groups-rings-fields

[coral spindle]

I will not be calling her laurent that's not her name

[delicate orchid]

stop saying shit like this bro

what would yo mama say

[coral spindle]

Now you can definitely say "she hopf on my fibration till I homotopy"

[tribal moss]

She'd tell you to go to #chill.

[chilly ocean]

she would laugh and go "ok TT"

[hollow mica]

😂 ok TT

[proud zenith]

Try this out in sage and stare at it for a few minutes:
sage: R.<x,y>=QQ[]
sage: R.ideal(y^2+5*x+2)
Ideal (y^2 + 5*x + 2) of Multivariate Polynomial Ring in x, y over Rational Field
sage: i1=R.ideal(y^2+5*x+2)
sage: R.quotient(i1)
Quotient of Multivariate Polynomial Ring in x, y over Rational Field by the ideal (y^2 + 5*x + 2)

[static yew]

How do I do polys in F7 instead?

[proud zenith]

I knew that'd be next, yes here we go:

R.<x,y>=GF(7)[]

?GF

[meager fractal]

does f irred in R[x] imply no roots in R

?

sick

[slim kayak]

Suppose it had a root in R, what can you do with f?

[meager fractal]

reduce it.

[slim kayak]

Factor, but yeah

[meager fractal]

shhhh

[coral spindle]

It does not: linear polys are irred

Humpty is crazy

[meager fractal]

deg>=2???

[slim kayak]

Right, irr f of degree more than 1 would've been right

[coral spindle]

x-1 is irreducible in R[x] for any ring R

[slim kayak]

You can factor linear polynomials, but you just get an unit which doesn't is addressed in the definition

[formal ermine]

x...

[slim kayak]

Still, you may wanna convince yourself why it holds for deg > 1. Consider if r is a root the polynomial f(x+r), how can you express it?

Huh?

[meager fractal]

waiiiit no linear polys irred iff they have roots in R?

or just for fields?

[meager fractal]

or in genera degree leq 3

[formal ermine]

#groups-rings-fields message

[meager fractal]

why do we not rly care abt non unitary rings? or maybe in my course we didnt

[formal ermine]

Z/pqZ need not have zero divisors

[south patrol]

It does for p q prime tho

[formal ermine]

who said that p and q need to be prime

[coral spindle]

Every linear polynomial is irred, full stop

[south patrol]

Context

[coral spindle]

Ah, one catch: I assume monicity

[slim kayak]

Polynomials of degree 3 or 2 are reducible iff they have a root in the field. Linear polynomials can't be proper products of polynomials of non-zero degree

[meager fractal]

ye

nvm

schupid

very easy to see

[slim kayak]

Degree counting time

[formal ermine]

chmtupid

[meager fractal]

but Z/pqZ is a ring right?

so

(p+q)X + 1 linear factors to (pX+1)(qX+1)

[slim kayak]

Yeah, it inherits it's ring structure from Z. pqZ just isn't prime, so your factor ring won't be integral

[meager fractal]

yes but then (p+q)X + 1 linear but not irred

so for any ring with zero divisors we can have linear non irred polynomials?

[slim kayak]

Yeah, by repeating your construction.

[coral spindle]

2x - 2 is linear but not irred in Z[x], despite Z having no zero divisors

In a commutative ring, a zero divisor is an element that divides zero. That is, an element x for which there exists an element y such that xy=0.

[meager fractal]

yes true

[coral spindle]

Z/nZ has zero divisors if and only if n is composite

Does that clear things up, humpty?

[meager fractal]

monic would make evry linear irred

[coral spindle]

yeah

Oh I'm not familiar

yes as you mentioned, I see

A condition for the polynomial ax + b to be irreducible in R[x] is that for all y in R, if ay = b then y is a unit.

Or hold on

that's not right, but it's close. Give me a moment to think.

Yeah I think you can't really express it much better than a, b have no non-unit common divisors

In particular, you get that nx + 1 is irreducible in Z[x] for free

[tribal moss]

That's a necessary condition, but it is sufficient for unrestricted rings?

[slim kayak]

x+4 is equal to (2x+5)(3x+2) mod 6

[meager fractal]

so again then, no zero div has to hold

[coral spindle]

Yeah, I think not

That's also a beautiful example kerr

Zero divisors really screw things up

[slim kayak]

Yeah, integral + monic + linear= irreducible

[slim kayak]

Isn't it related to this? ax-b has a root r then ar=b, so a divides b. If a divides b, then there exists some t s.t. at=b, but then at-b=0.

[tribal moss]

Well, your mod 6 example beautifully showed that it isn't sufficient in general.

[slim kayak]

Yeah you right, force of habit to use illegal extra structure

[coral spindle]

It is sufficient in any integral domain, and I think that's pretty much as good as we can do

There may be some subtle rings in which there are elements not of the form a+b where a,b are zero divisors but...

it's hard

And before you ask, yes integral domains are exactly the rings R such that, in R[x], deg(fg) = deg(f) + deg(g)

[tribal moss]

(Hmm, is it even obvious that 2x+5 is not a unit in Z[x] mod 6?)

[coral spindle]

Well ok let's work through this

[tribal moss]

(yes, it's obvious, plug in x=2 and you get a zero divisor out).

[coral spindle]

ok nvm

I was hoping for some result that would bound the degree of an inverse

But the substitution trick is cool

[slim kayak]

I was ready to just see what happens if you multiply with an arbitrary element of deg n lol

It really is neat

[coral spindle]

me too

[tribal moss]

General results about the degree of an inverse won't cut it: We have (3x+1)(6x+1) = 1 modulo 9.

[coral spindle]

well my hope was that we could bound the search at least

[tribal moss]

Oh, I see.

[coral spindle]

So I mean I think your example is nice

Because the inverse is also degree 1

are there any degree-1 units in (Z/nZ)[x] whose inverses are not of degree 1?

Obviously they need to be at least degree 1 (and I hope you agree it is obvious)

[tribal moss]

Yeah.

My example above is more obvious when written as (1+3x)(1-3x).

[slim kayak]

What would the "general result" have been? Inverses always being linear sounds like the ideal general result here

[coral spindle]

If ax + b is a unit, then there exists y such that ay=0 and the leading coefficient of the inverse must be y.

ah, some such y.

well whatever, I'll think about this more later.

[tribal moss]

In (Z/p³Z)[x] we have (px+1)(p²x²-px+1) = 1.

[formal ermine]

so Hom(Z^n, Z) = Z^n?

[coral spindle]

So my bet is that it's bounded in some way by the power of p. For general n I don't know.

[coral spindle]

Yeah, but this is easy to see by the free-forgetful adjunction. Z^n is the free Abelian group of order n, so Hom(Z^n, Z) is in natural correspondence with Hom({1, ..., n}, Z).

Ah to be clear

that latter Hom is in Sets

[formal ermine]

was able to guess

wait lemme think for a sec

[coral spindle]

You can think of it as just choosing where the generators go, freely.

[formal ermine]

yeah like choosing the coefficients from the linear combination sort of?

or?

[coral spindle]

Yeah that's right

[formal ermine]

yeah alright thanks

[coral spindle]

I choose where (1, 0, ..., 0) goes, I choose where (0, 1, 0 ...) goes

etc

[formal ermine]

yeah

[coral spindle]

Forgot to say that this is great stuff. Did you just search with brute force?

[tribal moss]

For general n, we can just pick the highest power of a prime in n, and CRT everything to be 1·1=1 modulo other factors.

[slim kayak]

Right, that work too. You don't even have to appeal to freeness (at least not openly), any way you map 1 of each coordinate into Z will uniquely determine your homomorphism

[coral spindle]

great catch

[tribal moss]

Combination of brute force, guesswork followed by algebra, and inspired generalization. :-)

[coral spindle]

And I guess for $\bZ/p^n\bZ$ we have $(px + 1)\sum_{i=0}^{n-1}{(-1)^ip^ix^i} = 1$ right

[cloud walrus]

Boytjie (never-to-be-glomed)

[coral spindle]

Do you reckon this is right?

This is inspired by your example ofc

[delicate orchid]

(-px)^i moment

[tribal moss]

Yeah. It definitely works, but I can't yet prove it's the highest degree that can be inverse of a linear poly.

[coral spindle]

Right so we have a lower bound at least

I think we can relate some kind of nilpotence index to the degree somehow

[formal ermine]

wtf is $\on{Spec} \bigoplus_{a \in \bZ^n} \bC \chi^a$ where $\chi^a$ is $T \to \bC^\times$ of the form $(t_1, \cdots, t_n) \mapsto t_1^{a_1} \cdots t_n^{a_n}$

[formal ermine]

[delicate orchid]

Find the prime ideals looser

[formal ermine]

$T$ is any $(\bC^\times)^k$

[slim kayak]

What does C underscript thingy mean

[formal ermine]

where?

[slim kayak]

The thing you want to find the spectrum of

[formal ermine]

complex numbers...?

[slim kayak]

The X looking Greek latter with an a as an exponent

[coral spindle]

@tribal moss I think I've got it. We look at the inverse of $ax + 1$ in $R[[x]]$ which is in R[x] iff $a$ is nilpotent.

[cloud walrus]

Boytjie (never-to-be-glomed)

[delicate orchid]

chi? Lol

[formal ermine]

ohh that's not in the subscript lmaoo

that's just chi

$A \chi$

[formal ermine]

it's just down like that

always

[coral spindle]

We can reduce to ax+1 because in general for ax+b to be a unit in R[x], b must be a unit in R.

[formal ermine]

definition is after the where

[delicate orchid]

Cleaver….

[coral spindle]

THANK U WEW

IT MEANS A LOT BC UR REALLY SMART

[slim kayak]

yeah sure but what does the notation mean

[formal ermine]

it lives in the coordinate ring of T

[slim kayak]

Alge awww moment

[coral spindle]

#algebraic-geometry >:(

[delicate orchid]

Ban

[formal ermine]

nooooooo

[coral spindle]

Ban

[formal ermine]

that's a scary place

[coral spindle]

can't deny that

[formal ermine]

people are always very mean to me there because I always ask stupid questions

[delicate orchid]

You do mean C[chi^a] though right

[slim kayak]

Your spectrum thingy has a topology on it. You shall be banished from here

[formal ermine]

no

[coral spindle]

seriously though, that channel is more likely to be able to understand + answer ur question

[delicate orchid]

Then I do not care

If you’re not taking the C span then ur a nerd

[slim kayak]

I still am curious what A\chi^a actually means lol. Does it have a name?

[coral spindle]

I reckon it's the weight space of A via chi

[delicate orchid]

You can split this into looking at the spectrum of each summand can’t you?

I can’t remember how they interact it’s been a few years

[slim kayak]

Ideals of direct sums are direct sums of ideals, so that works.

[coral spindle]

In fact I think this shows that we don't get anything like this mod 6

Nice :)

[delicate orchid]

It’s not that simple I know that much

[tribal moss]

Hmm, and the power of a needed to get to 0 determines the degree of the inverse? I'll buy that.

[coral spindle]

Yeah it does, it's not too bad to see actually

You can derive the inverse of ax+1 in R[[x]] for any ring, and then this is just a special case

[tribal moss]

Yes, I see that. Right-to-left long division.

[coral spindle]

Now just to generalise this to degree-n polynomials...!

[slim kayak]

My bad, your right. Maybe spec is a nice enough functor to map products to coproducts?

[delicate orchid]

Oh wait yeah Spec commutes with products contravariantly

It definitely maps products to coproducts but I don’t know if the dual has to be true

[formal ermine]

https://tenor.com/view/nerd-emoji-nerd-emoji-avalon-play-avalon-gif-24241051

[delicate orchid]

Probably is

[delicate orchid]

Mate this is YOUR question

[formal ermine]

wtf is contravariantly supposed to mean in this context

[delicate orchid]

What it normally means

[formal ermine]

I know what a contravariant functor is

[slim kayak]

Uh... what kinda course are you taking

[delicate orchid]

Maps whatever to cowhatever

[formal ermine]

newton okounkov theory

[delicate orchid]

Dude why

[slim kayak]

Contravariant functors seem like they deserve a passing mention lmao

[tribal moss]

A mod 6 solution would reduce to (linear)·(constant) = 1 modulo 2 and 3 separately, each of which is absurd.

[formal ermine]

we didn't do any cat theory

the only listed prereq for this was algebra 1

[barren sierra]

what

[delicate orchid]

How are you thinking about Spec then? Literally as the set of prime ideals?

[formal ermine]

but he just assumes us to know diff geo, category theory, and commutative algebra

[barren sierra]

those things are not alg 1

unless you've somehow gone through one of the most hellish alg 1 courses I've heard of

[formal ermine]

with le sheaf

[slim kayak]

Hm, what is the right way?

[formal ermine]

yeah well he also taught the alg 1 course that I took last semester and this is alg 2

[barren sierra]

[formal ermine]

right now we're doing toric geometry

he has started leaving out all of the proofs

I assume because they are too difficult or even impossible with this little theory

[tribal moss]

Oh wait, now I see what you were saying. There are no nonzero nilpotents modulo 6.

[delicate orchid]

there isn't really a right way in my experience, it depends on what you're doing

[delicate orchid]

he needs to be fired

[coral spindle]

Yeah

Your way is good too

[slim kayak]

Just wanna know what I might be missing out on :/

[delicate orchid]

asking the wrong dude, I prefer the functor way though

[formal ermine]

what does it normally mean

[slim kayak]

I do too, did you just mean that spec is a contravariant functor at all?

[delicate orchid]

I meant that yeah

and obviously a contravariant functor that commutes with products will send them to coproducts

that's what I meant

[slim kayak]

Okay fair, helps remember how Spec is supposed to act

Yeah, that for example

[slim kayak]

Yeah so, what you have to do is really easy. Show that spec is representable

[delicate orchid]

gulp

I didn't actually know that lemme think about this one

[formal ermine]

I know a bit of cat theory

just not a lot

[delicate orchid]

ok spill the beans, what's the representing object

is it Z I bet it'll be Z

[slim kayak]

Idk, I just remember that was a criterion under which certain functors preserve limits/colimits. I am not even sure Spec is representable

[delicate orchid]

.... banter

well it preserves limits/colimits because Hom does so that doesn't tell us much

[slim kayak]

Actually, what is the exact category Spec maps into here? Spec with it's usual structure sheaf stuff / affine scheme?

[delicate orchid]

I presume so

I'll do some googling

lol

and lol

[slim kayak]

Truly a dark day today

[slim kayak]

Apparently it is that simple...

[glossy crag]

Is there a way to efficiently define these through more elementary concepts? E.g. how polynomial rings are examples of monoids rings.

[slim kayak]

Let R be the direct product of rings Ri. Products of prime ideals are prime. Conversely let P be any prime ideal , P isn't proper so it can't contain the i-th basis element of the finite direct product for some i. Since the product of the i-th basis element with the k-th ( i neq k) is zero and our ideal is prime, it must contain the k-th basis element. So P contains the direct product of all Rk and {0}. Projecting P to its i-th coordinates gives us a prime ideal Pi, so P is then the product of all Rk with k neq 1 and Pi

[slim kayak]

It also fails miserably in the infinite case

As in, spec(R) of an infinite direct sum can't be a disjoint union

[delicate orchid]

apparently so....................

[white oxide]

is there any way to check the irreducibility of this polynomial over R[x]?

[chilly ocean]

what's the nice condition that's equivalent to irreducibility for small degrees

[white oxide]

no zeros in the field?

but then there are too many numbers to check

[chilly ocean]

you already know what the roots are

[tribal moss]

Hopefully you also know that a polynomial cannot have more roots than its degree in a field.

[white oxide]

oh yea

right

oops

thx

[south patrol]

You could also just resort to high school methods

I.e. discriminant is < 0

If you didn't know the roots already

[rustic crown]

🥔

[chilly ocean]

potato confirmed a highschooler

[south patrol]

Gtg rip

Yeah

No I have galois tomorrow and I use discriminant a lot

[chilly ocean]

i wish i didn't forget all of the highschool stuff i learned

i'm so university brained that i forgot all of the really simple ways of doing things

[white oxide]

don't remember that stuff sadly

[hollow fjord]

Hi guys! I how could i prove that f(x,y) is irreducible, i know that R is a UFD.

[summer path]

I wish I didn't forget everything I learned in the first half of my undergrad already

[celest furnace]

Okay silly idea

Can u plug in y = 2 and show that that's irreducible by eistenstein

And claim it holds in general cuz it should work for all y

[hollow fjord]

i was thinking that generalized einsestein should do it

[celest furnace]

Clearly? f(x, y) is NOT the product of p(x,y) * q(y) where deg q >= 1

Oh my bad this might work in Q but not necessarily R

[hollow fjord]

If we see R[x] is a UFD, then (R[x])[y] is a UFD, and i can try to use Einsestein proving that (y) is a prime ideal . That was my strategy

[delicate orchid]

yeah that's exactly what I was typing

[celest furnace]

What ive said is like completely false

Apparently the irreducibles in R are of degree only 1 or 2

[tribal moss]

I think "R" in this case means a general ring (which we know happens to be UFD), rather than the real numbers?

[hollow fjord]

Yes

wtf now i'm blue

[white oxide]

what do they mean by prime subfield?

[white oxide]

also how does every automorphism of a field leaving 1 fixed show that Q is fixed, i get that if 1 is fixed then Z is fixed but not sure about Q

[uncut girder]

Think about it

How do you form Q from Z?

[white oxide]

well i get that if you have a/b in Q then phi(a/b) = phi(a x b^-1) = phi(a)(b^-1) = phi(1 + 1 + \dots + 1)(a times) x phi(b^-1) but idk how to express b^-1 in terms of 1

[ivory trail]

phi(bb^-1) = phi(b) phi(b^-1)

so phi(b^-1) is determined by phi(1)

[white oxide]

ah

[ivory trail]

and if phi(1) = 1, then phi(b^-1) b = 1 meaning phi(b^-1) = b^-1

generally, ring maps preserve units

[delicate orchid]

maps everything to 0.. heh... nothing personell kid.....

[white oxide]

oh ok i see, thanks!

[ivory trail]

preserves 1

in the zero ring, 0 is a unit

[delicate orchid]

oh yeah??? how can 0/0 when undefined??? exactly

[ivory trail]

i always think about localizing with respect to 0

[noble hedge]

If I have a field tower $K/E/F$ and some $\alpha\in K$ algebraic over $F$, what exactly is $[E(\alpha):F(\alpha)]$, is it still the degree of a minimal polynomial of $\alpha$?

[cloud walrus]

(𒀭)

[uncut girder]

Min poly over F? Doesn't need to be

a could be a primitive element for K/F in which case K = E(a) = F(a)

[noble hedge]

$\deg_E(\alpha) \mid \deg_F(\alpha)$, right? Since either the minimal polynomial of $\alpha$ in $F$ (say $m_{\alpha, F}$) is still the minimal polynomial in $E$, or it can factor in $E$, in which case $\deg_E(\alpha) < \deg_F(\alpha)$, right?

[cloud walrus]

(𒀭)

[noble hedge]

So does that imply $[E(\alpha) : F(\alpha)] \mid [E:F]$ have I just done some weird logical parkour?

[cloud walrus]

(𒀭)

[noble hedge]

$E(\alpha)/F(\alpha)$ is an extension, but it can be a smaller (degree) extension than $E/F$, if $\deg_E(\alpha) < \deg_F(\alpha)$, right?

[cloud walrus]

(𒀭)

[uncut girder]

Wht does it have to divide

[noble hedge]

if $m_{F}(x)$ is the minimal polynomial of $\alpha$ in $F$, and similarly for $m_{E}$, then since $E/F$ is an extension, either $m_F = m_E$ or $m_F = m_Eg(x)$ for some $g\in E[x]$ with $\deg(g) < \deg(m_F)$

[cloud walrus]

(𒀭)

[noble hedge]

Either the extension is large enough so that the minimal polynomial can factor, in which case the degree should divide that of the original, or it still remains the minimal polynomial, in which case the degrees are equal so it divides trivially, right?

[uncut girder]

Why does the degree divide?

[noble hedge]

$m_F(x) = m_E(x)\cdot g(x) \implies \deg(m_F(x)) = \deg(m_E(x))\cdot\deg(g(x))$?

[cloud walrus]

(𒀭)

[uncut girder]

NO

[noble hedge]

...they're additive aren't they

ah

mb lmao

[summer path]

You can see that that doesn't work with a very very simple example

[noble hedge]

yeah

okay so that doesn't work

which now makes me even more confused about where $[E(\alpha):F(\alpha)]\leq[E:F]$ comes from

[cloud walrus]

(𒀭)

[uncut girder]

Its true

[noble hedge]

I was still right about $\deg_E(\alpha)\leq\deg_F(\alpha)$ right?

[cloud walrus]

(𒀭)

[noble hedge]

dividing isn't true, but it should still be at most the same, right?

[uncut girder]

Yes

[noble hedge]

is that related to how to prove this, or is that just going in the wrong direction?

[uncut girder]

Its related

Draw the diagram

[noble hedge]

there's a diagram?

Like the $K/E/F$ thing or is there a commutative diagram?

[cloud walrus]

(𒀭)

[uncut girder]

Draw E(a) and F(a)

With E and F

And the inclusions

And label the degrees

[noble hedge]

Like $K/E(\alpha)/E/F$ nad $K/F(\alpha)/F$?

[cloud walrus]

(𒀭)

[uncut girder]

Forget K

[rustic crown]

i wish i didn't forget stuff after one month of not studying it

[uncut girder]

Its a diamond

[noble hedge]

okay

[summer path]

Me too

[noble hedge]

This diagram?

[uncut girder]

Yes

Label the edges

[noble hedge]

can we do tikz-cd diagrams with the bot?

Oh we can

$$
\begin{tikzcd}
& E(\alpha) & \
E \arrow[ru, "{[E(\alpha):E]}", no head] & & F(\alpha) \arrow[lu, "{[E(\alpha):F(\alpha)]}"', no head] \
& F \arrow[lu, "{[E:F]}", no head] \arrow[ru, "{[F(\alpha):F]}"', no head] &
\end{tikzcd}
$$

[cloud walrus]

(𒀭)

[noble hedge]

So $[E(\alpha):E] = [E(\alpha):F(\alpha)]\cdot [F(\alpha):F]$ and $[E(\alpha):E] = [E(\alpha):E][E:F]$, is that how the diagram works?

[cloud walrus]

(𒀭)

[uncut girder]

Yes

Some of those degrees are deg_E(a) and deg_F(a)

Finish the argument

[noble hedge]

Wait did I mess up, should the left-hand side of the equalities be $[E(\alpha):F]$?

[cloud walrus]

(𒀭)

[uncut girder]

Yes

[summer path]

Given the diagram, you kinda just read the tower law straight off of it

[noble hedge]

yeah

[summer path]

[noble hedge]

in which I failed to do so lmao

is it a division argument?

$[E(\alpha):F(\alpha)]\mid [E(\alpha):F]$ and $[E:F]\mid [E(\alpha):F]$?

[cloud walrus]

(𒀭)

[noble hedge]

$\frac{\deg_E(\alpha)}{\deg_F(\alpha)} \leq 1$, so $[E(\alpha):F(\alpha)] = \frac{\deg_E(\alpha)}{\deg_F(\alpha)}[E:F] \leq [E:F]$

[cloud walrus]

(𒀭)

[noble hedge]

That should not have taken me as long as it did lol

[noble hedge]

Thank you for your help and patience with me. I do genuinely appreciate it, especially since this definitely wasn't one of my brightest moments lol

[uncut cloud]

Hi, for a project on Galois inverse theory I am studying the example of the Monster Group. It may be the wrong place, but I would like to extract 4 columns from its table ( (1A,2A,3B,29A) ). I have them in the Atlas, but for computations I would like an excel and for presentation a .tex file. Do you know how can I generate them ?

[delicate orchid]

for computations I'd like an excel
why are you using excel for group theory

[warped fable]

sigma grindset

[uncut cloud]

I have to sum on all irreducible characters the product of the value on my 3 classes.
So I need a computer. But yes if it is another file it will be okay

[delicate orchid]

I mean why aren't you using an actual computational algebra system

how did you even get the character table lol

[uncut cloud]

If you have a solution of this kind I will take it. But on Mathematica the table is unavailable. While in the book of Conway it is complete

[delicate orchid]

not sure how you'd actually generate the monster group in something like GAP but once you do the character table isn't that big, only like 200x200
although at that point I'd just copy and paste them from whatever this mysterious "Atlas" is

[uncut cloud]

The atlas is an old book by Conway with all the tables you want. Good reference but really hard to use as it is on paper 😅

[delicate orchid]

sounds useful tbf

[delicate orchid]

bro this is atlas is weirdddd

[solar glacier]

My alg prof said back in the day they had it as a HUGE book in the school libraries

[formal ermine]

@slim kayak @delicate orchid remember that problem from yesterday? the spectrum of that group algebra?

turns out it's just a fucking torus

the group algebra is isomorphic to the laurent polynomial ring

[warped fable]

i dont know what the problem is but sounds like a certified G_m classic

[formal ermine]

,, \bC[M] = \bigoplus_{u \in M} \bC \chi^u

[formal ermine]

wait ibsen do you know toric geometry

[warped fable]

vanishingly little

[formal ermine]

I missed you

it got a little bit more boring here without you and your hatcher set top pilledness

[formal ermine]

anyway yeah but the spectrum of this

an explicit description is

but that is just isomorphic to laurent

so it's just a torus

[warped fable]

chi irrep of C^*?

[formal ermine]

group algebra

but yeah it's like

an element of the coordinate ring of a torus

[warped fable]

what is M

[formal ermine]

which has the characters T -> C^times as a basis

[warped fable]

ah ok irrep of (C*)^n

yes thats the laurent poly ring

[formal ermine]

Hom(N, Z) for some N iso Z^n so it's Z^n

[warped fable]

gotcha

a simple thing, but said in a complicated way

like all of algebra

[formal ermine]

so true

well N is a lattice tbf

[warped fable]

maketh sense

[delicate orchid]

You didn’t say this before

[formal ermine]

I thought you knew that

[delicate orchid]

I didn’t recognise them as linear characters because you never told me what T was

[formal ermine]

oh...

[delicate orchid]

But yeah it makes sense it’s a torus

The funny Diag(e^dudes)

[formal ermine]

what

[ivory trail]

did you know the cogroup structure on the group ring gives you an algebraic group structure on the torus

[formal ermine]

wdym

[ivory trail]

if you're doing toric geometry you might want to show that an algebraic torus is an algebraic group

[formal ermine]

trivial

like literally lol

no?

except the regularity part

[ivory trail]

yeah...

well if you're looking at the closed points then it's easy

i don't actually know much about the other points but i assume there are some nontrivial ones

anyway group rings have a cogroup structure, part of the hopf algebra structure

[formal ermine]

what do those words mean

cogroup

[ivory trail]

it's like a group but co

so instead of multiplication $\mu\colon G \otimes G \to G$ you have comultiplication $\Delta\colon G \to G \otimes G$

[cloud walrus]

196883

[ivory trail]

and this satisfies "coassociativity" and the sort

you have a counit G -> I and coinverse (?) G -> G which also do what you might expect

I here is the unit of the tensor product

which is taken over the base ring (or field, like the complex numbers)

wait actually i think i might have something mixed up

actually nah it's fine

this isn't technically a "cogroup object" but it's part of the hopf algebra thingy

https://en.wikipedia.org/wiki/Hopf_algebra#Examples

[coral spindle]

The co-ordinate ring of a topological group is a cogroup in the category of commutative k-agebras

[ivory trail]

the group object is defined with a cartesian product usually

so i don't really want to call this a cogroup object even though it's opposite to the group object in the category of affine k-varieties

[coral spindle]

Uh no

It's defined using whatever monoidal product the category has

In the case of Set we typically use the cartesian product, yes

[ivory trail]

that's a monoid object

[coral spindle]

but the equivalent thing in algebras is the tensor product

[ivory trail]

but in general defining group objects you need to verify that the inverse works

and you don't have a diagonal map to do this with

[coral spindle]

Uh, again, no. A monoid object is similar to a group object, the only difference being that we do not require the existence of an inverse morphism.

If you like the definition of a group object is a monoid object and an inverse morphism.

[ivory trail]

so how are you going to require that a a^{-1} = 1

without a diagonal map

[coral spindle]

Perhaps I'm just misremembering, but monoidal categories do have diagonal maps

[ivory trail]

there is no way

[coral spindle]

And there is a diagonal map in the tensor product anyway

[ivory trail]

https://ncatlab.org/nlab/show/monoidal+category+with+diagonals this is the case

[coral spindle]

Well thank you for informing me about this

[delicate orchid]

bialgebras scare me

I stumbled upon them far too early I think

[next obsidian]

I’m not surprised you’re biphobic

[delicate orchid]

... wtf

a phobic that's also a cophobic

[ivory trail]

group rings are extremely hopf algebra, you can't avoid it

[wooden ember]

i must agree

our rep theory course was fine and fun and then the teacher took 30 minutes to define a hopf algebra

[upper pivot]

hopf algebras are cool

I am a quantum group enjoyer.

[ivory trail]

you don't have diagonals for the tensor product of R-modules because it's not linear (ax otimes ax = a^2 (x otimes x))

[wooden ember]

theyre cool but i dont understand them well enough yet to fully appeciate them

[wooden ember]

oh she gave a lecture on quantum groups too and i understood nothing cause she condensed everything into a single lecture

[delicate orchid]

I work very much with the group algebra and I've avoided it

[upper pivot]

oof i see.

[ivory trail]

oh wait i think it does work

i'm kinda dumb and forgot to switch the product for a coproduct

and the coproduct is of course the tensor product

so it is a cogroup object

i did this whole thing a while ago lol

[coral spindle]

I suppose for a cogroup object we'd need a codiagonal map

which makes sense

[ivory trail]

i think the comultiplication/diagonal works in the category of R-algebras for group algebras as a linear extension of g -> g otimes g on the basis

the "codiagonal" would be the multiplication of a group object

i'm not sure if you need the base ring to be a field

because everywhere i look people assume it to be a field

[coral spindle]

I don't think you do, but ofc that is the default

I think at most you need it to be a commutative ring

[upper pivot]

btw I havent learn much about these earlier outside of some talks I have attended and like a few chapters of a book. But this stuff is really cool imo, in particular, to extend pontrjagin duality to non-abelian LCH groups you extend to the category of locally compact quantum groups for example :)

quantum groups are just like, C* hopf algebras btw. now locally compact quantum group is quite a bit harder to define oof

[white oxide]

this sounds pretty naive and ignorant, but can somebody explain how studying categories would be relevant to gaining information about certain algebraic structures? maybe it’s because I’m very new to the idea, but I can’t seem to parse how making diagrams with dots and arrows and connecting them can in any way disclose information about the objects in question

maybe it’s a matter of mathematical maturity? I know that categories are useful for abstracting data and working there, but I can’t see how they would give us any more information idk if that makes sense

[long nebula]

They are a different way of presenting the same information

Some things are easier to see once you look at them under a different lens

The key is that the arrows have to respect the algebraic structure

The mantra of category theory is that instead of looking at the objects, look at how the objects relate to one another

[open sluice]

in linear algebra you have linear isomorphisms, functions between two vector spaces that are both bijective and linear
in abstract algebra you have ring and group isomorphisms, bijections that preserve addition and multiplication
in topology you have homeomorphisms, bijections that preserve the topological structure of the space

these all have something in common

[mighty kiln]

Perhaps it's easier to see when you learn more about what you can do with categories

[delicate orchid]

Especially homology/cohomology and alg top

[karmic moat]

as for understanding how commutative diagrams enclose information, you can often find statements about certain properties (like descending to a quotient in quotient spaces, isomorphism theorems, etc) using said diagram

helps put comm diagrams into a bit of context

imo

[cunning dust]

anemone does cat theory? goddamn

[karmic moat]

i dabbled a bit, havent looked at it in a long time

but i see plenty of commutative diagrams

[long nebula]

Oh no I'm doomed

[cunning dust]

isoke eric

[mighty kiln]

Elementary examples:
Initial and final object in Grp: trivial group
Initial object in R-Mod: R
Product in Grp, R-Mod, Top, etc: product in usual sense
Coproduct in Grp: combining generators and relations
Coproduct in Ab, R-Mod: direct sum
Right-split short exact sequence in Grp: semidirect product

[delicate orchid]

How

[elder wave]

I mean honestly i think you'd have to go quite far until category actually becomes useful as a subject applied to the one you're studying and not just a convenient way of stating things

[delicate orchid]

It’s also useful to have an abstract framework in which you can internalise concepts like direct/inverse limits in

[warped fable]

what do you need to internalize direct and inverse limits for

arrows go forward and you zoom

direct limits

arrows go backwards and you vroom

inverse limits

[warped fable]

i think i have not gone far enough then

its like an elaborate prank

"just a little further now, only a few months to go"

inb4 40 years pass

[formal ermine]

Eilenberg and Mac lane were the biggest trolls in history

[hidden haven]

They invented arrows what did you invent???

[devout dirge]

What is the ring of witt vectors of length 2 over Fp isomorphic to?

Is there a better understading of it?

[rustic crown]

@tiny jolt if you're thinking it in terms of actual elements, then the map is just re-parenthesizing. that is, send the element
(m⊗n)⊗p --> m⊗(n⊗p)
and extend by linearity

[delicate bloom]

Z/p^2Z if I understand you correctly

[rustic crown]

but i would rather think of it by showing that both objects represent the same functor, namely Z --> {tri-linear maps f : M x N x P --> Z such that f(ma, n, p) = f(m, an, p) and f(m, nb, p) = f(m, n, bp)}

[ivory trail]

if you know enough about groups, rings, modules, topological spaces, and the like, or even of a few of these, then you should be able to motivate categories pretty easily

[warped fable]

yes but what do i do with them

[ivory trail]

look at how the categorical constructions generalize constructions in each case

[warped fable]

and what do i do with that

[ivory trail]

and look at functors and and natural transformations

notice similarities

generalization

math

[warped fable]

and what do i do after noticing similarities

[ivory trail]

see if these generalizations have made it easier to spot patterns elsewhere

[warped fable]

math is specific not général to me

[ivory trail]

produce conjectures

investigate them

[warped fable]

what do i do with that

[ivory trail]

write papers

[warped fable]

boring

[ivory trail]

but before that, look up your questions in books and stuff

[ivory trail]

do you care about classifying all the objects of a certain type?

like for example, finite simple groups?

or finite dimensional semisimple lie algebras

or anything like that

[warped fable]

both the examples are boring

i care about classifying manifolds

[ivory trail]

there's a thing called "algebraic topology"

and "homotopy theory"

[warped fable]

and what do i do with those

[ivory trail]

get invariants of topological spaces

which could help you classify manifolds

[warped fable]

yes but where is the category theory here

nowhere

[ivory trail]

lmao

both those subjects are heavy in categorical language

[warped fable]

algebraic topology was invented by poincaré. if you asked henri if hes constructing functors hed think youre deranged

[ivory trail]

https://en.wikipedia.org/wiki/Eilenberg–Steenrod_axioms check this out

well poincare is dead

we're here

if you told euclid about polynomials he would think you're literally an idiot

[warped fable]

i dont even remember the es axioms

[ivory trail]

probably

[warped fable]

i can do algebraic topology just fine

[ivory trail]

do you remember the long exact sequence in homology

[warped fable]

yes

[ivory trail]

do you not care about when similar long exact sequences exist

[warped fable]

no

similar where?

[ivory trail]

for any cohomology theory

or even the analogous one for homotopy groups

[warped fable]

i dont know what a cohomology theory is

why do i care

[lethal dune]

this convo 😂

[warped fable]

i know what singular cohomology is

[ivory trail]

when are we going to use this in the real world, teacher

[warped fable]

my point is most of category theory is useless

[ivory trail]

do you know what simplicial cohomology is

[warped fable]

yes

[ivory trail]

do you know what de rham cohomology is

[warped fable]

yes

[ivory trail]

do you know what cech cohomology is

[warped fable]

yes

[wraith cargo]

Lmao rho getting trolled hard

[ivory trail]

nah i'm just blowing off steam

[delicate orchid]

It’s like the funny!!! 🤩

Do u know what uhhhhh the induced cohomological poopy functor from the brauer homomorphism is

[ivory trail]

i think this has essentially reached a reductio ad absurdum

[wraith cargo]

Why would I care what a universal property is

[warped fable]

this is what people say when they cannot carry their thought process to the actual reduction

cope

[lethal dune]

I agree with that

[next obsidian]

How sad to see someone so brilliant yet still so deluded

[ivory trail]

you're the tortoise

you're literally a tortoise

i'm not going to argue with a tortoise

[delicate orchid]

Brutal

[next obsidian]

[warped fable]

gish gallop

you have no point

also ad hominem

[wraith cargo]

[next obsidian]

One day your collaborator will mention mirror symmetry and then you’ll be forced to care…….

Just you wait…………

[warped fable]

i care about mirror symmetry

wheres category theory here

[next obsidian]

[warped fable]

tell that to an actual string theorist and theyll give you the looks

[delicate orchid]

I like mirrors because it lets me experience the joy of gazing upon my jaw droppingly handsome face

[warped fable]

exactly

[delicate orchid]

Google the whitehead tower nerd

[warped fable]

thats not category theory

[ivory trail]

no true scotsman

[warped fable]

its just a co-cell decomposition

[wraith cargo]

Yeah theyll see my pretty face

[delicate orchid]

Literally an exact sequence

[ivory trail]

but unironically actually lmao

[warped fable]

no thats a wrong headed way of thinking about it

postnikov towers are dual cell decomps

theyre not “exact sequences”

[delicate orchid]

Wait did you just say… dual??

[warped fable]

the problem with category theory is you see questionable similarities everywhere

[delicate orchid]

As in… the categorical dual?!??

[warped fable]

dual graphs

[delicate orchid]

I won you lost bye bye good night

[warped fable]

dual vector space

[formal ermine]

dual cone

[ivory trail]

can you define "dual" for me

nlab can

[warped fable]

sure, dual is when you shlonky inside to the outside

and things flip

[ivory trail]

so you hate bourbaki too

[warped fable]

of course

with a passion

im a morally russian topologist

[ivory trail]

i think k-theory has something to do with the eilenberg-steenrod axioms

[warped fable]

really? wow

k

[delicate orchid]

Bros trolling too close to the sun

[next obsidian]

Grrrrrrr

[ivory trail]

have you heard of k-theory

[lethal dune]

summon the #category-theory guys here

[ivory trail]

i think it has applications

[south patrol]

K theory

[south patrol]

So do like most cohomology theories lol

[warped fable]

ill stop trolling but i feel like you guys need to come up with more compelling examples to convince people

[ivory trail]

wow is this true

[warped fable]

simple, but compelling

doesnt have to be fancy

[ivory trail]

can't you do that yourself?

[warped fable]

yes i can

[ivory trail]

then do it

[warped fable]

this is the simplest i can do. theorem: let M be a compact n dim manifold, then homotopy classes of maps M -> S^n are classified by degree (preimage of a generic fiber, counted upto signs)

Proof: [M, S^n] = H^n(M), use Yoneda

i can expand more if needed

this is the first example that convinced me category theory is not just a language game

maybe there are simpler more compelling examples

[dim widget]

I need a real one to convince me that Morava K theory is not a language game

[warped fable]

too hard for me

when people say words, which categorists are prone to, my instant reaction is “why should i care”

its a shame because unlike most of mr Lurie’s student gangs, Lurie is actually a very good expositor

he knows how to communicate with people from other branches of math

all of the last few messages are non ironic

[fervent gyro]

I'm having trouble proving this exercise in Jason McCullough's notes Introduction to Commutative Algebra. I tried to tackle a) by proving both inequalities, and for $\le$ I gave a projective resolution of length n of M, and applied $A_{\mathfrak m}\otimes_A-$ to try to obtain a projective resolution of $M_{\mathfrak m}$, but that approach doesn't seem to be working, as I can't assure that the localization at $\mathfrak m$ of the projectives are still projective

[cloud walrus]

fargoth_ur7

I'm having trouble proving this exercise in Jason McCullough's notes Introduction to Commutative Algebra. I tried to tackle a) by proving both inequalities, and for $\le$ I gave a projective resolution of length n of M, and applied $A_{\mathfrak m}\otimes_A-$ to try to obtain a projective resolution of $M_{\mathfrak m}$, but that approach doesn't seem to be working, as I can't assure that the localization at $\mathfrak m$ of the projectives are still projective

[chilly radish]

Projectivity is preserved by extension of scalars. There's a multitude of ways to prove this. What are the equivalent characterisations of projectivity you know

[fervent gyro]

The functor Hom_A(P,-) being exact, each short exact sequence with P on the right being split and P being a summand of a free module

[rocky cloak]

$A_m\otimes-$ maps the free module $A$ to the free module $A_m$, and preserves sums and summands.

[cloud walrus]

jagr2808

[fervent gyro]

Thanks, that reasoning works great

[chilly radish]

There you go!

[rocky cloak]

Alternatively: $Hom_{A_m}(A_m\otimes P, - ) \cong Hom_A(P, Hom_{A_m}(A_m, - ))$ which is a composition of exact functors.

[cloud walrus]

jagr2808

[tiny jolt]

What does “extend by linearity” mean?

[elder wave]

Elements of the tensor product are finite sums of elementary tensors

So you can define a linear map by specifying where the elementary tensors get mapped to

[tiny jolt]

Ah I see

Thanks!

[rustic crown]

but if you work with elements, you'll always have to be careful with things being well-defined. the elementary tensors aren't a basis after all.

and the stuff you verify is basically what i wrote in the other description

check (m⊗n)⊗p --> m⊗(n⊗p) would be (\bZ-)linear in each variable and that you can juggle around a between m,n and b between n,p

[static yew]

Okay, I'm still trying to make sense of the logic on this wikipedia page and coming up empty

It talks about an elliptic curve over F_q, and then extending that to the algebraic closure F_q bar, and then applying the frobenius endomorphism to coordinates, so (x,y) -> (x^q, y^q)

Except as best as I can tell from poking at sage, if x is an element of F_q, then x^q in F_q bar is x
Which would make the endomorphism an identity transform for all valid inputs

Which would mean that the several paragraphs dedicated to the endomorphism are irrelevant and I'm pretty sure that they aren't

[next obsidian]

This is not true

x^q = x for x in F_q

Not for everything in the algebraic closure

[static yew]

But the transform can't "lift" an element out of F_q into a higher order subfield, right?

[coral spindle]

Ahdhahshdh

There are elements of F_q bar that are not in F_q

If x is in F_q bar and x^q = x, then in fact it must be that x is in F_q.

For any other element of F_q bar, x^q =/= x.

[coral spindle]

The logic you've applied here is like saying "every cat is an animal, and every cat has fur, therefore every animal has fur"

[static yew]

What I'm trying to apply is "I have a field of cats, and then algebraic closure of all animals, then starting with a cat c and frobenius phi, phi(c) is a cat"

[coral spindle]

Yes, and that says nothing about other animals.

[static yew]

It does say one thing

In order to end up with one of those other animals i cannot start with a cat

[coral spindle]

OK

[static yew]

It might turn a gnu into an emu

But it will never turn a cat into anything but a cat

Sage seems to want to call the elements of GF(9) { 0, 1 , 2, z2, z2+1, z2+2, 2z2, 2z2+1, 2z2+2}

Am I reading this right? What happened to z1

[coral spindle]

z = 2z^2 + 2 if I've gotten the polynomial right.

Again, this is a matter of choosing representatives.

I don't know why Sage chooses those representatives, but it has.

[lethal dune]

Prolly quotienting polys

[coral spindle]

Well no, because the 3rd cyclotomic polynomial (the one they quotient by) is of degree 2

So they're not performing the division algorithm here.

[lethal dune]

ah that's right

prolly z² is taken as a single variable?

[coral spindle]

There's presumably some clever trick they're using that helps when computing with finite fields, but I wouldn't know.

[static yew]

Ok that's weird
GF(9).gen() is z2
GF(27).gen() is z3

[coral spindle]

woah that's not crazy at all that's actually what you'd expect it to continue as

[lethal dune]

ok so I thing sage is just using z_i as generator

[coral spindle]

where i is the index of the field

[lethal dune]

they aren't powers but subscripts

[coral spindle]

And GF(3^4) is going to be z4 gasp

[formal ermine]

hey stevio

[lethal dune]

no different variables

[static yew]

Yeah sage uses ^ for exponents

[lethal dune]

tey GF(2^8) and see if it's z4

[formal ermine]

did you move from finite char 2 fields to finite char 3 fields

[coral spindle]

2^8?? D:

[lethal dune]

yeah different altogether

let's see if sage is smart enough

[static yew]

Fun fact: In 2001 i had to troubleshoot an inconsistency between two systems

The people who wrote the spec knew VB and wrote 10^n in their algorithm to mean 10 to the nth power

The people implementing the spec were writing the code in C

[lethal dune]

yeah I think zi, i is the exponent here

[coral spindle]

If we had just seen a list for GF(27) this would've been immediately apparent

[static yew]

Actually I'm doing ECC math so I moved to fields of prime order approx. 2^521

[coral spindle]

So you're doing characteristic 2

[static yew]

No

[coral spindle]

2^512 is a power of 2

Do you know what I mean by the characteristic of a field?

Or are you saying that it's just close to that

[static yew]

Yes and yes

[coral spindle]

Gotchya

Indeed 2^521 - 1 is a mersenne prime.

[static yew]

That's why I said approx. The crypto guys like to do Montgomery reduction so they have primes like 2^521 - 2^256 + 65537

Oh then its prolly 2^521-1

I wondered why they didnt pick a multiple of 8 for the exponent

0x01ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

Yep one less than a power of 2

[coral spindle]

If they chose a composite for the exponent, the result wouldn't be a mersenne prime

[static yew]

Curve25519 field is 2^255 - 19

Yeah they usually have 2^a -2^b + 2^c - 1

Where a is the number of bits they want (which is usually a power of 2, or at least a multiple of 64)
And b and c are chosen to minimize effort for the NSA to hack it 😉

Anyway what exactly is gen() supposed to return for non prime order (i.e. field size q is p^k, k>1)

For prime fields gen() is truly a generator of the field

[coral spindle]

A generator of the field

Every finite field has a generator

[static yew]

Hmm generates how?
For a group I can do a^0, a^1, a^2... a^(q-1)

[coral spindle]

Much the same way

Thinking of it precisely that way is unhelpful though

Let me elaborate

We can have elements a,b,c, ... of a ring R. We can 'generate' a ring <a,b,c,...> which we define to be the smallest subring of R containing a, and b, and c etc.

[static yew]

But F_q = p^k has characteristic p

So no additive generator

And multiplicative group is missing 0

[coral spindle]

Listen first

So we have to show that such a thing exists, but I won't do this here (maybe you could do this as an exercise)

But in any case this definition tells you exactly what it means

[static yew]

Such a thing being the generator?

[coral spindle]

The elements a, b, c, ... would be called generators for the group.

Ring* rather

Now moving onto finite fields

[static yew]

Wait a sec

[coral spindle]

I'm listening

[static yew]

When you say generators

Do you mean that they are each generators on their own

Or that there's some sort of combination of elements that generate subsets of the ring and collectively they generate everything

[coral spindle]

I mean together they are generators.

OK, moving on

Finite fields.

Let F be a finite field. Then F\{0} is a group under multiplication.

[static yew]

That's F with the zero element removed, right?

[coral spindle]

Yes

[static yew]

One sex

Sex

SEC

[coral spindle]

Now this is a very cool fact about finite subgroups of F\{0} for any field

any such group is cyclic

This means there is an element g in the group such that <g> is the whole group.

And of course, you're very familiar with this.

So in particular for a finite field there is an element z in F\{0} such that <z> = F\{0} (n.b. as a group)

We get now, trivially, that <z> (as a ring) is F.

This is because any ring must also contain 0, which is the only element not given by F\{0}

[static yew]

For rings of the form Z/nZ

R \ {0, k | gcd(n,k) != 1} is also the group, right?

[coral spindle]

We would usually write this as F_p[z] but I find this explanation simpler, so dw about it

[coral spindle]

No.

No group has an absorptive element (except the trivial group, but nvm)

also this isn't relevant

Anyway, the point is that every finite field has a generator.

[glossy crag]

This is trivial, but I want to make sure I have the details right: a finite-dimensional $K$-algebra $A$ is simple if and only if it is isomorphic to $M_n(D)$ for $D$ a finite-dimensional division $K$-algebra and $A$ is central if and only if $D$ is. One direction is clear, conversely $A$ is left artinian (because $\dim_K(A)<\infty$), therefore by Wedderburn's theorem there is a division ring $D$ such that $A\cong M_n(D)$ as rings. Because of this isomorphism we have $K\subset Z(A)\cong Z(D)$, thus we can define a $K$-algebra structure on $D$ and $A\cong M_n(D)$ as $K$-algebras wrt it. Since $A$ is finite-dimensional, so is $M_n(D)$ and so is $D$ (which embeds in to $M_n(D)$ via diagonal matrices). Centrality equivalence follows from $Z(A)\cong Z(D)$.

[static yew]

Basically you're saying "we can use <z> as shorthand for F by taking successive powers of <z> until we get {1} again, then add in {0}"

[coral spindle]

Yes, but n.b. only in this nice, special case

In general this is definitely not true.

[static yew]

Wait which part was special? The finite part?

[cloud walrus]

leave_no_norm

[coral spindle]

The part where I said that F\{0} was cyclic, which I believe is indeed only true when F is finite.

[static yew]

Yeah ok
That makes sense. Theres no way to pick any element of Q R or C and "generate" the entire thing

[coral spindle]

I may be off here, but I believe that there is a version of Wedderburn that specifies to k-algebras. But I think you're right in any case.

[static yew]

Actually I bet Goedels diagonal argument proves it impossible for uncountably infinite fielss

[coral spindle]

You don't need diagonalisation to prove it doesn't work

It's very easy to show that <z> is always countable

so we're done

In fact *adjusts glasses*

🤓 for any variety of algebras with a finite signature this is true. And indeed rings have finite signature.

Now who was it that was such a fan of universal algebra around here

[coral spindle]

Yeah in fact this can be shown easily. If F\{0} = <z> is cyclic and char F =/= 2, then z^n = -1 for some n and so z^2n = 1 and indeed F\{0} is finite. Now if the characteristic is two, then I don't know.

[static yew]

In char 2, z^2n = z^0 = 1

[coral spindle]

Uh no

we wouldn't be able to guarantee that n =/= 0

remember, -1 = 1 in char 2.

[static yew]

Oh right I'm mixing up my orders

Yes, that one I know well

[coral spindle]

I should have specified n =/= 0 in my proof

[static yew]

Actually I had to unlearn that + = - when extending beyond F2

Cuz I was so used to turning -s into +s when translating general expressions

Crud. Gtg

[coral spindle]

Dang I think it is literally possible for fields of characteristic two!

Let me think

F_2(x)?

No that's wrong.

If char F = 2 and <z> = F\{0} is infinite, then z can't be algebraic since otherwise it would be a root of unity. So ok, it's transcendental, so F is isomorphic to F_2(x) and we're done?

I think?

That's cool

[minor glen]

I am trying to find a commutative ring that isn't Z/2Z that has an endomorphism x |-> x^6 on it, but I seem to have exhausted my search. Does anyone have any inspiration regarding rings that are worth digging through?

[coral spindle]

Well, any field of characteristic two is an example

So for example the ring F_4 = (Z/2Z)/< x^3 + x^2 + 1 >

Wait is that true? Have I messed up there?

Ah yes. The field F_4 is an example, but in fact I think I was wrong about any field of characteristic two being an example

Man OK that isn't right either. The problem is that there are nonzero, nonunit elements lmao

[minor glen]

It is a tough one that's for sure

It has to be a non-integral domain ring I believe, but I have run through all the ones I know if I am not mistaken

[coral spindle]

And we know 2 = 64, so char R | 62

Ah in fact we know 0 = (-1 + 1)^6 = (-1)^6 + 1^6 = 2

So in fact we definitely know it's a ring of characteristic 2

[minor glen]

If it is an integral domain then yeah, we can then show that R is isomorphic to Z/2Z

Otherwise I don't think you can get 2 = 0

[coral spindle]

Nono, we can

0 = (1 - 1)^6 = 1^6 + (-1)^6 = 1 + 1

So we have 2 = 0

So let's try and find the least complicated ring that does this thing, right

If there is such a ring R, we can choose some nonunit z in R and surject (Z/2Z)[x] → (Z/2Z)[z]

So if we can find an ideal of (Z/2Z)[x] that contains what we want...

So we want (a+b)^6 + a^6 + b^6 = 0, but squaring is an endomorphism, so this is the same as ( (a+b)^3 + a^3 + b^3 )^2 = 0

Maybe we can assume for now that we have no nilpotents. I'll put a pin in that 📌

so we want (a+b)^3 + a^3 + b^3 = 0, or in other words 3a^2b + 3ab^2 = ab(a + b) = 0

If we have some generator x that we're choosing, we can say e.g. x(x+1) = 0. Hm! So x^2 = x.

Does this make it work? Is F_2[x]/(x^2 + x) an example? Let's check.

(x + 1)^6 = x + 6x + 15x + 20x + 15x + 6x + 1 = x+1. Nice!

So there you go @minor glen I think I found the simplest example of such a ring

I think you can experiment with adding nilpotents (see my pin!)

[delicate orchid]

x^2-x albeit

[coral spindle]

Char 2 baybeeeeee

[minor glen]

Okay thank you so much, I am going to mule over this and see if it sticks :))

[delicate orchid]

doesn’t make sense

Try again

[coral spindle]

I'm dying wew, I need your approval

[formal ermine]

I'm dying wew, I need your approval

[coral spindle]

Wew doesn't approve of imitation

[rocky cloak]

F_2[x]/(x^2 + x) is just isomorphic to F_2 x F_2 though, so not that exciting.

[coral spindle]

Good point

Haha yes indeed, any number of products of F_2 works. How dull.

Well it's certainly true that x^2 = x for any element, so... well if we know what a Boolean algebra is we are unfortunately very restricted.

Or is it called a boolean ring? I forget.

[zenith hollow]

you Germans wanna see my math exam?

Damn i f'd it up so hard that my brain still hurts

[delicate orchid]

Jawohl!!!!

[coral spindle]

Indeed it's a boolean ring I was thinking of.

[delicate orchid]

Block decomp of a Boolean ring or as I like to call it, a Boolean ring

[coral spindle]

If we want to find any interesting finite examples, we will need to have nilpotent elements :(

[rocky cloak]

What about F_2[x]/x^2 ?

Seems to work

[delicate orchid]

When u say Nilpotent do you mean x^n = 0 for some n

Because all I’ve seen so far are idempotents

[coral spindle]

yeah I think so

[delicate orchid]

Boss nobody here speaks German

[coral spindle]

@delicate orchid it's related to this up here. I just wanted to have the property that x^2 = 0 implies x = 0

[minor glen]

So would Z/2Z X Z/2Z also work?

[zenith hollow]

this chat was full of germans

[coral spindle]

Yeah, and as jagr pointed out, the example I gave was actually just that all along

Note that F_2 is just different notation for Z/2Z

[delicate orchid]

Ah ok

[coral spindle]

Which makes everything idempotent :)

[delicate orchid]

Z/4Z my beloved

[minor glen]

Very fair, and because the function does not do anything (is the identity function), it is automatically a homomorphism?

[coral spindle]

Yeah exactly

this is why I could conclude it was a homomorphism from showing (x+1)^6 = x+1 alone

[minor glen]

Boring but very neat haha

[coral spindle]

I knew already that 1^6 = 1 and x^6 = x

Yeah it's nice

And as it turns out, every finite ring such that x^2 = x for all x is (Z/2Z)^n for some n

[slim kayak]

We dont actually exist, all german math people are out of bielefeld

[tiny jolt]

One more question, how are they not a basis? Can't every element in the tensor product be written as a sum of elements?

[coral spindle]

Just because they span the space doesn't mean they're a basis

E.g. in k (x) k, the space is certainly spanned by 1 (x) 2 and 2 (x) 1

but they don't form a basis!

[delicate orchid]

Society if

[tiny jolt]

ohh

[barren sierra]

https://link.springer.com/article/10.1007/BF01160336

anyone know of an English translation of this?

[formal ermine]

just speak german

[barren sierra]

of course why didn't I think of that one

also this https://books.google.com/books/about/Finite_Groups.html?id=dfSguAAACAAJ

[delicate orchid]

what you want to know about permutation groups, nerd

[barren sierra]

REU stuff

[formal ermine]

wtf even is a twice transitive simple permutation group

[formal ermine]

I can translate it for you but I will probably lose motivation half way through

[coral spindle]

'twice transitive' aka doubly transitive means that the action on X, when extended to the diagonal action on X x X, has only two orbits.

[formal ermine]

aha

[delicate orchid]

n-transitive means you can map any n-tuple to any other n-tuple except the ones you obviously can't u stinkyface

[coral spindle]

so true!!!!!!!!

[barren sierra]

well the book would be more important lmfao

also I wouldn't ask anyone to translate it

I mean worst case I don't use this book but it'd be nice to have as a reference since it seems alot of the stuff I'm reading refers to it

[delicate orchid]

could u translate the chapter titles at least so I know what's in this mofo

[barren sierra]

https://link.springer.com/book/10.1007/978-3-642-64981-3
here this has chapter titles in German if someone wants to translate lol

[delicate orchid]

I'll use the powerful combination of my C2-tier german and guessing

ok nothing in here seems particuarly special

[barren sierra]

yea

just some proofs and definitions which are hard to find elsewhere

like quasi-primitive linear groups

[delicate orchid]

groupprops will have a definition

https://groupprops.subwiki.org/wiki/Quasiprimitive_group

lol

[formal ermine]

basics
permutation groups and linear groups
nilpotent groups and p groups
displacement and p nilpotent groups
rep theory
solvable groups

[barren sierra]

see this isn't what I want lol

[delicate orchid]

weird

[barren sierra]

I already looked there trust me

the definition I need relates to G-modules

I found it elsewhere

[delicate orchid]

ok so

[barren sierra]

but some other proofs are in there which would be nice

but no big deal

[next obsidian]

Maybe try #combinatorial-structures ?

[long nebula]

Finally, groups, 1

Hope this helps

[south patrol]

That'd be endlich smh

[long nebula]

I know potato smh you don't understand my humor

[south patrol]

OK lmao

[barren sierra]

Fuck man I was struggling with that

Thanks

🙏🙏🙏

[long nebula]

Lol

[formal ermine]

beat me to it

DEUTSCHER ALGEBRAISCHER STRUKTUREN SCHAT

[elder wave]

…

[glossy crag]

If I want to show Brauer multiplication is well defined via [A]\cdot[B]=[A\otimes B] does the following work:
Suppose A\sim A', B\sim B', we want to show A\otimes B\sim A'\otimes B'. Let D and E be division algebras with A\cong M_n(D), A'\cong M_n'(D), B\cong M_m(E), B'\cong M_m'(E), then A\otimes B\cong M_nm(D\otimes E) and A'\otimes B'\cong M_n'm'(D\otimes E), i.e. both are isomorphic to a matrix algebra over the same (not necessarily division) CSA C. This implies similarity (if A\cong M_n(C) and B\cong M_m(C) for a CSA C, then there is a division algebra D with C\cong M_d(D), hence A\cong M_nd(D) and B\cong M_md(D) and A\sim B).

Making sure I understand this correctly, is this because if [A]\in Br(K) is an equivalence class, then [A]=[D] for a central division algebra D (A is CSA, therefore A\cong M_n(D) for a central division D \implies [A]=[D]) and D is unique up to isomorphism ([D]=[E] \implies D\cong M_n(F), E\cong M_m(F) \implies n=m=1 and D\cong F\cong E), so Br(K) corresponds exactly to isomorphism classes of central division algebras.

[glossy crag]

Also, since #linear-algebra is keeping silent on this:
Quick tensor sanity check, if U\subset W is a subspace and V\otimes U=V\otimes W (where the LHS is embedded in the RHS in the usual way), then U=W, right?

[rocky cloak]

Assuming V isn't 0, then yes

[chilly ocean]

The group GL_2(R) of invertible matrices with real entries acts on R^2 by left multiplication. How many orbits does the set R^2 break up into under this action?

What is wrong with my reasoning? The orbits are of form GL_2(R)(x) = { Ax | A in GL_2(R) } and since A is a 2 x 2 matrix it has entries say a_1, a_2, b_1, b_2 and we are free to choose these real numbers so I thought there would be uncountably many orbits, but apparently there are only 2. What is going on here?

[glossy crag]

an orbit of G acting on X is an equivalence class wrt the relation x\sim y \iff \exists g\in G with gx=y

the term itself is very telling, think of planets orbiting the sun and the orbit of e.g. Mars as all the places where Mars can show up (it can't show up in the orbit of Venus, right?)

now one obvious orbit is {0} since every A\in GL_2(R) sends 0 to 0

take any x\neq 0, you want to describe all the places it can be sent to by GL_2(R)

[carmine fossil]

Ah, so the two orbits are {(0,0)} and everything else?

[glossy crag]

if y\neq 0 is any other vector, you can construct an invertible matrix such that Ax=y

meaning there is only one other orbit besides {0}, namely its compliment R^2\{0}

[slim kayak]

Rotate any point until it intersects the y axis, then scale it until it gets mapped to (0,1)

So yeah

Notice how this works just the same with SL(2,R), you just modify the scaling operation by scaling by an inverse amount in the x-direction

[formal ermine]

hey did you delete my thread

[hidden haven]

Literally 1984

[elder wave]

Yes it was incredibly cringe

Plus somewhat against rules but that wasn’t the main point

[delicate orchid]

it was really cringe

[south patrol]

Also you spelled it incorrectly

Strukturen Schat

[karmic moat]

i'm reading about "classical Lie groups" and it says "the map $\operatorname{Gl}(n, \bR) \times \bR^n \to \bR^n$ is given by polynomials in the coefficients of the matrix and the vector, and so is continuous

what exactly is this map?

[cloud walrus]

anamono

[karmic moat]

i don't really get it lol

i know that the map Gl(n, R) to R is the determinant

[delicate orchid]

u ever seen matrix multiplication before

[karmic moat]

yea

oh it's just multiplying the n x n matrix by a n x 1 vector?

oh wait it's R^n lol not R

oops

[delicate orchid]

yeah, there's no R involved lol

although the determinant is also continuous for precisely the same reason

[karmic moat]

yeah

[south patrol]

Hm feels a slightly weird justification lol like

Saying "polynomial" when really it's just that this is a restriction of a bilinear map $M_n(\mathbb R) \times \mathbb R^n \to \mathbb R^n$ right

[cloud walrus]

potato

[south patrol]

But maybe they say polynomial to make it more clear how it generalises lol

[next obsidian]

Well it is polynomial as a special case