groups-rings-fields

[wild solar]

oh okay

[next obsidian]

So okay, p(x) - aq(x)/c = r

And so now we can let h(x) = a/c, and r(x) = r

And we see that p(x) = h(x)q(x) + r(x)

This is kinda silly right now since h and r are constants

But this totally works

[wild solar]

sounds good

[next obsidian]

So what if q was a constant?

Say q(x) = c

Well now p(x) -aq(x)/c isn’t a constant

but

[wild solar]

h(x)=1/c p(x)

?

[next obsidian]

p(x) - axq(x)/c

Not quite

We need to bump up q’s degree

So it matches p’s degree

[wild solar]

o right

[next obsidian]

So okay p(x) - axq(x)/c = r

For a constant r right?

[wild solar]

yeah

[next obsidian]

So I’m gonna introduce some new notation which right now seems silly, but it’ll be useful since the general case will use this

[next obsidian]

So we’ll let p_1(x) = p(x) - axq(x)/c

And h_1(x) = ax/c

Note that p_1 is like, what’s left over from p after step 1

And h_1 was what we had to fudge q(x) by to complete step 1

I introduce this notation since the general process is basically just doing this over and over

But okay, well so p_1(x) = r

So p_1(x) - rq(x)/c = 0

Since q(x) is just c

So we’ll let h_2(x) = r/c

And r(x) = 0

Well now we see that
p(x) = (h_1(x) + h_2(x))q(x) + r(x)

Maybe verify that this actually works

Because if you can see why this works in this case, you’ll be able to understand the general case

Maybe it’s easiest if we write it like this actually:
p(x) - h_1(x)q(x) - h_2(x)q(x) = r(x)

[wild solar]

sorry why did you define h1 like this

you type faster than I think

[next obsidian]

Precisely so that this step here works

Let’s look at what happens here

Do this first subtraction first

p(x) - h_1(x)q(x)

This is p_1(x)

Oh actually induction is totally gonna help so sweet

But don’t worry about that right now

[wild solar]

ummm, sorry, I'm kind of lost

[next obsidian]

Right so

[next obsidian]

Going back here

I wrote down p_1(x) = p(x) - h_1(x)q(x)

Because I let h_1 = ax/c

And I defined p_1(x) = p(x) - axq(x)/c

[wild solar]

okay

[next obsidian]

So the idea was like

Okay, I want to cut p’s degree down by one

So I multiplied q(x) by something [this is what h_1 is] so that h_1(x)q(x) has the same leading term as p(x)

This makes sure that once I subtract them, I get something of lower degree

And I just called that thing of lower degree p_1(x)

[wild solar]

okay cool

[next obsidian]

Then I kind of replaced p(x) with p_1(x)

Then said “okay now I want to fudge q(x) by something so it equals p_1(x)”

And this was h_2

So what I ultimately did was, I found h_1 and h_2 such that
p(x) - h_1(x)q(x) - h_2(x)q(x) = 0

Because well I could just look at it this way

[p(x) - h_1(x)q(x)] - h_2(x)q(x)

But this is just
p_1(x) - h_2(x)q(x)

[wild solar]

you mean equal p_1(x)-r(x)?

[next obsidian]

And I know this is 0

[wild solar]

what happened to r(x)

[next obsidian]

r(x) in this case will be 0

Because q is a constant, I can’t have deg r(x) < deg q(x)

So it has to be 0

[wild solar]

ohhhh ok

[next obsidian]

So did the idea make sense?

It was like, fudge q(x) so that subtracting h_1(x)q(x) from p(x) gives something of lower degree

[wild solar]

kind of makes sense

[next obsidian]

Then just keep dividing what’s left over by q(x), then add t all@up

Okay so the general case

It will actually be pretty much the same idea

So the first thing to note is again, if deg q(x) > deg p(x) it’s easy

h(x) = 0, r(x) = p(x)

[wild solar]

uh huh

[next obsidian]

Okay so now let’s write
p(x) = ax^n + …
q(x) = cx^m + …

[the … is just the lower degree stuff, they won’t matter] And we know that m <= n

Oh wait

Okay now we’re good

Good so far?

[wild solar]

yes yes

[next obsidian]

Cool so step 1:
Let p_1(x) = p(x) - x^{n-m}aq(x)/c

This looks scary, but all I want you to do is pay attention to the leading term (also x^{n-m} makes sense since m <= n)

You’ll see that the leading term cancels each other out, so that p_1(x) has degree < n

[wild solar]

yeah

[next obsidian]

Awesome, so we’re almost done

Let h_1(x) = x^{n-m}a/c

So that above equation can also be written as
p_1(x) = p(x) - h_1(x)q(x)

Seems familiar right?

[wild solar]

yes

[next obsidian]

So this is where induction simplifies our life

Because p_1(x) has degree < n, we can use induction to get h_2(x) and r(x) such that
p_1(x) = h_2(x)q(x) + r(x)
where deg r(x) < deg q(x)

Does this make sense?

[wild solar]

yeah it does

[next obsidian]

Okay so let’s combine our two equations!

[next obsidian]

Taking this and the one we got by induction we see that
h_2(x)q(x) + r(x) = p(x) - h_1(x)q(x)

But now let’s move some stuff and we get
p(x) = (h_1(x) + h_2(x))q(x) + r(x)

So if we let h(x) = h_1(x) + h_2(x) we did it!

[wild solar]

wow

this is a bit long but I'll read it a few more times to memorize it

thank u so much

[next obsidian]

The key idea is really this:

1. Try to induct

2. just find something so that once you fudge q(x) by it, you can match p(x)’s leading term

This lets you subtract the two, and then you end up with something of lower degree, now you just use induction to complete

[wild solar]

Sounds great!

😄

[next obsidian]

Also hopefully you can see why we needed a field

In order to match p’s leading term we had to use a/c

But this means needing to be able to write /c

And that’s where the field-ness comes in

But this proof shows you that if q(x) is monic we can do this! Since c = 1

[wild solar]

then what about integral domains

[next obsidian]

Or really, all you need is that q(x) has an invertible leading coefficient

[wild solar]

does R integral domain implies R[x] is ED?

[next obsidian]

Won’t work since we can’t divide

Heck no

You can prove this

If R[x] is a Euclidean domain then R is a field

I can’t succinctly really tell you why

But the key is that if R[x] is an integral domain then so is R

So this tells you R is an integral domain at least

Then you can show that it has to have the property that every prime ideal is maximal

[wild solar]

I see

[next obsidian]

But since 0 is prime (since R is an integral domain) this says 0 is maximal

So R is a field

So yeah, idk it needs a bit of commutative algebra, it isn’t too hard

But it isn’t really doable with what you have right now

At least I can’t come up with a proof

[wild solar]

sounds good

this has been informative

3Q

[next obsidian]

[next obsidian]

Is this supposed to be some face?

[wild solar]

oh sorry

[next obsidian]

It kinda looks like a pair of lips kissing a balloon

Lmao

[wild solar]

in chinese its pronounced as "thank you"

[next obsidian]

Oh lmfao

[wild solar]

good imagination tho

[next obsidian]

Is that like shi-kyu or something

Wait no

Idk lmfao I’m tryna base it off of knowledge of Japanese but shi is 4 not 3

I guess my guess would be mi-kyu or something but that doesn’t sound right lmao

WAIT

Is it san-kyu

[wild solar]

yes

exactly

[next obsidian]

Woooo

[wild solar]

you learnt some chinese

[next obsidian]

That’s the same as Japanese

[wild solar]

wait rly?

[next obsidian]

Yeah

The number 3 is san in Japanese

And then Q was well

The letter Q in English is pronounced like kyu

[wild solar]

you do realize there's no other pronounciation of Q right

w8 actually

I'm not sure

[next obsidian]

I mean idk I thought like

When you romanize Chinese

Q usually is a ch sound

Like qi is chi

[wild solar]

hahahhahahahhh

[next obsidian]

I think

And x is a sh

Like xie is… shea…??

[wild solar]

that's how foreigners pronounce it

[next obsidian]

Is it more like

Sheh

[wild solar]

like they would pronounce shea shea ni for xie xie ni

[next obsidian]

Or something

[wild solar]

and think that's the right way to say it

[next obsidian]

Well I know it for sure as hell isn’t pronounced like ksie or something

[wild solar]

👍

also do u recommend any textbook on rings and fields

we're using Herstein but the exercises aren't very similar to what we get as homeworks or exams

[next obsidian]

Idk about rings and fields specifically but like

Dummit and Foote I suppose if you’re just looking for exercises

I’m not really aware of stuff that focuses on just rings and fields that isn’t like, a commutative algebra book which is definitely not what you want right now

[wild solar]

okay thanks! I have so many subjects this sem and I've been focusing on ode and probability for a while and lacking off on algebra and now I'm panicking...anyway I'll take a look on the book you said >___<

[patent ocean]

Yh it’s weird

[chilly radish]

Can someone walk me through why the maximal ideal in a commutative, local artinian ring is nilpotent

Actually, I find a proof in this thread
https://math.stackexchange.com/questions/2250105/given-an-artinian-ring-a-mathfrakm-show-that-mathfrakm-is-nilpoten

But I don't understand why from $(x)=(x)\mathfrak m^k$ then can conclude $(x)=(x)\mathfrak m$

[cloud walrus]

ShiN

[chilly radish]

I think I understand the rest of the proof

[terse crystal]

$(x)=(x)m^{k} \subset (x)m \subset (x)$

[cloud walrus]

Cogwheels of the mind

[chilly radish]

ah, of course, thank you

[chilly radish]

I have two questions. First of all, if I have a division ring $D$ and some finite rank simple \textbf{right} $D$-Module $K$, is the endomorphism ring of \textbf{right} $D$-Endomorphisms going to be isomorphic to $M_n(D)$ or $M_n(D^{op})$?

[cloud walrus]

ShiN

[chilly radish]

I know it's op in the case of left endomorphisms

Secondly, and somewhat related, i'm trying to prove uniqueness in wedderburn-artin. I have gotten to the point where I need to prove that $M_n(D)\cong M_{n'}(D'}$ implies $D'\cong D,n=n'$. It;s clear to me that once i prove $D'\cong D$ the rest will follow. The hint I have says to consider $\operatorname{End}_{M_n(D)}(I)$ for a minimal left ideal $I$, and I want to show that this is isomorphic to $D^{op}$. The natural way would be to define the endomorphisms via left or right multiplication by a scalar matrix, but in the case of left multiplication I don't see why this would be closed under $I$ (Since it's just a left ideal and this would amount to right multiplication), and for right multiplication I don't see why it's linear. I'm actually not sure about surjectivity either (Injectivity is clear) but i'll get there when I get there.

[cloud walrus]

ShiN
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chilly radish]

I is an ideal of M_n(D) ofc

[somber thorn]

is that from nlab

[leaden scaffold]

what is the wedge?

[stark sigil]

Semidirect product?

[sand cradle]

Anyone have a proof that ⊕i∈I V_i ≅ ∏i∈I V_i for an indexed collection of vector spaces {V_i|i∈I}

[lethal dune]

that's not true if the collection is infinite

[chilly ocean]

Have you tried latex before?

[lethal dune]

for finite case, just use dimension argument to show they are iso

[grand sigil]

a (0), (p) for p prime
b (0)
c (0), (3)
d (0), ideals that dont meet (3) ?
e (0), every complex number
f same as e?

i think i got like 2 out of 6 right

for d i remember localized ideals has elements x/s where x in Z and s in Z-(3) setminua

I know C is a field and C[x] is PID and because of algebraic closed every polynomial is reducible so every prime ideal might be a linear polynomial

and same reasonings for f

[grand sigil]

what is C_p and the other one

[leaden scaffold]

Cyclic group with p elements

I got it

The other index is specified to be an integer

p^2-1 is a perfect square

[grand sigil]

oh okay

i forget semidirectnproduct

i just know. its like direct product

but multiplication is based on conjugation action?

something like that

also im bumping sorry

https://media.discordapp.net/attachments/496784958430380033/982841722000265236/image0.jpg

a (0), (p) for p prime
b (0)
c (0), (3)
d (0), ideals that dont meet (3) ?
e (0), every complex number
f same as e?

for d i remember localized ideals has elements x/s where x in Z and s in Z-(3) setminua

I know C is a field and C[x] is PID and because of algebraic closed every polynomial is reducible so every prime ideal might be a linear polynomial

same reasonings for f

[next obsidian]

Why is the set of power-bounded elements closed under addition?

I can prove it under the assumption that R_0 is a k-subalgebra of R, but it is only defined to be a subring

[terse crystal]

a,b correct
c: (2)/(6),(3)/(6)
d: no, contained in (3) so (0) and (3)Z_(3)
e: (0) and (x-a): a from C
f: (0) and (x)/(x^2)

[terse crystal]

What book is it btw?

[grand sigil]

yeah

i made mistake

2 is one

[grand sigil]

geometry of schemes

yeah for e I meant that

i shouldve written that explicitly instead of way i stated it

[terse crystal]

Okay thanks

[grand sigil]

(3)Z_(3) have elements of form 3x/s?

where x in Z and s in 1-(3)?

I definitely need to review localization stuff

[terse crystal]

Yeah, s in Z \ (3)

[grand sigil]

yeah thanks for notation clearup

[terse crystal]

Just the unique maximal ideal of Z_(3)

[grand sigil]

oh bruh moment

ill think about f more

also unrelated question but what is a good computer algebra system?

im using sagemath atm

but idk what is considered mainstream i just looked it up

[grand sigil]

Im assuming the first step is looking at 15 mod 7

The next step is taking the result x and writing it as x/7

So the answer would be 1/7

Similarly at the point (5)

0/5

Oh okay this makes sense

oh bruh

i forgot

So first step is 15 mod 7

taking result x

x/1

x is 1

So f is 1

In case of 5

15 mod 5 yields x = 0

0/1 is value of 15

So 15 is 0 at (5) and 1 at (7)

No I am definitely messing something up

I was right the first time

1/7 and 0

Since the second canonical function is field of fractions

bruh im stupid af

[next obsidian]

😔

[grand sigil]

Im pretty sure i was right the second time

the canonical map is x -> x/1

what the hell

this is useless map though

nvm def isnt

So final answer is 1/1 and 0/1

[chilly ocean]

How do you define both? They should be precisely the same by definition

For finite family I

[terse crystal]

Correct

That’s true for any finite indexed objects in any abelian category finite product coincides with finite coproduct

Infinity case clearly false like category of C vector spaces, coproduct of Z many copies of C is strictly contained in product of Z many copies of C

[wicked zephyr]

The first answer is 1 (mod 7) and the second is 0 (mod 5). It is important that functions take values in different rings.

[grand sigil]

uh

i think its just 1/1 and 0/1

why adding mod 7

[wicked zephyr]

First, Z/7Z is a field

so the second arrow does nothing

[grand sigil]

it will always do nothing lol

[wicked zephyr]

not always, the thing is that here you are evaluating at maximal ideals

[grand sigil]

wait

in these cases

it should always do nothing

or wrong?

[wicked zephyr]

Not in the case where we evaluate at (0)

then the functions takes a value in $\b{Q}$

[cloud walrus]

Potitov06

[wicked zephyr]

Q

rationals, i mean

[grand sigil]

o yea

no im sorry

dont see how it matters

the functions should just be f/1 mod nothing

[wicked zephyr]

Z/0Z is just Z

[grand sigil]

yeah

i know it will take values in rationals

but so does everything sort of

[wicked zephyr]

You mean, why is it necessary to take values in the fraction field?

[grand sigil]

i dont see why that part matters

[wicked zephyr]

The thing is

[grand sigil]

unless we mod out by non primes

[wicked zephyr]

Well, you see the definition of rational functions

For example, the "function" 1/7 is a function in a subset of Spec Z

for example you can evaluate 1/7 at 5, and it gives 3 (mod 5)

(is the inverse of 7 (mod 5) = 2 (mod 5)

[grand sigil]

slow down

[wicked zephyr]

1/7 has also a value at (0), it is 1/7\in Q

[grand sigil]

from the book it says functions are elements of base ring

so 1/7 isnt element of Z

[wicked zephyr]

yes, but you will soon define rational functions

[grand sigil]

so how can it be a function

oh

[wicked zephyr]

And then you'll need that def

[runic basalt]

I've showed that H∩N is a normal subgroup of H, any ideas of a well defined hom from H/H∩N to G/N?

[proud bear]

try to find a map H->G/N with kernel H\cap N

[runic basalt]

thanks, I'll give it a go

[runic basalt]

pretty sure i got it thank you so much

[wooden ember]

am i being smol brain here? Cant i just take the splitting field of f to make it not irreducible in an extension? Here C is a conic in PG(2,F)

[hot lake]

Maybe C has several variables ?

[terse crystal]

What is a conic? I want to learn about it

And PG(2,F) is the set of 2-dimensional subspaces of F^3?

[wooden ember]

In this situation a conic is the zero set of a homogeneous degree 2 polynomial within PG(2,F)

[wooden ember]

And you’re right I’m an idiot

Can’t just apply regular field theory to polynomials with multiple variables lmao

[wooden ember]

It’s not just a set it’s a projective space

So points are 1 dimension subspaces and lines 2 dimensional subspaces

Don’t ask me about it though I’m absolutely clueless on the subject it just happened to come up today

[terse crystal]

So

[terse crystal]

Is the underlying set {one-dimensional subspaces of F^3}?

[wooden ember]

Yup

[terse crystal]

And what is it? A F-linear space?

I mean what structure it has

[wooden ember]

A projective space, ie it satisfies some extra axioms to a point-line geometry. To be specific it’s a linear space such that every 2 lines meet in exactly one point and there exists 4 colinear points. PG(2,F) is an example of such a space (I’m pretty sure you can define isomorphisms that show all projective spaces are of this type but don’t quote me on that)

And when I say linear space I’m not talking about vector space

I mean that any two points are contained in exactly one line

[terse crystal]

Oh I see

Since Any two one-dimensional spaces are contained uniquely in a two-dimensional space (generated by them)?

[wooden ember]

Yeah exactly

And dually any two planes meet in exactly one line

[terse crystal]

Oh i see, dual of span is intersection?

[wooden ember]

(You can dually switch around lines and points in any statement on projective space because of this)

[wooden ember]

I mean again here im speaking axiomatically, PG(2,F) is a model of this where we consider vector subspaces

So you can talk about projective geometry without referring to the subspaces at all

[terse crystal]

I see

So conic is homogeneous polynomial of degree 2, of 3 variables?

[wooden ember]

According to the stuff I saw they call it n-dimensional hypersurfaces, idk if they’d be called cubics

[terse crystal]

Sorry typed wrong

[wooden ember]

Oh yeah

[terse crystal]

I see, thanks

[wooden ember]

Zero set of*

They aren’t the polynomials themselves

And that’s you need the polynomial to be homogeneous or it wouldn’t be well defined

Here to evaluate the polynomials you take a representative of the 1 dimensional subspace, and it’s well defined if you require it to be zero

[terse crystal]

? Then what a conic being irreducible or reducible means?this corresponding homogeneous polynomial being irreducible or reducible?

[wooden ember]

Yeah

Irreducible in any extension of F

Which is where my original confusion came from cause I’m not used to thinking about polynomials in many variables

[terse crystal]

I see

Thanks

[wooden ember]

Making them irreducible allows you to get rid of degenerate cases

Like lines are technically conics

Cause (ax+by+cz)^2 is homogeneous

But not irreducible

Degenerate conics fit the euclidian geometric intuition better

[terse crystal]

Oh like because V(fg)=V(f) intersection V(g)?

[wooden ember]

Like every point has a unique “tangent line”

Ie a line that intersects the conic only once

[terse crystal]

Oh, I see

So interesting

[wooden ember]

Here I’ll send you the link to the blog I’m following

[terse crystal]

What book are you reading, about those

[wooden ember]

Just started today but it’s really cool stuff

[terse crystal]

Okay

[wooden ember]

https://anuragbishnoi.wordpress.com/minicourse/

[terse crystal]

Thanks a lot 😁

[wooden ember]

And if you want some applications of projective space they’re a natural setting to talk about elliptic curves I’m pretty sure

You can check out “Rational points on elliptic curves”

I’ve only read the first 2 chapters but it’s neat

[terse crystal]

Oh, that’s great, another thing I am interested in

Okay found it. Thanks

[wooden ember]

I think it’s supposed to be pretty important in algebraic geometry too so people knowledgeable on that can tell you about it

[chilly radish]

Gonna bump.this. I'm still kinda lost

[terse crystal]

Sure, thank you

Sounds like Morita equivalence

[chilly radish]

Well, yea, brauer equivalence is a special case of morita equivalence, but I'd like to prove uniqueness using the method I outlined

[agile burrow]

maybe I'm misinterpreting, but isn't it the case that the product decomposition of a semisimple ring into simple rings is unique? then uniquenss in wedderburn artin follows directly from that. i can outline this if you'd like

[chilly radish]

well, that's what i'm trying to prove

but I specifically want to prove the uniqueness of the representation of a simple ring as matrices over a division ring

[agile burrow]

yeah i don't really see how to do it following the hint you mentioned

but i can describe the way I've seen it done if you might find that helpful

[chilly radish]

I'd appreciate that

[agile burrow]

hang on, lemme test out the latex bot rq, i'm new here:
let $R = \prod_{i=1}^{m} = \prod_{j=1}^{n} R'_j$

[cloud walrus]

walter

[agile burrow]

oh bet, that's neat

[chilly radish]

Oh wait, if you mean that the simple components are uniquely determined, then i've already proven that, I just want to show that even further, they're uniquely determined up to isomorphism by the division rings and order of the matrix ring

the way I did it was by taking a simple component in one decomposition, intersecting it with all the simple components in the other and concluding by simplicity that it has to be equal to one of them/

where the intersection with at least one of the simple components has to be nonzero

[agile burrow]

ohh, ok

that's kinda neat actually

[chilly radish]

Is that not what you were going to do?

[agile burrow]

wait i think it might be the same, i use products instead of intersections

[chilly radish]

wdym

[agile burrow]

like you get R_i = \prod R_i R_j', and each factor is either zero or R_i. Not all of them are zero, so there exists some specific R_j' such that R_i = R_i R_j', but R_i R_j' is also a two-sided ideal in R_j', hence it equals R_j'. It follows that R_i = R_j'

[chilly radish]

ahhh I see

yea I think both methods work just as well

[agile burrow]

here's something that might help with what you're trying to prove: let D be a division ring, V an n-dimensional vector space over D, and R = End_D(V) = M_n(D). Then D is isomorphic to End_R(V)

i think i remember reading something about M_n(D) having a unique simple module, i'll have to think about it a bit more

[chilly radish]

so that is essentially what i'm after, but I can't seem to prove it

do you have a reference?

[agile burrow]

the statement i provided or the thing about a unique simple module?

i remember reading some of this in Farb and Dennis' Noncommutative Algebra (which I really need to work through someday)

[chilly radish]

the statement you provided

[agile burrow]

yeah that one is definitely in farb and dennis

[chilly radish]

k

i'll look into it, thanks!

[agile burrow]

but the proof for that shouldn't be too bad, there's an obvious injection from D into End_R(V) by sending x to the matrix corresponding to scalar multiplication by x. as for surjectivity, note that V is generated by any non-zero vector v under the action of R, so any endomorphism T is uniquely determined by its action on v. Then to show that Tv = xv for some x in D, just use the projection onto the subspace generated by v

[chilly radish]

not sure I understand the last part

[agile burrow]

let p in End_D(V) denote the projection onto the subspace generated by v. Then for an endomorphism T in End_R(V), we have Tv = T(pv) = p(Tv), which is in the subspace generated by v, hence equal to xv for some x in D.

or do you mean you don't get how V is generated by any non-zero vector v

for that, lets say for the sake of simplicity that v = (1, 0, ..., 0). Then for any vector w = (a_1, ..., a_n), consider the matrix M whose first column is (a_1, ..., a_n) and 0 elsewhere, so Mv = w.

[chilly radish]

It was this, but I understand it now

thanks a lot!

[agile burrow]

ofc, glad I could help!

i'm trying to remember how to show that D^n is the unique simple module over M_n(D)

well I guess a simple module over M_n(D) has at most n generators or else there's a faithful action of M_n(D) on a submodule of M, but a projection from D^n onto M has kernel in D^n, hence it must be zero since D^n is simple

something like that

[astral galleon]

Can anyone have a look at this: http://mathb.in/70886

Confused on what the author means by complement and complement stable

the example I constructed seems to be nice but i'm not sure what he means by those last two words (I highly doubt what I did is correct considering where the problem is coming from)

[agile burrow]

so if $M_m(D) \cong M_n(D')$, then $D^m \cong D'^n$, hence $End_{M_m(D')}(D^m) \cong End_{M_n(D')}(D^n)$, and $D \cong D'$

[cloud walrus]

walter

[agile burrow]

forgot a D' somewhere but I think that's the general idea

[agile burrow]

the matrices A_i act on C^2 by normal matrix multiplication, and the image of vectors under this multiplication form a subspace V. In particular, V is "stable" under A_i, meaning vectors in V are sent to other vectors in V when multiplied by A_i. If V is not all of C^2, then it has a complementary subspace W such that V \oplus W = C^2. It's not always the case that W is stable under A_i; that is, for any vector w in W, A_i w is also in W

[astral galleon]

ahh kk so the example I came up with works ? I'll have to generlize for the infinite dimesional case and actually do the problem

[agile burrow]

that's exciting, best of luck!

[astral galleon]

Oh so I did 🙂

[agile burrow]

oh i didn't check your example, sorry

didn't realize you were asking a question, i can try to take a look at it later though

[chilly radish]

@agile burrow I thought about this a bit more and my method would not work. In general you can intersect every component of a product trivially but not intersect the product trivially, that's why you need to take the product there, because it distributes nicely

[agile burrow]

ah ok that makes sense

[chilly radish]

what's the field that splits a CSA called? It's not called a 'splitting field', is it

[astral galleon]

ok reread your comment and did a bit of working things out I think this example can be ported over. https://math.stackexchange.com/questions/325020/prove-that-a-set-of-matrices-is-a-subspace. (easy to build a pair of matrices simular to this where it behaves like the identity) I imagine if V is indeed a subspace then one should be able to get the complementary subspace where you get C^{2} back and is indeed stable. The idea I have should work because the idenity matrix one can indeed get a subspace where V+W = entire space. Stability should be achieved via the ideanity matrix. I think the mistake in my first attempt was having the matrices being taken to a powr

Maybe I should just stick to the idenity matrix

Update looked in the thread where the problem comes from the idenity matrix does generate an infinite subgroup GL(C) 😩

[pastel cliff]

an element of D_8 is technically an operation right

that being a symmetry, and the group operation is composition

oh god he's here

[delicate orchid]

there exists a natural group action of D_8 on a square if that's what you mean

oh wait yeah ok

[pastel cliff]

i mean i was leading up to asking what a group action is

[delicate orchid]

well in that case that's true for any finite group (maybe infinite?!) cause you can just embed it into some S_n and then all of the group elements are permutations, which are functions

[fiery fern]

Hey can someone explain me what i have to do to find T(1) , T(T) , T(T^2),etc... i dont know where i have to plug those value

[pastel cliff]

#linear-algebra

[chilly ocean]

#linear-algebra

[pastel cliff]

gottem

[fiery fern]

Oh thanks didnt pay attention

[pastel cliff]

anyways ive seen symmetry defined as a group action that preserves the position of vertices (sticking with n-gons)

but "symmetries" are also the actual elements of the dihedral group?

chapter on group actions is later in my book so i could be grossly misunderstanding

[delicate orchid]

hmmmm

not sure I like that definition

the group action would be a group element acting on a corner to give you another corner of the n-gon but I think of the symmetry (the group element itself) as a "function" (as you said) and the group action "evaluates that function" at a certain corner

I'm not explaining it too well

[pastel cliff]

hmmm

perhaps i should wait for group action chapter then

[delicate orchid]

I mean your questions are very natural ones to have and understanding the distinction is important

[agile burrow]

Unfortunate, this is a really interesting problem though

[agile burrow]

Not sure if there's a standard terminology, I've definitely seen splitting field used for the smallest field over which a division algebra is isomorphic to a matrix algebra

So I wouldn't call it a stretch to just use splitting field for CSAs as well

[astral galleon]

Yeah in the case for r being finite dimensions when r is infinte it looks like the problem may still be salvageable. I did write that in my post but seems from the lack of responses in the twitter thread the infinite dimesional case may not be intrestring 😢

[astral galleon]

Oh @agile burrow where the problem comes from https://twitter.com/littmath/status/1533564609382305798

Daniel Litt (@littmath)

Here is an open problem I am very interested in. If you construct three matrices with the properties below, you'll have proven a conjecture of mine (and disproven a conjecture of Grunewald-Lubotzky)!

Likes

251

Twitter · Jun 5, 2022, 9:41 PM

[agile burrow]

heyy, no wonder this sounded familiar! I saw him give a talk and he mentioned this problem

[astral galleon]

yeah i've been trying to have a go at it the only case I found what he indeed describes is possible is when r equals infinity

But from doing a bit of googling around it seems the realy money is in the finite dimensional case

[agile burrow]

yeah i figure that's probably why he's very interested in it lol

[astral galleon]

when you solve a problem and its not the one people want 😩

@agile burrow would you like to help with it this weekend 🙂

Maybe we could try to show how its not possible in certain cases

[agile burrow]

not sure how much free time i'll have this weekend, i'll try to ping you if i know ahead of time

[astral galleon]

kkk also do you know any good examples of infinite dimesional groups with nice subgroups we can test the problem @agile burrow

[agile burrow]

not off the top of my head, i usually stick to finite dimensions but i can do some poking around

experience has shown me that matrix groups are usually the cleanest

i love linear algebra

[astral galleon]

which matrix group in paritcular do we need to consider would the projective linear group be a good punch to throw

[agile burrow]

PGL can be okay, i was thinking more along just GL and SL

[astral galleon]

Because after thinking about the problem it comes down to showing that our compleement stable does not exist under the A_i

also how would you get A_i to generate an infinite subgroup when for some r you have finitely many elements in your group @agile burrow when r=inf when r is finite its hard

oh wait SL(n,C) is the subgroup of GL(n,C) 😅

[agile burrow]

Yeah lol

Well actually I don't understand your question

[astral galleon]

@agile burrow that was just me misunderstanding the question I now get a better feeling of it

Yeah then it should not be possible when r=2 since upon mutiplying our matrices A_i the vectors are sent somewhere else no ? @agile burrow

[agile burrow]

That doesn't necessarily mean the image isn't in the same subspace

[astral galleon]

ahh kk 👍 we should investigate the case when n is indeed finite

[prisma shuttle]

can someone explain to me wut this sentence means

i am having a lot of trouble comprehending wut it is trying to saying

for broader context it is on page 2 in this paper

https://arxiv.org/pdf/2111.14573.pdf

but idt any of the stuff on page 1 or before it is really that practical

[cloud walrus]

Zophike1

[agile burrow]

it doesn't say infinite dimensional though, it just says infinite

the group GL(r, C) is infinite for all r > 0

i also don't see how it holds for infinite dimensional matrices, do you have an explicit example?

when you say finite dimensional, i presume you mean finitely generated? @astral galleon
which is odd because GL(r, C) is not finitely generated lol

[astral galleon]

as well as here: https://encyclopediaofmath.org/wiki/Linear_group

[astral galleon]

ahh okay makes sense 😅 must have gotton confused

reviewd the definitions its a set of 2x2 matrices 😅 I need to get some sleep

[agile burrow]

mm

[astral galleon]

yeah I see the mistake I made eariler I didn't actually bother to look at the other matrices withen the set 😢

I think we can cheese the question by looking at GL(C) and making sure we have a digonal matrices pluked from the set

@agile burrow friend me on discord so we can work on this 🙂

[agile burrow]

i accepted your rq

[opaque pilot]

I'm trying to show this isn't a homomorphism using the counter example a = (1 2) and b = (2 3). Would this be correct? Specifically how I calculated $\phi(ab)$ and $\phi(a) \phi(b)$

[cloud walrus]

Kurama

[agile burrow]

this looks correct. perhaps an easier way to see this is that a homomorphism takes the identity to the identity, which this map clearly does not do

[opaque pilot]

Yep, that makes sense. I actually wanted to practice using cycle notation since Im not familiar with it

But yeah, that is faster

[agile burrow]

good idea, cycle notation becomes very useful for understanding the symmetric group

[opaque pilot]

Are all dihedral groups symmetric groups?

[next obsidian]

Nope

Symmetric groups have order n!

But dihedral groups can have order 2n for any n

[opaque pilot]

So they are not related at all?

[next obsidian]

They definitely are

A dihedral group is a subgroup of a symmetric group, which on the one hand sort of says nothing, but this is particularly natural

So actually any finite group is isomorphic to a subgroup of S_|G| via a group action, but that probably doesn’t mean that much to you right now

[opaque pilot]

dont know what a group action is lol

[next obsidian]

In the case of D_2n though, this is like, symmetries of an n-gon

You can think about these as special bijections on a set of n-elements, namely thinking of what happens to the vertices

So D_2n ends up being a subgroup of S_n in a really natural way, you can think of S_n as every possible bijection on a set of n-points (in fact this might be your definition)

And D_2n are a special subset (or subgroup if you like) which preserves certain edge relationships

So if you drew out a pentagon and labeled the vertices 1,2,3,4,5; you can’t swap 1 and 4 because you always need 1 and 2 to be next to each other

I can try to clarify what I mean here if some of this is too opaque

[opaque pilot]

Just trying to take in what you said. A lot of it makes sense so far

[next obsidian]

Yeah so imagine if you labeled the top left 1, top right 2, bottom left 3, bottom right 4, but in an absolute sense

So these pictures describe what point goes where

So R_90 is like (2134)

Because A which was 1 now is in position 3

B which was 2 is now in position 1

…

[astral galleon]

I think the symmetric groups > diheaderal groups in terms of order

[next obsidian]

But you can see that nowhere on here will you find (13)

This is like if you fixed B and D, but only swapped A and C

So this is why D_8 is a proper subgroup of S_4 because we don’t get all possible permutations / symmetries

We only get the ones that preserve our square

So they’re super connected, just not the same (dihedral and symmetric groups)

[opaque pilot]

make sense. this was what I was trying to put into words.

[next obsidian]

👍

[opaque pilot]

ty 🙂

[next obsidian]

[tidal crag]

I have a quick question regarding mappings.

Is the mapping from R -> R+ bijective? I wouldn't say so since negative reals would map with the positive reals, disproving that it's one-to-one.

But is there a good way to show this?

[next obsidian]

First off I think this is better for something like #proofs-and-logic, but you never specified what your map is

There’s definitely bijective maps R -> R^+

[tidal crag]

It's a mapping of 2 groups, R, and R^+

[next obsidian]

This isn’t the real numbers?

[tidal crag]

It is

[next obsidian]

But again you didn’t specify what your map is

[tidal crag]

Give me 2 minutes

$\phi : \mathbb{R} \rightarrow \mathbb{R}^+$

[cloud walrus]

Jon123276

[tidal crag]

I'm trying to understand how this is a bijection, which would lead to showing that this is a isomorphism with (R, +) and (R^+, •).

[next obsidian]

Dude

You still haven’t said what the map is

You just gave it a name

Like what is phi(x)?

Is it e^x?

[tidal crag]

Oh wait, i can't read

[next obsidian]

If so, this maps legit has an inverse

[tidal crag]

Yea, now I understand.

Reading is hard...

[west sinew]

Can someone help me see why the ideal quotient (mZ:nZ) will be qZ, where q=m/hcf(m, n)?
for ref. https://en.wikipedia.org/wiki/Ideal_quotient

[next obsidian]

You know that every ideal of Z is principal, and is generated by the gcd of the elements in the ideal. If it wasn't q then you should be able to draw some contradiction

[astral galleon]

the idenity function could work as well for your mapping @tidal crag

[chilly ocean]

[south patrol]

No it couldn't because - 1 isn't in R+

Among other reasons

[red drum]

Hey everyone, I'm reading a paper about BMS group, but I am blocked at the page 9... If anyone have resources to understand this part to the end I will be gratefull ! It's about the BMS Lie algebra...

[west sinew]

What is the map j and why is it injective? (This is from Atiyah & MacDonald's Comm. Algebra)

[chilly ocean]

@west sinew it's the inclusion map, j(x) = x

I think?

[quartz quiver]

In general you can factorize any function into a composition of a surjective and injective function

[tidal crag]

I'm having a hard time wrapping my head around why the kernel is important...

[chilly ocean]

do you mean kernel

[tidal crag]

Yes

[chilly ocean]

what kind of kernel

Kernel of a map f, which we'll denote by ker(f) is defined as {(x, y) : f(x) = f(y)} and it "measures" injectivity of a given map f

if your map is actually a homomorphism of a group or something like that, then we can replace this with ker(f) = {x : f(x) = 0} where 0 is the identity element

the same is true, this map measures the injectivity of your map f

In general a map f:X to Y is not injective, but by quotienting X/ker(f) and setting p:X to X/ker(f) to be the natural projection, there exist a unique map g such that g(p(x)) = f(x) for x in X, and this map is injective

In algebra, injective homomorphism is usually an isomorphism onto its image, so if we assume that we are working with this kind of context we get what's called the first isomorphism theorem, that is X/ker(f) is isomorphic to im(f) = f(X)

[leaden pulsar]

For all (a_1, a_2) , (b_1,b_2)∈R^2 , (a_1,a_2)⚹(b_1,b_2) = (a_1.b_1,a_2.b_2). Prove that <R^2,⚹> is not a group.

help

[untold basin]

Maybe try to understand first why <R,*> is not a group

[leaden pulsar]

yeah since every elemnt need to have an inverse, 0 will not have it

breh was it this

[untold basin]

yes

[broken stirrup]

This looks life fairly simple proof but what I don't get is how do we claim that [\phi ^ G, X] implies that \phi^G is a character?

I know that for any two characters A,B the 'inner product' [A,B] is a nonnegative number

[agile burrow]

I've done the first three parts, but I don't see how to define H on morphisms. I'm trying to induce an appropriate morphism to use snake lemma but idek if this is the right approach

I can elaborate on the second and third points if it would be helpful
Edit: I wanna say the proof should be something along the lines of an arrow theoretic proof that cohomology is a functor. The exercise is given under the context that it is a generalization of an earlier exercise where we derived the long exact sequence in cohomology as a consequence of the snake lemma. In particular, there I believe we viewed A is the category of chain complexes and we define dA to be the category whose objects (A, d) consist of a chain complex along with a shift map, or something along those lines

[agile burrow]

first you should note that \phi^G is a class function. Irreducible characters form a basis for the complex space of class functions and it can be shown that a class function is a character if and only if it is an non-negative integral combination of irreducible characters. I can elaborate on this last point if you'd like

[broken stirrup]

thanks, its clear

[opaque pilot]

https://math.stackexchange.com/questions/526409/order-of-the-product-of-two-commuting-elements-with-coprime-orders-in-a-group

Why is that the two elements must commute important in this proof?

[chilly ocean]

(gh)^{rm} = g^{rm}h^{rm}

(gh)^n = g^n h^n is not true, in general, if g and h do not commute.

[opaque pilot]

But if they did h^{rm}g^{rm}, since the order of g is m, doesn't g^{rm} = e anyway? So we'd be left with h^{rm} even without commuting

[chilly ocean]

writing it as h^{rm}g^{rm} also works, but you're still using the fact that g and h commute to do so

[opaque pilot]

If we write it like this, then we didnt use the assumption of two elements commuting?

[chilly ocean]

let me say it more explicitly

If $g,h$ are commuting elements of a group, then $$(gh)^n = g^n h^n = h^ng^n$$ for all $n \in \bN$. This is not true in general if $g$ and $h$ do not commute.

[cloud walrus]

TTerra

[chilly ocean]

this is what allows the poster to write (gh)^{rm} = g^{rm}h^{rm} (which also happens to be equal to h^{rm}g^{rm})

[opaque pilot]

Yep, I understand the fact that G is Abelian makes this (gh)^{rm} = g^{rm}h^{rm} true.

[chilly ocean]

no, it's not assumed that G is abelian. that certainly makes it true that g and h commute, but it's only assumed that these particular elements commute

[opaque pilot]

But in the context of the proof, since the order of g is m, does it matter whether they write it like g^{rm}h^{rm} or like h^{rm}g^{rm}? Since g^{rm} = e, the identity. And we'd be left with h^{rm} even if we dont use that g and h commute.

[opaque pilot]

oh ok, I should be careful with my wording

[chilly ocean]

i am trying to tell you that you can only make the step that (gh)^{rm} = g^{rm}h^{rm} because g and h commute. this step has nothing to do with the orders of the elements

g^{rm} = e alone (i.e. without assuming commutativity of g and h) does not imply that (gh)^{rm} = g^{rm}h^{rm}, in general

you NEED commutativity to write (gh)^{rm} = g^{rm}h^{rm}, and no other assumption present in the proof will give this to you

[opaque pilot]

Sorry Im still not understanding. If we do e=(gh)^rm = h^{rm}g^{rm} =h^{rm}, this still shows that n|rm does it not?

[next obsidian]

You can’t say that (gh)^rm = h^rm•g^rm unless h and g commute

Because (gh)^rm is a priori just
ghghghgh…gh

But now you need to swap a bunch of g’s and a bunch of h’s

[opaque pilot]

ohhh that makes sense

I see, thanks 🙂

[wise igloo]

is C=R(i)?

I would think s

so

but I just want confirmstion

[devout crow]

yes

but it's not a hard check: C is definitely inside R(i), and you can't make anything in R(i) that's not in C

[next obsidian]

Chmonkey

[wise igloo]

true

alright ty

yes

chmonkey

[hot abyss]

asking about problem 3, i don’t need help with this proof, i just really don’t know how to notate it

i know the generalization would be like, $a_1a_2\dots a_n =e\implies a_na_1\dots a_{n-1}=e\implies a_{n-1}a_na_1\dots a_{n-2}=e…$

[cloud walrus]

quantum

[hot abyss]

i just don’t really know how i would notate something like this

it’s going in like a cycle

[delicate bloom]

I think that's fine personally

just say cyclic permutations like you said

you could be more precise, but I feel like it would be less clear

[wise igloo]

quanty doing group theory now? done with lin alg?

[hot abyss]

yes

[wise igloo]

niiiiice

[next obsidian]

Quantum you want good group theory problems?

[wise igloo]

I do

@next obsidian

anyway

my favorite ring is R_69

[lethal dune]

find all homomorphisms from R_420 to R_69

[wise igloo]

make me

[next obsidian]

This is actually not hard

There’s exactly 3 of them

If I did my maffz right

[wise igloo]

how tf

[agile burrow]

bumping this

[hidden haven]

There is no snake lemma involved

This is like showing that a map of chain complexes induces a map on homology

You restrict the map to the ker d in the domain

Show that this factors through ker d of the codomain

Then show that im d in the domain maps into im d of the codomain

So there is an induced map ker d/im d of domain to ker d/im d of codomain

[agile burrow]

so to make sure I understand this, given a morphism f: (A, d_A) -> (B, d_B), we have d_B o f o ker d_A = 0 hence f o ker d_A factors through ker d_B

[hidden haven]

yes

[agile burrow]

and for this one, is it analogously that coker d_B o f o d_A = 0, hence im d_A is mapped into im d_B? i'm kinda bouncing between intuition from a set theoretic level and universal properties here

[hidden haven]

yep

[agile burrow]

just trying to reconcile this with the notation in the exercise, i get that we look at the composition ker d_A -> ker d_B -> H(B, d_B) and now I just need to show that this composition precomposed with \overline{d}_A is zero, but this feels different from what we were doing before with the images

i think it stems from me not really understanding what the \overline{d} map is

[cloud walrus]

Moldilocks1337 ✓

[hidden haven]

You proved that you have this commutative diagram right?

im d_A should really go to A but I am just denoting the inclusion into the kernel by that

[cloud walrus]

Moldilocks1337 ✓

[hidden haven]

Here I just took cokernels of the maps on the left

cokernel is an additive functor so you get this induced map H(f)

Maybe you can phrase this as a composition of additive functors, so it is additive. You first map a differential object A to the pair (ker d_A, im d_A) in the category of object subobject pairs, and then apply the quotient functor to this which is also additive

Haven't checked the details in this approach but should work

@agile burrow

[broken stirrup]

Any hints? RHS is induced character from H intersection K to K, am i right?

[agile burrow]

ahh ok I think I get it now, tysm

[agile burrow]

yes, that is what the RHS is saying. as for actually showing equality, do you know the character formulas for restricted/induced characters?

[broken stirrup]

I know the definition, yes

[agile burrow]

the computations are kinda annoying but i think that should be a straightforward way to show that it is true

I feel like there should be an alternative way through the fact that induced representations work nicely with respect to towers of subgroups and Frobenius reciprocity

[charred flume]

im solving for the splitting field $x^{4}+1$ over Q now the answer that i get is $\mathbb{Q}(\omega, \mathrm{i})$ where $\omega=\frac{\sqrt{2}+i \sqrt{2}}{2}$. But there's an answer online where he gets $\mathbb{Q}(i, \sqrt{2})$. Are they equal?

[cloud walrus]

Ji

[broken stirrup]

i think so

if w is there then its conjugate is also there. Just add them and you get sqrt(2)

[charred flume]

ok i'll try later

thanks

[devout crow]

what's an example of a field extension of C?

I know it can't have finite degree

[chilly radish]

field of rational functions

i.e. C(t) where t is an indeterminate

[devout crow]

how do you show it's not iso to C

[chilly radish]

t is transcendental over C by definition

[hidden haven]

It is not algebraically closed because x^2 - t has no root

[chilly radish]

every element of a field is trivially algebraic over the field

that also works

[hidden haven]

It's algebraic closure is isomorphic to C though

[devout crow]

ahh yeah makes sense

thanks guys

[devout crow]

this is cool

[lethal dune]

really

[hidden haven]

Ye algebraically closed fields with some characteristic and transcendence degree are isomorphic

They have the same transcendence degree because same uncountable cardinality

In particular the algebraic closure of the p-adics is also isomorphic to C

[lethal dune]

[hidden haven]

I mean algebraic closure after you take p-adic completion

[lethal dune]

cool

[delicate orchid]

hmmm

[lethal dune]

ikr

[lament dawn]

What does it mean for a group G to act on X via isometries

I've seen this a few times

[devout crow]

d(g.x,g.y) = d(x,y) for all x,y in X, g in G

[lament dawn]

Does it mean $d\lrp{x_1,x_2} = d\lrp{gx_1,gx_2} \forall g\inG$?

Oop

[cloud walrus]

Migillope
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lament dawn]

Okay yes haha thanks

[devout crow]

is this hard to prove

[hidden haven]

Not much but you might need some machinery

You take a bijection between the transcendence bases

That gives you an isomorphism of the purely transcendental part

Now you need to extend this to the whole extension

You do this using Zorn's lemma

And to apply Zorn's lemma, you need to know when a field map F → E can be extended to a field map F[a] → E for a algebraic over F

It is exactly when the minimal polynomial of a over F has a root in E (by the first isomorphism theorem)

[devout crow]

what's the underlying field? the closure of Fp (the closure of Q if p = 0)?

[hidden haven]

F_p itself

[devout crow]

oh that makes more sense

[hidden haven]

It's the same thing

Transcendental over F is equivalent to transcendental over the closure of F

Though ye here you want to start building this isomorphic up from F rather than from closure

No actually it shouldn't matter

[devout crow]

hmm I don't know enough field theory yet I think

maybe I'll ask again when I know what I'm talking about

[hidden haven]

F

[lethal dune]

F_0

[pastel cliff]

this valid?

i should probably specify every element in G

[delicate orchid]

yeah looks right

[pastel cliff]

mμ₂rbius sweep

[delicate orchid]

we need more people like you in the world, nitezba

people I can actually help in here

[pastel cliff]

so simpletons

gimme like 3 weeks i'll get to galois theory

i will speedrun dummy and foot

assuming i dont fall asleep first

[delicate orchid]

I'm currently reading some galois theory notes

[broken stirrup]

I failed to show this equality using the definition 😦

[broken stirrup]

lol Morbius you know character theory maybe you could help me as well xD

[delicate orchid]

that looks like a semidirect product except you simply stop caring about everything

my first thought is frobenius reciprocity but I don't think that goes anywhere

[broken stirrup]

You get towers of induced characters looks complicated to me

[delicate orchid]

ok so $\phi^G(x) = \frac{1}{|H|}\sum_{g \in G} \overline{\phi(gxg^{-1})}$ as per usual, and $\phi_{H \cup K}^K(x) = \frac{1}{|H \cap K|} \sum_{g \in K} \overline{\phi(gxg^{-1}}$ and we want to turn one of these into another

fantastic start

[cloud walrus]

Wew "Morbeline" Tbh ⊗

[delicate orchid]

imagine it's correct
the overline means \overline{phi(x)} = phi(x) x in H 0 otherwise as per usual

first thing is we want to sum over K and we're currently summing over G so is there a way to change that?

hmm

[broken stirrup]

let's note that g \in HK

[delicate orchid]

so g is equal to some hk for h in k

so lets split $\phi^G(x) = \frac{1}{|H|}\sum_{g \in G} \overline{\phi}(gxg^{-1}) = \frac{1}{|H|}\sum_{h \in H, k \in K} \overline{\phi}(hkxk^{-1}h^{-1})$

my god my latex today

bear with me

[cloud walrus]

Wew "Morbeline" Tbh ⊗

[delicate orchid]

ok good enough

now if H and K were disjoint I wouldn't have any problems with splitting $\sum_{h \in H, k \in K}$ into $\sum_{h \in H}\sum_{k \in K}$ but we can't cause they're not necessary disjoint

[cloud walrus]

Wew "Morbeline" Tbh ⊗

[delicate orchid]

hmm

ok what if we try messing with $\phi^K_{H \cap K}$ now

[cloud walrus]

Wew "Morbeline" Tbh ⊗

[delicate orchid]

at some point the 1/|H| has to turn into 1/|H intersect K| so how does that happen

so many questions

[broken stirrup]

I feel like we'll have to use thee fact that |G|=|HK|=|H||K| / |H intersect K|

[delicate orchid]

oh SHIT yes of course

ok so 1/|H intersect K| = |G|/|H||K| right?

= $\frac{1}{|H|}\frac{|G|}{|K|}$

[broken stirrup]

yes

notice that when \phi=1 you get that equality :d

[cloud walrus]

Wew "Morbeline" Tbh ⊗

[delicate orchid]

now when I see the index of a subset in a group I immediately think of cosets

[broken stirrup]

and

[delicate orchid]

and indeed

[broken stirrup]

we've simplification of the formula for cosets

i mean sum over transversal T of right cosets

[delicate orchid]

right cosets of K?

hmm

yeah ok I follow now, and since they're disjoint we can split the sums!

[broken stirrup]

hmm yes but we don't have orders involving in that one so i try to avoid it

just definition looks more useful

[delicate orchid]

I still think the key lies in how they're disjoint

cause that was our main problem before

although I'll keep thinking about that defn, but we really would like the sum to be over K not the transversal

ah ok I think I've got it

nvm

is $\phi^G_k(x) = \frac{1}{|H|}\sum_{g \in G} \overline{\phi}(gxg^{-1}) = \frac{1}{|H|}\sum_{t \in T}\sum_{k \in K} \overline{\phi}(tkxk^{-1}t^{-1})$ valid? as the cosets form a disjoint cover of the group?

[cloud walrus]

Wew "Morbeline" Tbh ⊗

[delicate orchid]

cause if it is then we can just IMMEDIATELY do some trolling and get $\phi^G_k(x) = \frac{|T|}{|H|}\sum_{k \in K}\overline{\phi}(kxk^{-1})$

using the fact that phi bar is a class function

ffs I forgot the sum

[cloud walrus]

Wew "Morbeline" Tbh ⊗

[delicate orchid]

OH AND THEN THE SIZE OF THE TRANSVERSAL IS JUST THE INDEX OF K IN G

YESSSS

[delicate orchid]

Ok I'm saying we can do this because I said so

[broken stirrup]

wait im still not so sure about this let me think xd

[delicate orchid]

yeah if we know that for sure then I think the result follows

[broken stirrup]

so i think this is valid but...

[broken stirrup]

phi is class function only in H so it doesn't have to be class function?

[delicate orchid]

it's 0 outside of H so it doesn't contribute to the sum, so whenever it's non-zero it's a class function

[broken stirrup]

tysm

[slate mortar]

I want to prove that the wedge product on finite free R-module M is a perfect pairing, does someone know where I can find a proof of this?

[chilly ocean]

Given the vector space $\bR$ over $\bQ$ lets call it V, prove that the dimension of V is non finite.

Does the fact that the cardinality of R mod Q being non-finite suffice as a sufficient explanation?

[cloud walrus]

simp

[chilly ocean]

Does the fact that the cardinality of R mod Q being non-finite suffice as a sufficient explanation?

[agile burrow]

No because R^2 / R is isomorphic to R , which has non-finite order

[chilly ocean]

ty bb

[agile burrow]

you could use some version of a cardinality argument

perhaps something about the cardinality of Q that the cardinality of R does not satisfy...

[chilly ocean]

I know like a way to show it

I was just wondering if R mod Q also showed it

[agile burrow]

ah ok

[chilly ocean]

but tahnk you nonetheless

[young hollow]

Getting into some algebra after a while; just making sure I’m approaching this correctly before continuing with the remaining examples:

[slate mortar]

m+i could be greater than n

[young hollow]

Ah yeah I forgot the “(mod n)

[young hollow]

Can anyone look over the first portion of this one, please:

Shoot, that should be a “- 1” that got cut off there, not “+1”

[slate mortar]

Case 1 could be in case 2.b)

[young hollow]

How so?

Oh I think you mean 2a?

Yeah that would’ve been cleaner

[chilly ocean]

No, but you can use this argument: If R was finite dimensional, then |R| = |Q^n| = |Q| for some n.

[broken stirrup]

I understand that since D is countable, there is a line that doesn't intersect D. Where can i find the proof that there is an angle such that for any positive n, rotation by n\theta doesn't intersect D either?

[chilly ocean]

this is in the plane, right

[broken stirrup]

oh im so sorry

I'm trying to understand why S^2 and S^2\D are equidecomposable

[chilly ocean]

what's the origin then

do they mean like, origin of R^3, or maybe origin of of R^2 which we project somehow under the stereographic projection, or what

[broken stirrup]

R^3 i believe

yes indeed

[chilly ocean]

we can always take a plane which contains the line l

then if you look at this circle that you obtained, there is only countable amount of points from D

Suppose that for every theta there exist n such that after rotating n times by angle theta, we get an element d(theta) from D

There can be repeats, d(theta) is in the set of points generated by a countable possible amount of angles

so we get a map from a set of size continuum onto a countable set with countable fibers

this is impossible

so there must be theta with the desired properties

@broken stirrup

[broken stirrup]

i see

thanks 🙂

[wise igloo]

Collection of Questions

[pastel cliff]

any advice? the hint isn't terribly helpful

[delicate orchid]

SYLOW

nah I kid

[pastel cliff]

this is chapter 1 of D&F

[delicate orchid]

it's a special case of cauchy's theorem so lemme try and think though the proof of that one

[pastel cliff]

lagrange seems like it would help but it hasnt been introduced yet so cant use that

[delicate orchid]

I will use whatever I please

and this is a partial converse to lagrange so it's not that useful

[pastel cliff]

i mean contrapositive would kill it - if there is no element of order two, then the order of the group is not divisible by two so it cant be even, boom proven

[delicate orchid]

ahhh

no no no this is all massive overkill

watch this shit

|G| = 2n

pair up the elements into pairs like this {g, g^-1} with each element g only being in one pair

since we can put every element must be in a pair, as 2n is even, we must have the identity in a pair

say, {e, a}

hold on I need to think about how to phrase this next part

if a^-1 not equal to a, then a^-1 must be in a pair {g, a^-1} with g not equal to a (as a is already paired up with the identity), but g is (by definition of this pairing) in {g, g^-1} and this is the unique pair that contains g (this uniqueness can be guaranteed by the fact that |G| is even), so we must have a^-1 = g^-1 => a = g, which is a contradiction. Thus we must have a^-1 = a <=> a^2 = e

I think this holds

or: IT'S TRUE BY CAUCHY'S THEOREM NOW LEAVE ME ALONE!!

[pastel cliff]

[delicate orchid]

I'm going to flip my lid

[pastel cliff]

why are you wearing a lid

[delicate orchid]

I am a jar

ok I'm gonna go google my proof cause I remember seeing it somewhere

yeah ok good I remembered it right

[pastel cliff]

i would not have come up with this

[delicate orchid]

I wouldn't've either

see if you can generalise this idea to primes other than 2 though

[young fiber]

in what way

[pastel cliff]

dont spoil it bruv

[delicate orchid]

show that if $p$ divides $|G|$ then there exists an element $g \in G$ with the order of $g$ equal to $p$, cauchy's theorem basically

[cloud walrus]

Wew "Morbeline" Tbh ⊗

[young fiber]

hmm

[wooden ember]

Cauchy’s theore is far from easy to prove imo

It requires some clever ideas

[delicate orchid]

yeah I never said it wasn't a challenge

[wooden ember]

This is a weird way to phrase it

You can say things in a much simpler manner by simply saying that the set containing e is a singleton since it’s its own inverse

Hence there must be another singleton, ie an element that is its own inverse (is of order 2)

Now I’m trying to remember how the hell to prove cauchy’s theorem lol

I think you need to do it for abelian groups first and then do something in the quotient by the center or smth like that

Some kind of induction maybe

[delicate orchid]

||group actions on some weird cartesian product of the underlying group + orb-stab I think||

[wooden ember]

Ah i see

[delicate orchid]

||that's the proof I knew ||

||and I think it kind of follows from the partioning idea, we're working in G^2 there so maybe try working in G^p?||

[next obsidian]

You can deduce it from Sylow

[wooden ember]

Yeah frobenius ought to work somehow

[wooden ember]

That’s overkill lmao

[next obsidian]

[delicate orchid]

guys I'm gonna prove the trivial group is a subgroup of every group!
using the SYLOW THEOREMS!

[wooden ember]

And I can’t remember how to prove sylow either except some weird action shit

[delicate orchid]

I can barely remember the theorems thank god

[wooden ember]

I probably don’t remember them in full

[next obsidian]

How does Sylow proce the trivial group is a subgroup huh?

What if your group is infinite

[wooden ember]

Check mate

[next obsidian]

Morbius

[delicate orchid]

what?

groups cannot be infinite

[wooden ember]

Take the intersection of all subgroups

Huh actually if there’s no element of finite order is intersection of all subgroups necessarily trivial ?

[next obsidian]

Yes

Cuz trivial subgroup

[wooden ember]

What a fool I make of myself daily

And trivial subgroup aside?

[delicate orchid]

now we're talking

[next obsidian]

Then no

Subgroups of a cyclic group are totally ordered

So there’s a minimum non-trivial subgroup

[wooden ember]

Oh I mean never mind you can do it for finite groups yeah

I must be on sleep mode today

[subtle ivy]

your profile name and picture make me want to scream. just fyi <3

[pastel cliff]

that's among the nicer profiles he's had

[young hollow]

I posted the proof of the well-defined-ness of the operation before, but I’ll link it below since it’s related to the associativity somewhat. Is there any other way to do the associativity than how I did using cases?

I skipped some basic algebra steps (those elementary in R) and the justifications for some steps (due to space).

The key step in case II is noting that an expression of the form [x + [y]] = [x] + y for integral y (by definition)

[delicate bloom]

I want to say carrying happens to the left, so removing the integer part has no affect on when you do it, that way you can say it inherits associativity from R. Probably there's a more rigorous way of saying it though.

[young hollow]

Yeah I think this problem took me longer because I was trying to make it more rigorous for a while.. then I decided based on the way they wrote it, maybe it wasn’t what it’s looking for

Also, I guess the third case isn’t entirely “analogous,” but you use the same ‘trick’ of substituting [a+b] + [b+c] for [a+b] and cancelling the [a+b]s that remain using the property in my previous comment

[opaque pilot]

what about when the order is infinite?

[delicate bloom]

you could suppose it maps to an element of finite order and work out the consequences

[opaque pilot]

Like this, showing the |g| is finite, so a contradiction to the assumption that |g| = infinite?

[delicate bloom]

looks reasonable to me

[opaque pilot]

ty

[delicate bloom]

yw

[young hollow]

When you say “carrying happens to the left,” can you elaborate, please?

[delicate bloom]

i can not

[young hollow]

😂

Fair enough

[delicate bloom]

lol

[warm holly]

Yo I have a question on existence of orthogonal basis! Let <,>:V x V-> K be a billinear symmetric form over V x V where V is K vector space (characteristic not necessairily différent of 2) if there exist an element v in V such that <v,v> neq 0 can we conclude the existence of orthogonal basis of V?

[hidden haven]

Do you mean orthonormal

Otherwise <v,v> neq 0 is not needed anyway

Because everything would be orthogonal to everything else if all products are 0

[warm holly]

Nop I meant orthogonal. But if characteristic is 2 you can’t conclude it no?

[hidden haven]

What's the definition of an orthogonal basis? Is it a basis (v_i) such that <v_i, v_j> = 0 for i neq j?

[warm holly]

Yup

[hidden haven]

Then if the form is always 0, then you have orthogonal basis

Any basis is then orthogonal

[warm holly]

What I was looking for is weak condition to get an orthogonal basis in vector space over field of char 2