Mathematics - 2016-12-08 (page 3 of 7)

What is $K$?

what's ze question

@KajHansen Kellogs, :D

lol

I just won my 20th nice answer badge :D

cngrts

10:03 AM

@KajHansen K is the splitting field of $x^8 - 2$.

Oh this question again

@BalarkaSen prove that $Aut(\mathbb{K} / \mathbb{Q}(\sqrt(-2))) = Q_8$.

Yeah @KajHansen I left it for some days

I think I have the way to think about it.

The fact that $x^n - 2$ isn't always $D_n$ galois group is weird to me

Unintuitive

The roots of this thing form a regular $n$-gon in $\mathbb{C}$ which is flipped by complex conjugation

if you want to do it indirectly, show it's nonabelian then show it's not D_8.

not sure how to the second easily

That's hard @BalarkaSen

Due to the argument I'm making

And $\sqrt[n]{2} \mapsto \omega \sqrt[n]{2}$ always seems to be valid

A "rotation" of the $n$-gon

It seems like right there you have the generators of $D_n$

10:06 AM

just a sec I was thinking of building the $Aut(\mathbb{K} / U)$ directly.

>:(

no not really @KajHansen

@KajHansen He's working over Q(sqrt(2)), so you're looking at a regular 4-gon though, yes?

I know there's something wrong @Adeek

Oh wait, you're right

because you have to compute what is the action of $\omega$

10:07 AM

it's the roots of x^4 - sqrt(2) = 0 you care about

There was another question though recently where this wasn't the case

We were working over $\mathbb{Q}$

@KajHansen for example in your map $\omega \mapsto \omega^5$

ohhhhh I bet that's it

so the maps $\sigma^i$ as you defined won't be the regular generators of $D_n$

The root I'm thinking of corresponds to a root of unity that isn't primitive

10:08 AM

that is the issiue.

sheeeiiiit

2

yeah it is little bit tricky.

lol @ whoever starred that

isn't every subgroup of Q_8 normal? that ain't the case for D_8

so just show every subextension of that is galois

Yes @BalarkaSen

Clever

10:12 AM

ohh @BalarkaSen

very clever

iirc it's harder to build an extension over Q with galois group Q_8 tho

i dunno how to do it

That inverse galois

2hard4me

me2

It is some weird construction @BalarkaSen I think some page about it somehwere

https://data.stackexchange.com/math/query/2111/worst-accepted-answers
^^Funny as hell

10:16 AM

haha

@KajHansen math.stackexchange.com/questions/1611006/…

I am dying from this answer

haha

@Adeek L.M.A.O.

It's even funnier considering it was accepted

This is the MSE equivalent of someone walking into a marketplace and successfully transacting with a vendor despite all the other shopkeepers warning that his money is counterfeit.

@KajHansen I might be interesting to see a similar query on academia. There I havea feeling that often the OP accepts the answer that agrees most with what they hope to hear, even if that is not really the truth.

I agree @TobiasKildetoft, that would be interesting

@TobiasKildetoft data.stackexchange.com/academia/query/2111/…

haha

@KajHansen The solution is to make our prime minister and MSE moderator.

*an

10:28 AM

Hi everyone

Hi @Alessandro

@BalarkaSen, PM Modi and his mathematical undertaking

hi @Alessandro

@KajHansen Cool. I actually commented quite a bit on the third one on the list it seems

I should read more of Eliot; he's got nice stuff.

10:31 AM

How was the physics exam? @balarka

it was alright

today was chemistry, which went ok too

Does anyone know an automatic way to get a BiBTeX reference form arXiv?

SteamyRoot

tried arxiv2bibtex.org?

@SteamyRoot Thanks

Hmm, that does not actually seem to do this correctly

s.harp

If $M$ is an $R$ module, $N$ an $S$ module and I have a ring homomorphism $R\to S$, then I can view $N\otimes_R M$ as an $S$ module by pulling the scalar multiplication of elements in $S-R$ into $N$.

Does this construction have a name?

10:36 AM

You asked me about 2 dimensional manifolds a few days ago, but we covered the compact case in class before I could think about it... @Balarka the non compact case looks much uglier though

s.harp

I would expect something like "extension of scalars" or "restriction of scalars", except more accurate

@Alessandro I had the compact case in mind :)

Yeah, the noncompact one is a little bit more complicated

@SteamyRoot Unfortunately, it creates something that gets cited without mentioning the arXiv id and it does not respect capital letters in the title.

Who's excited for hats ?

I am!

s.harp

10:39 AM

hats?

The annual winter bash!! @s.harp

who are we bashing?

Not violent bashing. Celebratory bashing.

When will it begin?

No clue

10:40 AM

@Alessandro A small exercise: classify all compact 2-manifolds with boundary.

I'm actually not the biggest fan of some aspects of hats. It encourages, tbh, some degree of detrimental behavior on MSE.

By that I mean you're allowed to have points nbhds around which looks like the upper half plane (w/ the point on the x-axis).

Collection of those points is called a boundary of the manifold.

I asked Ted about those yesterday because I was annoyed by things like the Möbius strip not being manifolds

I'll think about the classification

OK :)

SteamyRoot

@TobiasKildetoft Oh, right. I changed my bibtex style to include Eprint fields.

Never had a problem with capitals not being respected, though.

I'm afraid it's the only working thing I know :/

10:45 AM

@SteamyRoot It just copied the title, so it would never be able to retain capital letters

@KajHansen what you mean with hats?

I just changed it to what I usually do, which is list it as an article, rather than misc, and list arXiv as the journal.

@Null, "hats" is an annual thing where users do tasks to earn hats, which are graphics of literal hats that can be placed on your profile pic

If you get a rare hat, you can show it off. It's a source of much ado and excitement in MSE chat.

@KajHansen what are those tasks? answering things or doing things?

Both @Null

There are particularly detrimental ones like "Go to a different SE and upvote things", which causes a huge flux of people upvoting indiscriminately just to get the hat as fast as possible

Joining a new SE gets you a hat. Answering a question gets a hat. Pretty much every task you can imagine earns one. Some are really hard to get though.

10:57 AM

can ${n \choose 3}$ be a square? I've eliminated all cases but $n\equiv 2 \pmod 6$

Mew

SUP DAWGS

@KajHansen and when does this start?

@Mew rofl :D

Very soon @Null. Not sure exactly when

19 December @Null

@KajHansen i found the vox populi badge nonsense. But i realized that it is very easy to spot bad questions. just look for downvoted ones, and look if the OP has tried anything, then autovotedown xd

(or vote up, in case he tried something)

Q: Clarification about Hasse's Theorem for Elliptic Curve over Finite Fields

LOL @Null

Oh, on that note

3

In the proof of Hasse's Theorem over finite fields, one considers the Frobenous endomorphism $\phi(x,y) \rightarrow (x^q,y^q) \in E(\bar{F_q})$ and then notes that this endomorphism fixes the elements of $F_q\times F_q$ but permutes the rest of the element of ($\bar{F_q}\times \bar{F_q})\setminus...

Give this guy an upvote. He put some effort into that question, and it's his first post

11:07 AM

@KajHansen don't even understand the question, but indeed strange first post :D

I like encouraging new members who have good questions as much as possible. They're too rare

meh, i mean noteworthy

My first post ever on here got a "Good Question" badge lol. But it also was really welcoming so I stuck around :P

@KajHansen My first question on MSE was pretty mediocre. My first on MO on the other hand (back before MSE existed) is sitting on 40 upvotes

@KajHansen i will vote up questions that are "how do i calc this without a calculator" simply because it is the right approach. The question can be easily googable, but i find one shouldn't stop someone who tries to be calculator free. Now in the end it comes down to the question of course...

11:10 AM

My first question went unanswered for months on MSE just to be answered after a few hours I reposted it to MO

Nice! @Tobias. I haven't ever had anything MO-worthy

@Alessandro, hahaha

@KajHansen is this question about cardinality?

Was MO more MSE-like back in the day?

Accepting any and all questions?

Or has it always been exclusively research-level?

That's good encouragement @Null

I think the definition of research-level is pretty broad in MO.

Yes. Hasse's theorem gives a bound on the number of points on an elliptic curve over a finite field

11:12 AM

@KajHansen that's all i got because of "#" haha

I asked only a couple of questions on MO and neither of them was even near to research level :P

I think the point is those are research-level in the sense of being questions which can be considered individual student-research, not always as high-falutin' as a thesis paper or something.

@KajHansen It has had the same scope always, but the precise interpretation has varied a bit. My first question on MO might have been directed to MSE nowadays (at least, I would have asked it here instead)

Cool

the people in MO can do whatever they want in any case

they are the actual mathematicians

11:16 AM

Whereas we are dirty MSE peasants

Well, maybe not you Balarka or Tobias. Y'all grace us with your presence :)

i am a random dude from the internet

Tobias is already an accomplished mathematician

~~internet~~ West Bengal

you don't know that's true.

too old

that ain't me

11:20 AM

That's what I thought too. Guy was tagged as such though lol

at least not how i remember myself

And indeed there will be time
For the yellow smoke that slides along the street,
Rubbing its back upon the window-panes;
There will be time, there will be time
To prepare a face to meet the faces that you meet;
There will be time to murder and create,
And time for all the works and days of hands
That lift and drop a question on your plate;
Time for you and time for me,
And time yet for a hundred indecisions,
And for a hundred visions and revisions,
Before the taking of a toast and tea.

I found your "Madhyamik Pariksha" scores

yikes

don't look

Need to spend some more time reviewing history ;)

haha, ok I'm done. I don't mean to harass / stalk you

nah i don't

i didn't feel harassed; there are better things to do than looking up my internet-history tho

11:25 AM

Indeed

I found a lot of photos of Kaj in the meantime tho

ah, Alessandro's doing one of the better things

@Alessandro huh if I were you I wouldn't share these google urls. They might contain a lot of information

@KajHansen I am not sure I was aware before now that you were at UGA actually

maybe it just says you're using firefox but there's probably much more

11:28 AM

LOL @Alessandro

@Tobias, yep. Had Ted for three courses

I know, but I couldn't figure out how much I could cut from it, using only the ?search= field it goes to google instead of google images

@KajHansen Neat. I don't think I actually met Ted while I was there.

Are you a professor @Tobias ? I've never asked your level of education

@KajHansen Postdoc (well, between postdoc positions right now)

Ah, what were you at UGA for?

11:30 AM

As a visitor during my PhD (for 6 months, visiting Dan Nakano),

@Sophie you mean information about the searcher?

Oh very nice! I had Dr. Nakano for my grad algebra course last fall @Tobias

Very good professor. Explains things quite well

cool

a field is a ring, whith closure and associativity of multiplication?

Good morning friends

meow-mix

11:37 AM

good morning

oh thats weird lol

The only thing different about a field w.r.t. a ring is that a field has multiplicative inverses for each nonzero element

Oh wait, and commutativity of mult

ok, one thing more i have to learn more

it is nice to know what one does not know.

That is an uncountable set in my case

@KajHansen dito

@KajHansen but i think mine is slightly more countable than yours, yet :D

Just 16 more hours until I can put a bounty on my question :)

11:41 AM

The set of real numbers that can even be described in a finite amount of time is merely countably infinite. E.g. $\pi$ can be described in a finite amount of time because it's the ratio of a given circle's circumference to its diameter

"All real numbers"?

@KajHansen Yeah, there was a terrible question related to that yesterday

A good IRL friend of mine thought of this a while back while we were hanging out. He calls numbers that cannot be described "unspeakable" hahaha

That's not a description ?

No @GFauxPas. I'm talking about referring to an individual element of the reals

11:43 AM

Oh

@KajHansen for me it stops at $\frac{\pi}{e}$

The infinity that is $| \mathbb{R} |$ is incomprehensible to my mind

(not using the descriptions of pi and e of course :P )

@Null, all of the algebraic numbers can be easily described

I thought the irrationality of that is a conjecture

I remember o

Some comment on SE

11:45 AM

$\pi + e$ is unknown

$e^\pi$ is known

That if it's rational it's God laughing at us

@KajHansen So is $\pi e$ (unknown that is)

Actually the above might've been $e^{\sqrt{\pi}}$

Don't remember

but we know that at least one of $\pi e$ or $\pi + e$ will be trancendental

meow-mix

unknown what? transcendentality?

if thats a word :P

11:47 AM

we don't know if $\pi+e$ is rational or not @meow

My jaw dropped when I read that for the first time

meow-mix

oh

@KajHansen if it's rational, there exists some integer n>0, such that $n(\pi +e)\in\mathbb{N}$?

yes, @Null

I can see why one of them must be irrational @tobias, how do you show that one must be trascendental?

11:52 AM

Of course @Null

\pi + e = x/y

pick n = y

@Alessandro, I was thinking about that too. Trying to get a contradiction with algebraic numbers and field closure axioms

@Alessandro use the fact that the algebraic numbers are algebraically closed and look at the polynomial $(x-\pi)(x-e)$

Ah

Nice

that's what I did do decide one of them has to be irrational, but actually you're right

hey @KajHansen I am trying to build a field extension with galois group isomorphic to $Z_4 \times Z_2$

Not too hard @Adeek

11:57 AM

really ?

Consider $\mathbb{Q}(\sqrt{2 + \sqrt{2}})$ let $\alpha = \sqrt{2 + \sqrt{2}}$ I am considering $\mathbb{Q}(\alpha,i)$

I think this works.

any ideas ?

I could be wrong but the trick to getting this is letting $K$ be the splitting field for a product of two irreducible polynomials, one with Galois group $\mathbb{Z}_4$, and the other with $\mathbb{Z}_2$

yeah

So let $f(x) = (x^5 - 1)(x^2 - 2)$, e.g.

yeah I think what I am doing works

why is it galois over Q though ?

Which?