Mathematics - 2015-12-15 (page 3 of 7)

Ted Shifrin

04:00

And well you should be, @MikeM.

Balarka Sen

@TedShifrin Point is to not make Anthony biased.

Mike Miller

@BalarkaSen: To be honest, I have no idea what you're saying.

Ted Shifrin

I know, @Balarka. Time is of the essence here.

Balarka Sen

@TedShifrin Fair enough.

Ted Shifrin

And I like Anthony, so I'm trying to help efficiently.

Julian Rachman

04:01

@TedShifrin hello!

Balarka Sen

@MikeMiller The usual advice. To try out a few examples, and then go for a proof/disproof.

Ted Shifrin

hi again, @Julian

Balarka Sen

@MikeMiller Yep.

Took naps in between anyway.

Ted Shifrin

10 minute naps don't count.

Anthony

@TedShifrin Sorry, where exactly did we discuss the easiest way to get normal extensions?

Ted Shifrin

04:03

Degree 2, remember?

Anthony

oh

shoot

D'oh

Ted Shifrin

So give me a degree 2 extension of $\Bbb Q$.

Anthony

fingers crossed for $\mathbb{Q}(i)$

Balarka Sen

Yes. $i$ is root of $x^2 + 1$.

Which is irred. Really is deg 2.

Ted Shifrin

Fine. Or $\Bbb Q(\sqrt 2)$. Either one.

Now can you give me a degree 2 extension of it?

Balarka Sen

04:05

Can you construct a normal extension of the degree 2 extension you wrote down?

Ted Shifrin

I wish Balarka would work on his calculus instead of meddling with me.

Anthony

we can consider $\mathbb{Q}(\sqrt{2},i)$ right?

Ted Shifrin

We can, @Anthony. Is that one a normal extension of $\Bbb Q$?

Anthony

yes

Ted Shifrin

Correct (I'll let you check later to be sure).

Balarka Sen

04:06

~~Why?~~

Ted Shifrin

Can you give me another degree 2 extension?

Anthony

cause it splits $x^2-2$

Ted Shifrin

Nope

Anthony

wooo

I meant $x^2+2$?

Ted Shifrin

Nope.

Balarka Sen

04:07

not right.

Ted Shifrin

It won't be degree 2!!!

Anthony

Okay, what am I doing

Balarka Sen

@TedShifrin Good reminder. I am going to, after a nap.

Anthony

Alright I clearly have some fundamental misunderstanding here. Let me ruminate for a moment.

Ted Shifrin

That's a degree 4 extension, @Anthony, so if it's normal, the polynomial it's the splitting field for had better have degree 4.

(That's not always true, but it's true in this case, by the way.)

Anthony

04:10

So what, the splitting field is smaller?

Ted Shifrin

What polynomial has splitting field $\Bbb Q(\sqrt 2, i)$?

Note it need not be an irreducible polynomial!

Balarka Sen

(relevant fact @ what Ted said about degree 4 : $K/E/F$ extension, then $[K:F] = [K:E][E:F]$)

Anthony

@TedShifrin Oh should I just take the product then? $(x^2-2)(x^2+1)$?

Ted Shifrin

Sure. Fantastic!

Balarka Sen

Nevermind me.

Ted Shifrin

04:11

It doesn't need to be @Balarka. Shaddup.

Go prove the Primitive Element Theorem and be quiet.

Balarka Sen

What was that? If $K/F$ is normal then $K$ is generated by a single element over $F$? i.e., $K = F[\alpha]$?

Anthony

okay, so backing up

Ted Shifrin

Go look it up, @Balarka.

Balarka Sen

Oops, normal + separable, I think.

Ted Shifrin

Yes, @Anthony, give me a different degree 2 extension of $\Bbb Q(\sqrt 2)$.

Anthony

04:13

I'm looking for a degree two extension of $\mathbb{Q}(\sqrt{2})$

I could append another square root

Ted Shifrin

Another root of whom?

Anthony

Shall I consider $3$?

Ted Shifrin

No, here's the key idea. Take something in the extension we already have and take its square root.

Balarka Sen

@TedShifrin Ok, finite degree separable. Tell me something, why do you geometers like this theorem so much? prof asked me to prove this about a year ago.

:P

Anthony

shall I take $\sqrt{\sqrt{2}}$?

Ted Shifrin

04:14

Not geometers. It's a super powerful result.

Balarka Sen

Yes.

Ted Shifrin

Bingo, @Anthony!!

So what is the extension we're looking at now?

Anthony

oh

Mike Miller

@Balarka: I sent you a note.

Anthony

You just meant $\mathbb{Q}(\sqrt{2},\sqrt{\sqrt{2}})$?

Ted Shifrin

04:16

Otherwise known as $\Bbb Q(\sqrt[4]{2})$, right?

Anthony

I don't know how to show that this isn't normal, but I can see why is wouldn't be

oh no I wouldn't

yes

Oh it just doesn't have the other roots of the minimal polynomial

Ted Shifrin

Perfect! OK. I'm done with you!

:)

Anthony

:D

Can I ask one more, hopefully shorter question?

Ted Shifrin

I hope you actually learned and understood what we just talked about :)

Yes?

Anthony

@TedShifrin I did. Thank you.

Balarka Sen

04:18

@TedShifrin Cool, I don't remember how to prove primitive element theorem (except vague sketches). Something to work with on the back of the head.

Anthony

Let me try to remember my confusion- I was looking I think at $\mathbb{Q}(\sqrt{2},\sqrt{3})$

Ted Shifrin

I'd rather you worked on calculus, but I thought it would give you something to shut you up now, @Balarka.

OK, @Anthony, just like $\sqrt 2$ and $i$ a minute ago.

Balarka Sen

It worked.

Anthony

And I was looking at the Galois group I believe.

@TedShifrin Oh I suppose.

Ted Shifrin

OK, @Anthony.

Anthony

04:19

In any case, I was reading somewhere that the elements of the Galois group need to fix the fourth basis element, namely $\sqrt{6}$

Is this true?

Ted Shifrin

Nope.

Anthony

Ah I remember where I was reading it, hold on. Let me find their statement.

Ted Shifrin

The action on $\sqrt6$ is determined by what the element does to the other two.

Anthony

When you take an algebraic extension, it permutes the roots- what's the statement in this case?

Should I look at the minimal polynomial with both those as roots, or something?

Ted Shifrin

Permutes the roots of an irreducible polynomial, yes.

Not if it's reducible. You can't send $\sqrt 2$ to $\sqrt 3$, only to $\pm \sqrt 2$.

You don't need to mess with the minimal polynomial. That is not illuminating.

But if $f(x)$ is irreducible and splits in a normal extension $K/F$, then any element of $\text{Gal}(K/F)$ must permute the roots of $f(x)$. Yes.

I made my students in my algebra class 3 years ago prove that on the final, I think.

Maybe @Balarka should remember how to prove that, too :D

Anthony

04:26

@TedShifrin Then how do I gain information about the Galois group for this problem?

The statement I read was that: "One of the elements of the basis of this field extension is not a conjugate of the roots of our polynomial, and so it must remain fixed by our automorphisms (namely $6\sqrt{6}$)." The polynomial was $(x^2-2)(x^2-3)$

Ted Shifrin

You have two automorphisms, @Anthony, one sending $\sqrt 2$ to its negative and fixing $\sqrt 3$, and the other doing the reverse. They generate a Klein 4-group. That must be the Galois group. Why?

Where did you read that statement?

It seems to be garbage.

Anthony

2

A: Galois Group of $(x^2-p_1)...(x^2-p_n)$

Think about the case where $n=1$, you have a quadratic extension and the galois group will just consist of the identity automorphism and the automorphism that sends $\sqrt{p_1}$ to its negative. This is isomorphic to $\mathbb{Z}/(2)$ (we can send the identity to $0$ and the other automorphism to ...

Ted Shifrin

@Anthony: Be very careful what you read. It's crap. Total crap.

I just put a comment on it and downvoted.

If $\sigma$ switches the sign on $\sqrt 2$ and fixes $\sqrt 3$ and $\tau$ does the reverse, then $\sigma\tau$ does fix $\sqrt 6$, but neither $\sigma$ nor $\tau$ does!

Anthony

I see. Then how would you go about investigating the Galois group of something like that?

I mean, even in our simply case.

Ted Shifrin

I have done that for you in my comments in here.

Anthony

04:32

oh

sorry

I forgot, let me read

Ted Shifrin

We get a group of order 4, and the normal extension has degree 4, so we must have all the Galois group.

Anthony

I guess it's easy to show that flipping the sign on the two roots gives homs? And then as you said we've found them all.

But so the idea of having an irred. poly and getting an extension, then looking at the permutations of the roots permissible feels comfortable.

When we starting gluing on random things to extend by, I grow worried.

Ted Shifrin

My comment was surely proved in your course and is essential.

Balarka Sen

@TedShifrin I was off. Prove what?

Ted Shifrin

What I just linked to ^^.

Forever Mozart

04:37

have you ever met a hippe mathematician?

Balarka Sen

Yah, sure.

Forever Mozart

hippie

Ted Shifrin

OK, @Anthony, I'm done with you for tonight. Good luck in the morning!

Clarinetist

@TedShifrin Quick question for you, what was that physics book that you recommended I read a long time ago?

Anthony

@TedShifrin OH

Ted Shifrin

04:37

heya @Clarinet

Anthony

Sorry.

Ted Shifrin

@Clarinet: Were we talking about a good first year mechanics book?

Balarka Sen

@Anthony knows it too, I guess. That's why the thing embeds in $S_n$.

Anthony

@TedShifrin Thanks. For some reason I kept saying minimal, or something. Any irreducible polynomial. Wonderful. Got it.

Clarinetist

@TedShifrin Sounds right

Ted Shifrin

04:38

Kleppner/Kolenkow @Clarinet

Anthony

@TedShifrin Thanks so much.

Ted Shifrin

You're welcome, @Anthony.

Clarinetist

Thanks @Ted, have a good night!

Anthony

Oh man, Kleppner.

Ted Shifrin

You're welcome, too.

Anthony

04:39

That's the first book I read in college.

Ted Shifrin

It's a fabulous book!

Once again, supremely good exercises.

Balarka Sen

I should get another hat anytime now

Ted Shifrin

Good morning, @Balarka, and bubye.

Anthony

Hasta luego, Ted.

Balarka Sen

Bubye!

Ted Shifrin

04:40

BTW, @Balarka, if you're still trying to get hats by commenting/downvoting, go deal with the garbage that Anthony linked.

Good luck, @Anthony. :)

Mike Miller

@TedShifrin I love the phrasing. "This is garbage, I'm afraid."

Balarka Sen

There's no hat for downvoting. And it's garbage, switch $\sqrt{2}$ and leave $\sqrt{3}$ put.

Ted Shifrin

Shall I rephrase? Probably too late.

Mike Miller

No, I think it's as polite as you could make it.

Ted Shifrin

I didn't say "garbage." I said "pretty much completely wrong." I was more polite :D

That's the trouble when people think everything on the internet is, like Donald Trump, the absolute truth.

Mike Miller

04:42

It's far more polite than I could make it!

Ted Shifrin

Thanks for the compliment — I think.

OK, I'm gone. Damn, I'm getting sick :( Ugh.

Balarka Sen

Does one need to be Lang to be a mathematician?

Mike Miller

@TedShifrin I guess Balarka gave it to you.

Balarka Sen

I think this is the highly upvoted question I am going to ask through android.

Mike Miller

lol

Anthony

04:44

@TedShifrin Noooo, get well, or not-sick soon.

Night.

Balarka Sen

@TedShifrin Not good, get well.

Anthony

@MikeMiller What gives, one of the hats is for asking a question today. I did that!

Mike Miller

How do you tex the circle integral?

Balarka Sen

\oint

Mike Miller

@Anthony: It's now the 15th UTC. But if you just posted your question, maybe it will still go through.

Anthony

04:46

@MikeMiller No, it was the testestsetsetes post I did that you saw.

I swiftly deleted it, but that wasn't against the rules!

Mike Miller

@Anthony: Probably doesn't count. Try upvoting something, that should be enough.

Balarka Sen

Horrendous service. I upvoted 10 questions/answers via android but I still haven't got the hat.

Anthony

lol

kevin Tah N.

What is the best elementary text on applications of matrices in physics

Mike Miller

@BalarkaSen Huh? Are you looking at... last year's hats?

Anthony

04:48

Wait @BalarkaSen, the Galois group is the same size as the extension?

Balarka Sen

If the extension is Galois.

Anthony

Oh.

Because there is that $n!$ thing. Somewhere. That's a number.

Balarka Sen

@MikeMiller winterbash2015.stackexchange.com/wireless

Mike Miller

Oh, I already forgot about that one. Patience.

Balarka Sen

Askew glasses looks nerdy.

Forever Mozart

04:55

I want a tophat

Balarka Sen

Why the hell is it taking so long? Same thing happened with the glasses.

Forever Mozart

what?

Balarka Sen

Hat-arriving.

Forever Mozart

oh I have 2 hats

but they are dumb so I dont have any

wear any

Balarka Sen

You should request an Alexander horned sphere hat.

Forever Mozart

04:59

hah

Balarka Sen

My meta answer here seems to be too neutral :(

Need controversial meta answer >:(

Forever Mozart

I usually delete my old questions that have no good replies

Julian Rachman

05:13

@TedShifrin hello again. We can finally have a conversation without disruption from my activities

Balarka Sen

@MikeMiller Sent stuff.

Anthony

05:43

@BalarkaSen You still there?

Balarka Sen

Yes.

Anthony

I probably should just search these theorems on my own

Actually

I don't know what to ask

I thought the galois group divided the $n!$ where $n$ was the degree of the extension

This is always true?

in the case of a galois extension, why is the galois group order $n$?

Balarka Sen

@Anthony What's your defn of a Galois extension?

Anthony

normal and separable

Rajesh Dachiraju

Hi @all

Balarka Sen

05:48

@Anthony Right, ok. Do you know the primitive element theorem?

Rajesh Dachiraju

Can some one explain this answer to me, mainly the part i mentioned in comment?

2

A: Placing delta's at maxima, Is there any smart equation based expression?

While I still agree with my comment that there might not be a nice formula, there is something. I just don't find composing a delta function with a function very nice. For a function $f$ with non-degenerate zeros we can naturally interpret $\delta(f(t))$ as $\sum_{a\in f^{-1}(0)}|f'(a)|^{-1}\del...

Anthony

@BalarkaSen I have both heard it,and should know it for my exam. It's just that any finite separable extension is simple, right?

Balarka Sen

Yes.

Now assume $L/K$ is finite (otherwise degree makes no sense) and Galois. Then it's also finite and separable. $L = K[\alpha]$ for some $\alpha$ algebraic over $K$ (the primitive element).

Equivalently, $L$ is the splitting field of the minimal polynom of $\alpha$, right?

Anthony

Mmmhmm.

Well, is that because the extension is normal?

Balarka Sen

@Anthony Well, yes.

Anthony

05:53

I see.

Balarka Sen

Now note that $[L : K]$ must be the degree of the min poly of $\alpha$.

Anthony

This is true, but doesn't that still just mean that the Galois group has order that divides $[L:K]!$?

Balarka Sen

Yes (as Gal acts on the roots of the min poly) but you can say something stronger.

I claim $|\text{Gal}(L/K)|$ is less than $[L : K]$.

Pick a $K$-aut of $L$, $f : L \to L$. $f$ is just determined by where it sends $\alpha$ though. i.e., there is a bijection $\text{Gal}(L/K) \to X$ where $X$ is the set of roots of the min poly of $\alpha$

Which is given by $f \mapsto f(\alpha)$.

You follow me?

Anthony

hold on, let me read real quick

Balarka Sen

You want an example?

Anthony

05:59

Possibly, although aren't you just saying that we're looking at permutations of the roots?

aren't there still $[L:K]$ roots?

Balarka Sen

There are $[L : K]$ roots. And no, we're not permuting roots.

Anthony

oh

Balarka Sen

We're looking at image of a single root by a galois automorphism.

Anthony

Where it sends $\alpha$, alone.

Balarka Sen

@Anthony Do you see why $f(\alpha)$ has to be another root of the min poly of $\alpha$?

Anthony

06:01

Perhaps not.

I don't think I do, actually.

Balarka Sen

Want me to reveal, or want a hint?

Anthony

Could you just reveal?

I'd think about it, but at this point there's a lot of stuff that I have jumbled.

Balarka Sen

I'll give the example anyway. Look at the simple extension $\Bbb Q(\sqrt{2})/\Bbb Q$. Let $G$ be the Galois group. With no priori knowledge of $G$, $\sigma$ be an arbitrary element of $G$. What is $\sigma(\sqrt{2})$?

Well, $\sigma((\sqrt{2})^2 - 2) = \sigma(0) = 0$.

Thus, $\sigma(\sqrt{2})^2 - 2 = 0$ (by homomorphism property + fixing of $\Bbb Q$).

So $\sigma(\sqrt{2})$ has to be a root of $x^2 - 2$ anyway.

Anthony

I see.

Balarka Sen

The same argument applies here.

$g$ be the min poly of $\alpha$. $g(\alpha) = 0$.

Anthony

06:06

Yeah, I think I see where the argument goes.

Balarka Sen

Thus, $g(f(\alpha)) = f(g(\alpha)) = f(0) = 0$.

$f(\alpha)$ is thus just another root of $g(x) = 0$.

Anthony

Wait, why were you allowed to switch the functions?

Balarka Sen

$g$ is a polynomial. Write it down explicitly. Then use the fact that $f$ is a field homomorphism.

Anthony

oh

Alright.

Balarka Sen

Are you clear on this?

Anthony

06:08

Yes.

Thanks, @BalarkaSen.

Balarka Sen

Ok, so $\text{Gal}(L = K(\alpha)/K) \to X$, $f \mapsto f(\alpha)$ is this nice map between sets. Why is it a bijection? Do you see?

Anthony

Let me think for a moment

Oh wait, you're just continuing?

Balarka Sen

Yes.

We were discussing an example above. Haven't proved yet why $|\text{Gal}(L/K)| = [L : K]$.

@Anthony Scrap bijection, first tell me why it's an injection.

Anthony

So on the one hand hand, a permutation is determined by its value on $\alpha$, on the other hand the value of $\alpha$ determines where it sends everything else, right?

I might have just said the same thing twice- but it's straight forward isn't it? Forgive me if I'm getting slower as the night goes on.

Balarka Sen

You should write the math down instead of saying vague statements. You need to prove the map above is injective - prove this from the definitions.

What does it mean to say it's injective?

Anthony

06:13

if $f(a) = f(b)$ then $a=b$

perhaps $f$ was the wrong choice of notation here

Balarka Sen

And what does it mean for that map $\text{Gal}(L/K) \to X$?

Yes, it was.

Anthony

If two permutations in the galois group mapping to the same root, they must be the same permutation

Balarka Sen

Not written in words, but a bit explicitly.

And there's nothing about permutations in my definition of the map.

Anthony

Yeah, sorry. Let's see: If $\sigma,\tau \in Gal(L/K)$ and $g(\sigma) = g(\tau) $ then $\sigma = \tau$ where $g$ is the map between $Gal(L/K)$ and $X$.

Balarka Sen

Look, here's the big picture. You note that L = K(a) for a primitive element a, you take the min poly g of a. You set for each element f of Gal(L/K) the root f(a) of g. So you're trying to figure out what the elements of Gal(L/K) are by trying to see where it sends a to.

@Anthony $g$ is again a wrong choice, as it was defined as the min poly of $\alpha$.

In any case, what was our map? What you wrote up there is not quite explicit.

Anthony

06:19

I'm not sure which map you were talking about, but I think I understand the gist of the argument.

Balarka Sen

The map Gal(L/K) --> X.

Anthony

You said what it was, didn't you? $f\mapsto f(\alpha)$.

Balarka Sen

Yes. Now what does it mean to say it's injective?

Anthony

lol, if $f(\alpha) = g(\alpha)$ then $f=g$

Balarka Sen

That's right. Do you see why it's true?

Anthony

06:22

Because the values the automorphisms in the galois group take on every element are determined by where they send the roots

I can't type properly

Frosty cake

Do we find the discriminant of $f=x^5+x^4-4x^3+3x^2+3x+1$ by calculating $Res(f,f')$? Is there a way I can do by hand?

Balarka Sen

Yes. Bit more explicitly, $f(\alpha) = g(\alpha)$ means $f(a_0 + a_1 \alpha + a_2 \alpha^2 + \cdots + a_{n-1} \alpha^{n-1}) = g(a_0 + \cdots + a_{n-1} \alpha^{n-1})$ for all $a_i$.

Anthony

Yeah, I was just writing something like that.

Thanks @BalarkaSen.

Balarka Sen

As every element of $K(\alpha)$ is like that, $f = g$ for all elts of $K$.

Anthony

Yeah.

So the map is injective.

Balarka Sen

06:24

But this just proves that $|\text{Gal}(L/K)| \leq [L : K]$, we have to prove $=$ :P

Anthony

Oh

I mean, you can consider such a map for each root.

Balarka Sen

So, any root of the minimal poly of $a$ is image of $a$ by some elt $f$ of $\text{Gal}(L/K)$.

Why is this true?

Anthony

I mean, isn't that just the case? If you give me a root, we can send $\alpha$ to it.

Frosty cake

@BalarkaSen Can you rephrase this? It doesn't make sense to me

Balarka Sen

Why can you extend that to an automorphism of $L$?

Anthony

06:27

But I suppose there's more, since I can't just send the roots anywhere.

Because everything in $L$ is just a polynomial in $\alpha$.

Balarka Sen

@Frostycake Don't know what to rephrase. Given any $b$, root of minimal poly of $a$, there is an $f$ in Gal(L/K) such that $f(a) = b$.

Frosty cake

@BalarkaSen $\sigma \in \text{Gal}(L/K)$ is a $K$-automorphism of $L$, and a $K$-automorphism of $L$ maps $\sigma(\alpha_i)$ to $\alpha_j$ where $\alpha_i,\alpha_j$ share minimal polynomial $f$

Did you ask him to verify why that is true?

($L$ is the splitting field of $f$)

Balarka Sen

Your formulation seems unparseable to me :)

Frosty cake

@BalarkaSen Ok :D

Balarka Sen

This is all I am asking him to verify.

Not the contrived thing you just wrote.

Frosty cake

06:31

Sure

Your question was different to my interpretation for the record

Balarka Sen

@Anthony Yes, it's not clear if you can get an automorphism.

Frosty cake

You are asking about the isomorphism extension problem

Balarka Sen

First time I'm hearing that name.

Anthony

Okay, given a $b$, if I take the function such that $f(a) = b$, then because everything in $L$ is some polynomial in $a$, we have that $f((p(a))$ = $p(b)$

Correct?

Balarka Sen

Function as in? What is $f$?

Frosty cake

06:34

Is $L$ the splitting field of $f$ in your discussion?

Anthony

I'm making a royal mess of all of this.

Balarka Sen

$L = K(a)$. $f$ is just a Galois aut, @Frostycake.

Frosty cake

Oh, no wonder the confusion

Anthony

What's the problem? We can make a function that sends $a$ to $b$ and extend it to everything in $L$.

Balarka Sen

Why can you extend? :)

It's not as clear as that.

Anthony

06:35

Because everything is a polynomial in $a$, right?

Balarka Sen

Yes. So?

Anthony

so some polynomial $p(\alpha)$ gets sent to $p(f(\alpha))$.

I mean, because we're looking for a hom

i feel like I'm saying the same thing over and over, and just wasting your time

Balarka Sen

@Anthony I am not sure if I understand you, but here's a proof. Note that $K(a) \cong K[x]/(g)$ where $g$ is the min poly of $a$ (hence also the min poly of $b$). But then $K[x]/(g) \cong K(b)$. Thus, we have an isom $K(a) \to K(b)$, which is the same as an aut $\sigma : L \to L$ such that $\sigma(a) = b$.

I don't think there's any easy way to do this by hand.

Anthony

That says there is an isomorphism between $K(a)$ and $K(b)$, but does that tell us $\sigma(a) = b$?

Balarka Sen

@Anthony Inspect the isoms $K(a) \cong K[x]/(g)$ and $K[x]/(g) \cong K(b)$ to see why.

Anthony

06:41

Alright.

So in this case you get equality.

Balarka Sen

Yes. $\leq$ and $\geq$ implies equality.

We have injective + surjective, hence bijection.

Anthony

Yeah. Thanks a lot for your time.

In general, why is it that the galois group divides factorial of the degree then? It must be a similar argument, right?

I mean, nix the first part.

Oh wait- primitive element theorem would give us the same kind of thing for any finite extension.

Balarka Sen

Galois group = degree, so automatically divides factorial of degree ... ?

Anthony

Sorry, I meant in the case of a non-galois extension

Balarka Sen

I think you mean why $\text{Gal}(L/K)$ embeds in $S_{[L:K]}$?

Anthony

06:46

Half of what I'm saying is nonsense. If we have a finite-non-galois extension, we can still apply the primitive element theorem, right?

Balarka Sen

Depends. If your extension is finite separable, then yes. I.e., non-normal is still good enough.

If non-separable, then you're screwed.

Anthony

Oh, duh.

@BalarkaSen Thanks for all the help.

Balarka Sen

No problem. Probably haven't been much of a clarifier though, mostly because I do not understand what I need to clarify.

Anthony

@BalarkaSen No you helped. I'm just caught in circles of nothingness. I'mma go to sleep. Thanks again.

Quintic

08:23

Just a question. How can I invite someone to my chat room?

Balarka Sen

09:37

hey @Huy

Huy

morning @BalarkaSen

want me to downvote you as well?

user174558

I just watched Hostel, very violent movie.

Huy

should have watched Star Wars

not sure what you didn't like about it

Samurai

Hello!

Balarka Sen

@Huy let's see. not right now. upvote on my self-answer though :D

have to get the lift hat

Huy

09:42

sorry I only downvote

I got a hat too, but it's hardly visible

we need larger avatars

Balarka Sen

ok, fine, downvote.

it'd be a good start.

thanks!

Samurai

I'm looking for a question on MSE, I saw it a few days ago. I want to add it to favourites.

Huy

always happy to help

Samurai

It was: Prove that $(n!)^{n+1}$ divides $(n^2)!$

Huy

@MikeMiller
* Unitary representations of compact Lie groups: Peter-Weyl theory, weights, Weyl character formula
* Introduction to unitary representations of non-compact Lie groups: the examples of SL(2,R), SL(2,C)
* Example: Property (T) for SL(n,R)
* Discrete subgroups of Lie groups: examples and some applications

can you tell me anything about any of these topics?

Balarka Sen

09:46

I have only heard of Khazdahn's property (T)

Samurai

Can anyone provide a link or tell me how to find it? I can't find it even after rigorous searching on MSE

Huy

what about it @BalarkaSen?

Balarka Sen

@Huy That it was some relation with number theory. I have to get my notes to get more detailed than that.

*has

Huy

The study of the Maslov index will allow students to combine methods from differential geometry and algebraic topology in order to understand this classical index introduced by Arnold and Maslov in the 60's.

The goal of this seminar is to understand and master the following concepts
- Lagrangian Grassmannian
- algebraic intersections
- axiomatic definition of the Maslov index
- explicit construction(s) of the Maslov index

or wat is this about @Mike

bolbteppa

@Huy That is the representation theory of Lie groups

Balarka Sen

09:51

I never understood the statement, @bolbteppa.

Wanna tell me?

bolbteppa

Which statement?

Balarka Sen

property T

that was what you were referring to, right?

Off-topic: Any good ideas on an HSM question? What about "Is it true that Steve Smale solved the h-cobordism theorem while sunbathing, sitting upon the Brazillian beach?"

Transcript for

$Mathematics$ Mathematics