groups-rings-fields

[fossil hazel]

whats this then?

[chilly radish]

That's field theory

What you posted is calculus

[fossil hazel]

close enough

[formal ermine]

lol

[fossil hazel]

which channel?

[chilly radish]

#calculus or use one of the help channels/help forum

See #❓how-to-get-help

[floral raptor]

let me differentiate your forehead pls

[fossil hazel]

dm me

[floral raptor]

ok babes

[fossil hazel]

ok slow down

[chilly radish]

Its coefficients are invariant under all maps F(a_1,...,a_n)->F bar which preserve F

Do you know that this implies an element is in F?

[chilly radish]

This is true since such maps.permute the a_i, and the coefficients are symmetric in the a_i

I think that's what they were going for

[rocky cloak]

f is the minimal polynomial of alpha over F, hence by definition it is in F[x]

[chilly radish]

Oh duh

I missed that part lmao

It factors in Fbar but the polynomial itself is still in F

[white oxide]

Hm ok

[chilly radish]

(Actually this might also not work here if there's just 1 repeated root but it doesn't matter cuz it's irrel)

[white oxide]

how is this representation in block matrix form well-defined? if you have any (v1, v2), then you can't just compute it as [phi_g^1 x v1, phi_g^2 x v2] since phi_g^1 and phi_g^2 are both matrices on their own if that makes any sense

the dimensions of the matrices are wrong

[stone fulcrum]

@white oxide
Think of 0 as a 0 matrix

So there's 4 matricies in this matrix

But we tend to just call that a "block matrix" and write them all as 1 matrix.

[glossy crag]

My question is still languishing in obscurity, please drop an upvote if you can:
https://math.stackexchange.com/questions/4726122/proof-of-the-double-centraliser-theorem-in-lorenzs-algebra-ii

[frigid lark]

Sorry, I don't have enough rep

[void cosmos]

yo if A --> B --> C --> 0 is SPLIT exact

then Hom(D,_) is split exac

the inverse map from the splitting of the first one induces one that also works

as a splitting

right>?

[next obsidian]

A should be 0

And it’s Hom(D,-) applied to this SES that’s split exact

But yes

[void cosmos]

yea thats what i meant

if its just a right ses then it doesnt work?

ik it doesnt if it doesnt split

[next obsidian]

Split exact is a term for short exact sequences

You’ll stay exact on the right if the last map has a section though

[void cosmos]

wait split exact doesnt mean split AND exact ? 😄

[next obsidian]

What does split mean

[void cosmos]

yea so splitting does not make sense

if the sqeuence is not exact

ig

[vagrant zinc]

Hi guys, could you give me a hint please?

[frigid lark]

||Well if a^n = e, then a^{-n} = (a^n)^{-1} = e, so the order of a^{-1} is at most n, then use symmetry to get equality.||

[next obsidian]

a hint

full solution

Make it make sense

[azure copper]

I have a question as well. My book doesn't say this so probably is wrong. If F: G => G' is a isomorphism, does this imply that |G| = |G'| ?

[frigid lark]

If a^n = e, what is a^(-n)?

[next obsidian]

It depends, but I’m assuming G is a group, in which case F is a bijection so yes

[azure copper]

yup

[vagrant zinc]

Ok I'm doing fine, I understand that part because I was doing it that way, now another question, when you refer to symmetry, what should I think or what do you mean?

[vagrant zinc]

ay :c I don't want to get confused bro

[frigid lark]

Say the order of a is n, and the order of a^-1 is m, then the argument above shows that m is less than or equal to n

[azure copper]

so, say I see something going from f: C -> R, or g: R -> Z I can already state that because of their cardinalities being different, f and g can't be isomorphisms, right?

[lusty marlin]

ℂ and ℝ have the same cardinality

[next obsidian]

R and Z have different cardinality so this works, but knight watch is correct

[vagrant zinc]

ooh |a|<|a^{-1}| and |a^{-1}|<|a|

[next obsidian]

In fact C and R are isomorphic as groups

[frigid lark]

But replace a with a^(-1) and perform the same argument, say a^(-1)^m = e implies a^m = e, you thus argue n is less than or equal to m, hence equality

[next obsidian]

<=, but yes

[azure copper]

oh right

[chilly ocean]

What does "coefficients of matrix" mean

[vagrant zinc]

ye, mb, simboly

[next obsidian]

You can produce a group isomorphism by noting they’re vector spaces of the same (first uncountable) dimension over Q, so they are isomorphic as vector spaces. But oh wait, that isomorphism is in particular a group isomorphism

[next obsidian]

The entries probably

[lusty marlin]

The proof of this fact is very beautiful too. Was great fun seeing the Hamel Basis construction and all that

[vagrant zinc]

Ok thanks, I'll try again!!!

[azure copper]

actually ye I'm dumb C and R are isomorphic, I kinda forgot cause my Book says | C | = | R^2 |

which is the same R

[lusty marlin]

The groups being isomorphic doesn't immediately follow from their cardinality though.

[azure copper]

well different cardinalities means I can't make a 1 to 1 map from group G to G', so my morphism is either not surjective or injective.

this was the idea in my mind

[lusty marlin]

But having the same cardinality doesn't imply that they are isomorphic.

[azure copper]

right

[lusty marlin]

You need to construct a basis for both of those groups over ℚ

[azure copper]

different cardinalities definitely means no isomorphism tho right?

[lusty marlin]

Yes

[azure copper]

I thought about this literally a few minutes ago when I wake up, den I wondered why my book never specified this. It woulda made my life easier. 😭

[next obsidian]

I forget what a Hamel basis is lol

Is that the normal basis from linear algebra

Or is it the infinite sum one

[lusty marlin]

Yeah basically

[next obsidian]

What’s the other one

[lusty marlin]

I am not sure why it has a special name though

[next obsidian]

Cuz in functional analysis you consider bases for infinite sums

With convergence and stuff

[lusty marlin]

Wdym infinite sum one

[next obsidian]

And need to distinguish them

[open sluice]

analysis

[next obsidian]

Like you want a countable basis for

A countable infinite product

But this doesn’t actually work with Hamel bases

[lusty marlin]

The Hamel basis isn't countable though

[next obsidian]

Yeah exactly

https://en.m.wikipedia.org/wiki/Schauder_basis

[summer path]

I remember the name Hamel basis but I don't remember what it is and I remember there was a similar basis but I also forgot what the name is

Oh

[lusty marlin]

I haven't studied functional analysis yet, sadly. So I haven't really seen what you are referring to

[next obsidian]

I see

I figured you did because no one really ever says Hamel basis unless they’re in a functional analysis context

They just say basis

BASIS

WTF AUTOCORRECT

[lusty marlin]

Lol wtf

[summer path]

What did I miss

[lusty marlin]

Something

[lusty marlin]

Ah. My set theory professor decided to use the term while proving the above mentioned group isomorphism. He didn't really explain why it needed a separate name though

[cloud walrus]

jonathanphan

[hexed mantle]

Here's a rough sketch of my sol.

Not sure if it's correct.

[summer path]

What was the question

[next obsidian]

It’s above

I don’t think the 3rd and 2nd from the bottom lines are correct

Like, x(xy) ≠ x

As far as I can tell

But it looks like you replaced x with x(xy)

Also like, you should put more parentheses I think

Because you’re unsure what the word for E is in English, I don’t know if * is associative

And the condition in the hypothesis seems to suggest it isn’t

So you need more parentheses

[hexed mantle]

#groups-rings-fields message

[hexed mantle]

Aight, just searched up the terms, Here's the question, rephrased.

Given an operation $\ast$ and a set $E$ such that $\forall \ \quantity(x,y) \in E^2,\ x \ast \quantity(x \ast y) = \quantity(y \ast x) \ast x =y$, prove that $\ast$ is commutative.

Hope it's easier for you to understand.

[cloud walrus]

jonathanphan

[azure copper]

again how do i prove that:

  : z         -> 5z```
is surjective?

[prime sundial]

i remember doing this one a while back. i think what you wrote assumes the operation is associative, but you can't assume that

if you want a hint ||consider x * x maybe a better hint is to notice that the equation is true for all x,y, so you can consider x = x * (x * x) and so on||

[ashen heron]

how is quasi group defined in the text?

[prime sundial]

do you know what surjective means?

[azure copper]

yup

[prime sundial]

also, is (Z, +, 0) the ring with multiplication mn = 0?

[azure copper]

issa group

[prime sundial]

ok

what have you tried

[azure copper]

ive wrote down the following:

for each y in 5Z, there exists x in Z such that 5x = y

which is the definition of surjectivity

[prime sundial]

yes. so take an arbitrary y in 5z. what would you expect the x in Z to be so that f(x) = y?

for example, if y = 10

what would you say x is?

[azure copper]

well y is in the form 5 * n, so x is n, in this case 2.

[prime sundial]

what operation did you do to 10 to arrive at 2?

in general, what operation would you do to 5y to arrive at x?

[azure copper]

i multiplied by 1/5

[prime sundial]

that's the whole idea. take an arbitrary y in 5Z

let x = y/5

show f(x) = y and that x is in Z

[ember field]

Guys is every S^-1A module M of the form S^-1X for X being an A module.

[wraith cargo]

No

every S^-1A module is of the form U^-1X, where S is a subset of U I believe

(Note I came up with this on the spot after thinking abt it for a bit but there could be a counterexample I'm not seeing)

[lethal dune]

I want to say yes, my argument is you can look at the $S^{-1}A$ module $M$ as $A$ module by restriction of scalars, call this $N$. Now If you induce again you get $S^{-1}A\otimes_A N = S^{-1}\otimes_A S^{-1}A\otimes M = M$

[lethal dune]

Now maybe someone has a different argument or answer, I would like to know

[wraith cargo]

counterexample:
Z_(3) is a Z_2 module
Here Z_2 is the localization at {1,2,4,8,...}

[lethal dune]

why does it not work? I don't see it. Z_(3) as Z module then you localize at {1, 2, ...}. Z_(3) is already divisible by powers of 2

don't you just get Z_(3) back?

[wraith cargo]

hm

Okay I see it now yeah then it is true

[lethal dune]

oh okay, let's see if anyone else has any other comments

I'm not sure myself if that's correct

[wraith cargo]

tbh this was my original ideal

every S^-1A module is of the form U^-1X, where S is a subset of U I believe
But now I can see that you can localize U^-1X at S and get back the same thing so it is of the form S^-1N

[south patrol]

I'm pretty sure restricting scalars should be fine

[lethal dune]

also categorically L \circ R naturally isomorphic to id

[summer path]

Mfw qual problems are easier than lang exercises

[rocky cloak]

Qual problems have a limited time frame right? Exercises you can ponder for weeks

[frigid lark]

Oh, they must be trivial, looks at ch5 ex 34

[untold turret]

$B_1,B_2$ are two bases for $\mathbb R^2$, given by $B_1={\begin{bmatrix} 4 \ 2 \end{bmatrix}, \begin{bmatrix} 5 \ 3 \end{bmatrix} };B_2= { \begin{bmatrix} 1 \ 0 \end{bmatrix}, \begin{bmatrix} 0 \ 1 \end{bmatrix} }$

I informally remember change-of-basis as involving the following procedure:
$\begin{bmatrix} a \ b \end{bmatrix}{B_1} \overset{\phi_1}{\mapsto} a\begin{bmatrix} 4 \ 2 \end{bmatrix}{B_2} + b\begin{bmatrix} 5 \ 3 \end{bmatrix}{B_2} = \begin{bmatrix} 4a + 5b \ 2a + 3b \end{bmatrix}{B_2} \overset{\phi_2}{\mapsto} \begin{bmatrix} 4 & 5 \ 2 & 3 \end{bmatrix} \begin{bmatrix} a \ b \end{bmatrix}$ but I'm trying to get the right formalization of change-of-basis. Don't $\phi_1, \phi_2$ have standardized names? isn't there maybe a nice commutative diagram of this somewhere?

[cloud walrus]

nrs

[untold turret]

i'm trying to remember how change of basis went in linear algebra because I need it for some module stuff, but I'm getting confused wrt. exactly what relationship is involved between these four objects: matrices/vectors in matrix-form in base 1, matrices/vectors in matrix-form in base 2, vectors in linear combination form in base 1, and vectors in linear combination form in base 2

[ember field]

you can replace free modules with vector space for your purposes.

Taken from aluffi pg 322. You can refer to it for detailed exposition of change of basis in free modules

[untold turret]

niiiiiiiiiiiiiiiiiiiiiiiiice! thank you so much!!!

[solemn dew]

If a group G is infinate and it's order is infinate, why does one say the order is infinate and not "order does not exist"? Is there a reason for this?

[rocky cloak]

Because the order does exist...

Like the group is infinite, that's the thing you want to communicate, so you say that

I'm not sure why you would say something else than what you mean

[coral shale]

The order is a cardinality

[chilly ocean]

it's the same with cardinalities of sets

[coral shale]

'infinite' is imprecise about the exact cardinality ofc

but it wont be stated if we dont care

[lusty marlin]

Also, a group being infinite is the same as its order being infinite. Not sure why you listed those two things separately with an 'and' between them.

[solemn dew]

@lusty marlin Because I am not sure what it is

But alright guys thanks

[lusty marlin]

The order of a group (G,*) is the cardinality of the set G.

[solemn dew]

Right

My thoughts were:

since g^(m-1) = e, then how is it that g^inf = e

[lusty marlin]

What is m here?

[delicate orchid]

the order of an element is different to the order of a group

[lusty marlin]

In an infinite group, there may exist elements of infinite order.

[delicate orchid]

specifically, if you take the cyclic group generated by an element - the order of that cyclic group is the order of the element
so if there doesn't exist an m such that g^m = e, that cyclic group will be infinite and so we say "g has infinite order"

[lusty marlin]

That is, elements g for which g^n ≠ e for all n ∈ ℕ

[solemn dew]

Oh okay I get it

thanks 👍

[frigid lark]

What does g^inf even mean?

[solemn dew]

I don't know. Been studying this for 3 days

[frigid lark]

Well your making good progress

[hexed mantle]

wait wrong link sorry guys

Here we go!

Let $(E,\cdot)$ be a set with an operation such that $\exists \ n \in \mathbb{N},\ n \geq 2$ satisfying
$$\forall \ \quantity(x,y) \in E^2,\ \quantity(xy)^n = yx.$$
Prove that $\cdot$ exhibits commutativity.

[cloud walrus]

jonathanphan

[frigid lark]

But g^inf doesn't have any meaning here.

[hexed mantle]

Here's my proof. Is this correct?

[delicate orchid]

how did you conclude the last bit

[frigid lark]

(xy)^n=yx

[delicate orchid]

yeah, for the left hand side sure

[frigid lark]

(yx)^n = xy

[hexed mantle]

Oh yeah, forgot to mention that $\cdot$ is associative.

[cloud walrus]

jonathanphan

[frigid lark]

But this seems incomplete, like it doesn't cover the case when x cannot be expressed as z^2 for some z

Does E have inverses?

[solemn dew]

So this is self study homework:

Is my proof valid or is it stupid?

[cloud walrus]

jonathanphan

[solemn dew]

@frigid lark Do you know a nicer way of writing I(n) -> I(n+1) for induction? Or is this okay?

[barren sierra]

Looks fine to me

[frigid lark]

I just handwave induction

[barren sierra]

Eh

[hexed mantle]

Just write $k = n $ and $k = n + 1$. They're sufficient.

[cloud walrus]

jonathanphan

[barren sierra]

Use words

Suppose now that {thing} holds for n. Now we show {thing} for n + 1

Look at how induction is written in textbooks

I'd actually write it out but typing latex matrices on phone 💀

[frigid lark]

💀

[barren sierra]

But words are good

[solemn dew]

Thanks a lot!! 🙂

Will write with words from now on^

[wraith cargo]

does the problem ever state that (yx)^n = xy?

it only says (xy)^n = yx

[delicate orchid]

I think we're trying to think about that as a relator

it's not a relator, it's a massive set of relators

[frigid lark]

For all (x,y) in E^2

[delicate orchid]

^

[wraith cargo]

ah so for any two elements

OK yeah

[delicate orchid]

which was my problem too

wait do we even have an identity

guess not

[frigid lark]

we import group structure from an abelian group of the same cardinality

[delicate orchid]

we have the little troll that x^(2n) = x^2

[hexed mantle]

||yes I noticed that lmao||

[delicate orchid]

no idea if it helps us

[hexed mantle]

trivial qed lol

[delicate orchid]

I'm so bad at these word problems where you have nothing to work with

like in a group this is obvious but noooo have to do it in a semigroup

[delicate orchid]

(xy)^n = yx => (xy)^2n = (yx)^2 => xy = y(xy)x lol

[frigid lark]

Don't you get that by considering n=2, like (yx)^2 = xy

[delicate orchid]

wdym, I got it from this

[frigid lark]

It seems like overkill

[delicate orchid]

nothing is overkill until we get an answer

[wraith cargo]

I got that x^2y^2 = xyx lol

[frigid lark]

Fair

[delicate orchid]

honestly? I don't believe this result

[lethal dune]

[delicate orchid]

we can't even work backwards and chain iffs together because no inverses

[frigid lark]

yx = (xy)^3 = x(yx)^2y = x^2y^2

[wraith cargo]

YXY = XY FUCK

[delicate orchid]

so we have
yxy = xy
xyx = yx
yxyx = xy
yx = x^2y^2
wonderful

[wraith cargo]

that's equivalent to yx = xyx

[delicate orchid]

how come

[summer path]

i feel like i remember a similar problem of this flavor that was in this channel a few months ago

[delicate orchid]

some courses like introducing group theory by having students prove things in obscure structures for god knows what reason

it's such dogwater pedagogy and I'm not just saying that because I can't solve this

[frigid lark]

Yeah torture them to make them appreciate how nice groups are

[wraith cargo]

yx^2 = x^2y
x^2 y = x(xy)(yx)x
x^2y = x(xy)x
x^2 y^2 = x(xy)^2
x^2y^2 = xyx

[summer path]

that seems to me like a waste of time tbh

[solemn dew]

Is my proof on (ii) okay? I was thinking that we could observe that inside this homomorphism is normal. Or is it logically wrong?

[delicate orchid]

oh have we just collectively agreed to set n = 2 now

[wraith cargo]

idk I used it here

[delicate orchid]

I was using general n

[delicate orchid]

it is

[wraith cargo]

ye it works

[solemn dew]

Thanks @wraith cargo

[wraith cargo]

I FUCKING GOT IT

[delicate orchid]

START SPITTIN BOY

[frigid lark]

I will spray 🔫

[wraith cargo]

Fuck lmao no I didn't I got too hyped I made a mistake
I showed that yx = x^2y and xy = y^2x
Which I thought were equal but I was mistaken

I hate this problem so much

[open sluice]

lol

[frigid lark]

Could you explain the second line?

[wraith cargo]

(yx^2)^2 = x^2y(yx^2) = x^2y^2x^2

[delicate orchid]

there's no way they actually wanted an undergrad to waste their time with this shite

get me in contact with your professor

we're going to court

[wraith cargo]

ur implying that a grad student's time is less valuable lmao

[delicate orchid]

correct

[frigid lark]

Isn't it (yx^2) = yx^2yx^2 = x^2y

[delicate orchid]

we had this at the start

[wraith cargo]

it is

[frigid lark]

Yeah my bird brain ain't following

[wraith cargo]

that's the next line

I mean currently the closest thing I got is yx = x^2y

[hexed mantle]

what the fuck have I brought into this chat lmao

[frigid lark]

Pain

[wraith cargo]

OK OK wait I might be close as fuck

I just need to coom with my brain to get a good idea

[karmic moat]

at least grad students get paid for it

[delicate orchid]

I'm going back to category algebras fuck this shit

I do not get paid

[wraith cargo]

calculating cellular homology is less tedious than this holy fuck

[karmic moat]

oh

[wraith cargo]

cope + seethe

[frigid lark]

What do you guys do, read book, read paper?

[wraith cargo]

1. reding is hard
2. I am not even in uni yet technically lol

[frigid lark]

At least you can fix homology groups with a gun

[wraith cargo]

I can point a gun to my head and hope that'll solve this problem

[karmic moat]

grwm — grad student edition:
10:00am wake up
10:30 go back to sleep
1pm: wake up again
2pm: read book
5pm: read paper
7pm: eat the only meal of the day
8pm: goof around
2am: sleep

[frigid lark]

Wait

I'm currently doing that

[elder wave]

you shouldn't

[wraith cargo]

same bestie!!!

[delicate orchid]

this except it's 11am

[karmic moat]

oh man here we go

youtube productivity guru

[frigid lark]

Yeah, my bird doesn't let me sleep past 10

[karmic moat]

guy who makes twitter threads about how to improve ur life

[delicate orchid]

and goofing around is finely mixed in with work from 1pm-1am due to adhd

[wraith cargo]

timo be like
"I'm the paragon of virtue"
ur a lawnmower stfu

[karmic moat]

ok dont slander lawnmowers

lawn mowing is an art form

[wraith cargo]

I hate mowed lawns 🚬

[karmic moat]

which one cannot truly appreciate until one mows the lawn

[frigid lark]

Lawns are a capitalist invention designed to oppress Americans

[elder wave]

Chat, am I a youtube guru for not wanting to have an unhealthy lifestyle

[karmic moat]

the lawn is a canvas and the lawnmower is my paint brush

[karmic moat]

yeah

[south patrol]

#chill

#❓how-to-get-help

[elder wave]

oh this is #groups-rings-fields wtf

my bad

[wraith cargo]

yeah we're trying to solve a problem go away

[delicate orchid]

timo solve this problem

[elder wave]

first step is eating more than one meal

yw

[delicate orchid]

you're competent

[wraith cargo]

I am starting to believe that this is some sick joke

[delicate orchid]

I said like 15 mins ago I don't believe the result

[elder wave]

what's the problem even

[south patrol]

What is the result

[glossy crag]

Quick sanity check, if $A$ is a $K$-algebra, $L/K$ an extension, and $B$ an $L$-algebra such that $A\otimes_KL\cong B$ as $L$-algebras, then they're isomorphic as $K$-algebras as well, right

[south patrol]

I will solve it

[cloud walrus]

leave_no_norm

[wraith cargo]

Also assume associativity

[south patrol]

Iso of L algebras implies iso of K algs right lol

[south patrol]

Hm

[glossy crag]

ikik, just for a second i doubted it for w/e reason

[elder wave]

is that true

[wraith cargo]

It's somebodies hw

I feel like it is
I managed to get that yx = x^2 y
And I think it can be reduced further

[delicate orchid]

restriction of scalars - extenstion of scalars adjoint isn't it

I forgorrrr

[barren sierra]

Don't you need to do something with the "order" of yx in addition to xy?

That's my first thought

[delicate orchid]

(yx)^n = xy

[wraith cargo]

also no cancelling

[summer path]

wow irony really hyped us up for nothing

[frigid lark]

And then exhibits commutativity is restricted to only elements in E^2, or it was a group all along

[delicate orchid]

yeah it's not cancellitive

[south patrol]

Yeah this seems weird lol

[delicate orchid]

everything in E commutes with everything in E^2
everything in E^2 commutes with everything in E

hmmm

[wraith cargo]

I got it

confirmed

[delicate orchid]

do those two conditions together with this (if this is true) imply the result?

no I don't think so

[barren sierra]

?

[frigid lark]

yx = (xy)^2n = (yx)^n = xy

[south patrol]

Yes though like extension of scalars is the left adjoint

[frigid lark]

For n = size of monster group

[delicate orchid]

https://media.discordapp.net/attachments/1025970402662563866/1123062919618367667/goofy3.gif

[south patrol]

Lol

[wraith cargo]

OK so we had these:
yxy = xy
xyx = yx

Now mutliply the first by x from the left to get
(xy)^2 = x^2y
yx = x^2y

Now multiply the second from the left by y to get
(yx)^2 = y^2x
xy = y^2x

Multiply this by x from the left
x^2y = (xy)(yx)
x^2 = xy

Hence xy = yx

[wraith cargo]

if this works I'm offing myself

[delicate orchid]

I get the intuition backwards in my head, image of the left adjoint is "easy to map out of" right

yeah that makes sense to me

[wraith cargo]

IT DOES NOOOOOOOOOOOO

Wow okay

turns out it's not a joke we were just idiots

[delicate orchid]

I'm offing myself

https://media.discordapp.net/attachments/1025970402662563866/1123062919618367667/goofy3.gif

[wraith cargo]

I managed to prove it in the most contrived way possible anyways 😎

[delicate orchid]

wait when did we get

waittttt

why is yx = (xy)^2n

[frigid lark]

Original identity

[south patrol]

Idk why you say image

Oh sure ye

[delicate orchid]

the original identity is yx = (xy)^n

[frigid lark]

R u trolling?

[delicate orchid]

there are fucking paragraphs of random xs and ys

[frigid lark]

Or is your brain that fried from this problem

[delicate orchid]

my SINCEREIST apologies for not following every little detail

[karmic moat]

it’s okay, i forgive u

[frigid lark]

Fair

[karmic moat]

water under the bridge

[delicate orchid]

thank you anamono

[karmic moat]

anytime

[barren sierra]

Wait what's the proof

There's so much shit above

[frigid lark]

^

[wraith cargo]

xy = (yx)^2n = (xy)^n = yx

[delicate orchid]

oh you swap it twice

[wraith cargo]

I think we should ban parrot I am angry now

[delicate orchid]

ban me please PLEASE

[static yew]

I'm not far enough through the book to know what the notation "F^*" means
also that's technically an endomorphism on the multiplicative group, right? but not on the field itself except for fields of characteristic 2 which is hmmmmm

but I think I see where this is going

for an odd prime field F_p, in F_p \ {0} the mapping x -> x^2 is always surjective, because x^2 = (-x)^2. This means that we consume two elements to create each perfect square, and there aren't any elements left to produce the other half of the field. this probably extends to odd prime powers greater than 1 but I haven't thought it all the way through

for an even prime field F_p, 0 is a perfect square, 1 is a perfect square, and there aren't any other elements cuz the only even prime is 2. poking at it with sagemath suggests that all powers of 2 are quadratically closed but I have no idea how I would prove that

[frigid lark]

Sorry irony

I am gonna go sleep and let real math happen

[wraith cargo]

I'm gonna go back to my Tors and Exts now

I don't wanna see another word in my life

[static yew]

real math == spending 500 pages to prove that 1+1=2?

[frigid lark]

Is that meant to be better?

[wraith cargo]

yes

[delicate orchid]

derived functors make you feel smart

[frigid lark]

Damn this was rough

[delicate orchid]

word problems make you feel like a fucking idiot

[karmic moat]

i be walking around saying “bijection” to myself sounding smart

i bet my bed thinks im a math genius

throwing math buzzwords like it’s wheel of fortune

[wraith cargo]

I am happy to hear other people do dumb shit like this too lmao

[summer path]

can we ban word problems

[rocky cloak]

So by F^* I mean the nonzero elements of F, which forms a group under multiplication.

What I was hinting at is that the kernel of the group homomorphism x^2 is 1 and -1. So this is injective iff 1 = -1, i.e. the field has characteristic 2. For a function from a finite set to itself being injective is the same thing as being surjective. So in a finite field every element is a square of and only if 1+1=0.

This happens exactly for fields of order 2^n

[barren sierra]

#1100503863586455632

[static yew]

Tubular Cat is working on an algebra problem. By applying the Goedel function to English words and mathematical symbols, he has defined two operators "+" and "*" on pairs of word problems, creating a ring of all word problems.

Exercise 1. Is Tubular Cat's ring a field?
Exercise 2. Are all word problem rings isomorphic? If not, how many distinct morphisms are there?
Exercise 3. Is this ring a field? If so, is it algebraically closed?
Exercise 4. Is the Frobenius endomorphism valid on elements of this ring?

needs more BS

[barren sierra]

Wait I'm actually dumb why is (xy)^2n = yx

[static yew]

what are x and y? what group/ring/field/whatever do they refer to

[barren sierra]

OH ITS THE SAME N FOR EVERYTHING

Wait no I'm still dumb I think it's too early to do math

[barren sierra]

#groups-rings-fields message

[static yew]

wait I don't think I've ever seen it written that way before
does the (x, y) \in E^2 just mean x \in E \and y \in E ?

[barren sierra]

Yes

Unpacking ordered pairs and such

[south patrol]

The set E^2

[barren sierra]

I genuinely do not see how (xy)^2n = yx

[static yew]

what if it's really the square root of the set E^4

so it's a vector space, or at least a module, right?

[delicate orchid]

it's a semigroup

[barren sierra]

This may be a symptom of me waking up not too long ago

[static yew]

for all x, y in E, a natural number n>=2 exists such that (xy)^n = yx

that means (yx)^n = xy

[barren sierra]

Yes

[static yew]

that means if you square both sides

(xy)^2n = xy

[barren sierra]

huh?

[static yew]

oh wait

lemme reread

yeah

actually it's not 2n

it's n^2, isn't it?

(xy)^n = yx
(yx)^n = xy
(xy)^(n^2) = xy

[barren sierra]

Ok thank fuck that's what I had but I thought I was going crazy

It is n^2 by a double swap

[wraith cargo]

it's 2n

[static yew]

where did you get 2n from?

[wraith cargo]

((xy)^2)^n

n is any number geq 2
So you can take n=2

[static yew]

where did the 2 come from

[south patrol]

What lol

[static yew]

that's not what it says, Irony

[south patrol]

It's "there exists n"

Not for all n

[static yew]

it says there is an n

[south patrol]

Otherwise it'd be trivial

[barren sierra]

You cannot just take n = 2

[wraith cargo]

Oh wow okay then
Looks like we were solving it even more wrongly than we thought lol

[delicate orchid]

no that's just you

[wraith cargo]

Noooo

[delicate orchid]

I warned you we needed a general n

[wraith cargo]

:(

[delicate orchid]

double swap idea should still work I think

[static yew]

personally I'm betting n is the sixth Fermat prime, F_5

[wraith cargo]

OK I leave now

[barren sierra]

Ok so it hasn't been solved right?

[wraith cargo]

parrot still solved it

[barren sierra]

ok then how does 2n work

[wraith cargo]

xy = (yx)^(n^2) = (xy)^n = yx

[delicate orchid]

yeah how do we get that first equality

[wraith cargo]

Oh wait

Okay

[barren sierra]

xy = (xy)^(n^2)

[wraith cargo]

I'm still thinking of stuff in the wrong way

ugghhhh

[barren sierra]

not yx

[wraith cargo]

It's so much easier if it can be any n thooooo

we should just pretend that's the case

[barren sierra]

that's the point

[wraith cargo]

Occam's razor

[barren sierra]

[delicate orchid]

I still don't see this
xy = (yx)^n = ... = (xy)^(n^2)

what are we swapping around

[barren sierra]

$(xy)^{n^2} = \left( (xy)^n \right)^n = (yx)^n = xy$ right?

[cloud walrus]

spamakin

[static yew]

wait does xy = (xy)^(n^2) mean that it's cyclic? barring trivialities like "xy=the same element for all x,y"

[delicate orchid]

you just keep restating the same thing

[delicate orchid]

not a group

[static yew]

not a group?

[delicate orchid]

I said it was a semigroup like

23 years ago

[barren sierra]

?

[delicate orchid]

how. does. ((xy)^n)^n = (yx)^n follow from the identity we've bene given

what two elements

are we swapping

[barren sierra]

(xy)^n = yx right?

[static yew]

I don't see how that's relevant here
(xy)^1 = (xy) trivially
(xy)^(n^2) = (xy) per the rules

that means that (xy) generates a set of at most n^2-1 elements

[delicate orchid]

thanks

I now believe you

[barren sierra]

but this is magic to me

[wraith cargo]

I think parrot was maybe doing it the same way I was

[delicate orchid]

ok? and?
how does this prove xy = yx for all x, y in E

I can take any group and generate a cyclic subgroup using an arbitrary xy it does not mean the group is commutative

xyxyxy = yx
yxyxyxyx = y^2x^2
yxyxyx = xy
xyxyxyxy = x^2y^2
hmm (n = 3 here)

[wraith cargo]

I almost said what if we use induction lmao

[barren sierra]

oh so it's wrong?

[wraith cargo]

yes

[barren sierra]

ok that makes me feel less dumb

[summer path]

I missed a bit, did someone get it yet

[delicate orchid]

nope

[summer path]

Oh

[lusty marlin]

(xy)^n = yx
y•(xy)^n•x = y²x²
(yx)^n+1 = y²x²
yx²y=yxxy=(yx)^n+1=xyyx=xy²x
Similarly, we can get x²y²=xy²x=yx²y
Therefore x²y²=y²x² for all x and y
All that remains is maybe showing that every element is a square in this set

[delicate orchid]

we had x^2y = yx^2 right at the start

good luck actually showing that every element is a square

[lusty marlin]

Haha I seem to have skipped over that whole conversation

[delicate orchid]

it's ok it was literally hours ago at this point

[lusty marlin]

Yup, we can't even tell if this is actually the case or not at this point

[rocky cloak]

Hmm, interesting problem. So if n=2, then

xy = (yx)^2 = (yx)(yx)

Thus yx = (xy)^2 = (yx)^2 = xy

If n is not 2, I'm not sure how to generalize

If the operation is associative, then xy = (yx)^n = (yx)(yx)^n-1

Thus (xy)^n = (yx)^n so xy=yx

But I'm pretty sure you need associativity

[white oxide]

oh ok thank u

[rocky cloak]

Maybe depends how x^n is bracketet

[chilly radish]

How do you conclude from that that (xy)^n=(yx)^n

What

[rocky cloak]

xy = ab where a=yx and b=yx, therefore (xy)^2 = ba

Or in the n>2 case: xy = (yx)(yx)^n-1 so (xy)^n = (yx)^n-1(yx)

But then I need associativity to conclude that (yx)^n-1(yx) = (yx)^n

[rocky cloak]

Ah, they said to assume associativity. Then it works no problem

[static yew]

wait what's the usual way for exponentiation in nonassociative?
is a^3 a(a(a)) or ((a)a)a?

I would have thought it would be the latter

[south patrol]

uh the way you've written brackets doesn't really make sense

a(aa) or (aa)a i guess you mean

[open sluice]

a(a)
a as a function of a

[south patrol]

I imagine you'd just not use exponential notation in that case

[static yew]

i'm getting unlambda vibes

[south patrol]

unless you really want to confuse people

[static yew]

if I didn't want to confuse people I wouldn't be anywhere near abstract algebra

[open sluice]

algebra without power associativity
why even live

[rocky cloak]

Well I just went with the former, but I need both to be equal, so you get problems either way

[real sparrow]

$$\left(^2F_4\left(2\right)'\right)^2$$

[cloud walrus]

s5

[real sparrow]

This is great

[delicate orchid]

le fischer group has arrived

[glossy crag]

If L/K is purely inseparable of degree n, then x^n\in K for all x\in L, right

[south patrol]

Yeah if n = p^M then each element x of L satisfies x^{p^m} in K for some m (possibly depending on x) and necessarily m<= M so that p^m | p^M

[glossy crag]

Yep, just double-checking, gracias

[south patrol]

np!

I had to work this out now lol didn't realise it was a fact so that's for teaching me lol

[glossy crag]

Yeah, it's not mentioned explicitly in any text I've seen (probably because it's an easy corollary)

Pretty useful though, I've seen it used in several proofs

[wraith cargo]

How do we justify that T is projective here (since Weibel claims that P_n (x) T is projective so I believe this is only true when T is projective)

is it something like that T is finitely presented somehow?

[south patrol]

If Q is a projective R-module then Q (x)_R T is a projective T-module

[wraith cargo]

hm
Okay

okay I see where I went bad
I was looking at stuff as if they were R-modules

[south patrol]

This follows p quickly from the like extension / restriction of scalars adjunction ig

[round sandal]

If we take a real vector space $V$ and consider the complexification $V \otimes_{\mathbb{R}} \mathbb{C}$ as a set what kind of elements does this have? They should be formal sums $\sum_{k,j} a_{k,j} v_k \otimes e_j$ where ${v_k}_k$ is a basis for $V$ and ${e_j}$ a basis for $\mathbb{C}$, but a basis for $\mathbb{C}$ is given by ${1,i}$ so the index $j$ goes only up to $2$ right? What kind of sums are we left with?

[cloud walrus]

estebanita

[south patrol]

Yes, you should say an R-basis for C though

But yes, one way to think of this is by changing notation slightly: write (slightly abusively i suppose) $v_k$ for $v_k \otimes e_1$ and $i v_k$ for $v_k \otimes e_2$. Then the point is we end up with $C$-linear combinations of the $v_k$ - you can check that what I'm saying actually works lol

But yes note that the actual construction of $V \otimes_{\mathbb R} \mathbb C$ may be different from what you wrote depending on source, like for example there are constructions which don't require a choice of basis which is nice

[cloud walrus]

potato

potato

[round sandal]

I'm getting slightly confused here. I would like to say that we end up with sums $$\sum_{k} a_{k,1}v_k \otimes 1 + a_{k,2}v_k \otimes i$$

[cloud walrus]

estebanita

[south patrol]

Sure, that's just rewriting it

[real sparrow]

Square of the Tits group

[round sandal]

Can we do something with the coefficients a_{k,1} and a_{k,2}?

[south patrol]

Not really, but the point is like

$V \otimes_\mathbb R \mathbb C$ is naturally a $\mathbb C$-vector space if you define $(a + ib) \cdot (v_k \otimes 1) = a ( v_k \otimes 1) + b (v_k \otimes i)$

[cloud walrus]

potato

[south patrol]

and so on

So if we write $i v_k$ for $v_k \otimes i$ and $v_k$ for $v_k \otimes 1$ then we can just write every element as a $\mathbb C$-lianr combination of the $v_k$ and indeed these form a basis over $\mathbb C$

[cloud walrus]

potato

[south patrol]

Which is sort of the whole point of the complexification

[glossy crag]

The centralisers of conjugate subalgebras are obviously conjugate, right

[rocky cloak]

Yes, if $zb = bz$ then $$x^-zxx^-bx = x^-zbx = x^-bzx = x^-bxx^-zx$$

[cloud walrus]

jagr2808

[glossy crag]

Intuition confirmation: if A, B are simple artinian algebras of length m,n such that A\otimes B is simple artinian, then its length is mn, right?

[glossy crag]

Does just taking 0\subset I_1\otimes1\subset\cdots\subset A\otimes 1\subset A\otimes J_1\subset\cdots\subset A\otimes B work?

[coral spindle]

Hmmmmmm

Over a field?

[glossy crag]

Yes

I_k\otimes 1=I_k+1\otimes 1 implies I_k=I_k+1

same with the other side

Only question is if every left ideal sandwhiched between these is of the form I\otimes 1 and A\otimes J respectively

[coral spindle]

Well I mean the quotients are just gonna be the composition factors of A and B right

That oughn't be hard to prove at least for the \otimes 1 side

Now what I'm worried about is that tensoring by A will kill off something in B

Would this be prevented by A being simple, is this your idea?

[glossy crag]

A is assumed simple, yes

[coral spindle]

Yes but does that mean that e.g. the map B -> A (x) B is injective

[glossy crag]

you mean b\mapsto 1\otimes b?

that's always injective

regardless of the properties of A and B

That's just embedding subspaces into tensor products

[coral spindle]

Oh right, because we're looking at a field

[wraith cargo]

is there a semi-sane way to show that Hom(Z[1/p], Q/Z) is isomorphic to the p-adics

[glossy crag]

So if $I'\subset I\subset A,J'\subset J\subset B$ are left ideals, do we have $(I\otimes 1)/(I'\otimes 1)\cong I/I'$ and $(A\otimes J)/(A\otimes J')\cong J/J'$ as left $A\otimes B$-modules?

[cloud walrus]

leave_no_norm

[glossy crag]

Actually nvm, I/I' and J/J' are not really A\otimes B modules

or are they?

[coral spindle]

Well I mean

I think they could be interpreted as them

like k (x) I/I'

[glossy crag]

if you make them into right B-modules and left A-modules respectively

[wraith cargo]

The isomorphism is that (A (x) J)/(A(x) J') = A (x) J/J' I believe

[coral spindle]

yar

[glossy crag]

what about the other?

[wraith cargo]

same,
You get I/I' (x) 1

[glossy crag]

What would that mean, 1 is not an algebra

[wraith cargo]

wait what is 1 here

I thought it was an element of B

[wraith cargo]

what I was thinking is that that quotient has the form i (x) 1 + I' (x) 1, then you can take out the 1 to get (i+ I') (x) 1?

[south patrol]

Inch resting hm

[rocky cloak]

Maybe just do it completely explicitly. If 1 maps to 0, 1/p maps to k_1/p, 1/p^2 maps to k_1/p^2 + k_2/p, etc. This exactly defines a p-adic integer.

Then for any rational q between 0 and 1 "multiply" the map by q, sending 1 to q

[formal ermine]

I think constructing the isomorphism explicitly is the way to go

[rocky cloak]

I don't see how the ring structure comes in naturally here

But maybe there's a way to do it

End( Z[1/p]/Z ) should be the p-adic integers with ring structure though

[formal ermine]

can't we do this p^n procedure for all n in Z?

[wraith cargo]

I asked somewhere else and was told that you can do it by writing Z[1/p] = lim-> 1/p^n Z, extracting this out of the Hom and getting the p-adics

[formal ermine]

the k_is should also all be equal because it's a homomorphism

[formal ermine]

but yeah wlog we just get a p adic integer

[rocky cloak]

$A\otimes J / A\otimes J' = A\otimes J/J'$ unless I'm tripping

[cloud walrus]

jagr2808

[rocky cloak]

So for I_1 < I_2 < ... < A, and J_1 < ... < B, I think the composition series should be I_1xJ_1 < I_1xJ_2 < ... I_1xB < I_1xB + I_2xJ_1 < ...

Guess this is what @wraith cargo was already saying(?)

[glossy crag]

If L/K is an extension, the base extension of M_n(K) is M_n(L), right?

just checking

[glossy crag]

I think I've finally put together the thing I was asking about the last couple of days, does this work?

[glossy crag]

Doesn't every algebra have maximal subfields (maximal subalgebras that are fields)? Start with K, then use Zorn. What's the big deal about these maximal subfield theorems.

[rocky cloak]

Which theorems are you thinking of? All I could find by googling are things classifying the maximal subfields or giving conditions for when they are isomorphic

[glossy crag]

Here's one example from the book.

Idk, it just felt like it was giving it the fact of existence more significance than it deserves (them all having the same degree is deservedly significant).

[rocky cloak]

I guess, but it should also be mentioned. But maybe it should have been outside the theorem in a remark or something...

[glossy crag]

a) Why exactly m>1, can't it happen that D\neq K and D has no subfields besides K?
b) The reasoning for <: the translates of L* cover D* and all translates intersect non-trivially, so |D*|<|D*:N||L*|, right?

[rocky cloak]

If x is not in K then K(x) < D is a field

[glossy crag]

And D\neq K(x) because we assumed D is non-commutative, right

[rocky cloak]

Well D could equal K(x), but then D is a field

Or K is the center, so I guess that's only possible if D=K

[glossy crag]

Man, Skolem-Noether is neat

[ripe basalt]

im reading a book that has a pretty category theoretic definition of an algebra. But part of the definition is they have a ring map $n: k \to A$ whose image is in the center of $A$ and a multiplication map $u: A \times A \to A$. A here is the vector space and k is the ground field. What exactly does n correspond to here?

[cloud walrus]

*-algebra

[ripe basalt]

like i see u is how you define multiplication of elements in the algebra

[south patrol]

To me this is just a standard part of the data of a k-algebra

[ripe basalt]

yeah what data is it exactly?

suppose my algebra is C for example

what is n(i)

err

[south patrol]

Algebra over what

[ripe basalt]

maybe not a great example

i was going to say as a C-algebra 🙂

but if you could just tell me what data n is that is enough

[south patrol]

I mean that in some definitions this corresponds to like itself, it is already part of the definition

Like in commutative algebra an R-algebra is just a ring S equipped with a map R -> S

[ripe basalt]

well maybe just tell me what n corresponds to

a description like "u corresponds to the multiplication of elements in the algebra"

[south patrol]

No I meant there is nothing it corresponds to lol

[ripe basalt]

yes there is

[south patrol]

Huh lol

[ripe basalt]

it means something intuitively

[south patrol]

Okay that's different lol

[ripe basalt]

this definition came after "older definitions" of algebras i am sure

like an older definition of an algebra wouldnt talk about a map u, they would talk about a closed binary operation etc

(same thing but slightly different flavor)

[south patrol]

Sure I mean one way to think about it is that it gives you the structure of a k-vector space

[ripe basalt]

how?

[glossy crag]

Yeah, what's the problem, the two are equivalent?

[south patrol]

Because now you can multiply by elements of k since they are in ur ring essentially

[ripe basalt]

okay yes this is what i thought but

how would you usually map from k into A?

[south patrol]

I mean to me this is already usual lol

[glossy crag]

If A is a k-VS that can be made into an algebra, then a\mapsto a\cdot 1 is a ring homomorphism of k into the center of A.

[south patrol]

What alternative definition do you propose

One is that k is already a subset of the centre of A

In which case the map would be the inclusion

[glossy crag]

Conversely, if there is a homomorphism of k into the centre of a ring A, then the ring becomes a k-VS and an algebra

[ripe basalt]

wait

can you just answer my question though

i dont know why you're asking me for alternate defns to propose etc

im just asking intuitively what this map is for

[south patrol]

Well this implies the existence of smth more usual

[ripe basalt]

how?

[south patrol]

But yes the key thing is a copy of k in ur ring

Which allows you to multiply by elements of k among other things

[ripe basalt]

i see..

[south patrol]

Which induces the vector space structure in particular

[ripe basalt]

so this is like "scaling a vector by an element in the ground field"

[south patrol]

Sure, that is part of it

[ripe basalt]

but how come it's not a map from k x A then?

[south patrol]

But this is stronger since you actually get a copy

Like you get an induced map k x A -> A by restricting multiplication

[ripe basalt]

like imagine some boring vector space, i would think then n: k x A --> A by n(k,a) = ka

[south patrol]

But you also get a copy of k you can use to add stuff etc

[glossy crag]

If you want yet another definition of an algebra, here's one using tensor products.

[ripe basalt]

that's what i am looking at ocean 🙂

[south patrol]

Like a nice example is smth like C[x] as a C algebra

[ripe basalt]

oh wait no it isnt

i have a field not a ring

[south patrol]

Like you get a C vector space structure

But you czn can think of C[x] as extending C in some sense

[glossy crag]

You can define algebras over general commutative rings the same way, just a little FYI for you.

[south patrol]

Since you have a copy of c in it

[ripe basalt]

oh yes then how would you define n: C --> C[x]?

[south patrol]

The natural inclusion

[ripe basalt]

hm

do you know an example that's not that basic?

[south patrol]

Usually it is just smth like that

Smth slightly more interesting sure would be

M_n(C) as a C algebra

By mapping C to the diagonal matrices

[ripe basalt]

like z --> diag(z)?

[south patrol]

But note a nontrivial ring map out of a field will always be injective

So you will get smth which often looks fairly trivial aha

But yeah this is probs more interesting

[ripe basalt]

mm

what's the whole importance of the thing where u get a copy?

yall mentioned something about that

[white oxide]

Bro gave a super thumbs up

[frigid lark]

Why is it uniquely determined?

Lang page 294

nvm, was proved later

[azure copper]

not sure if this fits here but intervals graphs are basically graphs where every vertex is an interval and i connect these vertexes if their corresponding intervals overlap right?

did i understand correctly?

[tender wharf]

it's not abstract algebra but yes appears so

[hidden haven]

Yes

[solemn dew]

Does anyone know how to start with this problem?

This is what I've done so far. Don't know what it means for something to have infinate orbits in addition to the zero vector

[wraith cargo]

what abt the multiples of the eigenvector?

[solemn dew]

That orbit is finite I guess

This is my argument so far

[hot lake]

that statement looks super false to me when A = identity matrix

[frigid lark]

Doesn't I have an eigenvalue of 1?

[hot lake]

yeah but all the orbits are finite

[frigid lark]

I thought it meant it has an infinite number of orbits

[hot lake]

C^n is uncountable and Z is countable so it's gonna have uncountably many orbits no matter what A is

[frigid lark]

true

[solemn dew]

I am a little confused of what you guys are discussing 😅

[hot lake]

I think the correct statement you might want to prove is

[wraith cargo]

The problem is saying that there are infinitely many different orbits in addition to the 0 orbit
Not that there is at least one infinite orbit

at least that's what I think it's saying

[hot lake]

Show that phi has a finite orbit other than {0} <=> there exist some eigenvalue of A such that lambda^k = 1 for some integer k >= 1

[solemn dew]

Oh shit

Yes

Wow I miswrote it

So sorry

Can the eigenvalue be imaginary?

Because if it only can be a real number, then there exists only one phi^k = 1

But I am not sure how this works in the complex field

phi^k = 1 must have infinate answers if phi is in C, since 1 = r(cos(theta) + i * sin(theta)), and theta depends on r

So doesn't this mean that the orbits of A^kv = phi*v is infinate and not finite?

[coral shale]

Let n = pq. p, q distinct primes.
Let R be a (commutative, unital) ring of order n.
I'm trying to think of an example of R with char p. So far I got the elements that aren't integer multiples of 1 have to be fractions, is that right?

I also wanted to check an example actually exists, and if there are many, before I continue

[south patrol]

P sure it can't have char p by Cauchy

[lusty marlin]

What result are you referring to here?

Ah fine, I get it

[south patrol]

Well like Cauchy's theorem for group, like we have a nonzero element of order q

[lusty marlin]

Yeah, char p shouldn't be possible

[south patrol]

And one of order p

[lusty marlin]

Due to this

[south patrol]

So the characteristic would have to be divisible by p and q

I mean also like

More cleanly

As groups this is isomorphic to Z/p x Z/q or Z/pq

So characteristic is n I suppose

But idk how weird the multiplicative structure can be

[formal ermine]

bro defined a vector space on a ring instead of a group through a universal property

[south patrol]

I mean that is the correct definition 🔥

[formal ermine]

hi potato

[south patrol]

Hi

[formal ermine]

how are you doing

[south patrol]

I just am getting up lol hru

[formal ermine]

same here lmao

some rando called me so I woke up

[south patrol]

Bruh

Yeah I still have alarms left over from uni I need to turn off

Lol

Today I am do ALGEBRAIC GEOMETRY

[formal ermine]

I need to cope with toric geometry today

[south patrol]

Bruh how come lol

[formal ermine]

my newton okounkov theory course is 5% newton okounkov theory 10% algebraic geometry and 85% toric geometry

[south patrol]

Bruh

Inch resting

[frigid lark]

https://tenor.com/view/donut-simpson-homer-simpson-yummy-nom-gif-12662023

[south patrol]

Fr

[delicate orchid]

You made all of those word’s up

[formal ermine]

it also has a bunch of prereqs like cat theory, rep theory, comm alg and what not even though it's algebra 2 lmao

[delicate orchid]

I still don’t believe you when you say it’s algebra 2

[frigid lark]

How can something like that share a name with a subject whose description is:

Rings, fields, ideals, number fields and algebraic extension fields. Coding theory applications of finite fields. Gaussian integers, Hamilton's quaternions. Euclidean Algorithm in rings.

[wooden ember]

Lmao what uni has an algebra 2 course that does Toric geometry

That’s mental

Our « equivalent » of an algebra 2 course does basic ring and field theory lol

[delicate orchid]

I don’t care

[south patrol]

Speak English

[frigid lark]

I don't understand therefore it does not exist

[delicate orchid]

I still don’t believe you

[formal ermine]

sprich deutsch du hurensohn

[wooden ember]

Ouch now it looks like Wew is being rude to me 😔

[delicate orchid]

We both know deep down that I would never do such a thing

[formal ermine]

sometimes alg 2 is scheme theory, sometimes it's comm alg, it depends on what the prof wants to do

[south patrol]

Bruder muß los

[formal ermine]

muß

[south patrol]

Yh I did that for meme

idk why

[wooden ember]

Shouldn’t that just be called « algebraic geometry »

Instead of algebra 2

[south patrol]

Spotted the French persin

[delicate orchid]

Yeah algebra 2 for me was like. Rings and ideals + basic number theory

[wooden ember]

What do you do in algebra 1 🤔 illum

[south patrol]

We didn't number them like that

[wooden ember]

And this is a second year course?

[delicate orchid]

Although my first AA course was in first year so who knows

[wooden ember]

Yeah we don’t either

We only number analysis courses for some reason

[delicate orchid]

Neither did we

[formal ermine]

groups, rings, modules, fields, galois, applications of galois

[wooden ember]

Damn

[formal ermine]

like third year

[delicate orchid]

I don’t buy any of this

[wooden ember]

That’s still kinda insane

[formal ermine]

I think I'm gonna take alg nt 2 next sem

[wooden ember]

Module theory here is third year algebra

[frigid lark]

You also only get to the good shit in algebra 808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,001

[formal ermine]

1 was fun this semester

the prof is so nice

[delicate orchid]

The warwick undergrad was about half as cracked as this

[formal ermine]

he said that we'll be so few people that it's gonna be more of a discussion and reading course

[frigid lark]

At least its taught

[wooden ember]

So a seminar

[formal ermine]

no presentations

[south patrol]

We do alg k theory in algebra 1

[formal ermine]

just like read X until Y then we talk about it

[south patrol]

tamiyaka shimura theory in algebra 2

[formal ermine]

I think it's gonna be on the global local principle

[south patrol]

Based

[frigid lark]

The Japanese names is when you know shit's getting real

[formal ermine]

isawa

[frigid lark]

Like if I hear a random European name in a math textbook, they could have been born in the 1500s for all I know, but a Japanese person recognised in western literature is like guaranteed to have contributed in the 1900's

Meaning cracked math

[south patrol]

Kinda true

[mighty kiln]

IUTT for algebra 3?

[south patrol]

Oh nah that's part of algebra 2

It's just a bit too trivial for me to mention

[wraith cargo]

taniyama shimura and you're probably thinking of Eichler Shimura theory :P

[wraith cargo]

wait Illum are you a grad student

[formal ermine]

I'm a high schooler lol

[wraith cargo]

I literally don't believe anything you say

[formal ermine]

[summer path]

We briefly touched on taniyama shimura in grad algebra 3

I don't really think it is feasible to make a good course covering all of the basic algebra stuff up to galois theory in one semester

[wraith cargo]

We had a hw problem with something Langlands related lol
But we never did anything that cool in class

[summer path]

Our homework in that class were easy computations and easy proofs

The lectures were nice though

[wraith cargo]

our hw were

our prof actually wanted to kill us I feel like

That first sem with grp theory was rough

[summer path]

I didn't take grad algebra 1 or 2

But audited grad algebra 2

[wraith cargo]

I took algebra 1 and 2 before I left uni
Alg 3 was field theory/ commutative algebra

[summer path]

Algebra 1 was hard rep theory

[wraith cargo]

wtf lol

[summer path]

2 was ag

[wraith cargo]

our algebra 1 was super intense grp theory

[summer path]

Grad algebra

[wraith cargo]

yeah

[summer path]

Ug algebra was the easy stuff

[wraith cargo]

Like our algebra class was take everything in UG algebra and crank it up to 11

[summer path]

That doesn't sound very fun

[wraith cargo]

cuz the prof was very much a pure algebraist
So he loves group stuff
We even did a bit of group cohomology and stuff like that

This is what our hws were like lol

We also did way too much Sylow stuff
I know it by heart now but I just can't anymore

Problems like this

[formal ermine]

more flexing time for me

this is easy compared to what we did

[rustic crown]

easy hw are uwu

[wraith cargo]

okay then
Solve it

show me it's easy

[agile burrow]

Sylow is cool

[formal ermine]

I'm at the dentist right now

but isn't it just like there's only one subgroup of order l lol

[spice whale]

ok prove that

put your mouth where your money is

eat the money

[tender wharf]

abstract algebra fight club

[spice whale]

I don't know where this was going to go

[wraith cargo]

no?

[south patrol]

Interesting

[hidden haven]

What illu said is equivalent to what you have to prove though

[south patrol]

Now prove it for finitely presented instead

[hidden haven]

And doesn't sylow say that no of sylow-l subgrps = 1 mod l and divides pq so divides p or q but that leaves only 1

[wraith cargo]

oh yeah lol I didn't see it said normal

[hidden haven]

Were you gonna prove the existence of a Sylow-l subgroup 💀

[wraith cargo]

no lol
I already solved this like 5 months ago when it was due lmao

[hidden haven]

wacky and uncharacteristic

[spiral scarab]

so g is fixed right?

[wraith cargo]

the argument boiled down to counting subgroups and then taking some quotients the reduced the problem iirc

[spiral scarab]

so all elements of the subgroup H multiplied by a fixed element of G?

or is it all elements of G?

[hidden haven]

So you didn't assume Sylow's theorems?

[spiral scarab]

or G/H?

[wraith cargo]

we did

[hidden haven]

Yes

[wraith cargo]

lemme open my overleaf

[hidden haven]

ok

[hidden haven]

gH is the set of all gh for h in H

And this is the definition of gH for all g in G

That's what the last "for all g in G" in the picture means

[wraith cargo]

This is my solution

[spiral scarab]

but then why is the number of cosets called an index? shouldnt there be an coset for every element in g?

or does it have to keep the subgroup structure

[hidden haven]

Different elements of G can give the same coset

The cosets are not subgroups

They don't usually contain identity

[spiral scarab]

oh yeah that makes sense

[spiral scarab]

then why do we care about them?

[tender wharf]

homomorphisms

[spiral scarab]

besides seeing if a subgroup is normal

[hidden haven]

The first isomorphism theorem

[wraith cargo]

essentially the problem boils down to showing whether there exists a normal p,q or l subgroup because at first you don't know which

[spice whale]

allows us to take quotient groups

[tender wharf]

lagrange theorem

[spice whale]

which are very useful

if not just for first iso and the correspondence theorem

[spiral scarab]

so like its part of the kernel? for the injectivity?

[spice whale]

no

[hidden haven]

The natural map from a group to its kernel maps entire cosets to single elements

And the preimage of each element is also a coset

[spice whale]

if the left and right cosets of a subgroup N are equal then you can make a new group G/N out of these cosets
where (gN)(hN) = ghN

[hidden haven]

I'll be honest I am not reading allat 💀

[spice whale]

the normal subgroups of G/N are in direct correspondence with normal subgroups of G containing N

which can be useful

[hidden haven]

Cosets of the kernel of f are precisely the fibers of f

Fiber meaning the inverse image of a single elements

[wraith cargo]

the argument boils down to:
Show that you can't have s_p, s_q, s_l > 1 (i.e. one has to be normal)
Assume wlog that ur p-subgroup is normal and the l-subgroup isn't
Get a contradiction

[hidden haven]

Oh damn

Is the issue in the fact that pq might be 1 mod l?

Thinking about why the argument I gave doesn't work

oh yeah if you take 2,3,5 💀

[spiral scarab]

I have no idea why quotient groups are defined like that honestly

[summer path]

I hate sylow so much

[spice whale]

how would you define them any other way

[hidden haven]

There's a first iso thread in this channel

[spice whale]

that's like
what they are

[hidden haven]

Where I have given one explanation

[spiral scarab]

isnt it supposed to be G without the subgroup N?

[spice whale]

no

[wraith cargo]

yeah
Tbh this argument took quite a while to come up with
my friend and I were working on it in the math building when the prof accidently saw us working and gave us a few pointers lmao

[hidden haven]

wow ok

[spice whale]

it's supposed to be G with N squished down into the identity

[hidden haven]

horrid

[tender wharf]

lmao, G \setminus N is no longer a group since you removed the identity

[hidden haven]

What

[spice whale]

the obvious way to do this is just say that N is the identity of the new group

[tender wharf]

what

[spiral scarab]

Oh right

[hidden haven]

Oh setminus

[spiral scarab]

right

[tender wharf]

yeah

[hidden haven]

[tender wharf]

I mean

[coral shale]

using \setminus rather than -

[tender wharf]

what's so sullyable about that

[spice whale]

quotients aren't removing elements they're squishing them together

across maths

[hidden haven]

why do you have 4 digit numbers in your name

[coral shale]

i think its bad. Anyways.

[tender wharf]

protesting discord change

[coral shale]

hes clearly meming

1337

[hidden haven]

[spiral scarab]

to preserve the initial structure right?

[coral shale]

anyways, uve all managed to confuse me on which way the quotient slash goes

[south patrol]

\

[tender wharf]

#groups-rings-fields message

[south patrol]

Depends