groups-rings-fields

[delicate orchid]

yeah there isn't one, because the evaluation maps for polynomial rings are part of a universal property, so if another ring has it it must be isomorphic to a polynomial ring

which clearly isn't the case here

but I think if you're mapping into a field you can do it analogously? maybe?

nah, cause if you pick x -> k for any k then wtf does the element 1/(x-k) get mapped to

or am I getting confused

[coral shale]

yh that is the problem

[crystal turtle]

Infinity duh

[delicate orchid]

so true....

[crystal turtle]

I mean that half-unironically. Freely adjoin infinity, and you can probably make it like a semi-ring homomorphism? or something?

[delicate orchid]

I think I'm gonna be sick

[coral shale]

wheely time

[south patrol]

😭

[white oxide]

can somebody please explain these equalities at the bottom to me? i can't understand it for the life of me... i don't understand why $\pi(\sigma) \pi(\tau)f(x_1, \dots, x_n) \neq \pi(\sigma)f(x_{\tau(1)}, \dots, x_{\tau(n)})$ and also the second equality

[cloud walrus]

okeyokay

[white oxide]

since we have $\pi(\sigma)(\pi(\tau) = f(x_1, \dots, x_n)) = \pi(\sigma)f(x{\tau(1)}, \dots, x_{\tau(n)})$

[cloud walrus]

okeyokay

[delicate orchid]

oh look it's a wreath product

[white oxide]

whatever fuck latex

[delicate orchid]

lemme read

you're applying $\pi(\sigma)$ to $\pi(\tau)f(...)$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

it's applied first

as is standard for wanting to show something is a group action you wanna show that g(hx) = (gh)x

so you first apply the outermost element and apply the second one

[white oxide]

ah i see so he did the left hand side of the equality first?

in g(hx) = (gh)x

[delicate orchid]

I think so yes

alternateively

[white oxide]

huh ok

[delicate orchid]

$\pi(\sigma)(\pi(\tau)f(x_i)) = \pi(\sigma)f(x_{\tau(i)}) = f(x_{\sigma(\tau(i))} = f(x_{\sigma\tau(i)})$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

not sure why they didn't just do this

[white oxide]

lmfao yea that's so much easier

what's the purpose of the $\pi(\sigma)$ lmfao

[cloud walrus]

okeyokay

[white oxide]

is it just to emphasize that you're projecting the permutations onto the indices or some shit

[delicate orchid]

I don't know

it seems like the authors are just allergic to defining what a group action is and are so skirting around the issue

[white oxide]

nah lang defined a group action earlier

[lost surge]

Someone told me to ask here so hope you can help me

Let $z\in\overline{\mathbb{Z}}$ an algebraic integer of conjugates $z_1,...,z_n$. Let $P\in\mathbb{Q}[X]$ also be a polynomial of degree at most $n−1$, and $p\in\mathbb{N}$ a prime number. Which of the following statements are always true?

All $P(z_1),...,P(z_n)$ are conjugates of $P(z)$. \

The conjugates of $P(z)$ are among $P(z_1),...,P(z_n)$. \

If $P(z)$ is an algebraic integer, then $P\in\mathbb{Z}[X]$ \

The sequence $z,z^2,z^3,...$ of the powers of z contains at least one term worth 0 or 1 modulo p. \

$\overline{\mathbb{Z}}$ modulo $p$, i.e. $\overline{\mathbb{Z}}/p\overline{\mathbb{Z }}$ is a field \

There are an infinity of algebraic integers two to two distinct modulo p.

[cloud walrus]

Joseph.P

[rocky cloak]

So the P(zi) will be all the conjugates of P(z), yes.

If P(z) = 1/2 z, and z=2, then P(z) is an integer even though P doesn't have integer coefficients. If you mean that P(z) is an algebraic integer for every z, then it must have integer coefficients.

[lost surge]

Is it all ?

[white oxide]

I'm a little bit confused as to this proof - for instance, wouldn't saying that a 3-cycle may come in the form of (rab) cover the case for a 3-cycle coming in the form (sab)?

[rocky cloak]

barZ/p will not be a field, and will be infinite.

I'm not sure whether z^n is eventually 0 or 1 mod p, but that seems plausible

[lost surge]

But for z=1+sqrt(2) I think it’s never congruent to 0 or 1 mod 5

[rocky cloak]

If you raise it to the 12 power you get 1 modulo 5

[lost surge]

,w calculate ((1+\sqrt(2))^{12}-1))/5

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by unmatheux

[lost surge]

That’s wrong

[delicate orchid]

modulo 5\bar{Z} I presume

[rocky cloak]

(1 + sqrt(2))^12 = 1 + 19600 + 138600sqrt(2)

[lost surge]

,w calculate (1+sqrt(2))^12

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by unmatheux

[delicate orchid]

,w 1 + 19600 + 138600sqrt(2)

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by wewladstbh

[delicate orchid]

you do seem to be slightly off but it doesn't matter, this is still 1 mod 5\bar{Z}

lets expand it out properly

,w expand (1+sqrt(2))^12

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by wewladstbh

[delicate orchid]

just a slight numerical error

[lost surge]

I tried the combination 1,3,4 but it’s wrong

[rocky cloak]

Yeah, I just wrote down one too many zeros

[rocky cloak]

What do you mean 1,3,4?

[lost surge]

The statements : the first one, the third and the fourth

[rocky cloak]

What do you mean you tried the combination?

[chilly ocean]

Why p=/=2^s+1

[lost surge]

In the question I have to check a box if they’re right so I checked the first, the third and the fourth

[rocky cloak]

Right, but the third is wrong

1,2 and 6 are right. And then we're unsure about 4

[lost surge]

Oh thought the 3rd was right

[rocky cloak]

1/2 * 2 is an integer even though 1/2 is not

[rocky cloak]

If s = rt with t odd, then

2^rt + 1 is congruent to 0 modulo 2^r + 1

[lost surge]

The fourth was wrong

[rocky cloak]

Coolio. Wonder what a counter example could be

[lost surge]

$\dfrac{1+\sqrt{17}}{2}$

[cloud walrus]

Joseph.P

[chilly ocean]

Tq

[white oxide]

can i get a hint for finishing this problem? i assumed p > 2 (the case that p = 2 was already proven in lang) and assumed for contradiction that [G: N_H] was not equal to 1. Since [G: N_H] = k for 2 \leq k \leq p, if 2 \leq k \leq p - 1 then that's our contradiction since [G:H] = p implies that we must have [G: N_H] dividing that index p which is impossible. however i'm a bit stuck on the case that [G:N_H] = p

<@&286206848099549185>

[topaz solar]

|G| = p|H| or smth right?

@white oxide

[white oxide]

yea what about it

lol ok time to bs some shit

[topaz solar]

|G| = p|N_H|

hrmmm

If the index is p, not 1, anyway

so what now homie

[white oxide]

no vlue

[topaz solar]

Try multiplying cosets in G/H by g in G

Group action

[white oxide]

does it involve some conjugation shit

im trying to follow this argument

[topaz solar]

Try left multiplication on the cosets

Or something like that, since the orbit of H would be p right?

[rocky cloak]

I guess a useful trick is that H is normal iff the action of H on G/H is trivial.

[topaz solar]

Permutation on p elements, kernel of index p or smth

[white oxide]

ok i'm still trying to understand this but yea

[topaz solar]

I guess that doesn’t as easily just say it’s H oop, but it’s contained in H

[white oxide]

i don't understand the conjugate and kernel part of the proof tho as well so i'll have to think about that

[topaz solar]

See that the kernel is a subgroup of H?

*of index p

[white oxide]

i guess so

yeah

yea makes sense

if g in G is sent to the identity permutation then gHg^{-1} = H so g is a subset of H or rather that doesn't make any sense

idk

i don't understand why the orbit would be p, there are p cosets of H including H, how does that imply that there are p copies of xHx^{-1} for x in G

and what the hell does it mean for G to operate on an orbit

that's saying G operating on Gs

or G operating on xHx^{-1}

like g(xHx^{-1})

[topaz solar]

I’d rather say g |-> gH

Which id think is better than conjugation shenanigans

Clearly hits all p cosets of H

[white oxide]

ok sure

[topaz solar]

And is (iso to) subgroup of S_p

[white oxide]

alr imma come back to this problem i'm starting to get frustrated lmfaoooo

thanks tho

[topaz solar]

Take the kernel of G’s operation on G/H

clearly a normal subgroup of G, and a subgroup of H

Since gH =/= H for g not in H

What’s the index of the kernel? What’s the index of the kernel in H

@white oxide to see either of these indexes, consider ||1st iso lol|| and ||lagrange||

[normal cypress]

is \Phi common notation in galois theory?

saying that for $x^2 + ax + b \in F[x]$, where $F(\sqrt{a^2 - 4b})$ is the extension, $\Phi(-a/2 + (\sqrt{a^2 - 4b})/2) = -a/2 - (\sqrt{a^2 - 4b})/2$

[cloud walrus]

Hsna¹

[terse crystal]

A subset yes but you should deduce what it is directly.
g acts on cosets of H, so g mapped to identity permutation iff gaH=aH for any a, meaning g is in intersection of aHa^-1.
Ker=intersection of aHa^-1, consider prime factors of |G/ker| (any prime factor q of [G:ker], can q<p? can q>p?)

[stone bane]

isn't D4 a subgroup of the group containing all of the symetries on a cube?

[quiet pelican]

Yes

(Well, isomorphic to)

Wdym exactly?

[stone bane]

forget that

[summer path]

Well it's not so difficult to determine the group of symmetries of a cube, I think

[stone bane]

the group of rotational symmetries of a cube is D4+90 and 270 degree roations in pitch and yaw

right?

[mighty kiln]

It is generated by those elements yes

[stone bane]

Wdym generated

[mighty kiln]

Every rotation is a combination of those

But not every rotation is one of those

Say rotation about an axis that passes through opposite corners

Or opposite edges

generate means you take some elements and combine them in every possible way to get more elements

E.g. the subset {6, 8} of the group of integers generates the subgroup of even integers

Since every even integer is a sum/difference of a bunch of 6s and 8s

[stone bane]

oh right those are counted as elements too

[stone bane]

is there an isomorphism between S1xR+ and RxR?

[coral spindle]

What group structure does R+ have? Did you mean to post in #point-set-topology? Add context please.

Oh, no I see now.

Hint: torsion

[stone bane]

uhh, the positive real numbers under multiplication?

the question is the same as "is there an isomorphism between complex numbers in polar form and rectangual form" (i think, that is why i'm asking)

[next obsidian]

Well not exactly, polar coordinates are degenerate at r = 0

But basically, yeah

[rocky cloak]

More like wheter the additive and multiplicative group of complex numbers are isomorphic.

There's a very natural homomorphism from one to the other, but it's not injective.

[stone bane]

at the pole they become redditors?

and those redditors make the mapping homo(morphic)???

[coral shale]

[open sluice]

hmm

[errant shadow]

What's an example of a non-normal subgroup of GL_2(R)?

proper subgroup

[delicate orchid]

the subgroup generated by (1,1)(0,1)

[errant shadow]

and non-trivial

[rocky cloak]

Non-trivial in what way?

[delicate orchid]

it's not... the trivial subgroup.....

[errant shadow]

as in not trivial subgp yea

[next obsidian]

But that’s normal

🗿

[delicate orchid]

so is every non-proper subgroup so w/e

[rocky cloak]

I would be more interested in an example where the trivial group isn't normal 😉

[errant shadow]

generated over matrix multiplication of itself?

[delicate orchid]

that tends to be how you generate subgroups yes

[rocky cloak]

It will consist of matrices of the form (1, n; 0, 1) for n an integer

[errant shadow]

But won't it have to generate the Id at some point to be a group then? and if it does then it becomes cyclic and hence abelian which would make it normal?

Or am I making a mistake somewhere?

[next obsidian]

You have to do M^n for all n in Z

So Id is when n = 0

And inverses appear as negative n

M being that matrix ofc

[errant shadow]

ahh makes sense

my bad

thanks

[rocky cloak]

Subgroups of abelian groups are normal, but abelian subgroups need not be normal.

[errant shadow]

oh yea fair, sheesh, what a mistake

thanks

[coral spindle]

No, this isn’t correct

R x R is not the multiplicative group of complex numbers

It is the additive group

[delicate orchid]

ceral2

anyway it's clear to see that S1xR and RxR cannot be isomorphic, the former has torsion

[coral spindle]

Upon the witnessing?

[delicate orchid]

theflamesofbabylon I have been trolled once more tearsofthenile

[coral spindle]

I want to see these emojis someday

[crystal turtle]

what the fuck could those emojis be

[coral spindle]

Fantastic 10/10

[chilly ocean]

Why in a finite ring A with an odd number of elements, 2 is in U(A)

[errant shadow]

How do you show that G acts on itself by left translation g.s:= gs?

[chilly ocean]

just check that that defines a group action

[errant shadow]

I am just confused about the definition of left translation, what exactly is it? rest I can figure out

[barren sierra]

It's what you wrote

Left translation of g is the function mapping s to gs

[errant shadow]

wait what? that's it?

woah okay

[barren sierra]

So then check that G acts on itself by translation, simply plug in the definition

It should immediately come together

[errant shadow]

gotcha

[chilly ocean]

Can someone help me with that

[coral spindle]

Hint: think about what you know about the group of units of Z/nZ

[rocky cloak]

What could the characteristic of A be?

[chilly ocean]

An odd number?

[rocky cloak]

Indeed, do you see why?

[chilly ocean]

Yeah from lagrange theorem on (A,+)

[rocky cloak]

And do you then see why 2 is a unit?

[chilly ocean]

Uh

This is where I got stuck

[coral spindle]

This may be the time to consider this, then

[rocky cloak]

So if the characteristic is 2n-1

[delicate orchid]

characteristic + 1 is e- ffs

[strong jasper]

1+1=2?

[rocky cloak]

What could a multiplicative inverse to 2 be

[strong jasper]

aña

[chilly ocean]

n?

Thanks:)

[delicate orchid]

this method is way smarter than what I had in mind lol

[rocky cloak]

More generally, for a finite ring A, n is a unit iff it is relatively prime to |A|.

[white oxide]

thanks, i'll look at this in a bit

i'm still trying to understand lang's example, or rather how the index of H in G being 2 implies that the orbit is of order 2

clearly one of the elements in the orbit has to be H

[topaz solar]

I’m surprised at least one is real

[delicate orchid]

they're all real

[white oxide]

so the other coset of H is of the form gH, and if the orbit is 2, that would imply that the orbit would look like {H, aHa^{-1}} for some a in G, i'm trying to see how this is true or implied

i'm also still uncertain as to what it means to operate on an orbit, i'm guessing he means that G operates on the set of subgroups under conjugation and happens to operate on H that way

[delicate orchid]

our action is conjugation right

[white oxide]

yea

[delicate orchid]

then H being index 2 implies H is normal so the orbit is size 1

[topaz solar]

I’d rather use a |-> aH

[delicate orchid]

sharp what pack were u smokin

[topaz solar]

I have said multiple times to use left multiplication

[white oxide]

nah im just tryna understand lang's proof

[topaz solar]

Also, the orbit is either 1 or 2 in that case

So you contradict

[delicate orchid]

me trying to infer the theorem statement from a dodgy proof

[white oxide]

let me post the proof/example gentlemen

[delicate orchid]

I'll scroll up for ONCE and only this ONCE

[topaz solar]

He’s trying to show smallest prime divisor index normal

[white oxide]

nah i got u

[summer path]

wew gone scrolling

[white oxide]

oh wait

i'm disabled

[delicate orchid]

I'm back from my pilgrimage

[white oxide]

oops

[delicate orchid]

I dislike this proof but I can't remember the one I like...

G/H = {H, aH} if aHa^-1 = H we're done, aHa^-1 = aH => Ha^-1 = H => H = Ha

[topaz solar]

The one I was thinking is G acting on left on G/H

[delicate orchid]

hmm

[rotund aurora]

why should aHa¡

[delicate orchid]

bro really hit me with the ¡

[rotund aurora]

lmao I couldnt see the keyboard well because dark and autohotkey wasnt activated

I was gonna edit but too late

I even opened minecraft education by accident, didnt even know that existed or was in my computer

[delicate orchid]

I ain't gonna question whatever the fuck your workflow is here

[white oxide]

"if [G: N_H] = p, then the orbit of H under the conjugation by elements of G on the set of subgroups is of order p, and G operates on this orbit. in this way we get a homomorphism of G into the group of permutations of p elements. since there is one conjugate of H unequal to H, then the kernel of our homomorphism is normal, of index 2, hence equal to H, which is normal, a contradiction which concludes the proof."

yeah yeah???

QED?

wait sharp ur right left multiplication on cosets is easier

wait

[white oxide]

WTF HOW DOES G OPERATE ON THE ORBIT IS IT JUST BY LEFT MULTIPLICATION OR CONJUGATION

[topaz solar]

The text you pasted literally says conjugation

[white oxide]

nah bro we got G operating on G operating on the set of subgroups by conjugation

that's crazy

anyways

what would this homomorphism even be given by

my guy lang waving his magic wand

[topaz solar]

Try left multiplication

You act on G/H by g |-> gH

[topaz solar]

(Also, g |-> g^-1Hg?)

[white oxide]

oh yea ur right

my fault

[white oxide]

aight

you should've suggested that earlier!

[topaz solar]

My guy

[summer path]

usually the actions are pretty natural, like if you're working with cosets you just slap it with left multiplication; but if you're working with normal subgroups, you slap it on both sides with conjugation

[topaz solar]

Well if it’s normal, the conjugation one is trivial

[summer path]

ok poorly worded

but im tired and im not fixing it

[white oxide]

wait my fault how would this be G acting on G/H if we're left multiplying H by g in G

[topaz solar]

Anyhow, kernel of this is evidently a subgroup of H

[white oxide]

g |-> gH

[topaz solar]

yeah tf you mean how is G acting on it?

[white oxide]

fasho

[white oxide]

like

g(aH) = (ga)H?

[topaz solar]

yeah?

[white oxide]

oh shit that's crazy

[topaz solar]

[white oxide]

anyways

[topaz solar]

What’s G:Kernel

[white oxide]

wait bro bro isn't this only if G is abelian, cuz h x (aH) = haH = H iff haH = ahH = H you feel me

[topaz solar]

tf u on about blud

Have you ever done a quotient group

Those aren’t always abelian

[white oxide]

nah, i've done a factor group tho

[topaz solar]

[white oxide]

anyways hmmm why is the kernel a subgroup of H....

OHH

[topaz solar]

[white oxide]

loll

[topaz solar]

Where’s timo’s 0 days when I need it

Anyhow

[white oxide]

don't get that nobody reference

anyways

[topaz solar]

K is a normal subgroup of H

[white oxide]

mhm

[topaz solar]

What’s G:K

[white oxide]

shittt maneee, it's greater than or equal to the index of H in G i believe

[topaz solar]

Try Lagrange

[white oxide]

ye it is

[G: K] = [G: H][H : K]

oops

i can smell a contradiction a brewing

LOL

i saw that Wew

[topaz solar]

What’s G:H

upon the witnessing

[delicate orchid]

What are we doing now

[crystal turtle]

theflamesofbabylon

[topaz solar]

waiting for okay to realize

[white oxide]

someone play the jeopardy theme

anyways it's p but i'm trying to reach a contradiction

[topaz solar]

No we aren’t

G:K = p * H:K

What’s H:K’s relationships

[white oxide]

ohh

wait i think i got it

nvm

[delicate orchid]

||G/H = {H, xH} then xH = H => x \in H so xH = Hx. Now assume x not in H, so xH \neq H. G/H is a partition of G so xH = G-H, likewise for Hx, so xH = G-H = Hx and so the left/right cosets are equal so H is normal||

this isn't related to ur action proof

such over kill for the p = 2 case

[topaz solar]

gg

[delicate orchid]

https://cdn.discordapp.com/attachments/487651208048410627/1153048752668553256/image.png

[rocky cloak]

Would be nice if you could generalize this strategy for p>2, but I don't think you can....

[delicate orchid]

neither do I

u have to do the action garbage

but even then it's a trivial consequence of schur's lemma ;3

[topaz solar]

Welp, when we eventually get that H:K = 1 then big W

[rocky cloak]

I guess the nice thing about p=2 is you don't have to worry about whether G is divisible by any smaller primes

[delicate orchid]

yeah you just get it immediately

cause |G| can't divide 0 😌

[topaz solar]

Other way around

[delicate orchid]

cereal2

[rocky cloak]

Do I want to know what this "cereal2" stuff means?

[crystal turtle]

Daybroken memes

[delicate orchid]

:theninecelestialsphereschimewithsuchasweethum:

[delicate orchid]

u know the old meme of the guy spitting out his cereal

it's that

[topaz solar]

What is that supposed to be

@white oxide have you seen it yet

[delicate orchid]

I made that one up

[crystal turtle]

yeah that one

[white oxide]

yea, i underestand that [H: K] = 1 implies that the kernel is all of H and hence is normal in G but my dumbass is still trying to get to [H: K] = 1

[topaz solar]

What’s G/K

Hint: 1st iso

[delicate orchid]

oh is our morphism the canonical surjection

[topaz solar]

Sure can’t be trivial

[topaz solar]

No, we’re permuting elements of G/H, a set of size p

I wonder what a helpful group homomorphism may be

[white oxide]

oh is it H

[delicate orchid]

map everything to the identity

[topaz solar]

why would you guess H

[white oxide]

idk rlly

[topaz solar]

You’re permuting a set of size p

Hmmm what’s a convenient group

[rocky cloak]

I prefer the proof where ||H acts on G/H by left multiplication||
||The orbit of gH can't contain H, so must have size < p, but must also divide the order of H||
||So it must have size 1, and hgH = gH, which implies Hg=gH||

[delicate orchid]

||I think that's what's going on here but they're so lost in the weeds the light of god cannot guide them ||

[topaz solar]

[rocky cloak]

Nah, sharps seems to be hinting that ||G/K is a subgroup of Sp||

[topaz solar]

I was going for ||K divides p!||

[delicate orchid]

I cannot follow any argument that's going on here ngl

[white oxide]

ok wait, so if we want to get [G: K]/p = p we want to get [G: K] = p or the quotient group G/K to be of size p

[delicate orchid]

I don't know the answer to this question and I'm tired of pretending that I do

[white oxide]

oh wait

G --> G/H

kernel K

G/K iso to G/H

which is p

[topaz solar]

huh

[white oxide]

?

[topaz solar]

No

[white oxide]

bruh

[topaz solar]

Try G -> S_{G/K}

[white oxide]

bruv

[topaz solar]

Since, ya know, it’s a group action?

That’s what they are

[white oxide]

nah not tryna do all that

but i'll try

[topaz solar]

First iso says G/K -> S_p

[white oxide]

hao

[delicate orchid]

can you prove this by appealing to the burnside ring of G or is that circular

[topaz solar]

I have no idea what the burnside ring is 😭

[warm saffron]

Usually, I'm helping answer questions, but I actually have a question I was hoping somone can clarify for me. Is the group U(12) a subset of the integers modulo n? I'm not sure if I should be putting equivalences symbols around my Cayley table. I feel like it is since otherwise 11^2=121 which would be outside the group if it wasn't a subgroup of Z_{12}

$U(12)={[a]:\gcd(a,12)=1,\ a\in\mathbb Z}$ is this correct?

[cloud walrus]

logician

[topaz solar]

Yeah but you could also list em

[delicate orchid]

yeah sharp I'm not buying this actually. S_p is p-transitive, how do we know the action of G on K is also p-transtitive?

it's definitely 2-transitive but we've done that

[topaz solar]

It’s not an isomorphism it’s an injection

[warm saffron]

I was seeing U(12)={1,5,7,11} but I imagine technically it's {[1],[5],[7],[11]}?

[delicate orchid]

right ok now I'm on board

[delicate orchid]

yeah it's 1,5,7,11

[topaz solar]

||so G/K divides p!, so it’s p||

[white oxide]

ok i'm not going crazy rightt his would imply G/K is of order p!

[topaz solar]

It does not

[white oxide]

bruh

what's the order of S_p then

the symmetric group on p letters

is it not p!

?

[warm saffron]

thanks for the clarification I imagine the equivalence classes symbols are only dropped for ease rather than constantly putting them around just like Z_n={[1],[2],...,[n-1]} in reality but most people just refer to Z_n as Z_n={1,2,3,...,n-1}

[topaz solar]

G->S_p isn’t a surjection necessarily my guy

G/K is the image

[white oxide]

dawg ur saying that G/K is isomorphic to S_p

[delicate orchid]

you did phrase it really annoyingly

[white oxide]

that implies they have the same order

[delicate orchid]

I thought that's what you meant as well

this entire proof is fucking stupid can I just add that on top as a little treat

[topaz solar]

I’m not a group theorist

[delicate orchid]

me neither

[rocky cloak]

It's just a matter of definition. Like you can define Z_n to be the set {0, 1, ..., n-1} where the group operation is a * b = (a+b) mod n

[delicate orchid]

it isn't actualyl I just think you're introducing it way too early

[rocky cloak]

Or you can define it as a quotient group, using equivalence classes

[delicate orchid]

the original post was only about p = 2

[topaz solar]

I’m not, since A -> B gives an injection A/ker -> B

[white oxide]

aight ok fair

[topaz solar]

The question he asked was because he had an assigned exercise for whatever p iirc

[white oxide]

ok so G/K has order less than p!

that's crazy

[topaz solar]

Yeah, and G:K = p * H:K

[white oxide]

and the image is a subgroup

[delicate orchid]

mb mb, as u were sarge

[warm saffron]

the books defines it as the positive integers less than that are relatively prime to 12. Since I was asked to write the cayley table for U(12) I was looking at 11 times 11 and that's 121, but since 121 isn't in {1,5,7,11} I figured 121=1 since leaves remainder 1 upon division by 12. Based on the fact that U(12) is a group and hence closed under the operation 11(11) must be in U(12).

[white oxide]

so it has ordering dividing p!

[topaz solar]

Yep

Now conclude G:K = p

Not as slick as jagr’s but it’ll get there eventually

Once you have that, H:K = 1 is immediate

[rocky cloak]

Yes, so 11*11 should be 1 in this group yes. Whether you want to think of that as the group operation being multiplication modulo 12, or you want to think of 11 as representing the equivalence class of 11 modulo 12 is up to you I guess

[topaz solar]

This where you finally use the assumptions

[warm saffron]

thanks for the help guys

[topaz solar]

No contradictions here too

[white oxide]

I wanted to take a moment to express my heartfelt gratitude for your invaluable assistance with a math problem that had been causing me quite a bit of frustration. Your timely help not only resolved my mathematical dilemma but also gave me a renewed sense of confidence in tackling such challenges.

Your willingness to offer your expertise and the patience with which you guided me through the problem were truly commendable. It is not every day that one encounters someone so generous with their time and knowledge. Your assistance went above and beyond what I expected, and I am deeply grateful for your kindness.

I believe that your name, Dragonslayer Sharp, truly reflects your abilities and dedication to helping others conquer their academic obstacles. You are a real-life math hero!

Please accept my sincere thanks once again. Your support has made a significant difference in my academic journey, and I will always remember your kindness. If there's ever an opportunity for me to return the favor or assist you in any way, please do not hesitate to reach out.

[topaz solar]

[delicate orchid]

hmmm... fusion system p-power index implies normal rizz? rizzed up?

[topaz solar]

Saturated with rizz, even

[delicate orchid]

actually wait I'm not even sure if that's true it might be weakly normal

[topaz solar]

Horrible

[topaz solar]

Why are you like this

[delicate orchid]

yeah ok "d"ragonslayer "s"harp I got a problem for you nerd

[topaz solar]

[white oxide]

idk i'm a very grateful person, my gratitude gets out of control sometimes

but fr thanks dude

[rocky cloak]

I think it's cute 🥰

Very heartwarming

[topaz solar]

Tears of the nile

[delicate orchid]

it's just like

we got S_{p^k} acting on it's Sylow p-subgroup S and I need the automorphism groups of different subgroups of S that this generates

[topaz solar]

Yeah that sounds horrific

[delicate orchid]

it's known for S itself and there's only one other subgroup I need for p \neq 2

[topaz solar]

0 shot that this is solvable in less than 3 pages

[delicate orchid]

remains to be seen

the other problem I have is way harder because it actually involves a proof rather than just computation

[topaz solar]

I mean this sounds annoying to compute and build the relevant intermediary results

How bad is the other problem

[delicate orchid]

the other problem can be summerised as "Clifford's theorem for fusion systems"

[topaz solar]

I don’t know that one so

[delicate orchid]

uhhh

[karmic moat]

yeah np

[delicate orchid]

$H \trianglelefteq G$ $\chi \in \text{Irr}(G) \Rightarrow \chi|H = e\sum{i=1}^t \psi^{g_i}$ for some $\psi \in \text{Irr}(H)$

[cloud walrus]

Wew Lads Tbh

[topaz solar]

Yep that’s rep theory I can’t help in the slightest

[delicate orchid]

the rep theorist is doing rep theory no way...

[topaz solar]

I don’t know anything about rep

[delicate orchid]

my discord status was something I was thinking about the other day

[white oxide]

wait would it even make sense to consider the kernel since the kernel is all elements that get sent to the identity, but G/H need not have a group structure since we don't know if H is normal yet right

or am i SMOKING

[delicate orchid]

it's the kernel of the group action right

iirc

so the kernel would just be all the points that fix G/H?

[white oxide]

oh shit that's crazy

[delicate orchid]

these are more commonly called "fixed points" but sharp is a nerd

no, sorry, Stab_{G}(G/H) or whatever

[topaz solar]

that is, the kernel of the action is the kernel of the homomorphism into the permutations

[delicate orchid]

u can either be an orb 🔮 or a stab 🔪

[delicate orchid]

uhh prove that for any action GxX -> X the set Stab_G(Y) = {g \in G : gy = y for all y in Y} is a subgroup of G for every subset Y of X @white oxide if u want another exercise

[white oxide]

ok i'll do this thx

[delicate orchid]

found a paper that does it for me (for k = 2...) hahahahahah I WON

[topaz solar]

W I guess

[white oxide]

here d and c are homomorphisms right

and i'm assuming we just send the basis x_i in F to some b_i in M

oh wait no

we send the x_i to whatever values the x_i are sent to under the map c

i believe

since b surjective

wait no

[topaz solar]

c(x_i) = b(y_i) for some y_i in M, yes?

[tribal moss]

maps are homomorphism of left R-modules
it says.

[crystal turtle]

reading is hard

[white oxide]

reading is hard

[white oxide]

that's exactly what i said!

like

put down on paper

not sent here i was spacing out

are u proud

[karmic moat]

here's what i have so far

i guess i'm just trying to show that linearly independent in M iff linearly independent in M/T(M)

where T(M) denotes the torsion submodule

idk how to show the other direction

do i just consider ${x_1 + T(M), \dots, x_n + T(M)}$ in $N + T(M)$? and then show that they're linearly indep and maximal

[cloud walrus]

ana(functor)mono(morphism)

[karmic moat]

something like,
[ x_1 + T(M) = r_2(x_2 + T(M) + \cdots + r_n(x_n + T(M)) = r_2x_2 + \cdots + r_nx_n + T(M) ]
implies
[ x_1 = r_2x_2 + \cdots + r_nx_n ]
contradiction to linear independence in $M$?

[cloud walrus]

ana(functor)mono(morphism)

[terse crystal]

On the right track but I think it’s not rigorous enough
Σ r_i x_i +T(M)=0 gives you
r Σ( r_i x_i ) =Σr r_i x_i =0 like this

It’s like max {m: m linear independent elements in M} =max {n: n linear independent elements in M/T(M)}, anything in the former correspondent a thing in the latter, so is the converse direction , so two maximum are the same. Things like this

[karmic moat]

hmm okay

imma try to fill in the details

im not quite sure how this concludes linear independence

is it just that \sum r r_i x_i = 0 \mod T(M) implies rr_i = 0 \mod T(M)?

[night onyx]

it's not mod T(M) it's that if \sum r_ix_i = 0 (mod T(M)) then \sum rr_ix_i = 0 (in R) for some r != 0, so use the fact that the x_i are independent and R is an integral domain

[karmic moat]

ohhh ok

that makes sense

[chilly ocean]

I have a problem: Show every element in a finite filed can be written as a sume of two squares

My try:

K finite field then |K|=p^k with p prime. Since K is a finite field then K is commutative

If p=2 then |K|=2^k then |K*|=2^k -1

So for every x in K* we have (x^2^(k-1))^2=x

So every element in K is a square and we are done

If p>2

Let phi:K*->K*, phi(x)=x^2

If phi(a)=phi(b) then a^2=b^2 then 0=a^2-b^2=(a-b)(a+b) (because K commutative)

Because K is a field 0 doesnt have non-zero divisors

So a=b or a=-b

What should I do here?

I dont think phi is injective.. then it will be a bijection so every element will be a square and this is false

[terse crystal]

It’s not , but you have this: inverse image has exactly two elements

I mean non-zero x^2 has exactly two inverse images x and -x

So any r in the field, you let X={a^2: a}, Y={r-b^2: b}. Consider what cardinality of X and Y are

[chilly ocean]

(|K|+1)/2

Right?

[terse crystal]

Yeah

[chilly ocean]

Thx

[terse crystal]

Np

[white oxide]

what are we inducting on? the order of a group?

[delicate orchid]

yeah

[next obsidian]

If you read the proof, you’d see what you’re inducting on

[chilly ocean]

What can we say about a ring A with 1+1≠0

[coral shale]

it isnt char 2...?

[rotund aurora]

it has >=3 elements

[delicate orchid]

GL_n over that field isn't simple

if it's a ring ur on ur own...

[white oxide]

dumb question, would this show that every free projective module is the direct sum of a free module? here since every module is the homomorphic image of a projective module, i would just take F/M' to be the homomorphic image of P (idk if this works lol) and M' a submodule of P, i feel like this doesn't work for some reason

using the fact that every free module is projective and i believe quotiening about by a module still preserves the module being free

[delicate orchid]

i believe quotiening about by a module still preserves the module being free
F(x)/(x^2)

[white oxide]

nvm then

[delicate orchid]

you have the right idea with that triangle though

just need to rearrange some things - remember you want P to be a direct summand - so your free module should probably go in the middle

and then you do a little trolling to find a section, hence the sequence splits

[white oxide]

OHH

i thought it meant like

oops

that it had F as a factor

in the direct sum

oops

[delicate orchid]

no no, it means if P is projective then there exists some Q s.t. P (+) Q is free

but u got that now

[white oxide]

wait maybe i misworded the exercise... so for the right direction of 1b i have to show that F = P \oplus Q for some Q right

[delicate orchid]

the second or the first thing

[white oxide]

proving that every proejctive module is a direct summand of a free module

[delicate orchid]

yeah

[white oxide]

ah ok thx

[late moth]

i have a notational question if thats ok

what is I(phi_2) and I_n-1(phi_2) here

this is from Eisenbud commalg

uh

which i did not read

ijust found this theorem but the book is massive

and i do not think i could search for the first instance of these notations in a reasonable amt of time lol

[fast stratus]

my brain hurts 🫠… too many G’s in one place

[crystal turtle]

Abhorrent notation

[delicate orchid]

I think that’s the worst thing I’ve seen all day

Firstly just write the triangle for G’_i normal in G_i

[crystal turtle]

oh no

[cloud walrus]

Ryx (Home for flowers)

[white oxide]

just got back to working on this problem

[white oxide]

this works right

here i took F to be the free-module with the property that P is the homomorphic of it lol

oops

pinged twice

my b

[delicate orchid]

Np

That’s the correct diagram and I think you’ve written that c is a section for g so yup!

[white oxide]

oh ok sick!

what's a section

[topaz solar]

One sided inverse to a surjection

Or epi morphism ig

[south patrol]

Aka right inverse

[delicate orchid]

Ah is that not standard terminology? ;3

[south patrol]

It is

[cold dust]

In general, if $G$, $H$, and $F$ are groups, is it true that $G*(H\times F)=(GH)\times F$, where $$ is the free product and $\times$ is the direct product?

[cloud walrus]

Eve

[terse crystal]

Probably not. If you take H={e}, you won’t have G free product F isomorphic to G product F

[crystal turtle]

I feel like usually two distinct operations on objects like this would rarely satisfy (X*Y)xZ=X*(YxZ) for distinct products * and x. Much more common to see some sort of distributivity I think? But you still shouldn't have that here I think

[cold dust]

Ah yeah, that makes sense. Thanks!

[topaz solar]

F x H ~= H x F

Consider how asymmetric that is

So that should be a big red flag

[small bramble]

I've been stuck on 13.2.9 in Dummit and Foote for a minute, specifically in trying to prove a necessary condition for an extension of the form $\mathbb{Q}(\sqrt{a + \sqrt{b}})$ being biquadratic (i.e. it being equal to $\mathbb{Q}(\sqrt{n}, \sqrt{m})$ for some $n, m, nm$ not squares). My guess is that the condition is $a^2 - b$ being a square and $b$ not being a square, mainly because we're asked to prove that this is the case if and only if $\sqrt{a + \sqrt{b}} = \sqrt{m} + \sqrt{n}$ for some $n, m, nm$ not squares earlier in the problem and not out of any genius insight. I've been able to do the rest of the exercise, including proving that $a^2 - b$ being a square implies biquadratic. "Just" having trouble with the converse of that. tysm i ❤️ u guys

[cloud walrus]

992qqoloy

[dense raven]

G is a residually finite group iff the intersection of subgroups of finite index in it is trivial. Seemingly, this property is preserved under quotients. However, the quotient of residually finite group is not necesarily residually finite. How is this possible?

The image of a finite index subgroup under homomorphism still has finite index, no?

[quiet pelican]

No
Consider Z -> Z x Z/2Z by x -> (0, x mod 2)
Then 2Z maps to the identity which is not finite index

Another thing where it could (and does, because (prove this) the image of a subgroup of finite index under a surjective homomorphism has finite index in the image) is that a subgroup of finite index could map to the whole group

[dense raven]

ah

[quiet pelican]

And taking intersections and images don’t commute

[dense raven]

quotients are surjective homomorphisms tho right

so finite index would be preserved under quotients

[topaz solar]

But can’t just move the intersection in & out of the map

[dense raven]

oh bruh

[void cosmos]

im lost at the second part of the proof

why does every submodule of I/S intersect pi(M) non trivially?

[summer path]

because you modded out by S, the largest submodule that was avoiding M

[void cosmos]

lmfao yeah obviously

cuz if not then itw ould be contained in S

by maximality

sorry mb

[summer path]

it's okay

[void cosmos]

why does

like

why is the chain nonempty

with zorn?

is it because

it has no essential extensinos so that there must exist a bigger module such that it has alteast 1 submodule not intersecting M?

[void cosmos]

@rotund aurora is the highlighted part the part where it uses what u just said?

thajt the category has enough injectives?

like that may not always be possible in general right?

[rotund aurora]

yes

[void cosmos]

tysm

[rotund aurora]

you dont even need to have injectives

[void cosmos]

:dw hta

wdym

[supple phoenix]

Any hints about the following quesiton

[rotund aurora]

In a general category I think you dont need to have injectives

[void cosmos]

yea ig

idk any cat theory let alone examples

but i trust you lol

[rotund aurora]

Im just learning about this

[coral spindle]

An arbitrary Abelian category need not even have projectives

[rotund aurora]

oh a nice example is the category of finitely generated abelian groups

[void cosmos]

yea ik that

its true in R-mod

[coral spindle]

If you restrict further to finite Abelian groups, there are neither projectives nor injectives

[rotund aurora]

you can prove that injectives in Ab are the divisible groups @void cosmos , which are not finitely generated

[void cosmos]

yea i can

[coral spindle]

No, that's not true. Finitely generated R-modules do indeed have enough projectives.

[void cosmos]

wait

yea ur right

i meant

like

no "projective hull"

or is that not true also

[coral spindle]

If I'm not mistaken, there is only one injective finitely generated Z-module up to isomorphism, namely 0.

[delicate orchid]

They’re called projective covers usually

[rotund aurora]

if you have two abelian categories A and B, then B^A defined to be the category where the objects are additive functors A-->B and the morphisms the natural transformations between them, need not be additive or anything right?

you do have zero object

and you can add natural transformations, but what you get need not be a natural transformation I think

wait I think it works

I was missing the bilinearity axiom

well idk if the rest of axioms are satisfied, I'll check

[terse crystal]

Is it true though? I proved K is either D_2n or a subgroup of <a>. When K has an element of the form ba^k, we can deduce that a^2 is in K, then K=D_2n.
Otherwise K is a subgroup of <a>. I don’t think it has to be <a>. When n=9, <a^3> is normal right

[delicate orchid]

indeed, there is a normal subgroup <a^d> for each divisor d of n

[terse crystal]

Yeah

[rotund aurora]

do you have kernels?

[balmy belfry]

What’s the first step in proving firs iso theo

[chilly ocean]

define the map

[prime arch]

Step 2: show its an isomorphism

[crystal turtle]

I mean

There's basically only one map it could be

[coral spindle]

you just do it you just fuckin do it you just prove the theorem that's all there is to it

[crystal turtle]

Step 0) open a fucking book

Step 1) actually fucking read it (literally impossible)

[prime arch]

Step 0) have intuition
Step 1) verify the intuition is correct

[crystal turtle]

Or step 1) actually go reread the explanation that Dami very politely wrote out for you that you evidently didn't read well enough

Altho actually he might not have had a proof there

[night onyx]

You could try to understand it just for sets, so if f : X -> Y is any set function, define the equivalence relation x ~ y iff f(x) = f(y). This is called the kernel equivalence with respect to f, and the equivalence classes correspond (bijectively) to points in the image of f. Specifically, if you have f(x) = y, then you also have f(z) = y for any z in [x]~. In other words, im(F) ~= X / ~

[rocky cloak]

This will in fact become an abelian category.

Prime example being if A is the category of finitely generated projective R-modules and B = Ab, then this is exactly Mod R.

[rotund aurora]

I could prove that Nat(F,G) will satisfy the axioms of an abelian group, but what if it's not a set?

I could prove that the category is additive I think, at least if I assume that Nat(F,G) is a set for all additive functors F,G, but how do I show that its abelian? For example, how do you show that kernels and cokernels exist?

every component of a natural transformation will have a kernel, but Im not sure how you can put all of these together and form a natural transformation

[next obsidian]

Taking kernels is functorial

[rocky cloak]

Yeah, you would need to assume either that A is essentially small, or deal with foundations somehow.

Once that issue is sidesteped you can compute kernels and cokernels pointwise. And check for yourself that this does indeed give you a natural transformation

[lunar brook]

nts if I should put this question under this channel or combinatorial structures but could someone help me with this question??

[coral spindle]

What's the Bose–Mesner algebra?

Hell, are they using the word scheme in an informal sense lol

[lunar brook]

From what I've learnt it's just that a matrix can be defined by <I,J,A> matrices

[coral spindle]

That's not really a helpful explanation. It sounds like this is some very specialist definitions that you will have to explain, or hope that another specialist is present

[rocky cloak]

From the definition I can look up, the algebra you're interested in is (for some enumeration of the group elements) generated by matrices D_i which have 1 in position (x, y) whenever xy^-1 is in C_i and 0 elsewhere.

Does the seem right?

[night onyx]

never heard of the bose-mesner algebra lol, but the center of C[G] is just C^n where n is the number of conjugacy classes of G, which just comes from representation theory, that C[G] is the direct sum of all matrix algebras End(V_{lambda}) where V_{lambda} are the irreducible G-modules, which are parameterized by conjugacy classes, no clue if that helps though!

[lunar brook]

Yes but we were introduced to it much earlier when we looked at strongly regular graphs so the way I saw it was that we can define some matrix with a bose mesner algebra which will be a linear combination of I J and A matrices.

[rocky cloak]

If so then just writing down what multiplication by Sum_Ci x should exactly give you the matrices D_i

[lunar brook]

damn ok I need to think about this for a second and then I'll get back to you

[rocky cloak]

So using the hint, you just consider each element of g as a permutation matrix (given by how it permutes g by left multiplication)

[lunar brook]

conjugacy classes are all disjoint right?

[rocky cloak]

Yes

[night onyx]

since conjugacy is an equivalence relation, yeah

[lunar brook]

Thank you for the help. I think this question is actually a fair bit easier than I thought.

my lack of knowledge in group theory is exposing me rn, the definitions helped a lot.

[rotund aurora]

can I get a hint in showing F(A oplus B)=F(A) oplus F(B), for an additive functor between additive categories?

[coral spindle]

There is a way of characterising direct sums in additive categories

That's the hint. You've probably seen this characterisation if you're following a book.

[rotund aurora]

you mean besides the definition right

[coral spindle]

Of course.

[rotund aurora]

I dont think I know, but I can try to figure out

if you write it, spoiler it plz

I mean ig its some sum of maps equaling zero or something, then you just take the F

[terse crystal]

Yeah. The split short exact sequence, under F they still form a split short exact sequence

[rotund aurora]

lol I did say spoiler it

but I realized

essentially, we are using that A oplus B is both a product and a coproduct right

[rocky cloak]

What you're using is that the biproduct can be expressed using a addition of morphisms

I.e. ||that the identity on A+B is the sum of the two projections||

[coral spindle]

I may be mistaken but although this does work in Abelian categories, I think the splitting lemma does not hold in a general additive category. This is nice but not what I was thinking of. Jagr has spilled the beans

[rocky cloak]

Yeah, you need the category to be weakly idempotent complete

Though I don't think cogwheels meant to use the splitting lemma

[terse crystal]

I thought biproduct exists in any additive category

Not abelian necessarily I mean

[delicate orchid]

ngl I just thought that was straight up the definition of an additive functor

[coral spindle]

This is the idea I had in mind. It says Abelian but it works in any additive category. I know because I wrote it.

[terse crystal]

I don’t know what splitting lemma is. I mean those morphisms i,j,p,q, pi=1, qj=1, ip+jq=1 thing…

[rocky cloak]

It's an interesting fact that between additive categories a functor preserves addition of morphisms iff it preserves finite biproducts.

[coral spindle]

I think a similar thing should be in whatshisname's intro to homological algebra

[delicate orchid]

oh right that's what we're doing

[coral spindle]

Rotman! Rotman's book.

[rotund aurora]

yeah

[coral spindle]

It might be an exercise, I don't recall

[rotund aurora]

that's the book Im using

[coral spindle]

Nice

[rotund aurora]

but I didnt read that lemma

[terse crystal]

The 10 chapters book?

[rotund aurora]

because it seemed convoluted

[coral spindle]

You should indeed have read the lemma

It's a great lemma

[rotund aurora]

and the next prop is that, so I figured to try to prove it on my own

[rotund aurora]

ik

I didnt mean to skip it for ever lol

[coral spindle]

Honestly I think I should quit math and take up soothsaying

[rotund aurora]

I did get the essential idea of the lemma before reading the lemma

[terse crystal]

Hi, what is this from? Is it from a textbook? If not what textbook you use for this course you are learning?

[rocky cloak]

Splitting lemma is basically if you have one side (i.e. i and p) then you can construct the other side (j and q).

But yeah for a biproduct you have both anyway.

[rotund aurora]

its page 304 in Rotman, Lemma 5.87, in case anyone is interested

[terse crystal]

Oh that. I see, thanks

[coral spindle]

Perhaps I'm not seeing it (I certainly can't visualise the diagrams right now) but I think you would have to assume that the functor has an image in every isomorphism class to prove that this is a coproduct

You would necessarily have the retract and section, that much is clear, but at least in my head I don't see this being sufficient

[coral spindle]

:uponthewitnessing:

See wew I'm cool

[rocky cloak]

I mean you don't need anything about the functor really. Once you have the section and retraction that sum to the identity, then you have a biproduct. And it's straightforward to prove that a biproduct is both a product and coproduct.

[coral spindle]

Right – that sum to the identity

I missed that

[delicate orchid]

this is the problem with learning stuff through osmosis I keep thinking things are the definitions of things but they're actually lemmas

[terse crystal]

Yeah you only need F is additive, F(0)=0, F(1)=1 I think. Nothing else

Oh so just additive …

And you said “additive F”

[lunar brook]

The content has been taken from these books

[terse crystal]

Thanks a lot☺️. All fields I want to read. I always want to read some algebraic combinatorics.

[lunar brook]

What are the rules for sharing pdf textbooks because I have the spectra of graphs book downloaded if you want me to send it over

touches on association schemes and designs in it but the other books in the reading list are definitely better for it

[coral spindle]

That's not permitted on this server, but who's to say what happens when you google "spectra of graphs type:pdf"

[rocky cloak]

I guess the server can't police what people dm each other in their own time

[delicate orchid]

I would be very upset if I recieved a direct message containing such content... I might even start crying....

[rotund aurora]

spectra of graphs, that sounds cool

[terse crystal]

Thank you! I have found links to all of them. They are probably all famous books so I found all.

[delicate orchid]

what's this matrix A here? Adjacency?

[rotund aurora]

I think so

[delicate orchid]

smells like the adjacency matrix

[rotund aurora]

yeah

[delicate orchid]

did not know the trace results

VERY swag

[lunar brook]

Yea just adjacency matrix of a graph

[rotund aurora]

absolutelly

[delicate orchid]

wonder what tr A^4 encapsulates

you're probably just doing homology at that point tbh

[lunar brook]

I'm doing my research thesis in graph theory so was nose deep in that book for a while

[delicate orchid]

ah yeah it is very homological - higher powers correspond to longer paths which in my deranged mind is the same thing as taking some sort of nerve

[terse crystal]

2n I think

Cycle of length n, corresponding n ways of selecting a vertex as beginning /ending point, and 2 orientations

[delicate orchid]

I buy it

[terse crystal]

What is this from? Is it graph theory ?

[rotund aurora]

spectra of graphs

[terse crystal]

Oh the one I asked about

I see, thank you. Very cool indeed

[lunar brook]

If you get through chapter 1 in that book you can solve the blackboard problems from Good Will Hunting with ease lol

my indicator for progress is if I understand the maths in movies

only true indicator

[minor dragon]

Hello, I’m struggling with an exercise. Let M be a module over a PID A. If M is a torsion module, there is just a finite number of primary components M_p that are not zero and the set of elements annihilated by p is of finite dimension over A/pA, show that M is artinian.

I don’t know how to show that the primary component M_p is artinian by using the fact that the set of elements annihilated by p is also artinian.

Does someone see how to do this ? Thank you.

[rotund aurora]

is the functor A-->Hom(A,-) always left exact

[delicate orchid]

do you just mean the functor Hom(A, -) or is something weird going on here

[rotund aurora]

you have a category C, and you send every object A to the functor Hom(A,-)

[delicate orchid]

ah ok so it is something more interesting

[rotund aurora]

so you have a functor C-->Set^C

[delicate orchid]

yur

[rotund aurora]

actually let me try writing things down

[delicate orchid]

yeah same

[coral spindle]

So what would it mean for this to be exact in an arbitraty category?

Since apparently you are not working in even an additive category C

[delicate orchid]

I presumed the category was at least exact

also

a natural transformation Hom(A, -) -> Hom(B, -) is just a function A -> B right

actually I'll verify it one sec

well you get a natrural transformation out of each one at least

[rotund aurora]

wait this functor is contravariant right

[delicate orchid]

ah wait you can show this with yoneda

nifty

[delicate orchid]

covariant

Hom(-, A) is contravariant

[rotund aurora]

Hom(A,-) is covariant

but you are looking at the functor A-->Hom(A,-)

[delicate orchid]

we should name this functor so I stop getting confused

let us call it "F"

[rotund aurora]

I think its contravariant

[delicate orchid]

lemme think

so f : A-> B gets mapped to a natural transformation f' : Hom(A, -) -> Hom(B, -) with f' = f \circ (-)

this is the action of Hom(-, whatever) on morphisms so should be contravariant yes

a lot of bifunctorality nonsense

So if we have an exact sequence 0 -> A -> B -> C -> 0
this gets mapped to uhhhhhhhhhhhhhhhhhh Hom(C, -) -> Hom(B, -) -> Hom(A, -) ? (cause we don't know if it's left exact yet)

[rotund aurora]

ye

[delicate orchid]

considering how similar the actions of F and Hom(-, A) are on morphisms could we mimic the proof?

only issue there is that Hom(-, A) is not left exact

[rotund aurora]

bruh the zero should be on the left right

if we want left exactness

[rocky cloak]

It maps colimits to limits. So should be left exact for any reasonable definition of left exact.

[rotund aurora]

🤓

[delicate orchid]

there's always something obvious i've missed

[balmy belfry]

Ty ❤️

[rotund aurora]

Rotman states this but with R=Z

are these equivalent? Or does the first imply the second (i dont immediately see why)

in Mitchell's book which Rotman references there is only the version for R-mod (and thats what I have seen in other places)

in the wikipedia article they define R explicitly

[rocky cloak]

This doesn't seem correct. It would imply that every ring is the endomorphism ring of some abelian group, which isn't true.

If you drop "full" from the requirement you can certainly do it.

[rotund aurora]

ok

yeah just before he states it without full

and so Mitchell does in another place of his book

but what an error

[rocky cloak]

Is this in Rotmann?

[rotund aurora]

page 316

[tiny pagoda]

the standard definition is just preserving finite limits/colimits

for non-abelian cats & non-additive functors

[rotund aurora]

what is the standard notation for coproducts

[coral spindle]

$\sqcup$

[cloud walrus]

Big Red Dog Theory

[coral spindle]

I'm used to this one ^

It's like a turned pi

[rotund aurora]

[coral spindle]

Similarly a product is $\sqcap$

[cloud walrus]

Big Red Dog Theory

[rotund aurora]

[coral spindle]

But it feels like everyone has their own notation

[rotund aurora]

why not just $\times$

[cloud walrus]

Croqueta

[rotund aurora]

thought times was the thing for products

[coral spindle]

[summer path]

What is Clifford theory about

[coral spindle]

reps of normal subgroups of a group

[rotund aurora]

lol I was just sully baiting 😭

[coral spindle]

kinda sucks when you want to do infinite products, and also \times is not the product in Ring, for example

But again it does feel like everyone has their own notation

[rotund aurora]

I am convinced with the cup/cap notation tho

[summer path]

I should import into my LaTeX

[tiny pagoda]

yeah this is the common one

well

$\coprod X_i$

[cloud walrus]

scrungler

[coral spindle]

See this makes me ick

[tiny pagoda]

is usually how I'd denote it

it is literally the upside down Pi

[coral spindle]

but yeah that's common

[tiny pagoda]

really?

I'd use sqcup for inline products

like $X \sqcup Y$

[coral spindle]

Idk I just dislike the fact it's just a turned pi for some reason

[cloud walrus]

scrungler

[tiny pagoda]

well that's kinda the point lmfao

[coral spindle]

yeah

[tiny pagoda]

it's the dual to product

[coral spindle]

Purely a silly personal preference

[tiny pagoda]

I like that notation a lot personally

[delicate orchid]

"for some reason" you mean the most intuitive reason on the planet?

and don't give me that \sqCAP garbage there is no such thing as a disjoint intersection

[rocky cloak]

Easy solution: only work in categories where the coproduct is $\oplus$

[cloud walrus]

jagr2808

[topaz solar]

$\amalg$

[cloud walrus]

Dragonslayer Sharp

[tiny pagoda]

this is too much serif for me

[karmic moat]

u probably celebrated when they took the serif away from I

[white oxide]

is this saying that if we write $F = S \oplus T$ then $S$ and $T$ are projective?

[cloud walrus]

okeyokay

[delicate orchid]

the second statement is saying that yes

[white oxide]

oh yea my bad

cool thx

pls have something to do with projections

[white oxide]

something like this??

wait no d is not from S to M

fuck

oh wait no i just have to switch up S and S \oplus T in the tower i believe

[delicate orchid]

biproduct moment

[white oxide]

would it look something like this where d is the map given by S \oplus T being projective

and i is the injection, c is any arbitrary map from S to M

i feel like i'm lost in the weeds

since we have i composed with c restricted to s is c

[white oxide]

wait this is not true huh

nvm i'm an idiot

[south patrol]

Hm may be easier to not worry about all the diagrams

[delicate orchid]

Or rather, the sequences

The coproduct diagram might still be useful

[delicate orchid]

Sorry I’ve had discord closed for a while. Have you seen the result that a direct sum of modules is projective if and only if each summand is projective? If not, prove it

[white oxide]

nah bro you're chillin

[white oxide]

well i just looked into hungerford and read that result and that was basically the solution ig lol

[delicate orchid]

Yeah but you can prove that result using the direct summand of a free module thing

[white oxide]

yea ig that makes sense

shouldn't it be $N = E \oplus M/(\Delta(M''))$, where we're trying to show $E$ is injective and we have the following diagram (with d not in the picture ofc)

[cloud walrus]

okeyokay

[white oxide]

and if i'm understanding correctly $\Delta: M'' \to E \oplus M$

[cloud walrus]

okeyokay

[white oxide]

also i'm assuming that i have to define a in some ay

way

bc in the given diagram it's only defined on M

[balmy belfry]

Is this channel modules or adv alg

[terse crystal]

Idk. It seems to me I can ask algebra questions in any one of them, and I choose the one that no current discussion is going on

[warm saffron]

was wondering if anyone is onboard with my idea that the symmetries of an regular n-gon (n>3 an integer) is 2n?

[terse crystal]

What are you talking about, this question?

tr(A^n)=(2n) (number of n-gons) thing?

[warm saffron]

Possibly? I was thinking of the symmetries of a triangle, square, etc. and thinking about their cayley tables and operations like rotations, or flipping about an axis... I'm talking about regular n-gons

I haven't gotten to tr(A^n) stuff yet so not sure if that's what I'm talking about

I came to 2n using a combinatorial argument

n places to fix vertex 1, then since vertex 2 must always be adjacent to vertex 1, it can go in one of two places hence 2n

Can anyone confirm?

pls tag/ping me

[terse crystal]

I confirm

[warm saffron]

thank u!

[fresh sun]

Hopefully this is the correct place to ask but. If you prove something to be true for the binary field can it be extended to all fields of characteristic 2?

[terse crystal]

You have to be specific about what property it is that you want to prove…

[fresh sun]

that (x+y)^2=x^2+y^2

[terse crystal]

You can directly prove it on any characteristic 2 field then

[fresh sun]

thank you

[void cosmos]

can someone give a counterexample

or more

an example

of two nonisomorphic R-modules having the same injective hull

?

ie a counterexample to the statement E(M)= E(N) --> M=N

[rocky cloak]

Any injective module is it's own injective hull, so just take any noninjective module

[void cosmos]

yes

[rocky cloak]

E.g. Z and Q both have injective hull Q

[void cosmos]

i am still not good with finding injective hulls in general

as in why is the injective hull of Z , Q

but yeah ig

it makes sense

its minimal

in a sense

tysm

[rocky cloak]

An abelian group is injective if it is divisible. So you can think about what it would mean for a divisible group to contain Z

[void cosmos]

yes

got it

tysm

king

[karmic moat]

let $R$ be a PID and $M$ a finitely generated $R$-module. Suppose $M = \langle x \rangle \oplus \langle y \rangle$, with $\text{Ann}(x) = aR, \text{Ann}(y) = bR$, with $a$ and $b$ coprime.

I'm not sure how to show that $M = \langle x + y \rangle$

[cloud walrus]

ana(functor)mono(morphism)

[karmic moat]

i guess what i'm stuck on is trying to mock a similar proof of the chinese remainder theorem