groups-rings-fields

[tribal moss]

(Not quite sure that reasoning still works if R is not UFD, though).

[frail zealot]

what's a good introductory book on representation theory

[chilly ocean]

ok

so this is would be the case for the polynomial ring being a direct product

but when expressed as a direct sum (1,0,…) is the identity

[white jackal]

since |Imφ|·|Kerφ| = |G|, then |φ(a)| divides
|a|. Why can you not make this leap? Can you guys help me out with this problem?

[chilly ocean]

so I still think A[x] is isomorphic to bigoplus_infty A as A-algebras

[chilly ocean]

take an example

[white jackal]

i know that if |g| is finite then |phi(g)| divides |g|

[gritty sparrow]

Do you mean as A modules? There is no direct sum for algebras

[white jackal]

do i just show that phi(e)=e?

[chilly ocean]

ive been told this already );

[gritty sparrow]

Ah sorry i didn’t see that this was part of an old conversation lol

[chilly ocean]

there exists sets with structure of infinite direct sum of rings

we just give different multiplication structure

[chilly ocean]

Like I understand in general for an A-algebra M, infinite direct sum of M isn’t necessarily an A-algebra

but is it possible to give inf direct sum a multiplication that allows it to be an A-algebra

because in case of A[x] as A-algebra

[gritty sparrow]

Yes, polynomial multiplication

[chilly ocean]

the multiplication is weird

[gritty sparrow]

Yep the same will work for any A algebra

[chilly ocean]

So is A[x] isomorphic to infty sum A, but with polynomial multiplication?

[gritty sparrow]

It isn’t really a well defined question unless you are asking about modules, in that case yes

And then you can define an algebra structure by transporting the multiplication over

[chilly ocean]

im trying to show A[x] flat A-algebra by showing each copy is flat

[gritty sparrow]

Lul that won’t work out in this manner

[white jackal]

can anyone help with my question?

[chilly ocean]

maybe atiyah macdonald has a typo im wondering

[chilly ocean]

have you worked with matrix rings?

[pastel cliff]

does an element of an algebraic structure being its own inverse imply that it commutes with itself

where's a wild moldi when you need one

[white jackal]

no our professor havent taught us that yet

[gritty sparrow]

You’re better off just showing it is flat directly

[gritty sparrow]

The question is correct

[chilly ocean]

the hint is the method i was approaching it at

it says use previous problem

but only works if polynomial is A mod

[pastel cliff]

referring specifically to the dihedral group here

[tribal moss]

Everything always commutes with itself.

[pastel cliff]

O

then is there an efficient way to do the last part of this

[white jackal]

for my question, do I have to show that phi(e) = e so that it |phi(g)| divides n?

[pastel cliff]

as in, without checking every pair

[chilly ocean]

also using A[x] tensor B iso B[x] is sorta cheating

[gritty sparrow]

Right, actually forget what i said about not using the direct sum and showing it directly, it can be done both ways in fact

Have you proven that tensor commutes with direct sum?

[chilly ocean]

yea

i was gonna do that approach

but im being told as A algebra A[x] is not infty direct sum

im sort of convinced it is though

[gritty sparrow]

It is as an A module

But as an A algebra it doesn’t really make sense

[gritty sparrow]

And this is all you need for flatness

[chilly ocean]

flat as A module implies flat as A algebra?

[gritty sparrow]

By definition

[chilly ocean]

flat as A algebra just means tensor preserves A algebra exact sequence?

[gritty sparrow]

No

[pastel cliff]

@tribal moss 👉 👈 only pinging bc ive been thinking abt it on and off all day and i got nothing

[gritty sparrow]

It has to preserve A module exact sequences

[chilly ocean]

wait wtf

[next obsidian]

Exactness doesn’t really make sense for A algebras

Exactness is all about module structure

[next obsidian]

eg kernels are not A algebras because they aren’t rings

[chilly ocean]

what do kernels have to do with anything

[next obsidian]

How can you state exactness as algebras

If the kernels aren’t objects in the category

This is why exact sequences of rings don’t make sense

[gritty sparrow]

Exact is kernel=image

[chilly ocean]

N->N’ injective implies N tensor R-> N’ tensor R injective?

[gritty sparrow]

Not in general

[chilly ocean]

im asking wouldnt this be a definition for exactness of A-algebra R

[gritty sparrow]

Yes that works

[next obsidian]

Or flatness I guess

[chilly ocean]

yeah my bad

i meant flatness

ig i havent looked at algebra homomorphisms enough

[gritty sparrow]

I think you looked at them too much

[chilly ocean]

but ig kernels not rings for same reason kernels not rings for rings

ngl i might be sleepy

[pastel cliff]

mood

chmonkey any chance u can scroll up please

would be quite of you

[next obsidian]

Just send a linky

[pastel cliff]

.

[chilly ocean]

wait wow

[next obsidian]

[pastel cliff]

i know, kinda cringe

[chilly ocean]

is there a name for condition of kernels and cokernels being objects in a category?

[next obsidian]

This problem looks like you’re realizing the dihedral group as a subset of matrices

Probably via a group action

[pastel cliff]

pretty much yeah

[next obsidian]

What’s the issue tho

Oh

[pastel cliff]

do i have to check all the pairs of elements of the dihedral group for commutativity or is there a more efficient way

[next obsidian]

That’s how I would do it

[gritty sparrow]

you might want to look up the definition of an abelian/additive category

[next obsidian]

I think you might be able to do some whacky shit like uhhh

I think things in the same conjugacy class

Should have the same number of things which commute

Namely if like uh

g = fhf^-1

[chilly ocean]

so weird that modules have this property but algebras dont

[next obsidian]

I think

The set of things g commutes with

[chilly ocean]

or maybe not

[next obsidian]

Is either like fSf^-1 where S is the stuff h commutes with

[chilly ocean]

maybe ring structures are meant to be unpleasant always

[next obsidian]

Or f^-1Sf

You’d have to check that

[pastel cliff]

FUCK we havent technically learned that

[next obsidian]

Just prove it omegalol

[pastel cliff]

i might have to just take the L and do it all manually

[next obsidian]

Anyway this lets you pick one thing from each conjugacy class

[pastel cliff]

prof is a cock

[next obsidian]

You could just program it

[pastel cliff]

i could do that

or i could be dumb

im also like 1.5 hours overdue

[next obsidian]

Monka

[pastel cliff]

making extensive use of my 4 hour grace period

[next obsidian]

Yeah I’d just bash it out

[pastel cliff]

but thank you chmonk

[next obsidian]

Sometimes you just have to

Do some shit

Rather than try to make that shit easier

[pastel cliff]

most of this homework is shit

[next obsidian]

[pastel cliff]

prof took a point off my last hw for not explicitly stating my claim

so im being obnoxiously rigorous in absolutely everything

[chilly ocean]

thatll show em

[pastel cliff]

does the fact that every element of D_8 commute with itself need justification?

and with the identity

[lethal dune]

D_8 is not commutative, wym?

[pastel cliff]

i know that

but certain pairs of elements do commute

[lethal dune]

yes

[pastel cliff]

im doing the last part of this

like can i just say that if f = g in D_8, the statement is trivially true

as well as if f = identity or g = identity

[lethal dune]

oh lol, yes

[pastel cliff]

or is it worth actually writing out

[lethal dune]

every element of <a> commutes,

here you can take a = rho

[pastel cliff]

yeah imm gonna show the ^2 and ^3 cases

[lethal dune]

so your question was whether $f\cdot f = f\cdot f$ or not \catthink

[pastel cliff]

i mean i know it's equal

and that it's aBeLiAn

[next obsidian]

Just prove it in any group lol

It’s as easy as

g•g = g•g

g•e = g = e•g

Ok there you go

[pastel cliff]

fair enough

[tall jay]

If H is a subgroup of the dihedral group D_4, and H has the elements {r0,r180,(x-reflection), (y-reflection)}, to prove it's not cyclic, I just need to check that each element of H does not have order 4, right?

[lavish nexus]

yes

also there are only 2 groups of order 4

cyclic or Z2xZ2

this is Z2xZ2

[tall jay]

So, I guess that begs the question: If I have a subgroup with n elements, to prove it's not cyclic, each element has to not have order n?

[lavish nexus]

yes

[lethal dune]

what if the group is Z

[lavish nexus]

if an element has order n then it generates a subgroup of order n

[lethal dune]

nvm you said it has n elements

[tall jay]

Would the converse be true? If a subgroup has n elements and at least one element has order n, it's cyclic?

[lavish nexus]

yes

[woven delta]

Yes

Try to construct the isomorphism

[tall jay]

Hmm, seems like a good exercise to prove

[lavish nexus]

This proves both directions

[tall jay]

"subgroup of order n" refers to the subgroup's n-elements, right?

[lavish nexus]

yes

[chilly ocean]

Isn’t it just H

Or am I tripping

[next obsidian]

No

Compute an example in like, S_4 or something

[chilly ocean]

can you walk me through this if you can

i actually dont know

so we setting G to s4?

what does S4 even look like

[lethal dune]

S4 looks like S4

[paper flint]

S4 is the group of permutations on a set of 4 elements

[chilly ocean]

thats like 16 things in it right?

[lethal dune]

!

[chilly ocean]

just trying to make sure if im seeing it right

[paper flint]

Oop

[chilly ocean]

[paper flint]

It should have 4!=24 elements as ryu said

[chilly ocean]

oh yeah

my bad

yes i knew that actually

4! elements

okay thats G

[pastel cliff]

here, does rho^3 commute with any element that has a tau in it??

i wouldve guessed that rho^3 would commute with rho tau but i dont think it does? unless im dense which is possible

[lethal dune]

no

otherwise rho^3 will lie in Z(D4)

[pastel cliff]

yea i just expected cuz rho^2 commutes w tau and rho^2tau it might also commute but ig not

those do commute right

[pastel cliff]

this was such a bitch to type

@next obsidian can u double check this is all of em please ❤️ im like 99% sure it is

[lethal dune]

why not use generator relation

[pastel cliff]

.

im being this obnoxious semi-on purpose

[lethal dune]

has nothing to do with generator relation

[lethal dune]

also WTF

[pastel cliff]

i wish i was kidding

[lethal dune]

[pastel cliff]

ignore the fact that i messed something else up

i asked him to double check and it was that the claim was missing

even though it was literally in the fucking question

does that look like all pairs tho

[next obsidian]

Fuck no

I do not want to bash through all that computation

[pastel cliff]

understandable, have a good night mr chmonk

[hidden haven]

Wait

You're still taking this course seriously???

[pastel cliff]

omg moldi is finally here to degrade me

i dont wanna wait

[hidden haven]

Lmao I would say the same thing I'm sorry

[pastel cliff]

it's either now or a year from now

,ti

[cloud walrus]

The current time for nitezba is 03:03 AM (EST) on Thu, 10/02/2022.

[hidden haven]

Like at some point you just gotta accept that you've done your best and leave it up to the prof to either be a good boy or an asshole 😌

[pastel cliff]

i am happy with my decisions i am happy with my decisions i am happy with my decisions

[hidden haven]

That's the spirit

[pastel cliff]

https://tenor.com/view/goku-anime-super-saiyan-dbz-dragon-ball-gif-16589862

[next obsidian]

Report the professor for causing you undue psychological damage

[pastel cliff]

id have to report most of my math department

[hidden haven]

[pastel cliff]

starting with my analysis prof for not teaching me any topology

[next obsidian]

Is there a nice criterion for when permutations commute?

You could try to find one then embed D_8 into S_4

But idk if that helps as much as it might make computation easier to do using a computer

[pastel cliff]

at this point ive turned my brain off and am bashing through computation

[next obsidian]

I was thinking something something cycles

But I think I only know like sufficient conditions

Not necessary ones

[hidden haven]

Maybe a cycle only commutes with its powers and nth roots

[next obsidian]

Yeah bht like

[hidden haven]

Or disjoint cycles

[next obsidian]

If you multiply 3 cycles

It could get really messy

[hidden haven]

Decompose into disjoint cycles first

Hopefully they don't interfere to make shit commute when it shouldn't

[next obsidian]

That’s my problem

I think it could

At least in S_4

Disjoint cycles are easy

[hidden haven]

Yeah I guess it is your problem

[hidden haven]

Peace

[next obsidian]

At worst you get like

[pastel cliff]

good time to use this

this should be a sticker

[next obsidian]

Product of two products of 2 cycles

[hidden haven]

Rip

[next obsidian]

Sussy butt baka

Okay this is so fake tho

Like

It isn’t just about disjointneas

[hidden haven]

Yes

[next obsidian]

Cuz anything commutes with itself

[hidden haven]

Did you see this

[next obsidian]

Okay I am

[hidden haven]

Good night chm

[next obsidian]

No

Not irl

I mean from this problem

[hidden haven]

Oh

Lmao

[next obsidian]

Lol

Okay I am gonna use my chmonkey sense rn

Either

This problem is solved in full generality

Or it’s insanely hard

And still open

[hidden haven]

[next obsidian]

Like no way ppl haven’t asked this question a million times

[hidden haven]

I feel like the people who have asked this question are profs giving this problem to ug first years before they've defined a group

Who else cares

[next obsidian]

Wait what class is this for?

Is this not a group theory course

[hidden haven]

Group theory apparently

[next obsidian]

But they haven’t defined a group?

[pastel cliff]

before you sully me ik this is an ugly way of writing it

[hidden haven]

But they are working with "algebras"

[next obsidian]

Yeah dude

I was like wtf is this shit

Wtf is an algebra

[pastel cliff]

I KNOw

[next obsidian]

No seriously

What is an algebra

[hidden haven]

Algebraic structure

[spice whale]

algebraic structure

[next obsidian]

A MAGMA?

[hidden haven]

Structure without relations

[spice whale]

at this point literally just say magma

[pastel cliff]

do i need to do subset both ways to prove that G_2 = A where A is the desired set

[next obsidian]

Are they literally dealing with magmas???

[hidden haven]

A magma is an example

[pastel cliff]

or can i just say it must be the smallest bc of the generating set

[hidden haven]

A set is an example

[next obsidian]

Wtf hahahaha

[spice whale]

wait a set??

[hidden haven]

A set with 5 operations and 2 constants is an example

[next obsidian]

This course is dick ass poopy butt

[pastel cliff]

guys please i know

[next obsidian]

Lol it’s doing it in a universal algebra way

[pastel cliff]

it's horsecock

[next obsidian]

Maybe you signed up for universal algebra

[pastel cliff]

pigshit, if you will

[next obsidian]

Not abstract algebra

[hidden haven]

Universal algebra is slightly more specific

[spice whale]

it only has one binary operation so it's a magma

[pastel cliff]

i'll send my syllabus later lol

[hidden haven]

In this case yeah

[pastel cliff]

"By definition the subgroup generated by a set is the smallest subgroup that contains that set"

is this enough proof that it must be the smallest

[hidden haven]

You want us to read that?

[pastel cliff]

[hidden haven]

Just omit the ∘s

And use powers for h ∘ h

[spice whale]

so, h is of order 4 and g is of order 2?

[hidden haven]

Will be a lot more readable

[next obsidian]

Yeah frfrfrfr

[spice whale]

oh wait yeah that's trivial

[next obsidian]

I even do that with maps when there’s too many

Like

Clearly fg = st or some shit

Ain’t nobody got time for •

[hidden haven]

Based

[pastel cliff]

fair

[hidden haven]

I also write fg for f then g

[next obsidian]

For f then g…

[pastel cliff]

would parentheses be acceptable to you sir moldi

[hidden haven]

Haven't done that in an assignment yet because I don't want TAs to cry

[hidden haven]

Parantheses for an associative operation?

[next obsidian]

I read cycles left to right

[hidden haven]

[next obsidian]

Because it’s what I did first

[pastel cliff]

hhhhngh

[next obsidian]

You can say

[spice whale]

sure

[next obsidian]

h^2

h^3

[spice whale]

you could do powers

[next obsidian]

It’ll make it a little less

[hidden haven]

Cycles should be read in the same order as functions

[next obsidian]

Hhhhnnnggghhh

And a bit more

[pastel cliff]

hngh

[spice whale]

also I don't think g and h commute

[next obsidian]

h^4n^3g^3h^3

[spice whale]

lol

wait do they

[hidden haven]

Maybe you should say that you are omitting ∘ intentionally

[next obsidian]

They totally do

[hidden haven]

I would not take any chances with this prof

[spice whale]

fair

[next obsidian]

I think

[spice whale]

so

[pastel cliff]

im doing this merci

ANYWAYS

"By definition the subgroup generated by a set is the smallest subgroup that contains that set" does this suffice as proof

im at my limit so i wanna say yes

and like it makes sense

actually fuck it

[lethal dune]

prove the definition

[pastel cliff]

this is correct

[lethal dune]

[hidden haven]

[spice whale]

i(-b-ai) = a-bi
g(i(a+bi)) = g(-b+ai) = -a+bi
they commute up to sign

[pastel cliff]

oh i dont have to do pairs of commuting elements for this thank god

[spice whale]

so they're not commutative

[pastel cliff]

it's isomorphic to D_8 so wouldnt be hard

but would also be extremely annoying

[spice whale]

gh = hhhg

yeah

[pastel cliff]

https://tenor.com/view/hng-gif-4501621

mood

[spice whale]

me when applied

[pastel cliff]

wait

is h^4 and identity element

it has to be right

yeah it is ok cool

[spice whale]

g has order 2
h has order 4

[pastel cliff]

fuck this shit

moldi and chmonkey and ryu i love you thank you mucho for shitting on my prof

[hidden haven]

Any time bro

[tall jay]

Would this be a valid argument, or would I have to do some sort of induction?

[hidden haven]

This is fine but you should also show that the identity is in the intersection

It's not implied by those 2 statements

[tall jay]

Oh, I used a theorem from a book that says if H is a nonempty subset of G, then H is a subgroup if (i) and (ii)

How come I still need to prove the identity is in there?

[hidden haven]

Then prove that the intersection is non empty

[steady axle]

what is a good reference for learning koszul complex

[next obsidian]

I liked the exposition in Serre’s Local Algebra

[steady axle]

ok i'll check it

[polar lotus]

on the context of commutative algebra, I don't know if that matters but, how can I show that A/<x> ~= A/ Ann(x), where A is a ring?

It appears on Atiyah's book solution from exercise 3.16 about flat algebras

[next obsidian]

I don’t this this is true?

[polar lotus]

ops

[next obsidian]

You should have that <x> ≈ A/Ann(x)

[polar lotus]

i missread it

yeaah

i just saw it

[next obsidian]

You have a surjevtion given by multiplying x

[polar lotus]

it was bugging me a lot

[next obsidian]

And the kernel is exactly the annihilator

[polar lotus]

yeah, that seems trivial now, thanks!

[next obsidian]

[wooden ember]

I don’t quite see why we can say that there is x_i in a such that xi not in pj for all j =/= i

Seems like a strong condition and I don’t quite see how it follows from the induction

Oh wait I see nvm

I misunderstood what the induction was trying to do

[coarse storm]

Proofs in Atiyah-MacDonald can be quite succinct, perhaps a little too much so as a first book in commutative algebra.

[inland otter]

So can somebody tell me how noether's normalisation theorem is like non-trivial?
Like it feels really intuitive that if $k[x_1,\ldots,x_n]$ is a finitely generated entire ring and $x_1,\ldots,x_n$ are algebraically dependent, we can just "take out" the elements which make it algebraically dependent and be left with r elements s.t. $k[x_1,\ldots, x_n]$ is integral over $k[y_1,\ldots,y_r]$

[cloud walrus]

dadaurs

[inland otter]

I feel like i'm missing something in the proof because the whole point of the proof is just constructing the right polynomials if x1,...,x_n are still alg. dependent

[hot lake]

"we can just take out the elements which make it dependent" is awfully vague and it can fail

for example if R = Z[2x, x]

and you decide to discard x instead of 2x

then R is not integral over Z[2x]

or if R = Z[2x,3x] then you have to be smart and find x inside it

[inland otter]

ah right that's what i was failing to understand completely about the proof

thanks for the response

[dusty sapphire]

Hello, how to count ring homomorphism from $\mathbb{Z}_m \cross \mathbb{Z}_n \to \mathbb{Z}_p \cross \mathbb{Z}_q$

[cloud walrus]

BLツ

[lethal dune]

wait $\mathbb{z}$

[lethal dune]

whatt

[dusty sapphire]

Hahaha

[lethal dune]

🙈

[dusty sapphire]

i.e. total number of ring homomorphism

[lethal dune]

you can count individually if you know how to count from Zm -> Zp

[dusty sapphire]

find possible values of f(1) like this but it could be hard some time

[prisma shuttle]

does anyone get why this chain is true

idk like why that's true unless ima missing smth obvious

[dense pumice]

that's true for any ideal

[chilly radish]

Isn't this always true for an ideal? Like if
I^2 = {i_1j_1+...+i_nj_n: i_i,j_i\in I}
Then clearly this is a subset of I, and you can keep going

[dense pumice]

since I is closed under multiplication by elements in R and I is in R

[dense pumice]

I^2 is a linear combination of elements of the form i*j for i,j in I, so their product, not sum

[chilly radish]

Right my bad

It should be finite sums of.oroducts of elements of I

[dense pumice]

right yea

[rotund aurora]

If $G$ is a finite group and $H$ is a proper subgroup of $G$ such that the cosets $g_iH$ form a partition of $G$, is the map that sends $g_i\to g_j$, $g_j\to g_i$, $g_i^{-1}\to g_j^{-1}$, and $g_j^{-1}\to g_i^{-1}$, where $g_i\neq g_j$ and $g_i,g_j\neq 1$, and leaves the other elements fixed an automorphism?

[cloud walrus]

Croqueta

[rotund aurora]

and you define the map to be an homomorphism (may not be well defined, but that's kinda what I'm asking)

[chilly radish]

Cosets always form a partition, in which case you need to choose representatives, and depending on your group and the representatives you choose this may not be well-defined

Say if g_i, g_j have coprime orders

This cannot be a homomorphism because the order of the image must divide the order of the original element

Simply because if g is an element of order n, f a homomorphism them
f(g)^n=f(g^n)=f(1)=1
So that
o(f(g))|n

[rotund aurora]

it must actually be the same if it is injective, no?

yes

Right, nice point

[chilly radish]

If it's injective then yes, because you can go the other way too, since if f(g)^k=1 then g^k=1

[rotund aurora]

Under what conditions is this a well defined automorphism? (if possible). What if g_i and g_j had the same order or if all elements in G had the same order?

[hidden haven]

Count how many from Z_m to Z_p, from Z_m to Z_q, from Z_n to Z_p and Z_n to Z_q, and multiply all 4 of these numbers

This is because maps out of a product of 2 abelian groups are in bijection with pairs of maps out of each group and maps into a product are in bijection with pairs of maps into each group

[lethal cipher]

Question: is this question trying to tell me any field homomorphism is 1 to 1?

[hidden haven]

Yes

[lethal cipher]

Hmm, okay. Then how might I go about that?

[hidden haven]

phi is injective iff ker phi is 0

What are the possible values of ker phi

[lethal cipher]

Umm, the only one that actually sticks out is 0. I'm not sure what possible values you are looking for.

[hidden haven]

Think of properties of kernels in general

Lol ok idk how to phrase this without it seeming like a trick question

But ||kernels are always ideals||

[chilly ocean]

Ideally you want to think about some subsets

[lethal cipher]

Oh and don't fields only have the two trivial ideals?

[hidden haven]

Yes 😌

[lethal dune]

🙈

[lavish nexus]

lol

[lethal cipher]

Ah,that's because if you take any x in I, then x x^-1=1 is in I, which generates the full field

[lavish nexus]

Anyways kernels in field maps are 0 or everything

[lethal cipher]

Ah, and the kernel is an ideal, because if you take any x in I and r if F, then p(xr)=p(x)p(r)=0 since it's a homomorphism

[hidden haven]

Ideals are defined so that they are exactly the kernels of ring homomorphisms 😌

[lethal cipher]

funny how that works

[lethal cipher]

Okay, final question, why can't phi just be the 0 map then?

[chilly ocean]

1 has to map to 1

[lethal cipher]

oh duh

[pastel cliff]

what is this

ive probably fallen for a trap

but i at least understand the induction sticker

[hidden haven]

It's called the first isomorphism theorem

You'll probably see it in your group theory course

[pastel cliff]

it's an abstract algebra class

[hidden haven]

You'll probably see it in your abstract algebra class if your prof actually gets to abstract algebra

The theorem is true for sets, groups, rings, modules and many other algebraic structures (all universal algebras)

It's a theorem about quotients

And a generalisation of it is the definition of a quotient in more general contexts

[pastel cliff]

uhhh

quotient groups are week 8 if that has anything to do with it

oh wow yeah

[hidden haven]

Yes that's it

[pastel cliff]

first iso theorem is beginning of week 9

[coral shale]

everything to do with it

[pastel cliff]

then week 13 is first iso theorem for rings??

[hidden haven]

Nice

[coral shale]

@hidden haven curious, whats your level of study/work/whtver rn?

[hidden haven]

MSc first year

[coral shale]

eh?

[pastel cliff]

oh it's cuz we're not gonna talk about rings until then...

[hidden haven]

In math ofc

[pastel cliff]

[coral shale]

uhhhh whassat?

is that like 4th year of uni for people

[hidden haven]

Masters in Science

3 year BSc and now 2 year MSc

Aka post grad

[coral shale]

right ok (I should know this)

[hidden haven]

Lol

[pastel cliff]

well ok

[coral shale]

damn wth, you know so much more than me

[prisma shuttle]

oh i see oops idk wut i was thinking

[hidden haven]

[pastel cliff]

what would a normal abstract algebra class be doing rn

im gonna have some time this weekend to self study

[hidden haven]

First isomorphism theorem for groups

Normal classes would do general algebraic structures after they've handled specific cases

[pastel cliff]

europe much

[coral shale]

Really get to know homomorphism and normal subgroup stuff I would say?

which then leads to quotient groups

[hidden haven]

India

[coral shale]

if you can

[pastel cliff]

oh right math server

shouldve guessed

[hidden haven]

He hasn't even been taught what a group is yet I think

[coral shale]

lol wut

[pastel cliff]

i know what a group is tho

[hidden haven]

Oh nice I see

[coral shale]

Theres a long road ahead then ig

[hidden haven]

It's a weird course

[coral shale]

how do you internalise stuff you learn? Lots of problems sheets?

[pastel cliff]

reteaching it

[coarse storm]

Examples and counter-examples.

[pastel cliff]

"lecturing" to myself too

[hidden haven]

ye tbh same that's the best thing to do

[coral shale]

that is a good way also but i dont think i have the chance to do that

[pastel cliff]

i like to just go to classrooms and pretend im giving really chaotic lectures

[coral shale]

I feel like I'm learning slowly =.....=

[hidden haven]

Learn cat theory, big acceleration guaranteed

😌

[pastel cliff]

what's the plato quote

[hidden haven]

I had not read a single book outside of my course work till 1 year ago

Then I read cat theory

And I can learn so much faster

It's really good

Every day I come here

Just to talk about how good cat theory is

[coral shale]

I don't have a chance to take it any more... what's a good book? 🤔

[hidden haven]

Mac lane and Riehl are both standard

Riehl is newer and more examples and easier to read

but has less content

[coral shale]

I'll probably go for that then. ty

[pastel cliff]

youre saying learning cat theory will boost other math learning?

[hidden haven]

Yes but you shouldn't learn it right now

[pastel cliff]

i wouldve listened to my cat more if that were the case

interesting tho

[coral shale]

You'll drown if you do cat before learning groups 😂

[pastel cliff]

im just gonna keep this in mind for now yeah

[coral shale]

yah sure. theres stuff before this

but yh

[maiden ocean]

learning category theory before doing any AT or anything like that is so broke

[pastel cliff]

and im gonna bother yall this weekend i will make something out of this fucking class

one way or another

[hidden haven]

Nice, I'll make sure to stay offline for the next 3 days 😌

[pastel cliff]

.

[hidden haven]

lmao

[pastel cliff]

im fighting for my life as it is

[rotund aurora]

This may be more of an english question than a math one, I don't know. Why are algebraic structures with some generating set called "free" (example: "free groups")?

[prisma shuttle]

i think its because you can basically generate "anyting" that is made up of those terms, so its the "freest" it could be

idk this is how i think about it

[rotund aurora]

makes sense

I was thinking maybe that it made reference to the generating set itself, because when you construct the, say free group, at first you are just given a set with no algebraic context, and maybe you would call it "free"

Wolfram says this

But I don't know what it exactly means tbh

[chilly ocean]

Have you seen group presentations before?

[rotund aurora]

I know what they are, but no

Free groups give the simplest presentations or something

[prisma shuttle]

its like a generator set

like generators in NT

generators are just the generator set of $\mathbb{Z}/p\mathbb{Z}$

[cloud walrus]

JustKeepRunning

[chilly ocean]

Yeah
Like Z_10 for example is not free because it has the presentation <x, x^10=1>
But if we remove that restriction of x^10 being one and have all powers of x being unique elements, then you get the free group on a single generator

[rotund aurora]

ah I see

so "free" in the sense that no restrictions like x^10=1 are imposed?

[chilly ocean]

Yeah you get it exactly

Free groups can also be characterized as the set of words you can make out of an alphabet of symbols, with the group axioms applied to them

[rotund aurora]

yes, but in the case of groups you need to define an equivalence relation between the words, no?

[chilly ocean]

Only as it concerns powers and inverses if we're talking about free groups

[rotund aurora]

In the case of monoids is way simpler I suppose

[chilly ocean]

Like how aaabbcaa is the same as a^3 b^2 c a^2

Less restrictive yeah

[rotund aurora]

But are the words tuples?

or nah

it seems a little bit weird to me having some random objects like a,b,c and writing abc. Idk

I guess is fine

Thanks to both

[chilly ocean]

can we use lagranges theorem on subgroups of subgroups?

[delicate orchid]

sure why not

[chilly ocean]

ye

Can I use that idea here?

im not sure what the numerator of the lagrange would be in this case

[delicate orchid]

the order of any subgroup in the intersection must divide both the order of H and the order of K as it is a subgroup of both by definition of intersection

and thus must divide the gcd

[chilly ocean]

yeah

okay so say if |H| = 5 and |K| = 3

[delicate orchid]

mm hmm

[chilly ocean]

so G doesnt even matter in this case right?

im assuming

[delicate orchid]

I think G is needed so the intersection is another subgroup, rather than just a set

[chilly ocean]

so the only thing that can divide the two

is 1

hence

|H && K| = 1

?

[delicate orchid]

so true...

[long obsidian]

I wanna show the free product of two nontrivial groups A and B has trivial center.

I was thinking if I just took w\in A*B that is reduced it's first letter should be in either A or B (without loss of generality say in B). So w=b_ r w_1 ... w_k for some reduced word b_r from B (assume w is reduced).

Then if I take a\in A nontrivial
Reduce(aw)=ab_r w_1...w_k since a and b_r does not reduce as they are from different groups.
Likewise
reduce(wa)=b_r ... That is it's first letter should be b_r since non of the letters from a could reduce letters from b_r as they are from different groups. Then as the first letters of both are distinct the two strings are not the same.

Hence any element of the free group is not in the center since it would have to commute with every word of A*B.

Am I missing anything?

[delicate orchid]

I feel like your explanation for reduce(wa) could be trimmed down a bit but otherwise that seems right

and the identity is still in the centre, so I wouldn't say "any element is not in the centre"

just that the centre is trivial

[long obsidian]

Thank you so much :)

[delicate orchid]

actually no on re-reading it your explanation for reduce(wa) is perfect

[chilly ocean]

What you think

this valid

[delicate orchid]

drop the "the" in front of |H \cap K| it's ... clunky

otherwise yeah that's fine

are you allowed to assume that the intersection of two subgroups is another subgroup btw?

just realised you might not have proven that yet

[chilly ocean]

umm

i mean

its an intersection

so its in both

how would i go about proving its a subgroup then?

[delicate orchid]

lagrange's theorem only applies to subgroups, H \cap K is only guaranteed to be a set

[delicate orchid]

prove it's closed has all da inverses, and the identity

[chilly ocean]

how would i do that lol there is no operator

[delicate orchid]

it's literally in a group

they are subgroups, there is an operator

[chilly ocean]

so to show closure exists in H and K

how would that go about

[dense pumice]

if a and b are in H and K, is ab in H? is ab in K?

[chilly ocean]

yeah it is

[dense pumice]

bam! ur done

for closure at least

[delicate orchid]

yeah joyes is correct

[chilly ocean]

i thought we had to show closure exists in H and K

not the original subgroups

[delicate orchid]

if and a and b are in both, then their products must be in both

because both are closed

[dense pumice]

so its in the intersection

[delicate orchid]

very similar for inverses

if a is in both H and K, then we know that a^-1 is in H and a^-1 is in K by definition of a subgroup

thus a^-1 is in the intersection

it follows very nicely from the definition

[dense pumice]

its a common thing thats done as u learn more algebraic structures

[delicate orchid]

yeah intersection is nice at preserving properties like this

unlike unions 🤢

[chilly ocean]

for closure ab has to be in H because H is a group right?

[dense pumice]

ye

[lethal cipher]

Hmm, I could use some help on this one.

[dense pumice]

probably show that any square must be > 0, then 1 is > 0 so -1 isnt

a^2 = (-a)^2 so any nonzero square is > 0

[chilly ocean]

damn writing all the the theorems and corollarys on a piece of paper before doing the hw has been so helpful \

its like a toolbox lol

[delicate orchid]

black box the theorems

[chilly ocean]

lol

[pastel cliff]

seems 2 work

[delicate orchid]

until you need to use part of the proof that appears half way in a different proof

[pastel cliff]

fake liberal media

[dreamy jewel]

If A is a finitely generated ideal, and suppose S is a possibly infinite set that generated A, can you always find a finite subset of S that generates A?

[hidden haven]

Each of the finitely many generators is a linear combination of finitely many elements of S

[dreamy jewel]

Why is guaranteed that the linear combination is finite?

Did I just make a linear algebra mistake

[hidden haven]

There are no infinite linear combinations in algebra lol

You need some metric for that to work

Or topology

[dreamy jewel]

Sorry I mean why is a generator necessarily a linear combination of finitely many elements of S

[hidden haven]

Because the ideal generated by S is exactly the set of linear combinations of S

[next obsidian]

Cuz they’re in I

And S generated I

[dreamy jewel]

…………………………..

[dense pumice]

A is finitely generated, it's by definition

[dreamy jewel]

Thank you

I am actually so dumb

[next obsidian]

Hi “actually so dumb”

My names Chmonkey

[hidden haven]

[dreamy jewel]

^_^

[chilly ocean]

celeste is a trash game!!

[pastel cliff]

@delicate orchid this post right here

https://tenor.com/view/programming-crazy-hard-typing-mad-gif-7866344

[delicate orchid]

baby rage baby rage cope seethe harder analysispilled tradbased copepilled seether, reddit chungus unwholesome downvoted reported anhilated police, fire department, coast guard, ambulance services, US military, Sudan Millitary, Croatian government have been contacted and your IP is being traced right now.

mad cause bad

add that in somewhere

[hidden haven]

[delicate orchid]

there's something about the field of algebra over the field of analysis that just attracts the most chaotic behaviour

it's remarkable

[long obsidian]

Is there a simple relationship between the fundamental group of A in X and the fundamental group of A^c?

[hidden haven]

That seems not very plausible

[long obsidian]

Hmm okay :(

[pastel cliff]

i forgot i have to grade homeworks by tonight

[hidden haven]

Wait by A in X do you mean relative homotopy group

[pastel cliff]

#abstract-chill

[hidden haven]

Me till high school

Now I'm

[pastel cliff]

3rd year of undergrad

[pastel cliff]

please i cant take it anymore im at my limit

[delicate orchid]

simply reverse the arrows and approach your colimit

[hidden haven]

Yooo

[delicate orchid]

moldi no

don't do it

[hidden haven]

I'm gonna send it

And no mods can stop me

[delicate orchid]

[pastel cliff]

@/mniip

[hidden haven]

Last time I jokingly said honourables can't be banned in #cats

Mniip showed up instantly

[pastel cliff]

dear god what has he become

[hidden haven]

And I had never seen him there before that

[delicate orchid]

moldi I have a... question

[hidden haven]

[delicate orchid]

what should I do in this situation

[hidden haven]

Did you make this

[delicate orchid]

yes

obviously

[chilly radish]

You go to #chill in this situation

[delicate orchid]

I'm in chill

[chilly radish]

Be more chill

[hidden haven]

I would be perfectly fine with committing that affront to nature

[pastel cliff]

permastudy

[delicate orchid]

a hot take

[pastel cliff]

doing my best to keep it on till the semester ends but my shitposting instinct gets the better of me

[delicate orchid]

I will post that in chill though

[delicate orchid]

I believe in the dream

[pastel cliff]

i just want 4 out of 5 A's

fuck statistics

[hidden haven]

✓

[delicate orchid]

stats

[pastel cliff]

some of the most mindnumbing lectures ive ever had to sit through, and i had to go through fucking rigorous algorithms class

[delicate orchid]

is it baby stats

or like

jan Niku stats

[pastel cliff]

for now baby stats

you best believe im gonna bother jan later tho

[delicate orchid]

poor jan

he's like the only person in this server who knows advanced stats stuff (that I've met)

[white jackal]

Prove that the group of rational numbers Q under addition has no finite
index subgroup other than Q itself. Can you guys help me with this

So its asking me to show that there are no finite cosets that exist right?

[dense pumice]

not that there arent finite cosets, its that there isnt a group which has a finite number of cosets

[white jackal]

So all the groups have infinite cosets?

except Q itself?

[dense pumice]

have an infinite number of cosets if thats what you mean yes

[white jackal]

To reiterate the question its asking me show Q under addition that the subgroups have infinite number of cosets except Q itself?

[dense pumice]

yes

[white jackal]

So lets say H is a subgroup of G

then how would i show H has infinite cosets?

[dense pumice]

well thats the problem now isnt it

hmmm perhaps you can pass to the quotient since Q is abelian so every subgroup is normal?

[white jackal]

Since H is a normal subgroup of G then G/H?

[dense pumice]

G/H has finite order

(idk the answer, Im just writing down thoughts)

[white jackal]

G/H means all the elements in G but not in H right?

[dense pumice]

no

thats G\H

have u learned quotient groups?

[white jackal]

we just covered it so im kinda new to it

[dense pumice]

yea so cosets form a group

if youve seen modular arithmetic, thats the main example to keep in mind

[white jackal]

so all the cosets of H form the quotient group

[dense pumice]

so lets say |G/H| = n, then any element of Q can be written as an element of G divided by n (consider the image of the number in G/H)

[dense pumice]

yes

ok so I think you can use this to show every element of Q must be an element of G: consider a in Q, a/n is in Q which implies that n*(a/n) is in G which means a is in G

yea thats a very weird proof if you arent yet used to the ideas

[white jackal]

So that is the proof?

[dense pumice]

yea

[dense pumice]

I mean if its hw, dont write that esp cuz this skips like 3 steps

[white jackal]

how does this show that Q under addition has infinite index subgroup other than Q itself?

[dense pumice]

it doesnt

you dont need to show that

[white jackal]

ohhh

so [q:H] = n means aQ is in H

[dense pumice]

oh wait I messed up lol

[Q:G] = n so nq is in G for any q in Q

[dense pumice]

and then do this

[white jackal]

so G is the group made up of elements of Q and H is a subgroup of G right?

[dense pumice]

there is no H, ignore that

G is the subgroup of finite index

[white jackal]

So G is the subgroup of Q

[dense pumice]

yes

[white jackal]

okay lemme try to put this all together

Let G be the subgroup of Q then [Q:G] = n then Assume a is in Q, a/n is an element of Q such that n*(a/n) is in G this implying a is in G

is this correct?

[dense pumice]

yes

[white jackal]

okay so this is the correct answer

gotcha

but im kinda confused on the question

Prove that the group of rational numbers Qunder addition has no finite
index subgroup other than Qitself.

so this answer shows that G is the only group that is finite

[chilly ocean]

its time to read the textbook bro

[dense pumice]

we started with the assumption G was a subgroup of finite index and showed G=Q, that proves the question

[prisma shuttle]

why is the red part true

like why do we have to show the jacobson radical is equal to R

i thought we just had to show it was equal to a nilpotent ideal.

[thorn delta]

If J = R then in particular, 1 is in J so I^n = 0 @prisma shuttle

[prisma shuttle]

oh i see

[next obsidian]

J is not the Jacobson radical

[prisma shuttle]

wait but is it necessary

[prisma shuttle]

huh

oh really

[next obsidian]

Yeah, I is

They defined J right before that

[prisma shuttle]

oh shoot

[next obsidian]

You want to show it’s equal to R

[prisma shuttle]

wait then what is J

[next obsidian]

Bro

[thorn delta]

oh i didn't see you misunderstood that

[next obsidian]

It’s defined in the line

Right before that

Hahaha

[prisma shuttle]

yea i know

but like wut is it

[next obsidian]

The set of x such that xI^n = 0

If it’s equal to R

[prisma shuttle]

does it have like a special name

[next obsidian]

Then you can plug in x = 1

No

Its special name is J

[prisma shuttle]

oh shoot bruh i got confused

[thorn delta]

its a special case of something called an "ideal quotient" but i don't think the specifics are super important here

[prisma shuttle]

ok thx so much

wait also one more question

do u get why we have that $J'=J+Rx$ for some $x\in R$ as they claim

[cloud walrus]

JustKeepRunning

[next obsidian]

If it wasn’t

Then you could find something even more minimal

If you required two generators besides J you could just omit one

[prisma shuttle]

oh i see

thats nice

[frank fiber]

is it true that every abelian p-group of order p is cyclic?

[thorn delta]

every group of order a prime p is cyclic

[frank fiber]

thanks

[white jackal]

for part b i said <a> intersect <b> will have the size of 2 iff p=q is this correct?

since p and q are the same then a^p and b^q will generate the same group and both have trivial group right?

[proud bear]

i don't think so. <a> intersect <b> is a subgroup of <a> (and of <b> too), so its size has to divide the size of <a>, and |<a>| is a prime so...

[white jackal]

ohh okay

[lavish nexus]

is Gal(E(a)/E(a^p)) isomorphic to (Z/pZ)*

I can for now assume E=Q

[oblique river]

what is a?

a specific element of E? or an indeterminate?

[lavish nexus]

take a to be the primitive (p^n)th root

for some n

[oblique river]

if n > 1 then that extension has order p, not p-1

if n = 1 then that's just Q(zeta_p) / Q which does have order p-1

but for n > 1 and you have something like Q(zeta_(p^2)) / Q(zeta_p) , that extension has degree p

and has galois group Z/pZ

(i use zeta_n to denote a primitive nth root of unity)

[white jackal]

since |<a>| is prime and by lagrange |<a>| / |<a> intersect <b> then that means |<a>| can be divided by itself or 1 right?

[lavish nexus]

I thought [Q(zeta_p):Q] = p

and the min poly is x^p-1

it's a Galois extension so shouldn't [Q(zeta_p):Q]=|Gal(Q(zeta_p)/Q)|?

but then I did prove that it is isomorphic to (Z/pZ)*
which has order p-1

[oblique river]

that's not true

x^p - 1 is reducible

[lavish nexus]

oh right nvm

🤦‍♂️

[oblique river]

this is correct

for n > 1, you do get size p though

[lavish nexus]

yes the min poly would be x^p-zeta^p

[oblique river]

that's right

or in general, it would be x^p - zeta_(p^k)

[proud bear]

yeah the possible sizes of <a> intersect <b> are 1 and p

[white jackal]

so its 2?

p and {1}

[proud bear]

yeah

i thought when you said 2, you meant that the size of the intersection is 2, not that there were 2 possible sizes

[lavish nexus]

ty
I was proving Gal(Q(/zeta_p^n)/Q) is cyclic for p odd prime and n<=3

Now that I'm done proving it I think it might be true for all n

since Gal(Q(/zeta_p^n)/Q((/zeta_p^n)^(p^n-1))) should be cyclic generated by \phi(/zeta)=/zeta^2 if (2,p)=1

[oblique river]

yes it is true for all n

[lavish nexus]

oh wait why didn't I just let this \phi generates the whole Gal(Q(/zeta)/Q)

why did I go through this tortuous argument of first modding by (Z/pZ)* lmao

[winter hound]

For any element a in G, where G is a group, and H is a subgroup of G, we define aHa^(-1) = {axa^-1 : x in H}. Prove that, for any a in G, aHa^(-1) is a subgroup of G, and that aHa^(-1) is an isomorphism with H.
I need some help

if anyone could

super lost

[lavish nexus]

prove ah_1a^(-1)(ah_2a^(-1))^(-1) is in H

notice that the inverse of axa^(-1) is just ax^(-1)a^(-1)

[winter hound]

im a bit confused

I'm at the part where i am proving its closed under operation

[delicate orchid]

take two elements, axa^-1 and aya^-1, and multiply them
see what you get

[winter hound]

i started off saying "let x,y be in aHa^-1. then, x,y is in H. So, axa^-1 and aya^-1 is in aHa^-1"

[delicate orchid]

yes that's a good start

a bit redundant but that's fine

[winter hound]

don't i introduce xy in H?

[delicate orchid]

you don't care about H, you're trying to prove aHa^{-1} is a subgroup

you already know H is a subgroup

[winter hound]

oh okay

[delicate orchid]

I take it back, I don't think that statement is true at all

[lavish nexus]

you know every element in aHa^(-1) is of the form aha^(-1) for some h in H

so just take axa^(-1) and aya^(-1)

then check their product and inverses are in aHa^(-1)

[winter hound]

okay, lemme try that

[lavish nexus]

incidentally this turned out to be the right thing to do because
there exists p for which 2 is not a primitive root

I am now less upset

[winter hound]

for the product, i now have axya^-1

[lavish nexus]

remember H is a subgroup

[winter hound]

mhm

[lavish nexus]

where is xy in

[delicate orchid]

so what can you say about xy

[winter hound]

xy is in H,and also in aHa^-1?

[lavish nexus]

yes

axya^(-1) is in aHa^(-1)

[winter hound]

so it's closed under operation

now we try closure under inverses

since H is a subgroup of G, x^-1 is in H as well

[delicate orchid]

hint: ||the inverse of (axa^-1)^-1 is ax^-1a^-1, think about why||

[winter hound]

isn't it from what i stated?

[delicate orchid]

we do not give a damn about H

stop looking at elements of H, look at elements of aHa^-1

while it is true x^-1 is in H, this comes at the end of your proof

[winter hound]

oh okay

[delicate orchid]

it's more like (axa^-1)^-1 = ax^-1a^-1, therefore because x^-1 is in H as H is a subgroup, (axa^-1)^-1 is in aHa^-1

you can't just write the middle part

[winter hound]

oh, i was starting off from the middle

[delicate orchid]

not the best place to start

[winter hound]

I totally skipped over the first step

true

lol

[delicate orchid]

and showing the identity is in aHa^-1 shouldn't be too hard

[winter hound]

i was told that if we show that it's closed under inverses, we already know there's an identity

[delicate orchid]

that is true actually, very good point

still though

aea^-1 = aa^-1 = e

[winter hound]

alrighty, i now have to show its an isomorphism with H

1. injective
2. surjective
3. morphic part

[delicate orchid]

morphic part I like it

[winter hound]

oops

yeah, that's what my professor likes to call it lol

can we turn aHa^-1 into a function?

[prisma shuttle]

it is lol

[winter hound]

like: f(x) = axa^-1

[prisma shuttle]

its called conjugation

[delicate orchid]

sure, looks like a function to me

[winter hound]

okay

[delicate orchid]

give it a try

[winter hound]

sorry, i mainly meant just for the f(x) part

[delicate orchid]

btw, instead of proving surjective/injective you can just find an inverse function

which is much easier in this case

[winter hound]

mhm, but i wanna show more work since he wants it in like a full proof format

but it'll be like a page long, which isn't much

[delicate orchid]

then write "a function is bijective if and only if it has an inverse"

[winter hound]

for injective, i got b=c (using these elements to make life easy)

cuz cancellation law

[delicate orchid]

aba^{-1} = aca^{-1} <=> b = c yup

[winter hound]

mhm

I'm assuming c is in H. and considering b = a^-1ca?

so i can get f(b)=c

[delicate orchid]

see this is why I think just finding an inverse function is so much easier

but, consider f(H)

by definition of f, f(H) = aHa^-1, so it's surjective

[winter hound]

it would be but we haven't really went through much of that

it's also why I'm taking the long route

[delicate orchid]

fair enough - it's good practice

[winter hound]

mhm

[delicate orchid]

do you understand the explanation I gave btw?

[winter hound]

yeah

[delicate orchid]

nice, now it's just the morphism part

[winter hound]

makes a lot of sense and much simpler

okay done!

[delicate orchid]

nice

f^-1(x) = a^-1xa btw
much cleaner imo

whatever works works though