groups-rings-fields

[median pawn]

How then would we get a unique group homomorphism from F to G extending phi?

[rocky cloak]

1 is not the identity in Z

[median pawn]

Yeah shit I was thinking rings for some reason, crap

[rocky cloak]

You don't, no set containing the identity is a free generating set.

[median pawn]

Correct, thanks

Why is {2,3} not a free generating set? phi(3) - phi(2) gives us phi(1), and once we have phi(1), we have phi(n)

[rocky cloak]

Let G = Z. And consider the function f: {2, 3} -> G such that f(2)=0 and f(3)=1

How would you extend this to a group homomorphism from Z?

[median pawn]

If there is an extension, call it g, it must satisfy g(1) = g(3) - g(2) = f(3) - f(2) = 1; but this forces g(2) = 2 and g(3) = 3 which is a contradiction

Fair

[delicate orchid]

Apologies if this comes in an hour late (spotty internet) but it’s because there’s a non-trivial relator between 2 and 3, mainly that 2*3-3*2 = 0, hence the group they generate cannot be free. Alternatively we could use a dimension argument, if {2,3} was a free generating set (aka a basis) then Z would have dimension 2 as a Z-module which is obviously nonsense

[topaz solar]

the lack of relations between the elements is a key point

as mentioned, they're like a basis

[median pawn]

ahh, yeah, makes sense!

[topaz solar]

and if there were a relation between them, then sending them to incompatible elements fails the free group property

[median pawn]

just {2} alone doesn't work either because if G = Z and phi(2) = 3 (or any odd number) then we've no choice for phi(1)

[topaz solar]

so there'll be a set where it fails

[median pawn]

the basis intuition doesn't work here tho

[topaz solar]

2 don't generate Z

[delicate orchid]

It does ^

A basis has to be a generating set

In fact, a module is free iff it has a basis

I know we’re doing groups here but shush

[topaz solar]

not as a Z module it don't, you wont get 1 as a multiple of 2

[median pawn]

yeah!

[topaz solar]

(2Z isomorphic but we don't care)

S will generate F if this is true, hence freely generated

consider lifting the map S->F

[median pawn]

thanks!

[topaz solar]

this makes a unique homomorphism F->F

where S->F->F = S->F

what is that homomorphism

[delicate orchid]

Yur we need them to be equal as sets. Just like how (1,0) isn’t a basis for <(0,1)> despite them being isomorphic

Can’t wait for this message to be sent at 6:43pm

[topaz solar]

quick exercise: suppose we had two elements in S such that some expression made out of them is identity in F, show it can't be freely generated by this S

[median pawn]

the identity

[topaz solar]

exactly

this shows you've generated it by S, but you can fill in the details further if you wish

[delicate orchid]

I’m going to start ranting about the connections to presentations now

[topaz solar]

(this should be a quick argument, could do it pretty immediately ||with like <S>)||

[topaz solar]

quotient time

[delicate orchid]

Given a generating set S that gives you a free group F(S), since it’s free there are no realtors between elements of S so we can write F(S) = <S| >.
However, given a map from S into a group G (say G has the presentation <S | R>) the induced map is the canonical map given by quotienting F(S) by exactly the normal closure of the subgroup generated by the realtors R in G, which I find to be a really good intuition for not just this map but quotients in general

[median pawn]

could you explain this some more?

[topaz solar]

S -> <S>

[delicate orchid]

So the universal property of the free group is basically just the universal property of the quotient but a little more general

[topaz solar]

this has a unique homomorphism

notice this is kinda also the same as S -> F

gg

See it?

theres a couple similar routes you can take but this is basically t he idea

[median pawn]

uhhh no wait. so suppose S freely generates F. the inclusion S -> <S> is extended uniquely to a group h.m. F -> <S>

[topaz solar]

there's exactly 1 map F -> <S> extending it

[median pawn]

what do you do with this extension? we want <S> = F, so an isomorphism

[topaz solar]

what do you know about <S>

a priori, what do you already know is true

[median pawn]

smallest group containing S in F

[topaz solar]

so <S> < F right?

[median pawn]

yes certainly

[topaz solar]

rewriting this, <S> -> F

F -> <S> -> F

see it now?

[median pawn]

yes!

[topaz solar]

So that composes to identity, and the latter is inclusion, so it was the identity all the way through

[median pawn]

the inclusion <S> -> F is the inverse of F -> <S> obtained by the univ property

[topaz solar]

do you see how to do this one?

[median pawn]

let me give it a thought

[topaz solar]

how would you immediately suppose to show this

first intuition

Using wew ladz's statement is kinda cheating in the sense that it's related to what we're proving in the first place, so I'd ask that you don't reference it

[hidden haven]

Never ignore wew's advice

You can get a lot farther by doing the opposite of what he says

[delicate orchid]

I never care about the details of the problem I just chat about cool shit

[topaz solar]

don't overcomplicate it, just assume some expression formed by some a, b will spit out the identity in F, and show that this must violate the universal property

[hidden haven]

Based

[hidden haven]

I think wew meant to reply to your message here @median pawn

The basis intuition works; it's the same universal property in a different category.

[delicate orchid]

I meant “the intuition does still work, {2} cannot be a basis because it doesn’t generate the whole space”

[hidden haven]

[topaz solar]

I will elaborate that this is essentially saying we want independence, though not exactly linear since ya know

not a linear space

[median pawn]

y'all i went for a coffee break so i don't think stupid shit

i'm back

[topaz solar]

the exact same idea for why a linearly dependent set ain't a basis works

[delicate orchid]

I wonder why

[topaz solar]

well presumably universal properties are a bit newer in material than basis-es

[hidden haven]

Full subcategory of Ab spanned by the Z/p modules

That's just Z/p-Mod lmao why did I say it like that

[topaz solar]

average category theorist

p torsion or smth

[hidden haven]

No they're not because before the universal property was discovered (not invented) people didn't know what a basis was

[median pawn]

i need a clarification first, why is phi (the unique map from F to <S> extending the inclusion S to <S>) the identity map?

[topaz solar]

this composes to identity yes

<S> -> F is the inclusion map

so i(phi(s)) = s, but i(phi(s)) = phi(s)

[delicate orchid]

Do we have to recite the “universal property characterises objects up to iso” meme or is this different

[topaz solar]

this is different

[median pawn]

ok why?

[topaz solar]

it extends S->F

there's only 1

and that 1 is 1

[median pawn]

no but we're extending (S -> <S>) to (F -> S) right

[topaz solar]

? no

F-> <S>

there's only one

and F->F there's only one

but we can compose F-><S> with <S>->F

that extends S -> <S>, and we compose with <S> -> F

[delicate orchid]

Get quiver up and draw the triangle

[topaz solar]

you've extended S -> F

[hidden haven]

I can't follow any of this discussion someone needs to draw some nachos

[median pawn]

nachos lmao

[hidden haven]

I'm hungy

[cloud walrus]

The great Sharp

[median pawn]

awesome so the map F to F is the identity

[topaz solar]

drawing like this

[delicate orchid]

See how easy it is with the funny Doritos

[topaz solar]

notice i:S -> <S> composed with i:<S>->F is just i:S->F

[median pawn]

AH I SEE all of it now

[hidden haven]

God

[topaz solar]

triangles unnecessary, just had to see inclusion is inclusion

[median pawn]

we use the univ property with F once, and then with <S>

[topaz solar]

local mathematician learns subset is transitive live

[delicate orchid]

Bullshit. I don’t believe it for a second

[topaz solar]

now do this

this is similarly "oh, duh" the second you open your eyes

[hidden haven]

Bro let the guy celebrate a W

[topaz solar]

bro is gonna open his third eye rq

[delicate orchid]

Oh right we’re doing this haha I like this

[coral shale]

meow

[hidden haven]

And then the furries showed up

[coral shale]

nacho neko desu

[topaz solar]

in fact, you can immediately generalize it by saying having any list a_1, ..., a_n having an expression which goes to 0 (nontrivially) cannot freely generate F

once you get 2 elements, getting n is immediate

[delicate orchid]

Hint, there’s a nice choice of G here

[topaz solar]

the dumb choice works

[delicate orchid]

In fact there’s two nice choices of G

[median pawn]

do we not show the surjectivity of the unique map from F to <S> at any point?

[topaz solar]

composes to identity -> bijection

this isn't exactly a sound derivation in general but uhh

[hidden haven]

Have you checked that they compose to identity both ways

[topaz solar]

it's iso here

[delicate orchid]

Don’t worry! It’s ok ✅

[hidden haven]

:reassured:

[topaz solar]

notably, <S> -> F is injection by inclusion map

and composing that with F-><S> is identity

[median pawn]

yep it follows from there ofc just wondering if checked it a priori

all good all good

[topaz solar]

any handy details about g such that g o f is identity?

[median pawn]

yes! g is surjective, f is injective

[topaz solar]

injective + surjective = bijective

gg

[hidden haven]

lol I was asking the sharp, idk what the problem is I am just here to shitpost

[topaz solar]

I didn't check because it's clearly surjective

presumably it's later on in the text, but showing F is unique (given S) up to isomorphism, and that such an F exists for every S, might be fun for ya

[median pawn]

i have something for this

[cloud walrus]

hausdorff

[median pawn]

Looks good?

[topaz solar]

I'd just say send a to (not zero) and b to zero, then p phi(a) = 0

[topaz solar]

this doesn't work though because uhh what if F not abelian

in fact, the only abelian free group is Z

[median pawn]

= 1?

[topaz solar]

(consider unique lifting to non-abelian things)

[median pawn]

yeah i'd have to fix the hypothesis, but that's it, right? (i.e., change from pa + qb = e_F to a more general expression)

[topaz solar]

doesn't matter what it sends to, so long as it doesn't go to identity in the end

1 isn't special here

[formal ermine]

good morning algussy

[topaz solar]

lets say I pressed a hypothesis of, say, aba^-1 b^-1 = 0

[median pawn]

ok minor glitch here

[topaz solar]

that is, that a & b commute

[median pawn]

it's 0 = some shit

not 1

[delicate orchid]

It’s 1

0 is for abelian categories

[topaz solar]

show it does not freely generate a group (in the sense of free groups, not free abelian groups)

[median pawn]

ok let me fix this thing, i got it

[median pawn]

this is confusing lol you're using 0 on the right but multiplicative notation on the left

[topaz solar]

Whoops

a b a^-1 b^-1 = identity

[median pawn]

are u suggesting a map from {a,b} to F?

[median pawn]

phi(a)phi(b)phi(a^-1)phi(b^-1) = phi(e) = e and setting phi(b) = e gives phi(a)phi(a^-1) = e which doesn't help bc it's always true

[topaz solar]

this is exactly asking phi(a), phi(b) to commute

[median pawn]

ah so just assign them to non-commuting elements

[topaz solar]

universal property says it lifts to unique homomorphism

whoops, it fails

what if I were to take some arbitrary expression $a^{a_1}b^{b_1}a^{a_2}...b^{b_n} = e$

[cloud walrus]

The great Sharp

[median pawn]

give me a moment

we gotta make cases though, when the coefficients a_i sum to zero and when they don't

similarly for b_i

[topaz solar]

that's a reasonable idea

if it sums to zero, what does that say about what you can do with it

or can't, perhaps

[teal vessel]

question: for most cases, it's probably best to check closure first when trying to qualify or disqualify a particular subset as a subgroup, yes? i.e. the inherent structure of the group that is a superset will usually satisfy most of the conditions of groupness on any subset, and if the two conditions of closure and inverses are met, then that directly implies the identity condition anyway.

[topaz solar]

Pretty much

[coral spindle]

most in what sense lmao

this just seems too vague to have a precise answer

[hidden haven]

if the two conditions of closure and inverses are met, then that directly implies the identity condition anyway.
The subset could be empty

[coral spindle]

Sure, that in some way is gonna go wrong a lot.

[topaz solar]

Can you show being a subgroup H<G is satisfied if a, b in H implies ab^-1 is in H?

• plus nonempty ofc

[delicate orchid]

How do you show inverses without identity omegaLUL

Oh subgroup

[hidden haven]

Owned

[topaz solar]

Very imprecise question, but I mean

[teal vessel]

it's not a precise type of question, just trying to save myself some mental energy by tackling the discriminants in order.

"is this the vibe" type question

[topaz solar]

This should be an iff because subgroups obviously satisfy the condition

[delicate orchid]

If an implication is trivial there’s no point including it lest you look the fool

[topaz solar]

Ok mochizuki

[delicate orchid]

Got DANG who are these mfs

[mighty kiln]

Why not include them lol

[teal vessel]

I mean, when dealing in definitions it is often important to use biconditionals ime. Not doing so can make for confusion when someone else reads it.

[delicate orchid]

Anyway yes I’d start with closure as it’s usually the most annoying one

[topaz solar]

The easiest is being inhabited but that’s uhhh probably obvious in context

[delicate orchid]

The easiest would be identity if subset isn’t given imo

[teal vessel]

I was just showing that the subset of ((Z x Zn)-(torsion group))U{(0,0)} is not a subgroup and went the dumb way and just checked if it was a group via the 4 properties like a numpty instead of recognizing the operation is already defined as a group-able operation.

[rocky cloak]

You can still define an inverse as a^(-1)ab = b I guess. But then it follows that a^(-1)a is the identity.

[median pawn]

lol yeah i'm stuck

because even if it doesn't sum to zero, there's a possibility that a^{\sum p_i} = 1 already

[topaz solar]

Ok, let’s say they dont sum to 0

[median pawn]

or b^{\sum q_j} = 1

[topaz solar]

What if I sent them to a convenient commutative group

That’s not gonna have any issues with commutativity

So then you can turn it into a variation of your original idea I suppose, but there’s other commutative groups

[median pawn]

let's just use Z

[topaz solar]

Name one that isn’t Z and should be helpful here

[median pawn]

okay why not Z

[topaz solar]

Because I say it’s a good idea

[median pawn]

the real question is why is Z a bad idea

[topaz solar]

We have 2 elements and we want them to be independent in our group

An abelian group

[median pawn]

Z x Z?

[topaz solar]

Yep

Send a to (1,0) and b to (0, 1)

Because our condition isnt a commutativity thing

And choosing Z in particular means we have no coprime things to worry about

We just care that being commutative doesn’t screw us immediately

This make sense?

(This also obviously generalizes for n element version)

[median pawn]

yes it does

[topaz solar]

The real tricky thing is, what if the a_n and the b_n both sum to zero

[median pawn]

i'm just in awe of the idea of choosing Z x Z

[topaz solar]

As said before, it’s just independence ye?

So let’s map them to independent elements (linear) when being commutative isn’t an issue

This may not be the “best” method, but I think it’s effective

[median pawn]

thinking

[topaz solar]

So, I’ll suggest that we can assume WLOG that there are no relations on any other elements of S, or that there are no other elements even

Whatever you happen to think of as usable

[median pawn]

yeah it's enough to assume S = {a,b} bc even otherwise the other elements of S never show up in the proof

the relation we have only contains a and b

[topaz solar]

Ok, so you know you have a relation on a, b but you cannot use a commutative group

But all you need is a group where the relation doesn’t hold for some pair a’, b’

[south patrol]

What is the question

[topaz solar]

If we have a relation on some generators then it doesn’t freely generate a group

(Universal property version)

It’s super obvious if you construct F(S) but I believe that’s later in the text

Or maybe it isn’t and that’ll give buddy here an idea

Well, if any relation kills it, then center is trivial

[median pawn]

correct

[topaz solar]

Since, ya know, commutator

[topaz solar]

The trick is, make it

You’d be hard pressed to have a group where every possible relation fails somewhere, I sure don’t know it off hand, so maybe make one for the relation

Well, no sane one

[median pawn]

ahhh okay

[topaz solar]

Or maybe you’ll big brain it and find such a group

You get the idea though

(And how it applies for n elements of S in an identical way)

@south patrol you see it?

[south patrol]

Idk I'd just use the free group lol

[topaz solar]

Lmfao that’s kinda ya know

What we’re showing in a sense

Did you see the earlier thing where if one element’s exponents summed to nonzero, we just use Z^n?

[median pawn]

yep if any one of the sums is not zero we just send stuff to Z^n

[topaz solar]

I quite like that one for making the independence thing obviously

[topaz solar]

So, they’re all zero, so you’ve gotta do something more specific to the relation

(Assuming all sums are 0 is non-essential, but Z^n is clean so I want it there for exposition)

[median pawn]

do i need to define a group using generators and relations? i'm not 100% comfy doing that rn lol

[south patrol]

ye i've done a similar exercise

[topaz solar]

Well, you could instead, say, build the elements of the group itself

A finite group will suffice

I think

Or at least a very simplistic group, anyhow lol

I’m pretty sure my reason why what I’m think is finite is a scuffed argument oop

[south patrol]

Wait so how are you defining "freely generating"

[median pawn]

universal property

[south patrol]

But you can't use F(S)?

or are we like

Idk I find that odd lol

[median pawn]

@south patrol see def 2.2.4 here

that's all we've got

[south patrol]

Sure

So we have the definition but haven't shown existence for every set, got ya

Use the adjoint functor theorem then /s

[topaz solar]

So what this is, is saying there are no relations in F, which I thought is a fun exercise from this point

*nontrivial relations

[south patrol]

Sure i think it's a good exercise

[topaz solar]

It’s trivialized by ya know

F(S)

But problem is you gotta build that

It’s kinda the uh

Definition of it lol

[south patrol]

depends how you build it ig lol

But i'd be interested in whether you can prove it from the universal property rather than by constructing an explicit model

[topaz solar]

But I brought in the Z^n thing in particular for the independence bit

[south patrol]

but i guess that is what we are doing lol

[topaz solar]

Oh yeah you 100% can (*pending my argument working)

I’ve got an argument but there’s certainly others

let me know if you want a hint in the direction I’m thinking in particular

[charred crescent]

i have a question

maybe im being dumb, but i am working through dummit and foote and encountered an exercise that asks to prove that Z(G) is a subgroup of N_{G}(A) for any subset A of G

[south patrol]

sure

[charred crescent]

Z(G) is the center of G and N_{G}(A) is the normalizer of A in G, A subset of G

[south patrol]

Okay so what definition of normaliser are you using

/ what characterisations of it do you know

[charred crescent]

N_{G}(A) = {g in G | gAg^-1 = A}

[south patrol]

Sure

Now if g is in Z(G), why is gAg^-1 = A

[charred crescent]

it is the collection of elements of G that "make" A a "normal subgroup"

[topaz solar]

gagged

I’m assuming Z(G) are elements which commute with all elements in g?

[charred crescent]

yes

[south patrol]

The centre

[topaz solar]

gag^-1 when g is commutative is…?

[south patrol]

That is what I was asking shook lol ye

though "g is commutative" i assume is a typo

[charred crescent]

honestly i was able to show the 2 conditions of a subgroup i just cant show it's a subset lol

[south patrol]

well show this

[charred crescent]

for some reason it doesn't click why every eleemnt of Z(G) must be contained in N_{G}(A) for any A subset of G

[south patrol]

It should be quite clear that gAg^-1 = A for g in the centre

[charred crescent]

okay hang on

[south patrol]

like if a is in A, what is gag^-1

[charred crescent]

i may have been thinking about this wrong

[topaz solar]

You were

[topaz solar]

You know what I mean, center-y

[charred crescent]

im a bit rusty

[charred crescent]

okay yeah it is clear now g in Z(G) => gag^-1 = a for all a in A => gAg^-1 = A

[south patrol]

Yes

But a like slightly nicer way imo is to rephrase it as like

[ornate vessel]

if H is subgroup of G and for a single g from G and for all h from H it is true that ghg^-1 is in H, does this mean gH=Hg?

it is clear that gH is subset of Hg

but the other way idk

[south patrol]

If H is finite, then you have equality since gH, Hg are the same size and one is contained in the other

[topaz solar]

gH = Hg should look like gHg^-1 = H?

[charred crescent]

okay thank you @south patrol im not quite sure why that one was giving me trouble

[south patrol]

np

[topaz solar]

Do you see why?

[ornate vessel]

yes

[topaz solar]

Now, surely you should be able to work from gHg’ = {ghg’ | h in H}

See a solution?-

[ornate vessel]

not yet 😄

[topaz solar]

Well, what do you know about ghg’

[ornate vessel]

it is in H

[topaz solar]

right, can you have two distinct ones map to the same element in H? @ornate vessel

[teal vessel]

meanwhile, I'm over here trying to prove that for H,K < G, H U K < G iff (H subset of K) OR (K subset of H)
Got the subset => subgroup direction, but my workings out in the subgroup => subset direction have lead me to need to prove that H-(H intersect K)={} or K-(H intersect K)={}, but I feel like I'm going about it slightly wrong.

[ornate vessel]

if gag' = gbg' then a = b

is every element in H of the form ghg'

[topaz solar]

well, clearly each h maps to a different point in H

so clearly gHg’ < H at least

[hidden haven]

Draw a Venn diagram of H and K inside G. You'll find that if neither is a subset of the other, all regions in the most general Venn diagram that you can draw are non empty. Try multiplying elements from different regions and seeing where the products land using the fact that H, K, and their intersection are subgroups

You will find that there are 2 regions such that multiplying an element from one with one from another can't be in the union. You should be able to guess which 2 regions these should be from our assumptions.

[topaz solar]

What if we did like, g’gHg’g = g’Hg?

H = g’Hg we would also expect too, what can you say about this

take h and look at g’hg

[ornate vessel]

isnt this almost the same as gHg' 😄

[topaz solar]

It’s similar yep

Is it possible for g’hg to not be in H

[ornate vessel]

right

[topaz solar]

If you answer no, then that means you’d get your desired equality right?

[ornate vessel]

oh

true

g'hg is always in H

or wait

[topaz solar]

If you use a similar argument to your original with g’ you could get it that way too maybe?

[ornate vessel]

yes but

[topaz solar]

If we had x in G and g’xg in H, then x is in H

[teal vessel]

so supposing there are some elements a,b uniquely in H,K respectively, ab existing in either of them would necessarily imply that both a and b must be in both, right? i.e. put ab in one group, multiply it by b inverse and you necessarily get a, which drags a into the group. So in order to have closure on the union of H and K, you need ab to exist for all a,b in the union, and in order for that to happen, ab has to be a member of H or K originally, which necessarily implies that both a and b must be part of that same group by closure. Therefore our assumptions safely "absorb" all of one group into the other group.

side question: doesn't this setup also directly imply that H<K or K<H (I don't have the leq symbol and I'm lazy)?

[hidden haven]

Yes this proves that if the union is a subgroup, then one must contain the other

[teal vessel]

I mean particularly that one is a subgroup of the other, not merely subset

[hidden haven]

That fact is a lot more basic

[teal vessel]

it seemed so, but sometimes I need a sanity check

[hidden haven]

Sure

[ornate vessel]

from ghg' in H for every h, doesnt it follow that gHg' = H 😄

[hidden haven]

It follows that gHg' is a subset of H

[ornate vessel]

if we map from H to H

h -> ghg'

we know that for every h, ghg'

is also in H

[topaz solar]

This map G->G by conjugation has an inverse

g’ x g

If g’ x g is in H, x is in H

[ornate vessel]

if x is in H not always g'xg in H?

[coral shale]

prove forall g in G, gHg' sub H => forall g in G, gHg' = H

[topaz solar]

well, I’m saying the inverse might be helpful to look at

[ornate vessel]

this will be true if gHg' sub H for every g ?

what if gHg' sub H for only one element g

[coral shale]

im saying to prove it.

added modifiers to be clear

[hidden haven]

You want to prove that gH = Hg, why don't you check both inclusions? Take an element of the form gh, and show that you can write it in the form hg for some other h, and vice versa

This is very direct

[ornate vessel]

gHg' sub H => gH sub Hg; H(g')' sub (g')'H => Hg sub gH

[coral shale]

uh

[delicate orchid]

what lol

[delicate orchid]

do this

[solemn dew]

Any hints?

[coral shale]

i dont get how u get Hg'' sub g''H

[karmic moat]

use first isomorphism theorem

[hidden haven]

How second line

You should be multiplying both sides by g on the right

[ornate vessel]

yes

mb]

[coral shale]

thats a typo but idk where 3rd

[coral shale]

oh nvm i do know

yeah be a bit more explicit

[ornate vessel]

yes but

here im using

[ornate vessel]

that gHg' sub H for every g in G

[rocky cloak]

Are you able to construct a group homomorphism N -> NM/M?

[solemn dew]

i am not

[ornate vessel]

thats why from Hg' sub g'H we have Hg sub gH

[night onyx]

Yeah basically if ghg' in H for all h in H, then for any such h, gh = kg for some k in H. This shows gH \sub Hg. Since both have the same size (g being invertible) gH = Hg.

[ornate vessel]

my original question was

[karmic moat]

think of a typical element in NM/M

[delicate orchid]

there's a fairly obvious one, all elements of NM are of the form nm with n in N, m in M

[solemn dew]

first time i am using this theorem

[rocky cloak]

Do you know what NM/M is? I.e. what the elements look like?

[solemn dew]

so really hard for me to think a typical element in NM/M

[rocky cloak]

Do you know what NM looks like?

[solemn dew]

NM/M is NM * M?

[delicate orchid]

it shouldn't be any different to thinking of any other quotient

[solemn dew]

Yes NM = {nm, n is in N and m is in m}

NM should be a subgroup

[coral shale]

gHg' sub H
=> (gHg'g = gH sub Hg) and (g'gHg' = Hg' sub g'H)
wlog Hg sub gH
=>
gH = Hg

[rocky cloak]

Are you able to construct a homomorphism N -> NM?

@solemn dew

[solemn dew]

i believe so

[karmic moat]

ye

[coral shale]

so without relying on the forall quantifier i dont see how you achieve equality

thats the wlog bit

[ornate vessel]

yes that was my origianl question

when we dont have "forall"

[karmic moat]

notice that nM is an element in NM/M

[rocky cloak]

Here you seem to have constructed a map N -> NM/M. Which is good, that's what we wanted

[coral shale]

should be able construct explicit example

[rocky cloak]

Now what is the kernel of this map

[karmic moat]

have u learned first isomorphism theorem, jonathan?

[coral shale]

in particular if gHg' sub H but g'Hg isnt for some g

[solemn dew]

i've memorised it but i didn't understand the proof of it

but it states G/ker(phi) is isomorphic to H in this case?

[karmic moat]

yeah

[charred crescent]

iso to the image of phi right?

[solemn dew]

i don't understand how phi : N -> NM is a element in NM/M

[delicate orchid]

it isn't, that's a map?

[solemn dew]

NM/M = NM * M = NM ?

[delicate orchid]

no?

what?

do you know what a quotient group is

[solemn dew]

assume H is a subgroup of G, then a quotient group is a*H, where a is in G

[delicate orchid]

normal subgroup but yeah

[coral shale]

hmm about that loose notation

[delicate orchid]

so what does an element of NM/M look like, knowing that elements of NM look like nm

[coral shale]

if ur just starting to learn it

G/H = {gH : g in G} yes

(usually N not H to indicate H is normal)

[solemn dew]

why does it have to be a normal subgroup?

[delicate orchid]

because otherwise the quotient isn't a group

[karmic moat]

shuri ur spoiling it :(

[solemn dew]

Is it that G\H =H\G?

[coral shale]

cus gH = Hg is necessary

think about the identity axiom for the quotient group

[karmic moat]

first off, \ usually denotes set difference; second, in terms of quotient groups, not always

[coral shale]

g . H = H . g = gH is necessary

[delicate orchid]

it appears there are some colossal holes in your understanding

I'd go back and review

[coral shale]

eh sorry what

[karmic moat]

im jk

[rocky cloak]

The notation H\G is often used for right cosets.

[karmic moat]

my bad

[coral shale]

its ugly

[karmic moat]

ive never seen that notation before oops

[coral shale]

thank god i rarely talk about cosets

[night onyx]

Consider the map f: N -> NM/M given my the rule f(n) = nM. Show that the kernel is N \intersect M (this is pretty obvious if you write it out) and the map is surjective (also not super hard to show), then use the first isomorphism theorem.

[solemn dew]

if right coset = left coset it is a quotient group?

[rocky cloak]

Fits in with the notation for double coset H\G/K

[coral shale]

correct.

gH = Hg for all g in G

[solemn dew]

gotcha

[coral shale]

if and only if H normal

[delicate orchid]

that's an awful justification

[karmic moat]

i have also never seen the double coset notation LOL

[coral shale]

who cares, everything commutes

[delicate orchid]

I have, but it doesn't justify writing H/G, it could justify writing H\G but that's set minus

[solemn dew]

So NM/M = NM * M = M *NM then?

[karmic moat]

i always just use aB for left coset and Ba for right coset

[rocky cloak]

Why? Having nice and consistent notation is a good thing no?

[delicate orchid]

no, NM/M is not, and never will be NM

[coral shale]

ngl i never use set minus tbh

just use -

[delicate orchid]

except it isn't consistent, the slash is going the other way

[summer path]

I have seen it as shorthand for two left cosets

[rocky cloak]

?

[topaz solar]

M=*

[coral shale]

issue is its overlap with existing elementary notation

[delicate orchid]

elements of NM/M look like nmM = nM
now can you construct a map from N to NM/M or do you need more hints

[summer path]

Well it's obvious that it isn't a setminus in this context

[solemn dew]

I think I can continue from here

[charred crescent]

just to make sure i understand correctly, we can say an element of AB/B is just (ab)B = aB right

[karmic moat]

yeah

AB = {ab | a in A, b in B}

[coral shale]

god i find this confusing to read monke

[karmic moat]

AB/B is the set of cosets

[charred crescent]

cuz (ab)B = aBbB = aBB = aB ? (only in the case of normal subgroup)

[delicate orchid]

nope, b is in B

(ab)B = a(bB) = aB

[charred crescent]

oh how so? not my goal

[delicate orchid]

it wasn't confusing dw

[coral shale]

ive rarely had to read such myself

i didnt do much alg dw

[open sluice]

if you like A - B why not also use A + B and AB and f(A)
(tbh I only like the last one)

[coral shale]

wut

[karmic moat]

i hate A - B and i hate A + B

ugh ur uhu hasodfj alsk;djf

i hate it >:(

[coral shale]

whats fA

[delicate orchid]

A + B for coproduct of sets is based

[karmic moat]

nuh uh

[delicate orchid]

and who doesn't use f(A) lol

[karmic moat]

\amalg supremacy here

[delicate orchid]

I don't even know what that is

[karmic moat]

$X \amalg Y$

[cloud walrus]

ana(functor)mono(morphism)

[open sluice]

probably something like image(f)

[delicate orchid]

oh so \coprod but for weirdos

got it

[karmic moat]

wait i didnt even know \coprod was a command

LOL

[coral shale]

is A+B meant to be union. pretty unfortunately

[cloud walrus]

jagr2808

[delicate orchid]

yeahHHHHH

now we're talking

[karmic moat]

if A + B is coproduct then A - B is product

so true so true

[rocky cloak]

It's really big for some reason

[topaz solar]

Disjoint Union

[hidden haven]

\coprod is for the big indexed coproduct like \sum. For the binary operation you should use \amalg

[topaz solar]

Or, + like a sane person

[white oxide]

$\coprod$

[hidden haven]

You add morphisms not objects please

[cloud walrus]

okeyokay

[white oxide]

Sick

$\prod$

[cloud walrus]

okeyokay

[white oxide]

Wow

[rocky cloak]

Should be \coprod and \bigcoprod 😠

So it matches \oplus and \otimes

[hidden haven]

They should make it \plus and \bigplus instead of + and \sum

[coral shale]

,,\bigplus i

[cloud walrus]

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

[coral shale]

shame

[karmic moat]

i've actually seen some notation where it uses like

a big + instead of \sum

[coral shale]

we should do that

[karmic moat]

e.g. indexed disjoint union

it drives me mad

[hidden haven]

Has no one every defined a \bigminus

Mathematically

[coral shale]

its unfortunate when your plus on paper isnt quite big enough and confuses ppl tho

[karmic moat]

emdash

[hidden haven]

no but

mathematically

what does it mean

[formal ermine]

you guys need help

[coral shale]

make it commute first

[karmic moat]

indexed set minus LOL

idk

[hidden haven]

you too bby

[charred crescent]

can we discuss the natural projection pi : AB -> AB/B ? (when B is normal in AB)

[karmic moat]

if you have {A_n} n \in N then \bigminus_n A_n = A_1 \ A_2 \ \cdots \ A_n \ ...

[coral shale]

AB/B = {aB} as established earlier right?

[charred crescent]

yes

[coral shale]

= A/B or nah?

[rocky cloak]

Is there something to discuss?

[solemn dew]

how do I even find the kernel here? Since phi : n => nM

[charred crescent]

the "restriction to the subgroup A"

[solemn dew]

My question is: for do I find phi : n = nM = e ?

[hidden haven]

Is it surjective? Discuss.

[charred crescent]

so creating from it the map phi : A -> AB/B

[rocky cloak]

You should start by finding out what the identity in the group NM/M is

[solemn dew]

i'm on it 👍

[coral shale]

^

i feel like im missing something

[delicate orchid]

B might not even be a subgroup of A

[coral shale]

suppose it is

it might not even be normal?

[delicate orchid]

then AB = A trivially

[charred crescent]

i want to understand the motivation for why from the map pi : A/B - AB/B we "restrict to the subset A" and create the map phi : A -> AB/B

[coral shale]

ok i should think this shit up properly lmao

[delicate orchid]

almost like the 2nd isomorphism theorem tells you something about the intersection of A and B

[coral shale]

what wew says cleared it up for me

[charred crescent]

and what it actually means to "restrict to a subset"

[delicate orchid]

there's no restriction? who said this

[charred crescent]

dummit and foote

[summer path]

"A/B" is AB/B if you want it to be well defined

[delicate orchid]

more like dumb and foot

[karmic moat]

dummit and foote more like dumb and foot

[delicate orchid]

just take the map a -> aB

[karmic moat]

aw shit

[delicate orchid]

what's the kernel of this map

then first iso

[charred crescent]

A intersect B

[delicate orchid]

ta da

[white oxide]

CANONICAL EPIMORPHISM

[delicate orchid]

2nd iso proven

[charred crescent]

i get that part but it's this intermediate step that i can't find the motivation for

[delicate orchid]

which intermediate step

[charred crescent]

namely why we consider the natural projection pi : A/B -> AB/B and then "restrict to the subset A" and create this new map, phi : A -> AB/B

[hidden haven]

The second and third isomorphism theorems are built into the notation in Ravi Vakil's diagram calculus for abelian cats and using it means that you never have to manually apply these theorems

[delicate orchid]

look ignore what they're saying

[charred crescent]

okay

[delicate orchid]

literally just

a -> aB is a homomorphism from A to AB/B

kernel is the intersection, so use first iso

[charred crescent]

but don't we need to know that it is surjective to use first iso

[delicate orchid]

it's pretty obvious that it's surjective I'm ngl

[hidden haven]

Every map is surjective onto its image

So you can always use the theorem to get an isomorphism of the quotient with the image at least

[rocky cloak]

Every map is surjective.
FTFY

[delicate orchid]

yeah ok so that's why they're doing the restriction thing

to get around actually showing it's surjective

"let aB in AB/B, then a in A maps to aB in AB/B" shows it's surjective

[hidden haven]

Jagr set theory

[solemn dew]

ok i completed it

(not written nicely, just scratch workt)

i self study

but you guys are my teachers

thanks 😄

[delicate orchid]

ok now for the third iso

[hidden haven]

Where's my tuition fees

[charred crescent]

but i would have never thought of creating such a map

[delicate orchid]

oh well

comes with practice

[karmic moat]

dont forget to coauthor me in your first paper

i want second author status

no less

[delicate orchid]

see if you can use a similar idea to show that (G/K)/(H/K) is iso to G/H as practice

[charred crescent]

so third iso?

[solemn dew]

no worries, i'll make sure my cousin knows you helped me!!!

[delicate orchid]

yeah

[karmic moat]

does your cousin have one kajillion usd

[charred crescent]

okay i will try to prove it

[delicate orchid]

as for YOU @hidden haven

[charred crescent]

thanks @delicate orchid

[rocky cloak]

Your name starts with a, so I guess you should be first author

[delicate orchid]

you can help me with this nonsense

[hidden haven]

[solemn dew]

negative kajillion usd

same absolute value in R

[delicate orchid]

I know what these stupid characters are but I cannot be BOTHERED to actually do the calculations

"ooohghh induce from <a, b^p>" fuck offfffff

[karmic moat]

undergraduate math "research" in a nutshell

[hidden haven]

Lmao what are you even talking about

wew

[karmic moat]

professor says "I have these things but i dont want to calculate it. do it for me"

[delicate orchid]

I'm venting because I have to do this garbage

[south patrol]

Calculations are important

[delicate orchid]

https://tenor.com/view/soyjak-spinning-spinning-soyjak-soyjak-jazz-hands-excited-spinning-wojak-gif-25231392

[hidden haven]

sucks to be you

[coral spindle]

I feel this 🫂

[teal vessel]

So I have to show that the intersection of arbitrary many subgroups is itself a group, and I feel like it's fairly obvious, but it said not to assume that the number of intersections is countable, which spooked me. I know if it's countable I can merely prove it inductively, but for uncountable, is it sufficient to show that the structure of an arbitrary intersection must contain identity, inverses, and have closure? If it didn't that would imply that there is some inverse or product present in one member of the intersections but not in others, which disqualifies it from being a group.

[delicate orchid]

they're ordinary characters at least. The day I have to compute a brauer character is the day I drop out

[coral spindle]

They mean you should use the general form of the intersection instead of assuming a certain size.

It's really not that complicated.

[delicate orchid]

well those are the 3 properties of being a subgroup so yes that would be sufficient

[hidden haven]

Induction proves things for finitely many things not for countably many things

[coral spindle]

Wew help me I need to compute blocks in order to calculate some l-local Schur indices I want die

Yeah forgot to mention that

You CANNOT show facts about infinite things with induction

(ordinary induction at least)

[delicate orchid]

what's the group algebra boss

[coral spindle]

Can't remember can I lad

[teal vessel]

Sometimes I need to remember I'm not doing set theory where transfinite induction just exists

[delicate orchid]

well get to thinkin

[coral spindle]

I'm lazy, wew

Nah I just need to work out the partition of the ordinary characters

[delicate orchid]

if it's just a little one I'll bash it through the Brauer homomorphism with you

[hidden haven]

Transfinite induction exists everywhere but then there is no reason for it to stop at countable

[coral spindle]

It should really be said: You do not need transfinite induction to prove this statement!

[teal vessel]

I know I don't. I also don't need Fermat's last theorem to prove ³√2 is irrational, lol.

I'm mostly kidding.

[hidden haven]

[formal ermine]

regular flt ong

[delicate orchid]

although that being said, I think you know more block/modular memes than me so idk how much of a help I'd be

[coral spindle]

nah I don't know shit about modular reps

Hence why I am not™️ looking forward to it™️

[delicate orchid]

is ur field algebraically closed at least

[coral spindle]

Nah

[delicate orchid]

🚪 🚶‍♂️ _ _

[hidden haven]

How about you people learn some presentation theory before representation theory

You really need it

[coral spindle]

Honestly it shouldn't be that bad. I just need to understand a particular ring of integers

[delicate orchid]

I know presentation theory as well

oh the O thing

[coral spindle]

The O thing

[formal ermine]

isn't representation theory just presentation theory but done again????

[summer path]

But how else are you going to prove cbrt(2) is irrational without using flt /s

[delicate orchid]

yeah you look at characters over an algebraically closed field with positive characteristic and then translate that back into the ring

I'm assuming your thing is local

[formal ermine]

just pull a kummer and pretend every prime is regular

[coral spindle]

Yeah it's a local ring

It's not too bad

[delicate orchid]

yeah, that's the standard setting

[coral spindle]

Totally char zero too

[formal ermine]

(was it kummer?)

[coral spindle]

it's no biggie

[delicate orchid]

wtf do you call the field u get when you quotient a local ring by it's maximal ideal

[coral spindle]

residue field?

[delicate orchid]

whatever that is, is the field of char p that you do the brauer nonsense over

[delicate orchid]

that's the one

[coral spindle]

Two ordinary characters are in the same ell-block if their values differ by an algebraic integer in the local splitting field I'm working with

So it's fine I really just have to do that

[delicate orchid]

interesting

[hidden haven]

I had so much respect for you before I knew you were into rep theory borty

[formal ermine]

wtf

[delicate orchid]

I tend to work with W-blocks instead of L-blocks

[coral spindle]

Um uh it's just what my supervisor tells me to do

[coral shale]

I respect wew for being british and based

[coral spindle]

🙋‍♂️ 🇬🇧

[hidden haven]

oh then carry on may you recover

[coral spindle]

I didn't at all sign up for it

[coral shale]

wait u werent talking to wew lmao

[formal ermine]

I'm scared because of my exam tomorrow

I feel like I know NOTHING

[hidden haven]

I literally named the person I was talking to in my message

[coral shale]

i missed it

i saw rep theory i assumed wew

[hidden haven]

I respect wew he is one of the best musicians in the world

[karmic moat]

im scared because of nothing tomorrow

[coral spindle]

Yo that means ur really smart coz sock rates

[karmic moat]

aka i am not scared

haha >:)

[formal ermine]

I BARELY OWN ANY SOCKS

[karmic moat]

tomorrow i will eat breakfast >:)

then i will take a nap >:) then eat lunch >:) then take another nap

[formal ermine]

HELP WHERE DO I GET SOCKS FROM QUICKLY

[delicate orchid]

HOW

[coral spindle]

I am also a little replet ackshually

[formal ermine]

I ALWAYS RUN OUT OF THEM SO I JUST WALK AROUND BARE FOOT 😭

[karmic moat]

how tf do you run out of socks

do you like

throw them away

do u think socks are one time use

[formal ermine]

I just don't have many

[delicate orchid]

wOAH!! What a correct and true statement! As an unbiased third party I highly agree!

[formal ermine]

ok but seriously what do I do

I feel like I don't know anything

[coral spindle]

bit late honestly idk

[delicate orchid]

I'm hardly even a rep theorist I do weird shit that makes people's eyes go wide at conferences

[coral shale]

i see rep theory in this chat i assume wew

[coral spindle]

Don't make me rep all over you

[coral shale]

[delicate orchid]

"dude what if like... class functions but like...... there's even more conjugation dude.... like... just add MORE in..."

[coral spindle]

I proved a teensy tiny theorem about Schur indices a few weeks ago

[coral shale]

Schur

[coral spindle]

Schur

[formal ermine]

what is the intuition behind multi gluing besides glUING

[delicate orchid]

the most rep-theory thing I've done is explicitly compute character values for the sylow p-subgroup of S_{p^k}

[hidden haven]

I attended a series of talks by Henning Krause this summer

[coral spindle]

nerds lmao

[hidden haven]

He did rep theory

[formal ermine]

he doesn't exist

he's a "professor at bielefeld"

lmaooo

what a bunch of idiots

bielefeld isn't real guys

[teal vessel]

Are you allowed to do presentations on representation theory? Or just representations?

[hidden haven]

Please no sheeping in this chat

[delicate orchid]

combinatorics jumpscare

[rocky cloak]

Pretty sure that officially makes you a representation theorist

[coral spindle]

it was a joke

[delicate orchid]

this took me 13 hours straight to derive and I finished at 3am and started violently shaking from the adrenaline dump

[hidden haven]

wtf

I thought you ripped it off of google images

like a sane person

[delicate orchid]

idk what this means, they're two different fields

[hidden haven]

It's so much funnier now

[summer path]

If you do a presentation twice, does that mean you did a representation?

[delicate orchid]

this is the number of irreducible characters of C_p \wr C_p \wr ... \wr C_p with k wreaths of degree p^x

[coral spindle]

Huh

interesting

gross but interesting

[hidden haven]

I think this is the fifth time this joke is being made in this chat in the last 15 mins

[coral spindle]

That group probably has derived length k

[summer path]

I don't scroll up

[coral shale]

ive seen it for the first time

[delicate orchid]

u will never guess what

[hidden haven]

Whatever you say bro

[delicate orchid]

worked for C_5 \wr C_5 \wr C_5 good enough for me

[coral spindle]

I'm surprised there's partition stuff going on there

[delicate orchid]

it's the sylow subgroup of S_p^k why the fuck are you surprised

[coral spindle]

Ohhhh that makes sense

[hidden haven]

wew told me its true good enough for me

[formal ermine]

ok on an unrelated note, but do any of you guys happen to have adobe xd?

[coral spindle]

No xD

[coral shale]

adobe

[delicate orchid]

my humble abode

[teal vessel]

Language joke. Less funny than tubular cat's

[formal ermine]

I need to convert an xd file to after effects (I exported it from figma)

and FOR THE LIFE OF ME

I can't find someone that owns adobe xd

I even asked on their official discord

[delicate orchid]

it's actually kinda scary how simlar GL and S_n are in terms of their local p-structure

[coral shale]

bruh who uses that shit

[coral spindle]

Ooooh it's the Weyl group 👻

[hidden haven]

Bruh at least we were talking about algebra before

[delicate orchid]

this similarity is made manifest by the fact that S_n is GL_n(F_1)

and group actions are F_1-reps

[coral spindle]

Shut up F_un is a myth

not real not real not real

[hidden haven]

F_une, fields are feminine

[coral spindle]

F_une

based

[formal ermine]

or do YOU know two covariant quasi inverse functors between the category of adobe xd files and that of after effect files????

[delicate orchid]

it's so funny you just set p = 1 in all the formulas for GL and you get the formulas for S_n

[formal ermine]

NEIN

KÖRPER SIND MASKULIN

[hidden haven]

[hidden haven]

Bielfield mf spotted in chat

[coral spindle]

We all know illum isn't real

[hidden haven]

projective == illum?

[delicate orchid]

el nino character table

[coral shale]

yes hes tryna be cute

[coral spindle]

Uh-huh it's a change in brand

[coral spindle]

Wew I love and support u in all endeavours but u need to change the way you do tables in TeX

[hidden haven]

He had me tricked, I almost sent him all my savings

[coral spindle]

https://anorien.csc.warwick.ac.uk/mirrors/CTAN/macros/latex/contrib/booktabs/booktabs.pdf

^ booktabs package

[delicate orchid]

wait till he sees the oh no no no no

[coral spindle]

instant pretty tables

[formal ermine]

I am your grandson

I had a car accident

I need money

hi bruno

[delicate orchid]

oh no no no he is unaware

[formal ermine]

boytjtjie also needs money, his 23 kids are starving

[median pawn]

free groups stuff

[wraith cargo]

Message deleted 💀

[coral spindle]

Oh don't worry I fed them all ||to my man-eating plant||

[wraith cargo]

Tf did the bot delete my message lmao

[hidden haven]

Hausdorff we are just shitposting

You don't need to go to a thread

[coral spindle]

Sorry Hausdorff

[stuck fiber]

Those arrays scare me

[wraith cargo]

scary

[median pawn]

LMAOO

[hidden haven]

Who made this into a thread

[coral shale]

shitposting is the main topic of this channel

[delicate orchid]

simmer down in the back

[summer path]

We are having a good time shitposting since nothing else is happening here

[formal ermine]

uhmmm guys how do I get the grad+ role 👉👈

[summer path]

[delicate orchid]

apply for it

[formal ermine]

I actually asked a question but everyone ignored it ):

[coral shale]

hints in the name

[formal ermine]

I got denied

[delicate orchid]

cope

[summer path]

Nothing else is happening here

[formal ermine]

AAAAAA

[wraith cargo]

tf is multigluing

[coral shale]

idk what glue u using

[formal ermine]

gluing more than two affine varieties

[coral shale]

try using superstick

[formal ermine]

glUING N VARIETies wheRE N > 2

[delicate orchid]

society if G \cong \widehat{G} for all G

[summer path]

Your shift button needs help

[hidden haven]

Is widehatG the presheaf cat on G

[coral shale]

Whenever I see illum spouting words i dont understand I tell myself he's pretending to understand it to cope

[hidden haven]

Isn't that the category of right G-sets

[delicate orchid]

no it's the pointy yang duel

but yes it is also kinda the category of right G-sets

there's a nice connection there actually between the burnside ring and the rep ring

[hidden haven]

How is that different from the pontryagin dual

[formal ermine]

omg ponrtytagoing duality my beloved

[delicate orchid]

it isn't

[hidden haven]

[delicate orchid]

if it was true it would just mean all irreducibles were linear basically

[hidden haven]

oh I used the Burnside ring a lot while homotoping theory

[delicate orchid]

makes sense

[summer path]

I don't think I've ever seen someone spell poetry aging wrong

[delicate orchid]

golden quip

[formal ermine]

so true

what prereqs does the second chapter of neukirch have

@wraith cargo

[wraith cargo]

do u know algebra

[formal ermine]

I pretend I do

[wraith cargo]

that's enough

[formal ermine]

it's called improvised bullshitting

[coral shale]

u have to be an adult.

[summer path]

If you have taken undergrad algebra

[formal ermine]

I hate you

[summer path]

You can read neukrich

[coral shale]

haha

[wraith cargo]

do not lie

[summer path]

What

[formal ermine]

I took grad alg nt 1 1/2 this semester and I wanna take 2/2 next sem and wanna be prepared

[coral shale]

[wraith cargo]

You can't read Neukirch without grad algebra lol

[summer path]

What is the difference

[formal ermine]

the under

[summer path]

Like what do you consider to be necessary from grad alg

[wraith cargo]

what a noetherian ring is for example
localizations
category theory

idk if I went and looked I could probaby find more stuff

[formal ermine]

the only advanced thing we used was like localization, tensor products, limits

[wraith cargo]

like you need to know A LOT of Galois theory for Neukirch lol

[summer path]

Oh

Well I guess I had a bunch of that covered in topics classes

[wraith cargo]

Neukirch as a book is very deep but at the same time it's not tame

[formal ermine]

ping?

och ne timo

[elder wave]

I think you'll be fine illum

[coral shale]

[elder wave]

I've never heard anyone say Neukirch needs all these prereqs and it's used here to teach a course following basic algebra +galois theory and some commutative algebra

[delicate orchid]

ok done

it was <b> not <a, b^p> though so I had to do it twice

[formal ermine]

only took you two hours

[delicate orchid]

I only started like 30 minutes ago

[formal ermine]

I said

only took you two hours

[summer path]

Only took you a quarter of what illumi said

[white oxide]

can somebody give me a hint to find the kernel of gf

i mean i guess i could look at the image of a homomorphism A --> B under the exact sequence but im not sure what it is

[solar vessel]

what do we know about g

[white oxide]

it's a monomorphism, so the kernel of it is trivial but f is an epimorphism and ig that's where i'm having trouble

oh wait

oh it's just kernel f oops

i think

[solar vessel]

yes

[white oxide]

thanks you a g

what's the best way to proceed here? induction on the length of the sequence?

[white oxide]

referencing this

oh nvm i guess i can just consider an arbitrary exact sequence i think?

nvm i think induction is better, idk what's the better strategy for this kind of proof

[delicate orchid]

there's a really really nice diagram you can draw

[white oxide]

oh right i didn't even consider diagrams lol

weww.... glad for that

[untold turret]

is there an injection from the ideals of a unital subring to a unital parent ring? edit: consider Z -> Q, clearly not

[chilly radish]

Exact sequences are reasoned about visually almost always

[delicate orchid]

lmk if you want some hints

[white oxide]

ye imma struggle for a little bit but i'll probably need some soon lmfao

[median pawn]

what does the stuff in blue mean? compatibility means?

A* is the set of words and F(S) = A*/~ where ~ is the equivalence relation that takes care of cancellation with inverses

[coral spindle]

x ~ y means phi(x) ~ phi(y)

That’s compatibility

[median pawn]

do you mean phi(x) = phi(y)? that makes more sense

[coral spindle]

No.

[median pawn]

similarly for phi*

[coral spindle]

I do mean phi* tho

[median pawn]

phi*(x) is an element of G though? and there's no known equivalence relation on G. G is literally any group @coral spindle

[white oxide]

ok yea i have no clue where the fuck to start lol

[untold turret]

suppose A is a unital subring of B, and B is a commutative unital ring but not a field. is there an injection from the ideals of A to those of B?

[delicate orchid]

the long sequence A_n is exact

[white oxide]

oh ok i'll consider that thanks

[median pawn]

• not sure what "compatible with generating set of ~" means

[white oxide]

oh so basically i just have to describe the short exact sequences which make up that diagram

i guess that's pretty clear

i think??

well the trivial maps always exist so yeah that makes sense

[delicate orchid]

yeah so you use the fact that the image of f_i is the kernel of f_i+1

[white oxide]

right ok i'll try that out, tysm

[white oxide]

dumb question; could we just append trivial maps like this

because they always exist right

[delicate orchid]

that's really REALLY lame but I can't see anything wrong with it immediately

this is what I wanted you to do

hopefully it's clear that this commutes and the diagonals are exact

[white oxide]

LOL hey if it works it works

you might say it's canonical

😹

[delicate orchid]

it's not canonical, what I did it the canonical construction

[white oxide]

nah canonical means easier

[delicate orchid]

there will be some subtle reason why what you did doesn't work

oh right yeah, short exact sequences require 5 elements

but then you have 0 -> A_5 -> 0 -> A_3 -> 0 -> A_1

[white oxide]

wait isn't like 0 --> Ai --> A{i + 1} --> A{i + 2} --> 0 five elements

[delicate orchid]

how are you getting out to the zero

oh it's the zero from before

[white oxide]

yea