Mathematics

I happen to be rather careful about it, myself, and I don't understand how it can be difficult nowadays... Every computer has a reasonable spell-checker.

Eric Gregor

12:41 AM

i judge people's self worth by their measure of others' self worth

2

Peter Tamaroff

@EricGregor I forgot to add use of punctuation, too, and how they spell "succinctiness". JK

Any cool foreign word I can learn?

Danny Rancher

Anyone else try and trick people with large vocabulary?

Eric Gregor

@PeterTamaroff paralelepípedo

Fortuon Paendrag

Try Jugendtraum @PeterTamaroff

Eric Gregor

means cobblestone in portuguese

Danny Rancher

12:43 AM

Sending texts to numbers with "And don't worry, your looks definitely don't detract from your lovely personality".

Peter Tamaroff

@DannyRancher Just fooling around. You know, right?

Danny Rancher

No man, I'm for real.

Peter Tamaroff

@EricGregor Spanish is my mothertounge. Maybe a cool French word.

t.b.

@PeterTamaroff I once spent an evening memorizing and learning how spell and pronounce Muvaffakiyetsizleştiricileştiriveremeyebileceklerimizdenmişsinizcesine. Needless to say that I didn't succeed.

Peter Tamaroff

@DannyRancher You're serious about joking.

@tb What is that!?!?!?

Danny Rancher

12:45 AM

"Hey, where's your smile gone?! I've done nothing but garland you with florid adulation since we met."

t.b.

@PeterTamaroff the new longest Turkish word. Formerly it was, well, follow the link.

Peter Tamaroff

@tb Yeah, just read it. And I thought German had long words.

@DannyRancher Where's the trickery there?

Danny Rancher

@PeterTamaroff Ahh no trickery on the second one.

t.b.

@PeterTamaroff yes, but we can't construct that complicated things. How about: Donaudampfschiffskapitänsmütze?

Danny Rancher

"You're the definition of pulchritude." is funny.

Brian M. Scott

12:48 AM

@Danny: You now have an answer to your order question; basically you had it right.

Danny Rancher

@BrianMScott, I am understanding the answer at this exact moment! Thank you for your help.

Brian M. Scott

You're very welcome.

Peter Tamaroff

@DannyRancher It sounds like it is a negative connotation! In Spanish the word "Pulcritud" is not really used. What would be the adjective, as in "you are ___________" (In Spanish we use "Pulcro")

@tb I only get the Captain part, if it isn't anything else.

Danny Rancher

Yes, its especially exciting when they think they have been insulted.

I guess it does share the same sounds as sepulchre (tomb).

Eric Gregor

stillicide is a good word

Peter Tamaroff

12:52 AM

@tb Google split it in two: Donaudampfschiff Kapitänsmütze

Brian M. Scott

@PeterTamaroff 'cap of the captain of a steamship on the Donau'

Peter Tamaroff

Right.

When will THC start?

Brian M. Scott

@PeterTamaroff If you load a question, it now displays a message claiming that the THC is going on right now, with a link to it. My clock says that it's actually about 6 minutes away yet.

Danny Rancher

@PeterTamaroff I believe the adjective is pulchritudinous.

t.b.

@PeterTamaroff It might be that they now allow splitting those words into pieces. I grew up with the old version of the orthography, and I won't ever be able really learn the new one.

Peter Tamaroff

12:54 AM

@DannyRancher LOL

Brian M. Scott

@DannyRancher Yep.

Danny Rancher

@PeterTamaroff Yes dripping with irony.

Peter Tamaroff

I invite everyone here to THC, starting right now.

$Mathematics$ Math Town Hall Chat

meta.math.stackexchange.com/q/2348

Brian M. Scott

I'm giving it a miss; I'd rather read the transcript.

Fortuon Paendrag

I wonder where Gigili's gone..

Brian M. Scott

1:05 AM

@FortuonPaendrag Town Hall Chat, I assume.

Fortuon Paendrag

No, she isnt there either @Brian.

Brian M. Scott

Well, I think that she did say at one point that it would be a bad time for her.

Peter Tamaroff

@BillDubuque THC STARTED

@robjohn THC STARTED

Fortuon Paendrag

ah, that would explain it.

Brian M. Scott

@FortuonPaendrag Let me think; she's 7.5 hours ahead of me, so it's 0430 her time.

And robjohn might be late getting back from his birthday dinner.

Fortuon Paendrag

1:07 AM

oh, wow. its only 3 AM here in Budapest.

Brian M. Scott

Gigili's UTC+0330.

Fortuon Paendrag

I always thought she was from somewhere in South America, for some reason.

Brian M. Scott

In her candidacy she mentioned being able to cover Asian time fairly well.

Fortuon Paendrag

Ha. I should pay more attention to details, then.

Brian M. Scott

@FortuonPaendrag I was in Budapest for about 10 days once, back around 1980 or so. Big topology conference.

Fortuon Paendrag

1:12 AM

Its a beautiful city, is it not?

Brian M. Scott

I thought so.

Though there were still some pretty grim reminders of unpleasantness in a few places.

Fortuon Paendrag

Right. That was still back in the day..

Brian M. Scott

Yep. Different world then.

Fortuon Paendrag

The hour grows late. I wish you a good night, Brian! I will retire to bed now, for I have class early tomorrow!

Brian M. Scott

Good night!

(I always hated early classes.)

Fortuon Paendrag

1:16 AM

Also, you asked me to remind you of the sequence problem, which i just remembered myself.

Haha, same feelings here!

Brian M. Scott

I've a feeling that that's going to be hard to repair, but I'll give it a bit more thought.

Fortuon Paendrag

Thank you :)

t.b.

I just figured out why I was rather unhappy about answering here lately. Of my last 10 answers only 2 were accepted and most of the others had absolutely no feed-back from the OPs whatsoever. The accepts are basically the only thing that really counts.

Since I'm often the only answerer of a question this always leaves me wondering what's wrong...

Dylan Moreland

@tb :(

user19161

@tb Maybe the questions and answers are too difficult for the general population?

Dylan Moreland

1:24 AM

I don't so much have this problem, since the questions I answer are usually easy, but the site has lost some charm.

user19161

@tb This is a sign that you should forget about the site and do more real math!

t.b.

@ClarkKent I'm not complaining about lack of votes, I got quite enough (though for the wrong answers, as usual, I believe). It's the lack of feedback of the askers that I find really strange.

David K.

Does anyone else think that the town hall chat would be the ideal place for Jordan to complain about his academic life?

t.b.

@ClarkKent maybe.

user19161

@DavidK Sorry, but no.

t.b.

1:27 AM

@DavidK I'd rather love to see this :)

David K.

That would be awesome!

user19161

@tb I chuckled when I saw that question too.

user19161

@DavidK He invented the canonical form after all.

David K.

;)

t.b.

Okay, I think I'll go have a peek into the "town hall"...

mixedmath

1:32 AM

come on over ;p

user19161

I'll come too.

Danny Rancher

Since f is linear, it is an injection.
Can the inverse be used to show a function is not an injection.
Since f is non-linear, it is not an injection.

Dylan Moreland

What

t.b.

was this deliberately "shuffled"?

it seems like some non-s moved where they shouldn't have and some conclusions are just wrong.

Dylan Moreland

The lack of context is heroic, but certainly not all linear maps are injective.

2

Danny Rancher

1:37 AM

So, is this answer incorrect? math.stackexchange.com/a/140285/30011

user19161

@DannyRancher You need to look at an answer in the context of the question. You cannot just ask out of the blue "since f is linear it is an injection".

Dylan Moreland

@DannyRancher Bill is correct as always, but also terse as always.

You could ask for clarification there. But it is a general pattern that any map $f\colon \mathbf R \to \mathbf R$ of the form $f(x) = mx + b$, with $m \neq 0$, is injective.

If you restrict such a thing to a subset such as the domain in your problem, it's still injective.

Does that make sense?

I'm really surprised that "The Chaz" is not participating in this.

David K.

Can someone help me with yet another Galois Theory question?

Dylan Moreland

@DavidK I'm always up for Galois theory. Did you ever sort out that last one?

user19161

@DavidK It might be better to post on the main site to get more exposure.

David K.

1:45 AM

No. Not entirely.

@DylanMoreland I haven't yet determined if that Galois group of $$\Bbb{F}_{2}(x^{3})\subseteq\Bbb{F}_{2}(x)\subseteq\Bbb{F}_{2^{3}}(x)$$ is $C_{3}\times C_{3}$ or if it's $C_{9}$.

Where $C_{N}$ is the cyclic group with $N$ elements.

Dylan Moreland

Well, again, I don't see a way to believe that it's $C_9$ if you believe that it has two distinct subfields of degree $3$. I'll try to set aside some time for it.

It is pretty shameful to be constructing $\ell$-adic Galois representations and yet not be able to nail something like that down.

David K.

I agree that it isn't $C_{9}$.

Yes. Shameful indeed.

Dylan Moreland

And those are the only two groups of order $9$.

Hm. What's your new question?

David K.

Ok. Suppose that I have the set of distinct roots $\alpha_{1},\ldots,\alpha_{\ell}$ of the monic irreducible polynomial $f$ over $\Bbb{Q}[x]$.

Then form the product $g_{n}(x)=\prod_{i}^{\ell}(x-\alpha_{i}^{n})$.

I'm trying to show that $g_{n}$ is a positive integer power of an irreducible poly over $\Bbb{Q}$.

Oh and $f$ is not cyclotomic.

I feel confident that I can show that $g_{n}$ has coefficients in $\Bbb{Q}[x]$ using Newton's formulas for symmetric polynomials. But I can't figure out how to show that $g_{n}$ is a positive integer power of an irreducible poly.

I tried it out using a CAS and some example irreducibles, and it seems that $g_{n}$ is always either a polynomial to the power 1, or to the power $\deg\{f\}$.

Dylan Moreland

One way to go about this would be show that the $\alpha_i^n$ are all conjugate in some Galois extension of $\mathbf Q$.

t.b.

1:57 AM

what a bummer this THC...

Dylan Moreland

Right? I think that works. You really just want to show that the roots of that polynomial are roots of a single irreducible polynomial, and that happens iff you can find some automorphism over $\mathbf Q$ taking any given one to any other.

And from that perspective it seems clear. Look at the splitting field of $f$. The Galois group acts transitively on the $\alpha_i$, hence on the $\alpha_i^n$.

David K.

Hmm... Yes indeed. I hadn't thought of it quite like that.

But honestly I've been having trouble seeing the forest for the trees on this one.

@DylanMoreland Is this because for $\sigma\in\operatorname{Gal}(K/\Bbb{Q})$, $\sigma(\alpha_{i}^{n})=\sigma(\alpha_{i})^{n}$ ?

Dylan Moreland

Exactly.

mixedmath

The townhall is ending, and the candidates are likely to come over here.

robjohn

we're being kicked out :-)

Math Town Hall Chat

Transcript for

$Mathematics$ Math Town Hall Chat

$Mathematics$ Mathematics