groups-rings-fields

[last spoke]

Last year of undergrad

[delicate orchid]

[granite topaz]

which is quite funny, because I'm teaching a second-year about group theory 😄 we just passed by normal subgroups and are heading to group actions

[delicate orchid]

that's sad, why is algebra so under-represented in these unis

[granite topaz]

I must say I learned more about group theory that I taught

[last spoke]

Dude don't ask questions about my university, topology is a grad level course

[delicate orchid]

well at least they're consistent

[spice whale]

get a semigroups course in there

[upper pivot]

In america you have to take grad classes bc a lot of these UG classes cover what id consider a week or 2 of material of a grad class oof

[delicate orchid]

I get that it's your new hyperfixation but stop banging on about semigroups

[last spoke]

I mean it took my galois theory lecturer 4-5 weeks to get to galois maps

[dim widget]

They study a lot of semi groups in group theory ring theory and field theory

[delicate orchid]

YOU DID GALOIS THEORY BEFORE GROUP ACTIONS?!?!??!

[granite topaz]

heh? you had galois? and topo is a grad course?

[last spoke]

Oh we did group actions but brushed up on them

[granite topaz]

wtf is this uni?

[last spoke]

Galois theory is grad level too, we've just gotten to the group action bit of it and we're over halfway through the galois theory section

[delicate orchid]

:jawdrop:

[granite topaz]

TIL not all grad schools are the same

[last spoke]

When do you expect the Quintics have no roots bit to come up, it hasn't been discussed yet

[granite topaz]

You need Galois correspondence for that one

[upper pivot]

when I was at ross, me and my friends did a 10 hour lecture covering all of a galois theory course.

it was around hour 4!

[granite topaz]

With this pace, maybe before Christmas?

[upper pivot]

I think after galois correspondence makes sense

I like making it a motivitating example

[granite topaz]

proving Galois correspondenc would be a pain tho

[upper pivot]

like state galois correspondence and prove abel-rufini, and then go through proof of galois correspondence

[last spoke]

We have 3 weeks of galois theory left

[granite topaz]

Lmaoooo

[upper pivot]

honestly my hot take is this: the proof of galois correspondence is a waste of time in classes

dont cover it

[last spoke]

For an assignment due last Friday, the latest question from the textbook was proving x^4 - 10x^2 + 1 is irreducible in Q[x]

[granite topaz]

it's really technical, from what I can recall

[upper pivot]

yeah its not really insightful

[granite topaz]

Eisenstein?

[last spoke]

Yea Eisenstein into gauss' lemma

[granite topaz]

Can't guess of anything else. I didn't take Field theory

[last spoke]

I think

Oh we did fields in like 3 weeks lmao

[granite topaz]

Kekw, wtf could anyone cover in three weeks?

[upper pivot]

I mean this one you can kinda bash out lol, as you know the roots

[last spoke]

Oh yea I got that one fine

[glossy crag]

Lorenz in his Algebra 1 (which assumes only linalg at the German undergrad level) covers fields and Galois extensions before he does the concept of a normal subgroup, and introduces actions only in chapter 10, so it is perfectly possible to do this.

[last spoke]

The question before it was "Show if a_0 + ... + a_nx^n is irreducible in F[x] then do it a_n + ... a_0x^n" except that ain't true lmao

[granite topaz]

Lmao what? He really did prove Wantzel before introducing what a normal subgroup is?

Nahhh bro, this can't be right

[upper pivot]

Here are some fun exercises:

1. Find Gal(R/Q).

2. Prove that the Galois group of a 3rd degree irreducible polynomial over Q is A_3 iff the discriminant is a perfect square, and S_3 otherwise.

3. Prove the elementary symmetric polynomials generate all symmetric polynomials over any field.

[glossy crag]

What's Wantzel?

[last spoke]

^

[granite topaz]

Gauss-Wantzel theorem

[last spoke]

Nope

[delicate orchid]

:jawdrop:

[upper pivot]

oh the construction by straight edge thing

honestly who cares about that, i feel like thats the least interesting "application" of galois theory

[glossy crag]

This is page 77.

[last spoke]

Okay tbf when I took groups/rings/fields it was my 9am course

[granite topaz]

iirc, the proof starts with showing that constructible by ruler and compass is a field

[glossy crag]

He simply didn't need the concept of a normal subgroup before this point.

It's certainly unconventional, but I see the logic of it perfectly well.

[upper pivot]

I do think its fine to do rings/fields before group

[glossy crag]

His is a constructive approach, he introduces concepts as he needs them.

[granite topaz]

it either works spectacularly well, or it's a catastrophy

[upper pivot]

in which case normal groups wouldnt really show up until galois

[glossy crag]

First time I heard it called this.

[glossy crag]

The book? It's definitely fantastic, I recommend all to check it out, even people who already know the material.

[granite topaz]

There are two of them. Wantzel theorem about constructible number, and Gauss-Wantzel theorem about constructible angles.

[glossy crag]

Yep, it's more of a classical approach, I feel.

[granite topaz]

Wantzel was a student from Polytechnique, and French ppl do take a huge pride on that. Maybe that's why his name shows up

[upper pivot]

french moment

[last spoke]

Also calling something gauss' theorem doesn't really work

[granite topaz]

French ppl are definitely weird. I often know theorems by two different names

[glossy crag]

Russian texts are like this, they give a personal tame to every single result.

It's really annoying

E.g. the theorem critical point => derivative 0 is called Fermat's theorem iirc

[upper pivot]

we should number all of gausses theorem to avoid confusion!!!!

this can be gauss theorem 241!

[granite topaz]

yeah? try Euler's theorem

[upper pivot]

lol

[granite topaz]

rumor has it, that Cantor discovered uncountable number while trying to do that to Euler's theorems

[granite topaz]

What other ppl call Picard-Lindelof theorem, French ppl call Cauchy-Lipschitz theorem

God knows why

[granite topaz]

I vote for something more unorthodox to motivate students

Everyone knows Abel-Ruffini. It's a meme now.

[glossy crag]

Russians do too.

[upper pivot]

yeah but its standard for a reason

like, abel rufini is a very good showcase of how converting between fields and groups is useful

[last spoke]

Tf is Abel ruffini

[upper pivot]

quintic cannot be solved by radicals

[last spoke]

Oh that's what it's called

[glossy crag]

Surely there's more interesting stuff, but imo it's a very convincing proof at an elementary level of how useful field theory can be, multiplicativity of degree alone gives you answers to questions the Greeks were pondering 2500 years ago.

[granite topaz]

actually Abel-Ruffini is weak

It only shows that there's no general formula

[upper pivot]

maybe ocean but I always found this application quite boring, and a lot of my cohorts did too

[granite topaz]

What Galois shows, is that even for a particular quintic, it can be impossible

[upper pivot]

honestly to me the most interesting application when i first did galois was like

Prove the elementary symmetric polynomials generate all symmetric polynomials over any field.

this one

[granite topaz]

My favourite application is to show that for a polynomial, complex roots appear by conjugate pairs

I only appreciated its power when ODEs show up

[solar vessel]

uh

[glossy crag]

You mean a real polynomial?

That's it?

[granite topaz]

I should do more Math. I don't know math.

[solar vessel]

is this not obvious

[glossy crag]

That's your favourite application?

come on bro

[solar vessel]

like you can just conjugate

[upper pivot]

I mean ig this is the same idea as automorphisms rotating roots

but its kind of a trivial case oof

[glossy crag]

BTW if you're looking for applications to present in your class, you could talk about transcendental numbers. Lorenz has a chapter on the transcendency of pi.

[upper pivot]

transedantal galois theory

[solar vessel]

anyway the best thing we had in our Galois theory course was the classification of finite fields

who tf is abel and ruffini fr

[upper pivot]

oh yeah that shits good too

[granite topaz]

you can prove transcendency of pi without galois

just need a lot of willpower and analysis

[glossy crag]

I know, but the subjects are connected

• you can show C is alg. closed w/o appeal to analysis

that's another neat application

[upper pivot]

the galois theory proof still uses some analysis

[glossy crag]

all you need is that odd degree real polynomials have roots

[glossy crag]

obviously, because the structure of C is analytic

[upper pivot]

you cant truly avoid analysis, the reason C is algebraicly closed is analytical afterall

[glossy crag]

what i mean is it's simple

[upper pivot]

yeah fair

[glossy crag]

shrimple in fact

[upper pivot]

https://aareyanmanzoor.github.io/2023/08/07/Gauss-Sums-and-Kronecker-Weber.html

QR proof that uses some galois theory kek

(shameless plug)

[granite topaz]

Ehh, idk, analysis is my beloved

algebra is eww

[glossy crag]

Did you take this from Ireland-Rosen?

[upper pivot]

The gauss sum proof is quite popular, it might have been in ireland rosen

[glossy crag]

I mean, they're not the only source for it

[upper pivot]

I just motivate it using galois theory

[solar vessel]

I can't exactly say based but this is a lot of energy and I respect that

[glossy crag]

There's a really clean one using algNT, that's the first time i got it conceptually.

[upper pivot]

yeah I had to think hard about it bc I had to teach it. Like what is the motivation for this

[granite topaz]

I learn algebra out of spite, just so that I can shit on it

Also I'm teaching it now. Karma is a bitch

[glossy crag]

Exactly, I never get this or the standard proof, it's just cleverness.

Meanwhile the ANT proof is perfectly clear in how it works.

If anyone doesn't know it, check out Samuel's book or Brian Conrad's ANT notes, they have a section on this (it's standard material afaik).

[upper pivot]

Well what do you consider the standard proof kek. I guess the one of gauss counting the -

Or my blog!

[glossy crag]

it's Eisenstein's iirc

the one with counting points and lines and shit

hate it

[upper pivot]

I m psure its the same idea, its quadratic kronecker weber basically

oh lol yeah

[granite topaz]

I remember there's another proof of it, very beautiful

I know exactly where to find the book containing it in the library, but it's closed till next week

[glossy crag]

What kind?

[granite topaz]

it was included in the first chapter, which was intro to modern NT, so the author played with quadratic forms over integers a lot

[last spoke]

You want more cursed facts about my uni? Real analysis isn't needed for a math degree

[glossy crag]

And you're not in America? Truly cursed.

[granite topaz]

Maybe he's in Canada

[last spoke]

I'm in upside down canada

[granite topaz]

Can't think of anywhere else this cursed

So... Australia?

Brazil? Idk

[glossy crag]

Crikey!

[last spoke]

Australias canada

[glossy crag]

Aha

I see you engage in intimate relations with sheep

[granite topaz]

Where tf is that? New Zealand?

[last spoke]

Ye

[south patrol]

how do you do this using galois theory?

i've seen this used as a lemma in galois theory

[granite topaz]

Man... should have applied for math Master/PhD there. Why did I choose to stay in France?

[last spoke]

I could also skip linear algebra and multivariable calc

[glossy crag]

You can have Taika back btw, we don't want him anymore

[last spoke]

Taikas aight, fuck Thor 4

[glossy crag]

Fame has gone to his head and he's a hack now

[granite topaz]

So you do math, without knowing math

perfect

[glossy crag]

I contend that Jojo Rabbit was garbage too

[granite topaz]

Imagine I could graduate studying only Analysis

[glossy crag]

At least, in no way deserving of the massive praise it got.

[last spoke]

I only did real analysis after I graduated with my bachelors

[glossy crag]

Hunt for the Wilderpeople is good tho.

Sam Neil is a treasure.

Is Crowe officially yours or AUs?

[last spoke]

Idfk

[glossy crag]

How does it feel to hail from "LOTR Tour: the Country"?

[last spoke]

I've never seen LOTR

[glossy crag]

No true Kiwi

[last spoke]

Only Harvey weinsteins producer credit lmao

[glossy crag]

Fucker tried to stop the other 2 movies from happening, can you believe it.

[last spoke]

He shares my birthday

[glossy crag]

Truly, a distinction

I hope your privates aren't like his.

(supposedly his are gangrenous)

[last spoke]

[dim widget]

The chain Q(X_1,…,X_n)/Q(invts.)/Q(elem) is what we want to understand. The total extension is Galois with galois group S_n. However by definition so is the smaller extension

[upper pivot]

Basically take the field F(t1...tn), take s1...sn to be the symmetric polys. Look at the extension F(t1...tn)/F(s1..sn). Prove this is Galois, Galois correspondence says the fixed field of S_n is precisely F(s1..sn), which is the result we want

[south patrol]

okay sure

nice

[upper pivot]

sniped oop lol

[glossy crag]

But that just shows symmetric rational functions are rational in s_i, you still have to show this for polynomials.

[upper pivot]

hmm

[glossy crag]

Also, the proof I know of this already presupposes the symmetric polynomial theorem.

[upper pivot]

I think it shouldnt be too hard to get the poly out of this, let me think

[glossy crag]

I mean the proof I know that F(s1,...,sn) is the fixed field uses the theorem.

[upper pivot]

Sure, but the galois theory gives this to you much more easily

[glossy crag]

So how do you show it's the fixed field w.o the theorem

[upper pivot]

Galois correspondence ofc

it says the fixed field of the galois group, S_n, is exactly the base field

[glossy crag]

Proof I know: let L=K(x1,...,xn), S_n is a finite subgroup of AutL, hence by Artin L/L' (where L'=L^S_n) is Galois with group S_n. K(s1,...,sn) is obv contained in L', for the converse use the SymPol theorem.

Idk how to do it w.o SPT.

[upper pivot]

I mean you literally just look at K(x1..xn)/K(s1..sn). That this is Galois follows from the fact that x1...xn is easily roots of a polynomial

Each permutation of the xi is obviously an automorphism, and fixes K(s1...sn).

The galois correspondence tells you that they all fix nothing bigger

[glossy crag]

You need to use that the degree is at most n!, no?

[upper pivot]

Its the splitting field of a n degree polynomial, so that follows. Ig that shows the galois group is atmost S_n and what i said shows the galois group contains Sn

[glossy crag]

Yep

Still, that just shows a sympol is rational in s

[upper pivot]

I think for that you just say, a polynomial in K[x1..xn] is integral over K[s1...sn]

which follows bc the extension here is integral

[last spoke]

So uhh, how many hours would one expect galois theory to finish from Kummar theory

Where the last lecture started with proving Gal(E/F)/Gal(E/B) is isomorphic to Fal(B/F)

[chilly radish]

Ours took about 5 weeks

3 hours a week

And we moved at a decent pace

But you might do different or less topics in-between

Even more maybe

[wicked harbor]

I've got a question. Since $\sigma$ extends $\tau$, isn't it meant to be the case that $\sigma(i(a)) = i'(\tau(a))$ for all $a \in F$, where $i: F \rightarrow E$ and $i': F' \rightarrow E'$? Why then is it written here that $\sigma(a) = \tau(a)$?

[cloud walrus]

sunnyside1

[wicked harbor]

Also, it is defining an extension $E/F$ to be normal if every irreducible polynomial $f(x) \in F[x]$ with a root in $E$ splits into linear factors over $E$... Is this just the same as saying that $E/F$ is normal if $E$ is the splitting field of every irreducible polynomial $f(x) \in F[x]$? But the splitting field is defined to be the minimal field with that property, so would it rather be defined to be normal if $E$ contains the splitting field of every irreducible polynomial $f(x) \in F[x]$?

[cloud walrus]

sunnyside1

[chilly radish]

This is definitely a mistake or a misprint because tau(a) and sigma(a) are not even in the same field

[chilly radish]

No, every splitting field contains E

Because E is the smallest

So it's the intersection

[last spoke]

That's what we're looking at, 6 weeks, 3 hours a week but with 2 lectures missed. How long did it take to get to the Quintic shenanigans tho

[chilly radish]

About 2-3 weeks

[last spoke]

We haven't done that yet

Took 4 weeks to get to galois maps and the stuff Sunnyside is discussing

[chilly radish]

I meant from the point you stated

After the fundamental theorem and the quotient thing

[last spoke]

Oh

What about 2 hours

[wicked harbor]

Yeah, this is how it is structured for me:
Week 1: Review of rings, basic stuff about fields
Week 2: Irreducibility tests, field extensions
Week 3: Minimal polynomials, algebraic extensions, tower law, algebraic closure, embedding theorem
Week 4: Splitting fields, normal extensions, automorphisms, primitive element theorem
Week 5 and 6: Fixed fields, Artin's theorem, Galois extensions and groups, Galois correspondence, fundamental theorem of Galois theory, fundamental theorem of algebra, solubility, radical extensions, Galois' great theorem

Then they cover modules up to the fundamental theorem of finitely generated modules over a principle ideal domain for the second half of the semester.

[wicked harbor]

I guess that it'll take 5 weeks or something to get to quintic shenanigans in my case

[last spoke]

We've said that splitting fields is normal extensions but we haven't gone into that yet

That looks like we might get done if my lecturer speedruns like he's never done before

[wicked harbor]

I just got to a result that a finite extension E/F is normal if and only if E is the splitting field over F of a polynomial f(x) in F[x]

[wicked harbor]

What other stuff do you have in the topic? It seems like it's half fields

[last spoke]

Idk man it's called galois theory and number theory. Number theory is after this so he mightve been building some stuff in for that

[wicked harbor]

Also, the whole Galois correspondence and fundamental theorem stuff looks insane... Don't know how I will do with understanding them... But they look very beautiful, potentially will be up there as my favourite theorems in undergrad

[last spoke]

This is grad level for me lmao. Have fun

[wicked harbor]

Yeah, this is probably the crux of the theoretical stuff for your theory, and the rest would be applying it to number theory problems

This is final year undergrad, all dedicated to fields and modules

[last spoke]

I mean this guy is a number theorist so it makes sense why it's half and half

[wicked harbor]

This topic is fun. I'm also doing some diff geo of surfaces and stochastic processes... The stochastic processes stuff is cool, but then they add in matlab and I haven't ever coded, so I'm dying

[last spoke]

Oh yea fuck Matlab, schocastic processes is dope though

I'm currently doing matroids and PDEs because half the grad courses here need PDEs

[wicked harbor]

Yeah, we're covering Kolmogrov DEs and birth-death processes this week. Pretty neat

[wicked harbor]

oof, sounds intense

[last spoke]

Kolmogrov DEs can go fuck themselves

[wicked harbor]

When I hopefully get to masters, they offer stuff like category theory, algebraic topology, differential geometry, and stuff like that. Pretty cool stuff

[last spoke]

Idk what I'm doing, I'm just vibing in postgraduate courses rn. Since I decided to go a more discrete path through undergrad I couldn't complete masters in time without picking up more pre-reqs

[wicked harbor]

What do you think you want to specialise in?

[last spoke]

I have no fucking clue, I'm not smart enough to think about that

The stats department likes me because I'm like the only math grad that took statsy courses, but discrete math is fun

[wicked harbor]

Stats sucks

But probability is alright

Studying pure stochastic processes would be cool

[last spoke]

Yea probability is fun, some stats is aight but not my favourite

[wicked harbor]

I'm wanting to learn some commutative algebra after this topic though. Interested in cracking into algebraic geometry eventually

[last spoke]

Algebra is okay, better than anything with derivatives

[wicked harbor]

I used to think that I would do analysis, but algebra is cool imo

I've done some basic functional analysis, and that was cool though

[last spoke]

I got an A+ in real analysis though I did it after I got my BSc

[wicked harbor]

This is such a nice proof and a really cool exercise too!

[tough raven]

How can I show for any $a,b,c,d$ in any commutative ring $R$ that if
\begin{equation*}
a^2 + b^2 = c^2 + d^2 = ad - bc = 1, ac + bd = 0,
\end{equation*}
then $a = d$ and $b + c = 0$?

[cloud walrus]

Raghuram

[tough raven]

Source: a textbook essentially implies in an exercise that $$\operatorname{SO}_2 \cong \left{ \begin{psmallmatrix} a & b \ -b & a \end{psmallmatrix} ;\vert; a^2 + b^2 = 1 \right}.$$

(It doesn't assume characteristic not equal to $2$ or anything and doesn't seem to assume that we are working over a field, just a commutative ring.)

[cloud walrus]

Raghuram

[solar vessel]

did they define it some other way

[glossy crag]

The splitting field of t^n-x_1t^{n-1}+...+(-1)^nx_n over K(x1,...,xn) is again a function field in n variables. Is there any chance this is always like this, i.e. every finite extension of K(x1,...,xn) is again K(y1,...,yn)?

[rocky cloak]

So you have
$a^2d - abc = a$

Which gives
$a^2d + b^2d = a$
So a=d

[cloud walrus]

jagr2808

[rocky cloak]

You also get a(b+c) = 0 from this

[solar vessel]

[rocky cloak]

And -bc = b^2

Thus b^2(b+c) = 0, and thus

(b+c) = (a^2 + b^2)(b+c) = 0

[tough raven]

matrices X st XX^t = Id, det(X) = 1
ie a,b,c,d satisfying my hypotheses

[tough raven]

Nice, thanks!

[teal vessel]

I love doing two proofs at once. "The center of any dihedral group must be {1, r^(n/2)}, if they exist (because reasons not included for brevity). 1 always exists, and r^(n/2) exists iff n=2k for some integer k, therefore, the center is 1 if n is odd, and {1, r^k} if n is even."

is it cheap to assume existence? maybe, but it gets wrapped up nicely in the end.

[slim kayak]

Is this the channel for combinatorial/geometric group theory?

[delicate orchid]

depending on the level, either here or #advanced-algebra

[slim kayak]

Thanks

[untold basalt]

Is there an infinite amount of fields?

[chilly ocean]

Yes, even just finite fields.

[untold basalt]

actually, of infinite fields*

yes I know about Z_p, I was wondering about infinite ones

[chilly ocean]

Yes, should be true as well

[untold basalt]

But is there a finite amount of "ways to construct a field"? For example all Z_p are "the same kind of field". Is there a finite amount of "kind of fields" in this sense?

[brisk bison]

"Ways to construct a field" is a bit too vague to answer this question. I assume you mean to include taking quotients as a way to construct a field. The answer seems like "yes" for some appropriate notion of construction, but not for any mathematical reason, and moreso by the fact that humans only know a finite amount of mathematics.

[slim kayak]

No, you at the very least have one Z_p for each p, of which there are infinitely many

But it's still an interesting question. The category of fields sucks, or more concretely, building new fields isn't very nice and more complicated than for other objects

[formal ermine]

Q_p lol

[topaz solar]

We do have ultraproducts, which you might consider the different ultrafilters used as different “ways” to construct idk

[slim kayak]

What does the ultraproduct tend to "look like" ?

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

This is uhh, not an integral domain probably

(Also, assuming none are empty but that’s irrelevant since fields)

[slim kayak]

Depends on your operation, idk

[topaz solar]

So uhh, the operations are pointwise here

It’s the usual product

[slim kayak]

So the F_x as the ring product?

Gotcha

[topaz solar]

But we want a field

So… take a maximal ideal containing things which are cofinitely zero

That is uhh maximal ideal containing all X-sequences which are nonzero on finitely many inputs

[slim kayak]

Wdym, one of those two (0,1,1,...) and (1,0,0,...) is zero

Or good enough of approximation

[topaz solar]

As in, (1, 0, …) ~ 0

[slim kayak]

I was still at the "integral domain" bit lol

[topaz solar]

Multiply those together and u get 0

[slim kayak]

Cofinitely 0?

[topaz solar]

$X-{i\in X|x_i = 0}$ is finite

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

That is, it’s nonzero on finitely many points

Have you seen the uh

Restricted product?

[crystal turtle]

Product of integral domains is not an integral domain
I'm not sure if the category of integral domains has products actually

[topaz solar]

It has ultraproducts tho

[slim kayak]

Not really an achievement tbh

[topaz solar]

real

[crystal turtle]

based

[slim kayak]

Consolation prize

[topaz solar]

This makes it a field again though

And uhh it looks like an averaging out of the F_x

[cloud walrus]

Dragonslayer Sharp

[crystal turtle]

integral of fields

[topaz solar]

A first order statement is true in this iff it’s true almost everywhere for the ultrafilter \mu

Where’s our ultrafilter? You might ask

[slim kayak]

No disrespect but this kinsa sounds like someone desperately trying to explain how checking "field have products" wasnt a mistake on the exam

[topaz solar]

Well, we took a maximal ideal, and exactly one of 1_A or 1_{X-A} is in our maximal ideal right?

So there’s an ultrafilter, and it’s nonprincipal too

In particular, you can get some fun results

[slim kayak]

Oh, is this choice of maximal ideal canonical in some way?

[topaz solar]

Not in the slightest

That’s just how maximal ideals are

[slim kayak]

Or how is this distinct from saying that you take ring product and then look at some random quotient field

[crystal turtle]

It will probably not be functorial, if that's what you're asking lmao

[topaz solar]

$1_A \cdot 1_{X-A} = 0$

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

It is functorial if you start from an ultrafilter on X tho

And has universal properties

Iirc

[crystal turtle]

oh cool, even the quotienting?

That's kinda surprising

[topaz solar]

Well yeah it’s a filtered colimit

[crystal turtle]

fair

I do just be talking without thinking sometimes

[topaz solar]

It’s a prime ideal so the quotient has to contain at least one of these

And you can’t have both

Since ya know

1_A + 1_X-A = 1 (in our ring)

[brisk bison]

The properties of the quotient depend not just on the terms in the product, but also the ultrafilter used.

[slim kayak]

Where do we get the diagram from?

[topaz solar]

And the fact that it’s maximal means the one we have left is a unit

So it’s just 1

[crystal turtle]

Sets in the ultrafilter, ordered by inclusion iirc

[topaz solar]

Lmao

[topaz solar]

Reverse inclusion

[crystal turtle]

to make it filtered

from filters being downward directed I guess

[topaz solar]

$$\lim_{\rightarrow\mu(S_0)=1} \prod_{s\in S_0} X_s$$

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

Something like that

The maps between them are gonna be ur projections and all

[topaz solar]

True, but the fact that it’s a field doesn’t!

And that it’s still got Łoś goin

[slim kayak]

The category of fields has colimits at least, if you ask very nicely

[topaz solar]

In particular, as you can see from the screenshot I posted, the ultraproduct of the algebraic closures of finite fields is just C

Regardless of your choice of maximal ideal

[topaz solar]

The whole thing is done in set for all we care

Łoś’s theorem pulls mad weight

Thinking about it categorically isn’t the most enlightening

[slim kayak]

Sure, it was kind of how this started going

How the category of fields category kinda sucks

[topaz solar]

Categorical approach just generalized it somewhat, the real leg work is just set models

Anyway, your ultraproduct averages out your field characteristics

Averages out your algebraic closed-ness

it has characteristic p>0 iff we have 1_A not in the maximal ideal and F_a for each a in A has characteristic p

Algebraically closed with the same condition, but replacing characteristic

[slim kayak]

It accepts any first-order formula 💀

[topaz solar]

Yep

Including ones asking if something is a ring

Or a field

[slim kayak]

Set theorists try not to make megalomaniac theorems challenge (impossible)

[topaz solar]

what part of this is megalomaniac

[slim kayak]

Did it have some use to answer some big weird problems about models and stuff like that?

[topaz solar]

It’s just like

Yeah bro induct of formulas

(And use a choice principle on the \exists part)

[topaz solar]

It’s a nice construction, and kinda immediately gives you arbitrarily large models of theories

And if your ultrafilter has nice combinatorics? Your theory has nice properties, such as k-saturation

I honestly don’t know what part sounds megalomanic to you

[slim kayak]

It just seemed very grand, thats all

[spice whale]

what is an ultraproduct

i keep seeing the word

[dim widget]

@split cipher

[topaz solar]

@spice whale

Or ya know

[spice whale]

it's measure theory?

that's unexpected

[topaz solar]

$\prod_{x\in X} A_x$ as a set, modulo $x\sim y$ iff ${i|x_i=y_i}\in U$

[cloud walrus]

Dragonslayer Sharp

[spice whale]

where U is an ultrafilter on some set i assume

[topaz solar]

\mu being a characteristic function of U, which is a finitely additive measure on P(X) where \mu(X) = 1 and exactly one of A and X-A has size 1

It being a finitely additive measure falls out from the requirements of the ultrafilter

[spice whale]

what are x_i and y_i

[topaz solar]

On X, here

and x, y are in \prod_X A_x, so they’re functions in X

[spice whale]

oh i see

[topaz solar]

Ye

[spice whale]

so i is an element of X?

[topaz solar]

Modulo “almost everywhere equivalence”

[topaz solar]

Yep

[spice whale]

and what's A_x

[topaz solar]

Whatever you desire

[spice whale]

so X is just an index

[topaz solar]

Łoś’s theorem guarantees this ultraproduct satisfies a first order formula

iff

It’s satisfied for some collection of A_x for x\in A \in U

That is, iff it’s true almost everywhere

Iff it’s false almost nowhere

Hence the averaging sort of statement and the integral symbol

[topaz solar]

Yep

[topaz solar]

This sort of construction works in some subcategories and all (think fields as a subcategory of rings), and you can formulate Łoś as some statement about pretopos functors into set or smth, it’s all about the same though any time set-looking things are involved

They also just kinda behave nicely

[wooden ember]

just to be sure im not saying something stupid, we always have $\operatorname{Hom}_K(L,K^{sep})=\operatorname{Hom}_K(L,\overline{K})$ right?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

(having fixed an algebraic closure and taking the separable closure within that algebraic closure)

for L/K separable i mean

cause something on the right is uniquely determined by its corestriction to K^sep

[chilly ocean]

Do finite fields always have a square root of -1?

[teal vessel]

so if we remember that I'm dumb and overthink things, is this sufficient to show that for a subgroup H of G s.t. |H|=2, the centralizer and normalizer of H in G are the same?

1. it is known that C(H) ≤ N(H) since any element that commutes with all members of H also conjugates H to itself.
2. Suppose we have some element x in N(H), with which we wish to conjugate the non-identity element of H (let's call it h). There are two cases: that xhx^-1 = 1, or xhx^-1 = h (because of membership of N(H)).
3. In the case of mapping to identity, we get the following:
   xhx^-1 = 1
   x^-1 (xhx^-1) x = x^-1 x
   h = 1
   which is a contradiction. The calculation for the other case is omitted, but does not yield such contradiction.
   Therefore, any element in N(H) must commute with h, which satisfies the definition of membership to C(H)

wow that's a lot of alliteration in my second step lol

[wooden ember]

no this only happens for cardinalities congruent to 2 or 1 mod 4

[wooden ember]

it's fine but you didn't need to say "the calculation for ... such contradiction"

you showed you cant have xhx^-1 = 1 and thus xhx^-1=h => xh=hx for all x in N(H). That's all that matters

[teal vessel]

habit from professors who told me not to assume such elements exist.

[dim widget]

No, what if L is \bar{K}?

[dim widget]

Ah yes

Being separable over a subfield is an intrinsic property, so a separable field cannot map to an inseparable one in a way which is compatible with the K-algebra structure

[wooden ember]

Yeah that was important lol

[wooden ember]

Yeah that was my intuition too thanks

[wooden ember]

If no such elements exist then N(H) is empty which isn’t possible as it always contains the identity

But saying something « won’t lead to a contradiction » is not a nice habit to get into imo

[teal vessel]

technically I omitted the "non-identity" stipulation for the x element, but yeah.

[teal vessel]

more of "check all cases before making assertions"

[glossy crag]

All GF(2^n) have this property trivially, so not just cogruent to 2.

[little shadow]

Is Z(N) normal in G when N is normal in G?

[coral shale]

N normal in G iff (forall g in G) gNg' subset N

Hence for any subgroup of N...

[solar vessel]

no

[coral shale]

heck did i do

oh crap

[coral shale]

ah so i guess not necessarily

or wait no, yes necessarily

[coral shale]

im unsure any more and dont wanna think

[solar vessel]

I would guess not necessarily

oh wait

should be yes

the centre is fixed under automorphisms

and conjugations induce automorphisms on N because it's normal

implying that the centre of N is fixed by conjugations and is thus normal

(when I say fix I mean it is the image of itself)

@little shadow

so yeah it is

[formal ermine]

center commutes with everything

gZ(N)g' = Z(N)gg' = Z(N)

Wait

center of a subgroup

[solar vessel]

ye hence ya gotta be careful

[indigo plinth]

If z in Z(N) and g in G and h in N, then
g'h(gzg') = (g'hg)zg' = z(g'hg)g' = zg'h =g'(gzg')h
cancel to get
h(gzg') = (gzg')h
So gzg' commutes with all of N, meaning gzg' in Z(N).

[indigo plinth]

But this is quite nice

[next obsidian]

Characteristic in normal => characteristic

Normal in normal =/> normal

[warm wyvern]

I like this

damn I can't believe smth lumin posted made me laugh

[solemn garden]

If R mod Z is like a circle, how do you visualize R mod (Z + sqrt(2)Z)

[coral shale]

R mod (a + rt2)

In the same way Id visualize R mod Q probably

Mark Z + rt2Z on real line

other equivalence classes are from shifts by real numbers not in Z + sqrt(2)Z

===
ig im taking a look at how this quotient group acts on itself, I believe

[coral shale]

R/Z works as a circle is more to do with topology. Z divides R into equal chunks, so it 'works' nicely

[delicate bloom]

I think you could even build on that in steps from the circle picture and mod sqrt(2)Z on that

[coral shale]

oh 3rd iso?

[delicate bloom]

idk, I'm just thinking about the cosets

x + Z + sqrt(2)Z also looks like it came from (R mod Z) mod sqrt(2)Z

idk is that the third iso? I don't have them memorized

[coral shale]

R/(Z+rt2 Z) = (R/Z)/((Z+rt2 Z)/Z)

I think thats 3rd iso

hmmcat. feels somethings up to me

Z+rt2 Z mod Z is looking dense in R mod Z

but ig that makes sense

[delicate bloom]

yeah, we can then imagine breaking up the circle into N equal parts and then distributing multiples of sqrt(2) around it, once you have more than N, by the pigeonhole principle you know that there are two that are within 1/N of each other.

we use this kind of thing to arbitrarily approximate any irrational number by a rational

[coral shale]

Very cool each R/(Z+rt n Z), n squarefree, dont touch each other except at one place

[delicate bloom]

yeah you can do any irrational, R/(Z+piZ) for instance

to elaborate a bit, so like let's say you take multiples of some irrational number a. you can then place the N+1 multiples of a around the circle so your multiples are 0 <= m_i < m_j <= N. Then you can keep track of what integer you subtract by to get it in the range [0,1), so now you know by pigeon hole that |(m_i a - n_i) - (m_j a - n_j)| < 1/N now you can rearrange to make |(n_j-n_i) - (m_j-m_i)a| < 1/N and so if you call those |x-ya|<1/N you can then show y < N and so we have |x/y - a| < 1/y^2.

[coral shale]

analysis techniques being used. devastation

[solemn garden]

the reason why I asked was that I realized a + bsqrt(2) could approximate any number if a and b were integeres

so it would have to be something different than a circle

but not a really small circle either

[coral shale]

🙄 readup from the top

[solemn garden]

what does this mean

[coral shale]

the circle we re talking about is the same R/Z

rather than the one youre thinking of which would need to be infinitely small

We're embedding the quotient you wrote into R/Z

[coral shale]

using 3rd iso

[solemn garden]

what does (Z + sqrt(2)Z)/Z look like

[coral shale]

that thing mod Z

so fractional part of multiples of sqrt of 2

[coral shale]

which as you observed, can approximate any real number

so this set is dense in R/Z

[static needle]

Guys quick question: is there a general method for finding quotient groups?

I get the definitiob

But im looking for like a method

[coral shale]

what does find mean though

[solemn garden]

find the set of cosets of the normal subgroup you’re quotienting by, then work out the multiplication table of the cosets

@delicate bloom were you typing something?

[static needle]

I mean calculate i guess

[coral shale]

Z + sqrt n Z
for n squarefree
dont touch each other except at integers

[coral shale]

thonk, gib example

[solemn garden]

oh

[coral shale]

i think eogs told you though

[static needle]

Great! What if theres no group isomorphic to the set of those elements in the mult table? Or like I cant compactly write it? Do I just express the entire set?

[solemn garden]

they have to form a group

[coral shale]

Give example

Cayleys theorem is a thing

[static needle]

So they are guaranteed to be isomorphic to some group which has compact notation?

[solemn garden]

there is a group isomorphic to them, namely themselves, since they have to form some group

[static needle]

Ayt

[delicate bloom]

yeah, basically I outlined a proof that there exists integers x, y such that |x-y sqrt(2)| is arbitrarily small, and since those are both 0 mod Z+sqrt(2)Z, we can squeeze anything arbitrarily close to any point we like

[coral shale]

For large enough groups it may be complicated

to express as a product

[solemn garden]

Cayley’s theorem guarantees that they are isomorphic to a permutation group, I guess you could call that “compact”

[coral shale]

i think

Also all groups are the quotient of some free group

so theres a presentation

[solemn garden]

ah, I see

[coral shale]

in any case

[static needle]

Ayt i cant believe i didnt know cayley's theorem

Im screwed in alg top

[coral shale]

this is probably the more useful one in at

[solemn garden]

and what do you mean by “compactly write”

[static needle]

Arent those abelian groups?

[coral shale]

no?

[static needle]

Notation wise I suppose

[coral shale]

Also, G/H

[solemn garden]

free groups are not always abelian

[coral shale]

is pretty compact already in terms of notation

[solemn garden]

Can you elaborate or provide an example?

[static needle]

hmm

R/Z i suppose?

[coral shale]

yes you keep it as R/Z

thats it.

[static needle]

Neatly, it's isomorphic to S1

[coral shale]

uhhh

[solemn garden]

if you have to write the elements, you can write [x] meaning the equivalence class of x

[wooden ember]

Hurb yeah forgot about those

[coral shale]

S1 under complex multiplication?

[static needle]

Ayup

Yeah complex mult

[coral shale]

well but like

[static needle]

Theres a homomorphism

[coral shale]

R/Z looks good too

[static needle]

From R to C

[static needle]

Yeah it does but I love writing it out as S¹

[coral shale]

[static needle]

Purely to confuse people

[coral shale]

topologists dont care about R/Z as a group (only) anyways

[static needle]

Buuuut if they know, they shouldnt be lol

[coral shale]

as a topological group probably?

[static needle]

Mhmm

[coral shale]

but also the ring structure?

so topological ring

[static needle]

S¹ is a lie group

Both a group and manifold

[coral shale]

idk what that is

ic

[static needle]

One can construct it via the manifold R quotient with all points that are in Z

[coral shale]

R/Z is a field right, ive forgotten

nvm no

[static needle]

Except Z is now a topological subspace

And we get a quotient topology

While now here in group theory we construct a homo from R to C

Via the map exp(2iπx)

I think

[coral shale]

homomorphism from R, + to C*, *

[static needle]

Whoops yeah i meant that. I was sorta implying that you alr knew those two as groups

Wrt addition and mult respectively

[coral shale]

yh well, just to be clear

[static needle]

Yep

Another lie group, I think, is the torus

It's now $\mathbb{C}/\mathbb{Z}[i]$

[cloud walrus]

messyinterval

[coral shale]

wth is this object

[static needle]

A torus!

[coral shale]

surely that is not a torus

[static needle]

Donut

[dim widget]

A donut

[coral shale]

it is?

[dim widget]

🍩

[coral shale]

damn gotta think

oh ok it is nvm

[static needle]

Yep!

I think the explicit map involves cosine and sine but eh

[dim widget]

The frosting on top is an arithmetic structure and the sprinkles are complex multiplication

[static needle]

No, fellow topologist... CW complexes!

The frosting are CW complexes!

[dim widget]

Yes it is a donut with cw complexes on top!!

[static needle]

Yosss!!!

Also we're getting way outta line this is a group theory channel lmfao

[dim widget]

An elliptic curve is a group

[static needle]

Anyway, neatly enough we can express T² as R²/Z²

[coral shale]

the topologists have come

[static needle]

Time to get more cursed

SO(n) is a lie group represented as a n-ball

You heard me

S-O-en

The matrix group

[dim widget]

Wait SO(n) isn’t an n-ball though

[static needle]

Isnt it?

Wait

[dim widget]

It’s an iterated sphere fibration

[static needle]

Hmmm

[static needle]

Ah—

I have only a rudimentary understanding of fibrations—

[dim widget]

SO(2) is a circle and SO(3) is S^3

After that you get an iterated fibration of odd spheres

[static needle]

Yep i meant that i guess. I was jumping to conclusions with the whole 'SO is a manifold' thing

[dim widget]

But you always have SO(n-1) \to SO(n) \to S^{n-1}

[dim widget]

It is a manifold! Just not that manifold

[static needle]

Might wanna latex that?

[static needle]

Is there a general uhh expression for it?

[dim widget]

I do not 😄

[static needle]

Like SO(n) is what as a sphere?

[static needle]

Eh i sorta understand it? Why the arrows tho

[dim widget]

It is an SO(n-1) bundle over an n-1 sphere

[static needle]

Errr

Yes...

[dim widget]

It’s a standard way of writing a fibration “like an exact sequence”

[static needle]

Ahhhh

I think I remember a lecture by niles johns (i think? Jones?) On fibrations and I saw him write out a sequence using arrows

[dim widget]

This just means it has a smooth map to S^{n-1} which is a submersion and all fibers are SO(n-1)

[static needle]

Immersion?

[dim widget]

And there is an action of SO(n-1) which acts transitively on the fibers

[static needle]

Never heard of the term "submersion"

[dim widget]

No submersion

Submersion is a map of manifolds which is surjective on tangent spaces

It is the counterpart to an immersion

[static needle]

Ah

It's a map from a manifold to a tangent space that is surjective?

[dim widget]

No no

An immersion is a map of manifolds which is injective on tangent spaces

A submersion is a map of manifolds which is surjective on tangent spaces

[static needle]

Man i never got this deep into topology so sorry if im illiterate

[dim widget]

No worries 🙂

[static needle]

Damn

Ok so

[dim widget]

Anyway all matrix groups are Lie groups SU(n), Gln, etc

[static needle]

Good to know!

Even O(n)?

(Well duh, right?)

[chilly ocean]

Is here any typo

[static needle]

Ok what

That isnt a quotient group?

Oh wait those are fields

Damn

[coral shale]

K/F is a field extension, notated K : F in some texts

[coral shale]

idk what typo u think there might be

[chilly ocean]

F must be a fixed field of Aut(K/F)

[coral shale]

Why do you think that could be a typo

(idk, btw but)

I feel like separability comes into play surely and hasnt been mentioned. No idea...

Is the characteristic infinite from context

[rocky cloak]

SO(3) is projective space, so it's only SO(2) that's a sphere I think

[rocky cloak]

The fixed field is the set of elements fixed by the group action. F is necessarily fixed, but the fixed field could be bigger

For example say F=Q and K = Q(cuberoot(2)). Then Aut(K/F) is the trivial group, so the fixed field is K. And indeed

|Aut(K/F)| = 1 <= 3 = [K : F]

[rocky cloak]

Yeah, F will be the fixed field iff the extension is seperable and normal. That's called being a Galois extension.

[coral shale]

Ah ok.

So fixed fields dont exist in non-separable

thonk

[rocky cloak]

Well if you have an extension that isn't seperable, you will have some element that is the only root of its minimal polynomial. So it would have to be fixed by any automorphism

[static needle]

I prefer the K:F notation. Jeez.

[coral shale]

The reason why it wont cause confusion is because fields dont have any interesting ideals

nothing to quotient in the ring sense

[dim widget]

Grothendieck preferred the other one, guess who wins?

[coral spindle]

It’s nice to have the notation [E : F] which mirrors the index notation used elsewhere

[static needle]

WHAAATTTT

NOOOOO

but but but quotients ;-;

[dim widget]

they're not quotients they're slices!

[static needle]

Ermmm in the field sense... Im...not following you

[dim widget]

A slice category $\mathcal{C}/y$ is a subcategory of objects $x \in \mathcal{C}$ equipped with a morphism $f: x \to y$. So for instance $L/K$ means $L$ considered as an object of the slice category $\text{Field}^{op}/K$. Where the op is for complicated reasons.

[cloud walrus]

Topos_Theory_E-Girl

[static needle]

Oh uh

Nice

[coral spindle]

There is only one subfield of any field that forms a quotient.

[static needle]

Ah...nice

[coral shale]

is the trivial field a field

wouldve thought 2 but ok

[coral shale]

Is there no clash in notation in category (absolutely no clue)

with regards to quotient things

[coral spindle]

0 is not a subfield of any nontrivial field, because it does not contain the identity

[coral shale]

{0, 1}

not an ideal right

ok gj me

[coral spindle]

Not an ideal.

[coral shale]

aaa it just feels weird you dont have a smallest thing to quotient by

F/1 = F

F/F = 1

kinda thing

[gilded torrent]

Yo nerds have some fun

Math is lame

Let's talk about something fun

[coral shale]

@gilded torrent Can you stop please. #chill for shitposting.

[gilded torrent]

Kk nerd

I'm so sorry it was a dare

😅😅

Have fun doing math

[coral spindle]

The smallest thing to quotient by is the zero ideal, but this is not a subfield as I said

[coral shale]

Right I was confusing myself. math makes sense again...

[dim widget]

$0 \times 0 = 0$

field with one element discovered 😱

[cloud walrus]

Topos_Theory_E-Girl

[coral shale]

feld

rng

[dim widget]

what part of 0 * 0 = 0 do you not understand???

[coral shale]

i understand felds but i dont have to accept them do i

[dim widget]

those who do not accept the gospel of 0 = 1 will be judged during the acharit ha-yamim

[coral shale]

f(a+b) = f(a)+f(b)
f(ab) = f(a)f(b)
f(1) = 1

wait sht

huh weird, so we kinda have to insist 1 not in kernel

[coral shale]

no wait this is completely irrelevant

We insist a subfield has to have the same 0 and same 1

[coral spindle]

I never did think Schneerson was the one

[formal ermine]

wait what

[dim widget]

0 * 0 =…..

……

……

0

[formal ermine]

ughhhhh idk??

[crystal turtle]

what

[south patrol]

"Non trivial field"

[coral shale]

its clearly not trivial if chats debating it

[south patrol]

i mean there is no trivial field

[formal ermine]

Q

[topaz solar]

F_1 geometry

[chilly ocean]

Not sure if this is the right place but I have a question. As we add more numbers, we change what we're working with. For instance, the addition of $i$ to $\mathbb{R}$ makes $R[i]$ or $\mathbb{C}$. If we go to the next group, we get the Hamiltonians, which lose the commutativity property. But why does it lose that? What is it about going to the next set that makes it not commutative?

[cloud walrus]

Dark Angel

[timber blaze]

This might be a really stupid question but how can cyclic groups be infinite? In finite cyclic groups the inverse of a^k is a^n-k, then how does an inverse exist in an infinite cyclic group where |a| is infinite?

[dim widget]

$\mathbb{Z}$

[cloud walrus]

Topos_Theory_E-Girl

[open sluice]

cyclic group means every element is a power of some element
that includes negative powers

[timber blaze]

Ah thanks

[south patrol]

So a nice definition of cyclic group is that G is cyclic if there's an element g such that G = <g> i.e. the smallest subgroup of G containing g is G

So for example, any subgroup of Z containing 1 must contain -1 (since the subgroup must be closed under taking inverses!)

And <g> is simply set of powers of g anyway, but the above hopefully motivates it more lol

[timber blaze]

I see

[south patrol]

Here's an idea for you though: show that any infinite cyclic group is isomorphic to Z

So Z is "the only" example

[timber blaze]

I haven't studied isomorphisms yet 😢

[south patrol]

Oh don't worry then

[open sluice]

every time you use the cayley dickson construction to produce an algebra with doubled dimension, you lose a property, because symmetry becomes more and more difficult to achieve
you lose ordering by moving from R to C (so you no longer have a negative half to match the positive half)
you lose commutativity when moving from C to H (so you can’t mirror products anymore)
you lose associativity when moving from H to O (so your algebra is now basically useless)
you get zero divisors when moving from O to S (so your algebra is properly useless)

[chilly ocean]

Why do you lose a property though?

Also, why are the octonions before the sedenions?

And if it's useless, what's the point of a construction which basically just generates useless algebras after not that long?

[open sluice]

have you tried reading math stack exchange

[chilly ocean]

I haven't been on stack exchange in ages because I didn't find it that useful

[dim widget]

People came up with all of these algebras separately. C, H were both extremely useful, and O has some niche applications in Topology

The construction is just interesting because it unifies a bunch of previous constructions for finding, eg, group structures on spheres

[chilly ocean]

What happens if I apply the CDC to $\mathbb{S}$?

That's a vector space $V: \mathbb{S} \times \mathbb{S}$, isn't it?

Whatever it is that you get from applying it to the sedenions

[cloud walrus]

Dark Angel

Dark Angel

[chilly ocean]

Isn't every function field in one variable over a field k isomorphic to k(t) ?

[dim widget]

What do you mean by “in one variable?”

[chilly ocean]

What everyone else means? Is this not a standard phrase?

[dim widget]

Not really

[chilly ocean]

I think it means an indeterminate symbol 't' in this context.

What are other reasonable interpretations of that phrase?

Where it makes sense to say "all" instead of just "the"?

[dim widget]

It’s hard to understand a sense in which you mean it in which your question isn’t a tautology…

Like for instance is F(x, y) where y^2 = x^3 - x “in one variable” because y is algebraic over the transcendental field F(x)?

[chilly ocean]

k is also algebraically closed in this context

[dim widget]

If what you mean is are the fields F(t), where t is a transcendental indeterminate, all isomorphic no matter what t is then yes, and the isomorphism just changed the name of t

[chilly ocean]

So I'm guessing that rules out interpreting k as F(x) ?

The full sentence is "For every function field in one variable over k, there is a unique curve with function field K, up to a unique isomorphism."

[dim widget]

A more interesting theorem is that any subfield of F(t) is either F or isomorphic to F(t)!

[dim widget]

Aha yes this is what I was thinking

So this is different from the thing about F(t)

In this case “in one variable” means “having a transcendence basis consisting of only one element”

So for instance my F(x, y) from above is another example

[chilly ocean]

Hmm OK, thanks for clearing that up

[dim widget]

Intuitively having only one transcendental element means being one dimensional. Whereas being isomorphic to F(t) means being basically affine line

[celest cairn]

Does any sum of nth roots of unity raised to the power of n give a rational number?
For example, does $(\zeta_{5}^2+\zeta_{5}^3)^5$ equal a rational?

[cloud walrus]

Sapphire

[prime sundial]

let JG be a group ring, and define phi : G -> JG by phi(g)(h) = 1 if g = h, and 0 otherwise
my book says this is product preserving. does that mean phi(gh) = phi(g)phi(h)? because that doesn't make sense to me

[chilly ocean]

What is phi(g)(h)? Is it supposed to be 'phi(g,h)' ?

[prime sundial]

no. JG is the set of maps from G to J with finite support

so phi(g) = g*, where g*(h) = 1 if g = h, and 0 otherwise

[topaz solar]

g* • h* will again be a function G->J

However, like a polynomial, you have to compose the indices

so yes $\phi(gh)=\phi(g)\phi(h)$

[cloud walrus]

Dragonslayer Sharp

[prime sundial]

i just realized i was thinking of the regular product

[topaz solar]

Polynomials are a sort of uh

Monoid ring

On N

There’s ur intuition

[prime sundial]

perfect

[topaz solar]

$ax^g \cdot bx^h = abx^{gh}$

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

It’s polynomials

Extend linearly

[prime sundial]

but yeah i was tihnking of literally (g^(x)h^(x)), instead of (\sum_{y\in G}g^(y)h^(y^{-1}x))

[cloud walrus]

maximo

[prime sundial]

i dont follow

[topaz solar]

You can write terms in J[G] as polynomials with exponents in G right?

[prime sundial]

i was not aware of that

[topaz solar]

x^g = \phi(g)

[prime sundial]

i see

[topaz solar]

Polynomials are N->R of finite support

The multiplication is the same

[topaz solar]

This is essentially what’s going on I think though

Idk if it gets screwier for noncommutative rings though oop

[topaz solar]

Apply this same formula, summing over N

And you should see the polynomial-ness

(Or, if you want negative exponents, Z)

This make sense?

[prime sundial]

i'll have to come back to this polynomial thing later

not really

let me play with the sum first

so i understand that this is indeed product preserving

ok duh this makes sense

g*(x) says g = x, and h*(x^-1y) says h = g^-1y -> gh = y

so it's clearly product preserving

still don't get the polynomial thing though

[topaz solar]

$\sum_{i} a_i x^i \cdot \sum_j b_j x^j$

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

What’s the formula for this

[prime sundial]

are you saying this is what we're emulating

[topaz solar]

Yes

What’s it called uh

Covolution

[prime sundial]

ah

i see the resemblance now

let me reread what you sent with this in mind

[topaz solar]

$\sum_i \sum_j a_i b_{j-i} x^j$ or something

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

I’m not sure how important your choice of g(y)h(y^-1 x) order for y is, but you should see it here

The sum in i is summing it over the basis of \phi(x) = x^g

$\sum_{i\in G}\sum_{gh=i} a_g b_h x^i$

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

Identifying the polynomial with the function of its coefficients

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

I meant x^g but you get the idea

[prime sundial]

i am processing all of this, thanks sharp

[topaz solar]

Consider how this works, without any change, for polynomials by substituting in N^(whatever)

Which isn’t a group but cmon it’s clearly related

[prime sundial]

ok i do see it now

[topaz solar]

Ye

This is literally just polynomials, but not coefficients in N (or Z)

[prime sundial]

is that the inspiration for the notation J[G]

[topaz solar]

So you can get some weird cycling (consider how cyclic subgroups make zero divisors)

Or very nonlinear setups

[topaz solar]

Probably

It’s literally polynomials in g \in G

[chilly ocean]

Hello I have a problem

[prime sundial]

thanks again sharp

[chilly ocean]

N subgroup of G and N included in Z(G). Show N normal and if G/N cyclic then G abelian

I just started 1 month ago group theory. I know it is a trivial result

The part to show N normal is obvious.

And for the second part when G/N cyclic I said let xN be a generator of G/N.

[topaz solar]

Idk any rep theory so idk how to motivate it, but symbol pushing is my domain

[chilly ocean]

x should be in G\N right?

[prime sundial]

yeah that's fair. im learning this for the sake of knot theory. i am very out of my element here

[topaz solar]

Well, unless G/N = {1} but if that’s the case obviously it’s abelian

[chilly ocean]

Yes cuz all elements would be in N which is in Z(G)

So in the case where |G/N|>1 we need to have x in G\N.

All elements in {x} union N commute right?

Now I am stuck

[topaz solar]

Well, here’s a quick exercise: if G has torsion then R[G] has a zero divisor

[topaz solar]

So, let’s pick x to be a generator of G/N

Or uh, x such that xN is ofc

So, for every y in G, yN = x^n N for some n

Can you do anything here

[chilly ocean]

Oh

Every element y can be written as x^some power × n with n in N

So the set A= {x} union N generates G and because all elements in A commute between them, G abelian

Thx

[topaz solar]

Might could do with some formal work since you’re starting out

Ya know, to ensure careful work going forward

But I think you get the idea, which is the essential part

[chilly ocean]

Yes

[prime sundial]

this is easy once you know phi is a group homomorphism haha

[topaz solar]

Yeah lmao

Give an example for when |g| = m

Once you do

Relate it to something you know

[prime sundial]

ok sharp

[prime sundial]

i take this back

but

i will say

the rabbit hole you took me down

is starting to look like Z[x]/(x^m)

or generally

[topaz solar]

Do you want a solution to torsion -> zero divisors?

[prime sundial]

J[x]/(x^m)

[prime sundial]

nono

i want to work it out

i like this a lot

[topaz solar]

Well, not everything can be generated by one element

[prime sundial]

ok right

[topaz solar]

But yeah, consider R[G] as R[F(G)] mod some multiplication relations maybe

Or well, you can maybe not should

[topaz solar]

Well, if you want a hint of the |g|=m one, consider R[g]

Since obviously nothing else is important for that particular one

[prime sundial]

i think i have a good educated guess

[topaz solar]

Try it

[prime sundial]

we take
1 + g* + ... + g^(m-1)*
and (1-g*)

is that at least on the right track

[topaz solar]

Yes

What does that multiply to

As a polynomial in g

[prime sundial]

1 - g^m*

[topaz solar]

Yep

g^m = 1

[prime sundial]

awesome

[topaz solar]

Consider

What does this look like

[prime sundial]

not sure what you're getting at, it looks like a lot of things

[topaz solar]

Well, x^m - 1 = 0

And you get a cyclic group of order m

It’s a ||primitive root of unity||

[prime sundial]

ok fair i should have seen that

[topaz solar]

Ye

[prime sundial]

wow yeah that's like hitting me in the face now

[topaz solar]

I don’t know the rep theory angle, but relations in the group are gonna look like being algebraic over R[whatever]

Polynomial root lookin

[prime sundial]

mhm for sure

this is a nice link

[topaz solar]

Especially with commutative groups ofc

[prime sundial]

yeah

[topaz solar]

Non commutative would be a lil screwier

[prime sundial]

lemme see if the book even talks about noncommutative stuff

[topaz solar]

Well, you can make this kind of thing work for semigroups

So this stuff works fairly general

Ask ally for that angle more than me though lmao

[prime sundial]

yeah this is all talked about in full generality. i don't know if they'll tie any of this to actual algebra, it's like the preamble to free calculus

[topaz solar]

I have no idea what that means

[prime sundial]

nice

that's a first

you maybe do just not by that name

[topaz solar]

I don’t know any use cases of these so probably not

[prime sundial]

it's doing derivatives on these group rings

it looks a lot like basic tensor stuff which is interesting but yeah

[topaz solar]

Ah, so like Kähler

How would that work for a random G hm

What text?

[prime sundial]

this is what i was talking about resemblance wise

introduction to knot theory crowell and fox

there's no like actual form stuff tho, at least not right now

[topaz solar]

What’s t

[prime sundial]

the trivializer

[topaz solar]

Also weird is it’s D(v1)t(v2) + v1D(v2)

Trivializes only on the left

[topaz solar]

Is this a weird thing or a typo

[prime sundial]

no im pretty sure this is intentional

[topaz solar]

Intriguing

[prime sundial]

https://www.maths.ed.ac.uk/~v1ranick/papers/crowfox.pdf
text if you ever feel curious enough to read it

[topaz solar]

So it’s a derivative like a vector field

[prime sundial]

it looks that way