text{disc}(\mathcal{O}_K)$

hmm I'm not sure how to derive it
oh wow it's Eisenstein at 2 and 3 simultaneously!
$(x-1)^3-2 = x^3 - 3x^2 + 3x - 3$
wow this is magical

@LeakyNun Going from $p$ not dividing $|\mathcal{O}_K : \Bbb Z[\alpha]|$ to the original mysterious statement, you mean?

4:58 AM
Hm

why on earth should $1 + \sqrt[3]2$ have $3$-valuation $1/3$
oh because of some $1^3 + \sqrt[3]2^3 = 3$ nonsense I suppose

Ah I think I understood that. It's divisible only once by $3$ in $\mathcal{O}_K$, is what that means, yeah?

something like $a^3 + b^3 = (a+b)(a^2-ab+b^2)$
yeah that's what it means
wait no that isn't what it means
it's divisible a third times

Some power shit. $(1 + \sqrt[3]{2})^3$ is divisible by $3$ exactly once.
Maybe?

$3$ is totally ramified
$3 = (3, 1+\sqrt[3]2)^3$
this is trippy

5:04 AM
Ah.

I don't see the 3-ness in $1+\sqrt[3]2$
except in the identity I wrote earlier
so $1 + \sqrt[3]2 = 3/(1-\sqrt[3]2+\sqrt[3]4)$
whatever I'm just speaking nonsense

If minimal polynomial of $\alpha$ is Eisenstein at $p$, I think that implies $p$ is totally ramified in general

indeed it is
and $p = (p,\alpha)^n$

Very cool
Yeah

but like one wouldn't expect $\Bbb Q_3(\sqrt[3]2) / \Bbb Q_3$ to be totally ramified
or maybe watch out because we're taking cube root
I'm not being rigorous here
but $2$ has no $3$-ness
ok what's next

5:10 AM
Not much now
That's all I got
Should get back to the grind with Dummit-Foote and field theory
I don't know why suddenly I am doing number theory

well they're not unrelated

Oh I know I was saying that in the sense that
topology has gone up in the air

ah the key is that
$f(x) = x^3-2$
$f'(\alpha) = {\Huge \color{red}3} \sqrt[3]4$
that's where the mysterious $3$ comes from

Lmao

no seriously

5:13 AM
How is the derivative relevant? Something something Hensel?

the different is generated by $(f'(\alpha))$

Oh ok

and the different measures ramification

Spooky

it is

5:15 AM
Where do I learn this stuff from, if I ever want to?
I was toying with the idea of reading Serre's local fields

Neukirch

Wikipedia says that nikola tesla considered women to be superior in everyway to men, but it also says that in his elderly years he claimed to love a injured pigeon "The way a man would love a woman" I mean, if he held women in such high regard then why was he nailing a trash dove

@LeakyNun Thanks

@BalarkaSen and also the course I'm in: math.mit.edu/classes/18.785/2019fa/lectures.html

Haha nice plug
I'll try these out
Damn these lectures look very good

5:20 AM
if $f(x) = \prod (x - \sigma(\alpha))$ then $f'(\alpha) = \prod_{\sigma \ne 1} (\alpha - \sigma(\alpha))$

Darn these are good problems

so something like $\operatorname{Tr}(f'(\alpha)) = \Delta(f)$
recall that the discriminant is the product of the difference of the roots
so that's how $f'(\alpha)$ is related
to the discriminant and hence to the different

2 hours later…
7:39 AM
Is stack exchange the only domain that can administer Wikipedia editing privileges ?

8:07 AM
@LeakyNun Hi leaky
@TedE Second Ted are you here?
I think ill just ask the question here haha
the definition of denumeable set is that either finite or has same cardinality as N
also that we can list the elements
most of the times however, it is not easy to find a bijection in a simple way
how does one proves some set is denumerable ?

you use lemmas like product of two countable sets is countable and subset of a countable set is countable

putting the elements in a list such A = { .... } is not satisfactory to me, maybe because I dont understtand why
Yes but that comes from accepting the defintion of denumerable
is it standard defintion ?
or is there a better one , more tangble

well you just said another one

@LeakyNun some of these ideas are intuitive but yet hard to grasp

it's either finite or has same cardinality as N

8:21 AM
yes
so can we try to prove that, every infinite subset of a denumerable set is denumerable?
I have made some progress on it
but want to discuss it with someone
you down for it? @LeakyNun
I just notice what i wrote is wrong
denumerable is just bijection with N
coutable is finite or denumerable
mixed up the names ^^

@LeakyNun What's an algebraic number field that isn't totally ramified at any prime?
There are no unramified finite extensions of $\Bbb Q$ right? That's some result I remember

8:58 AM
it’s at least degree 4
maybe your favourite quartic polynomial @BalarkaSen

$x^4 - 10x^2 + 1$?

yeah

I'm becoming an expert at this

lmao
let’s check
you can perform the computations XD

Basically Eisenstein never applies to that polynomial with whatever shifts, because it's always reducible mod $p$, I think.
Eisenstein-at-$p$ polynomials are irreducible mod $p$

9:01 AM
x^n - p is eisenstein

Ah, except those dumb cases
I should be careful

every eisenstein is x^n mod p
actually
wait no nvm
idk

Hm.
lmfdb says $2$ has ramification index $4$
I did $x^4 - 9x^2 + 1$ and suddenly $2, 7, 11$ are the only ramified primes, none of which have ramification index $4$

cool

9:28 AM
Seifert surface of the Trefoil (almost made a three-turn mobius strip)

Lol
Nice

9:40 AM
@LeakyNun So $x^4 + 10x^2 + 1$ is also an example. The splitting field is $\Bbb Q(\sqrt{-2}, \sqrt{-3})$ which has discriminant $2^6 \cdot 3^2$, so $2$ and $3$ are the only primes which could be ramified.
How would I show by hand that neither has full ramification index?

factor it mod 2 and 3

So mod $2$ that's $(x + 1)^4$ and mod $3$ that's $(x + 1)^2(x - 1)^2$
Let me think why this helps (don't tell)
Rambling out loud: If $2$ totally ramifies then $(2) = \mathfrak{p}^4$ in $\mathcal{O}_K$.
If it was a PID I would be able to say $\mathcal{O}_K/2\mathcal{O}_K$ has $4$-torsion
If $\alpha = \sqrt{-2} + \sqrt{-3}$, then $\alpha^4 +10\alpha^2 + 1 = 0$, so reducing mod $2\mathcal{O}_K$ on both sides says $(\alpha+1)^4 \in 2\mathcal{O}_K$ I guess

10:10 AM
Wait. So $(2, \alpha+1)^4 = (2)$, right?
@LeakyNun isn't that the prime decomposition of $(2)$

@BalarkaSen yeah it is

I misunderstood the $e$ down below here I think
It's some local field shit

that’s strange
oh no
it isn't monogenic
is it
that's what goes wrong

@BalarkaSen I thought it was monogenic

10:21 AM
So is my thing correct, and $4$ is indeed the ramification index, and their $e$ is something different?

no it's 2
their $e$ is the $e$

Hm, how so

@TedShifrin I saw Mathein today at uni and spoke to him :)

@BalarkaSen I can't compute anything
but basically lmfdb says so so it is so lol

Yikes

10:32 AM
seriously I don't work with non-monogenic number fields
go learn pari/gp

Hahah yeah I can see why it can be a pain
I need to learn some actual number theory first

Comment: can't you determine the ramification over $\Bbb Q(\sqrt{-2})$ and $\Bbb Q(\sqrt{-2},\sqrt{-3})/\Bbb Q(\sqrt{-2})$ separately?

Index is multiplicative, yeah?

yeah

(Thinking covering spaces again)

10:36 AM
oh right
lol what am I doing
ok so $(2) = (\sqrt{-2})^2$

Ramification indices and the other English word are both multiplicative
Trägheitsindex

inertial degree

yeah hahaha

flexing

#ForgettingWordsInYourMotherTongue

10:37 AM
weird flex but ok
and then $x^2+x+1$ right
wait that isn't right
oh yeah that's right

Why's $\sqrt{-2}$ prime in $\mathcal{O}_K$

@BalarkaSen ok so $(2) = (\sqrt{-2})^2$ is the factorization, and $\mathcal O_K/(\sqrt{-2}) = \Bbb F_4$

How did you compute that so fast lmao

because $x^2+x+1$ is irreducible
no I computed it in $\Bbb Q(\sqrt{-2})$ first
which is a PID (irrelevant)

Yes, that's fine.

10:41 AM
and then the second step is $\sqrt{-3}$
which we all know is $\zeta_3$
i.e. $x^2+x+1$

But why doesn't $\sqrt{-2}$ further split up in $\Bbb Q(\sqrt{-2}, \sqrt{-3})$?

we're talking about $\Bbb Q(\sqrt{-2})$ first
right?
and $(2)$ obviously splits in that way in $\mathcal{O}_{\Bbb Q(\sqrt{-2})}$

because $A[\zeta_3]/(\sqrt{-2}) = A[X]/(X^2+X+1, \sqrt{-2}) = (A/\sqrt{-2})[X]/(X^2+X+1) = \Bbb F_2[X]/(X^2+X+1)$
this is the "reciprocity" you used before

Doing the whole $\mathcal{O}_K/\mathfrak{p} \cong \Bbb Z[X]/\text{Blah blah blah}$ thing is very common

10:44 AM
it works for monogenic extensions only

Trüth

Of course, yeah, makes total sense.

the "Dedekind--Kummer theorem"
6.14
@ÍgjøgnumMeg /tRyt/

Although you have a similar trick if $\mathfrak{p} + \mathcal{F}_\theta = 1$ where $\theta$ is a primitive element for your number field and $\mathcal{F}_\theta$ is the conductor

$\Bbb F_3[X]/(X^2 + 2)$ is $\Bbb F_3 \times \Bbb F_3$ though, so $3$ has two prime factors, both of index $2$?
@LeakyNun Thanks

10:49 AM
@LeakyNun any chance you have links for the course before that course, and course before that one?

@BalarkaSen corresponding to the fact that there are two entries for $3$

Yeah

like I need problems in that vein but dialled back to those involving just basics

OK, so I guess all of this can be summed up to conclude that $x^4 + 10x^2 + 1$ can never be translated to satisfy Eisenstein at any prime
Although there are likely elementary ways to see that, but fuck it :P

Also: just contacted a prof. to supervise my master thesis o.o I have fear

10:52 AM
Nice, @Ig

I mean, hopefully he says yes; he's a big player in Iwasawa theory so it'd be cool if he'd supervise me

I got fingers crossed for ya

@BalarkaSen did you watch the video

Nope, sorry :P

it's not as viral as it should be, rawr
it's just 30 secs go watch it

10:55 AM
A'ight

His doctoral supervisor was Coates, who also supervised Wiles rofl

Ah damn

bit intimidating but w/e

@LeakyNun Jesus that took 3 seconds

ikr

10:59 AM
Usually professors refuse to supervise a thesis just because they don't have time (too many students already or they just have stuff to do), apart from that they won't eat you @ÍgjøgnumMeg

@BalarkaSen I guess this beats the record:

chessbrah oy

but it isnt as awesome

@Alessandro yeah, he often supervises master theses though (I've heard) and I don't know how many students he currently has, also Mathein mentioned that he's a really nice guy

I see, good luck then

11:12 AM
The stuff you guys discuss is much more advanced than I have ever learnt, and judging from what you are saying you are all very qualified, are my problems just too boring for people at your level to consider attempting? (I'm not going to be offended)

Idek what problems you're doing but I'm definitely out of my depth at my current level rofl

11:37 AM
no but things like field theory and rings etc like I have been reading the material for the past few years but it's a long way off being second nature do you understand what I mean? Like I know that it's all very relevant somehow to the things I work on and the questions I post, but you guys seem to be doing this with ease, that's why I ask
like I have to put things in first order sometimes and im not sure if its a good habit to get into or its obsessive compulsive disorder, but it's generally how I understand something, like this morning I had to get Lagrange's theorem out in formal and I known its ridiculously long, but the statements on the wiki didn't make the point that it makes no sense unless you are doing arithmetic modulo p
like the upper bound on the number of incongruent solutions only holds when you restrict $x$ and $y$ to ${\{0,1,2,...,p-1}\}$
in other words
put it this way, if that MIT assignment was due this week, and I was in that class, i would attempt one of the warm ups, then go on a 3 month bender in which i am permanently banned from campus then be able to get a decent score on it 3 years later
ie it would not be handed in on time in summary

11:57 AM
ergh "throughout this whole thesis we assume the conjectured finiteness of the Tate-Shafarevich group since this is crucial to computing the coefficients of blah blah"

1 hour later…
1:20 PM
@Adam The stuff that I have seen you post is usually one of 1) too unrigorous to even interpret, 2) contrived and unmotivated, 3) really basic. I think you should just work through the exercises in an algebra textbook like Dummit and Foote?

Dummit and Foote represent

1:38 PM
Ok thanks Ted is there a chance you submit an answer to one for me?
or perhaps give an example where rigor was lacking, and provide an example of what you would expect to make the question answerable or even interpretable?
But sure I will look for the text book you mentioned and go through some exercises
I'm not sure your second point is worth any salt tho, kinda seems like the type of point motivated by sporadic thyroid

2:05 PM
You always need to have motivation for any particular question. Even if the motivation is "I am curious about this topic"

2:23 PM
In my Linear Algebra class, my professor talked about something called "Slotovi slices" which are these matrices that generate all characteristic polynomials. Does anyone know where I can find more about this? I tried Googling "Slotovi slice," but found nothing. Also, I think I am probably misspelling "Slotovi" which may be the reason why I can't find anything on it.

I have never heard of these things. What does generating characteristic polynomials mean?

Basically, the way my professor explained it is that a "Slotovi slice" is a set of $n\times n$ matrices such that each matrix in the set has a unique characteristic polynomial, and that for every polynomial of degree $n$ with leading coefficient of $1$, there is a matrix in the set with that polynomial as its characteristic polynomial.

Sounds like the set of companion matrices is an example of such a thing?
There is a bijection between monic polynomials of degree $n$ and $n \times n$ companion matrices.

OK, I googled "companion matrix" and this seems to be exactly like what he was talking about. Thank you!

Strange name.

2:31 PM
Oh...I was spelling it wrong. It is "Slodowy slice." arxiv.org/pdf/1509.07764.pdf

Huh

2:46 PM
@Adam I'll point them out to you in future. I'm not motivated enough to dive back through your history to classify them.

@BalarkaSen have you proved that the two notions of separable degree are equal

@NobleMushtak If you want to read about these, there's a book on semisimple and nilpotent orbits that gives you the backgroun.

it’s like a cool but useless result

David .H. Collingwood "Nilpotent Orbits In Semisimple Lie Algebra"
You can think of this as being a geometric/topological interpretation of Jordan normal form if you're looking at GL_n(C).
For other Lie types, more work must be done.

3:48 PM
0

Let $G$ be a locally compact group. The union of countably many open $\sigma$-compact subgroups of $G$ generates an open $\sigma$-compact subgroup. For each $n \in \Bbb{N}$, let $L_n = \bigcup_{j=1} K_{n,j}$ be an open subgroup, where $K_{n,j}$ is compact. I was able to prove that $\langle \... @user193319 is it countability of compactness that you doubt? Ummm I think so; it's just not clear that the number of sets in the double union is countable. do you know that$\Bbb N^2$is countable? Yup. and$\Bbb N^3$? 3:56 PM Yup and$\Bbb{N}^t$for any$t \in \Bbb{N}$. so$\Bbb N^{2k+1}$? Sure. oh, it should say$\Bbb N^t$not$\Bbb N^k$anyway a countable union of countable sets is countable Okay. But a countable union of compact sets isn't necessarily compact. Isn't that a problem? but it's$\sigma$-compact! yeah so your$K_{n,j}\$ are wrong?
where you gettin' those open compact subgroups from

4:27 PM
@TedE ah buddy your initial statement carries a requisite of having classified the content I have posted don't you have bongs to smoke with mark Walberg?

4:40 PM
@ÍgjøgnumMeg Good to hear!

@TedShifrin he's just super busy with his dissertation
but he was in today holding a seminar lol

Did you go to it? :)

i didn't know it was on! But it was about rigid analytic geometry anyway, about which I have no idea

1 hour later…
5:57 PM
let it never be said that i was a lurker
hi chat, i'm still alive, how've y'all been

03:00 - 18:0018:00 - 23:00