groups-rings-fields

[delicate orchid]

ok it's kinda like field extenstions if you simply do not think about it

[barren sierra]

and I can't think of a good map

it is?

[delicate orchid]

my question is why does the trivial centre matter

that must have something to do with the answer

[barren sierra]

I presume it helps with showing H injects into H

or showing that the image of said injection is the kernel of the surjection from G to K

[delicate orchid]

sounds... first iso-y

[barren sierra]

it does but for that i need a map

[delicate orchid]

tried the canonical map?

[barren sierra]

and I can't even think of one from H to G or G to K

wdym

[delicate orchid]

G -> G/H by g -> gH

or I suppose in this case

K -> G/H by k -> (trivial automorphism, k)H

might work

it's just an idea

[barren sierra]

isn't that the wrong direction?

I need G -> G /H = K

[delicate orchid]

hmm ok

[barren sierra]

and some sort of injection from H to G

[delicate orchid]

I'm gonna ignore this bit cause the idea of equivalence classes of automorphisms is horrifying

wait

[barren sierra]

well that's what outer automorphisms are

[delicate orchid]

yeah which is why I really don't understand outer automorphisms

anyway!

[barren sierra]

ye I'll try this later I guess

[delicate orchid]

try playing around with the canonical map a bit more

I swear there's something there

[barren sierra]

what is the canonical map lmao

[delicate orchid]

_ _

[barren sierra]

oh

[delicate orchid]

wait a minute, Inn(G) are isomorphic to G/Z(G) right?

so Inn(H) is isomorphic to H

might be able to work with something there

[barren sierra]

is it?

hm

[proud bear]

for H->G does mapping x in H to (phi_x,1) where phi_x is conjugation by x work? theta(1)=Inn(H)=[phi_x] i think

[delicate orchid]

yeah cause the centre is trivial

[barren sierra]

lemme think on this

[delicate orchid]

wait I think it does, we just established that Inn(H) = H so mapping the conjugation/inner automorphisms back to their elements should give us something nice

[barren sierra]

yea I think I'll play with this and see how far it gets me

I completely forgot G/Z(G) = Inn(G)

[delicate orchid]

it's a useful one

[barren sierra]

this class moves so fast that I forget these things

we've done so much group theory and now we're going over ring theory

https://faculty.math.illinois.edu/~rezk/500-sp22/lecture-notes-500-sp22-1.pdf
all this in 5 weeks

(they are very nice notes)

[delicate orchid]

that's some dense material

[barren sierra]

ya

[proud bear]

are you guys going to do modules or field/galois theory after rings too?

[barren sierra]

first grad class also so I'm

ugh

dying

[barren sierra]

ye

[proud bear]

oof

[barren sierra]

[proud bear]

good luck bro

[barren sierra]

ty

i figure if I can survive this

I can survive many things

[proud bear]

[barren sierra]

this reminds me of the pace of my honors intro proofs course

which went from like "what is a proof" to "we are formally constructing the integers and rationals" in a sem

easier material than this but similar pacing

[delicate orchid]

Wtf I’m jealous

[barren sierra]

I'm not

I'm scared lol

[delicate orchid]

I’m having to self teach myself Galois theory slowly

[proud bear]

your prof is not good or you are learning it before you take the course?

[delicate orchid]

There ISNT a course

[proud bear]

[delicate orchid]

Thanks to my uni axing stuff like that my algebra skill set is rather uhhh let’s say “concentrated”

[proud bear]

there almost isn't a course for that at my uni either. i think 6 people are taking it this term

[barren sierra]

want my prof's notes?

[delicate orchid]

Nah it’s fine

[cloud walrus]

grist bundle

grist bundle

[echo orbit]

I'm really not sure what is meant by "a coset in the quotient C that is fixed by G"

I thought C was a G-module, not a quotient...

[cloud walrus]

grist bundle

[echo orbit]

Why is it saying "a coset in the quotient C"?

[echo orbit]

I do, I have to learn this as well. I've seen the topics before but need a refresher

[barren sierra]

https://faculty.math.illinois.edu/~rezk/500-sp22/lecture-notes-500-sp22-3.pdf

[echo orbit]

Thanks

man

theres no way galois worked all this out on his own

[upbeat swift]

Does anyone have an idea as to how to solve this? I'm a bit confused

What it means by Exercise F^^

I understand that

2 and 5 are prime factors of 10

But is that all I have to say?

Or is there more? Just need clarification

[echo orbit]

2 and 5 being prime factors of 10 just means theres an element of order 2 and another of order 5 in G

by cauchy's theorem

it doesn't mean anything else

but that's only the second sentence of exercise F

you actually have to use the rest of the reasoning

note that e, a, b are all distinct elements

in fact

let a have order 2

let b have order 5

then e != a != b != ab != b^2 != ab^2 != b^3 != ab^3 != b^4 != ab^4

those all form the 10 elements of G

[upbeat swift]

What's != ?

[cloud walrus]

grist bundle

[upbeat swift]

Oh

Is that all I'd have to say?

[echo orbit]

yeah

[upbeat swift]

Thank you

[ember field]

How many group homomorphisms are there from Z/n to itself?

[echo orbit]

n

[small bison]

shouldn't it be like phi(n) or something

[hidden haven]

I have never once tried to figure out any computations on any assignments that I have graded i doubt any graders do lol

[echo orbit]

hm interesting

[hidden haven]

That's automorphisms

[echo orbit]

^

[ember field]

i think this true for homomorphisms from Z to Z/n

[echo orbit]

this is endomorphisms

[hidden haven]

It's also true here

Hint

[wise igloo]

really

[echo orbit]

INTERESTING DIAGRAM!

where'd you draw that?

[hidden haven]

It's a sticker lol

[echo orbit]

are you just banging out tikzcd ad hoc

lol

nice

[wise igloo]

high schoolers complain about letters being introduced in math, wait till they see this

[echo orbit]

lol

[wise igloo]

"it's algebra bro" "NO THAT FUCKIN ISNT"

[echo orbit]

moldi do you group cohomology

[hidden haven]

I know what group cohomology is now but nothing beyond that

[wise igloo]

I know what a group is!

[hidden haven]

I gotta read clerk's thread

[ember field]

You mean like every homomorphism from Z to Z/n gives unique homo from Z/n to Z/n?

[hidden haven]

Yes

[ember field]

but how do I know that is all?

[hidden haven]

Because you can also go back

[ember field]

no you go back

[hidden haven]

A map from Z/nZ gives a map from Z

[lavish nexus]

problem from yesterday

[ember field]

ok thanks very much

[lavish nexus]

I still don't get the second part
how do you prove x^p-r^p doesn't split when r is not a root of unity

(assuming p is the smallest integer s.t. r^p is in F, otherwise I don't think it is doable at all

[hidden haven]

What do the roots of x^p - r^p look like

[lavish nexus]

r*\zeta?

[hidden haven]

Don't need that, since this talks about splitting, not irreducibility

[hidden haven]

Ye, and F is real

So the only case left to handle is p = 2

That should follow from the discriminant stuff

[chilly ocean]

idk what its supposed to represent

i did presentation on it

[lavish nexus]

I'm dum

I kept thinking about splitting into (x-1)*something

1 isn't a root

tyty

[chilly ocean]

do you know how group cohomology comes to use?

also another question very simple

what do maximal ideals look like in polynomial rings

[lavish nexus]

(x)

[hidden haven]

See the group cohomology thread on this channel lol

[chilly ocean]

lol woops

[lavish nexus]

actually wait (p,x)

[echo orbit]

@chilly ocean my only conception of its use is that it can be used to study the fixed points of G-modules

[chilly ocean]

thats H0 tho

also i learned about localization today

🙂

false

i learned about its applications to alg geometry

[hidden haven]

Applications

[chilly ocean]

we worked with example looking at polynomials A^2

also correct me if im wrong

A is just k[x] for some field k?

or is it always assumed k is C

the type set A for affine space

[hidden haven]

As a variety, it is k²

[chilly ocean]

yeah

[proud bear]

hi moldi

[hidden haven]

Hello stain

[chilly ocean]

oh wtf

i also learned weird stuff

in alg nt

like these different operators on this ring Z_p[[T]]

Z_p power series

and we just finished going over a bunch of operators on this space

the point is that we use this thing called a Mahler transform and connect it back to iwasawa algebra on Z_p

so all the operators on Z_p[[T]] corresponds to some change of a measure in iwasawa algebra

its v confusing atm for me

the last operator we covered is called a logairthmic derivative

oh last thing i learned today but confusingly was about correspondence between Lie Algebras and Lie Groups

[proud bear]

what's Z_p[[T]]? power series with coefficients in p-adic integers?

[chilly ocean]

yeah

[proud bear]

damn

[chilly ocean]

and there is a topology on it also

[hidden haven]

Power series in power series

[chilly ocean]

afaik Z_p are just p-adica with norm <=1

but i actually dont know

i just black boxed most the proofs

[hidden haven]

Ye or they are a quotient of the power series ring over F_p

[chilly ocean]

oh yeah

we went over that

oh ig the types of operators we were going over have a name

called Coleman operators

and they have a heavy relationship to this tower of number fields Q_p(u_p^n) where u_p^n are primitive p^n roots of unity

in each of these fields the unit group have inportance but i dont really remember

we focus a lot on these things called norm compatible sequences in these fields

all in all today was loaded

i wish i had a more directed courseload

[prisma shuttle]

can someone explain this

where do they get the $\sum (\text{dim} V_i)^2$ from

[cloud walrus]

JustKeepRunning

[prisma shuttle]

i get that they get the mapping from the jacobson density theorem (the analogy of CRT for irreps of A) but after that i get lost

and also i don't really get how the finally inequality implies the problem

[chilly ocean]

what are itreps

[prisma shuttle]

irreps?

they are basically finite dimensional irreducible representations

[chilly ocean]

oh

no clue what irreducible representation is

i think i know a representation is a homomorphism from group to GLnR

[prisma shuttle]

an irreducible representation is one that has no proper nontrivial subrepresentations

[prisma shuttle]

sort of but you can think of a representation as a k algebra (where k is a field) along with an algebra homomorphism to EndV

[chilly ocean]

is a subrepresentation a restriction of a representation?

[prisma shuttle]

its a subset of the representation where mutliplication by any element of the algebra is closed

sort of like ideals in ring theory

[chilly ocean]

oh

how can i think of a representation as k->End(V)?

how is k being formedd#

do you mind answering these questions

[prisma shuttle]

no k is the algebra

but you also have an "action" associated with it

[prisma shuttle]

no lol its fine

[chilly ocean]

i dont want to ask a bunch if you dont want to answer

ok ty

[prisma shuttle]

its like

how for a ring

you have an additive group structure along with multiplication

like its the two parts that make it up

[chilly ocean]

yeah

[prisma shuttle]

so you have the k-vector space V (the algebra)

along with an action (aka. a homomorphism)

if you want to think of it more intuitively

the homomorphism is basically a matrix you associate with the algebra

[chilly ocean]

so i thought a representatiom of a group G is r:G->GL(n,R) a group homomorphism, but you are saying it can be viewed at a k-algebra with multiplication defined by k->End(V)

oh

are you making it more general?

[prisma shuttle]

no no like the two definitons mean the same thing

like wut are saying is same as wut i am saying

just like how a group action is just a homomorphism to the symmetric group

[chilly ocean]

yeah

[prisma shuttle]

your homomoprhims is the "action" i am referring to

[chilly ocean]

oh i think i see it

[prisma shuttle]

wait really

how

like u see why the dim(v_i)^2 term is there

[chilly ocean]

oh lol no clue as of now

[prisma shuttle]

yea rep theory has been pretty hard to learn

[chilly ocean]

idk where square coming from

oh wait

notation is weird idk that V_i^oplus V_i is

[prisma shuttle]

oh that's just direct summing

honestly ima kinda confused about that notation too

[chilly ocean]

yeah but what is superscript

[prisma shuttle]

exactly

i think it just means it repeated in the direct sum a certain number of times?

[chilly ocean]

i mean i might have idea where square term coming from?

[prisma shuttle]

but idk why they have to dirrect sum that too

cuz if it was just raised to the power they wouldn't need the \oplus symbol there

[prisma shuttle]

do u get how we get from the inequality to concluding the problem statement

[chilly ocean]

yea

maybe its typo

[prisma shuttle]

cuz idk how this relates to irreps

[chilly ocean]

surjection makes it believable

[hidden haven]

They get dim V² from end V

[chilly ocean]

oh

[prisma shuttle]

could you elabroate

[chilly ocean]

end(v,v) are nxn matricies

bad phrasing ig

but if u have vextorspace V

the endmorphisms are square matricies with same dim

[hidden haven]

Oh ok they're phrasing it differently

[prisma shuttle]

oh i sorta see

[hidden haven]

V^⊕dim V

[chilly ocean]

yeah lol

wtf is that

y is the plus there

[hidden haven]

This has (dimension V)²

[hidden haven]

This is standard more notation for repeated direct sum

[chilly ocean]

seriously

how does this make anything less ambiguous

[prisma shuttle]

wut is a repeated direct sum

[chilly ocean]

seems redundant

[hidden haven]

If you remove the plus it is repeated direct product

[chilly ocean]

oh true

oh lol

wait nvm

[hidden haven]

V + V + ... + V lol

[chilly ocean]

ive seen this betorw

[prisma shuttle]

wait but if the dim was finite then direct product and direct sum are the same right

so it wouldn't matter?

wait but here some of the dim could be unbounded

[hidden haven]

Ye

[chilly ocean]

such bog

good notation

[prisma shuttle]

tocyem your idea about the matrices being nxn was pretty nice

is it valid for this problem

[chilly ocean]

??

its moldi

[prisma shuttle]

no like the thing about the size of End(v)

[chilly ocean]

but if you treat everything as a vector space then it makes sense

[prisma shuttle]

yea it does

idk if that works here though

i feel like the inequality basicaly boils down to the fact that $(a+b+c+d)^2\geq a^2+b^2+c^2+d^2$ if you think about it intuitively

[cloud walrus]

JustKeepRunning

[prisma shuttle]

like for an arbitrary number of letters a,b,c,d,etc

depending on the dimensions

[chilly ocean]

i mean idk

i thought it follows immediately from surjection

[prisma shuttle]

oh wait yes u are write i am overcomplicating

[chilly ocean]

like taking dimension of that surjection gives you take inequality immediately

the part that looks weird is just the notation but we understand now

[prisma shuttle]

also do u know why they use the two headed arrow

[chilly ocean]

two headed is shorthand for surjection

[prisma shuttle]

ohhhhhhhh

oops i didn't know that

thx so much y'all

[chilly ocean]

lol idk shit

i have no clue what most those objects really are even though you explained

looks interesting though

i think i know a bit tho

and def want to get more invested

[prisma shuttle]

yea you should read napkin

its pretty good

[chilly ocean]

evan chen

ye

i remember seeing that in HS

also another cool thing i learned were how localization makes a DVR and valuation represents number of times another polynomial intersects

[weary cloud]

What is the relation between inductively ordered sets and induction? I don't think we can do the usual induction on this set, as we are only guaranteed upper bounds on each chain. So what kind of induction can we do on this set? What are some examples of proofs/applications of such kind of induction? If there is none, why are they called inductively ordered? If someone could point to a resource, that'd be helpful too.

Yours gratefully

[white jackal]

What subgroups of Q8 correspond to the subgroups of Q/H under the Correspondence Theorem? H is {1, -1}

can u guys double check my work

[prisma shuttle]

i took a quick look over it and it seems correct o me

also can someone explain the notation here (what does the "opposite multiplication" thing mean?)

like division or smth?

[kind temple]

a groups opposite has the same underlying set, but the definition of composition on the opposite group is a x b = b * a if * is the operation on the original group

[chilly ocean]

Do people remember all those propositions

Does someone have some intuition or vid for cosets?

I'm having a hard time grasping what they are and connecting all the dots in relation to quotient groups

Could someone walk me through an arbitray group... (Z/m8Z, +)?

The quotient groups in this case would be the trivial groups, and what?

[kind temple]

try looking at cosets of R^2 when viewed as an additive group, and the sub-spaces of R^2 as subgroups. you can actually draw these cosets, and its a good way to think of them in general

[chilly ocean]

I haven't learned what a sub-space is yet

[kind temple]

its just a subset of a vector space that is also a vector space

[hidden haven]

You could check out the first iso thread

[lethal dune]

there's a first iso thread

[chilly ocean]

que es un first iso thread

[hidden haven]

Quotients of compact spaces are compact, does that apply here?

[foggy merlin]

well if it only was asking for compact this would be a simpler question. but I'm asking this for a friend and since I haven't had any algebraic geometry idk what a variety is

and if quotienting by such a group preserves that property

[hidden haven]

Yeah I'm not sure when a variety is called compact, but if it is when the underlying topological space is compact then yes

Since it's just a topological space with extra structure

[foggy merlin]

I tried to look it up but didn't really get very far as these words in exactly this order don't seem to be used commonly

ig cuz he may have translated from french

[foggy merlin]

but does that quotient operation also preserve the property of being a variety in that case?

also what I found kind of weird that this was asked in differential geometry and not some other subject, I'd expect that they would just ask this for manifolds there. maybe he means sth else idk?

[sharp sonnet]

arent varieties noetherian and thus always compact?

[hidden haven]

[sharp sonnet]

the algebro-geometric replacement is completeness

no idea if this is related to the question though

[foggy merlin]

wait what type of variety are we talking about now

[sharp sonnet]

the type that is studied in algebraic geometry?

[foggy merlin]

wait but then what topology were you talking about, because that seemed kinda wrong for the euclidean one? or is it not?

[sharp sonnet]

oh ok, yes

i am thinking of zariski topology

[foggy merlin]

yeah, since my friend was asking this apparently in diffgeo, I assume they talked about the euclidean one? maybe?

[sharp sonnet]

hm ok but wouldnt you then use the word manifold

[foggy merlin]

otherwise the question itself would be worded redundantly

[sharp sonnet]

(and ask in #diff-geo-diff-top)

[sharp sonnet]

my original comment was in reply to this ^and you saying the question is translated

thinking about it in french the word variety is used both for varieties and manifolds

[foggy merlin]

yes, also I asked there before and no one answered, so I thought I'd ask here today. I already mentioned this seemed kind of odd to me

[foggy merlin]

OH

yeah I don't speak french lmao

yeah ok now I know why no one answered lmfao

thanks for clearing that up

[simple valley]

I'm asked to show that SES of modules over K[G] always split if G is a finite group and K is a char 0 field

A little lost...

Is this to do with the relationship between free/projective/flat/torsion-free modules?

[rustic crown]

that's called Maschke's theorem

the group algebra k[G] is semisimple

[delicate orchid]

Maschke's theorem 😌

cool to know it works for any char 0 field and not just algebraically closed ones

[rustic crown]

i can never remember these hypothesis 😵‍💫

the proof was like averaging the splitting map

so yea it should work as long as char k doesn't divide |G|

[delicate orchid]

I've got 2 different proofs, both 2 pages long

[simple valley]

that sounds awful

[lavish nexus]

had an abstract exam

complete destruction

Given ζ 25th root of unity

I got |Gal(Q(ζ)/Q)|=20

and there was a 4 cycle

But how do you know if it’s abelian

[prisma shuttle]

wait really

in napkin it says

Let $G$ be a finite group, and $k$ an algebraically closed field whose characteristic does not divide $|G|$. Then $k[G]$ is semisimple

[cloud walrus]

JustKeepRunning

[prisma shuttle]

although not being algebraically closed sounds like a nightmare to work with

[next obsidian]

You prove it by taking an avg

The reason you need char 0 IIRC is just so you can divide by |G|

Which is why you can instead assume the characteristic doesn’t divide |G|

At least to my memory

[lavish nexus]

wait is |Gal(Q(ζ_n)/Q|=φ(n)?

[chilly ocean]

yeah

[lavish nexus]

Fuck so it’s just Aut(Z_n)

[chilly ocean]

uhh idk about that

[delicate orchid]

I've only looked at it through the lens of representation theory so I've only ever dealt with algebraically closed cases as they make the representations SO much nicer

[lavish nexus]

I was thinking about the order 5 element in that group

I wrote ζ->ζ^6 but somehow calculated it wrong and thought it wasn’t order 5

rip me

[chilly radish]

What does average mean in this context

[waxen hedge]

Sum over the group and divide par |G|

[chilly radish]

Ah

[barren sierra]

we can never have that the n x n matrices over a ring R can have an identity but R itself has no identity right?

[small ingot]

All rings have identity

Why is true that every map from a torsion group to a free group is trivial?

[barren sierra]

no?

[prisma shuttle]

it depends on what book you are using

[barren sierra]

for example the ring of even integers

[prisma shuttle]

some authros do assume all rings have identity

[barren sierra]

D&F

[prisma shuttle]

and are commutative

[barren sierra]

strange

[prisma shuttle]

yea D&F doens't

D&F basically assume bare minimum

no commutative

no identitty

[barren sierra]

I see

[ember field]

is a subgroup of a divislbe group divisible ??

[viscid pewter]

nah

consider the trivial subgroup

wait no

it still shouldn't be

ok consider the integers as a subgroup of the rational numbers, there we go

[ember field]

does divisible for multiplicative group mean every element has an nth root ??

[delicate bloom]

yes

[prisma shuttle]

why is the red statement tru

[terse crystal]

He just means that you need to determine all χ(g) to define χ, and G=S_3 is just an example that he will discuss about…

[prisma shuttle]

oh really

thats a weird way to word it lol

like "we just have to specify..."

[rose shard]

I started the course groups and rings and this week was my first week and I’m alr struggling bcs the course is in English and I’m native Dutch. Can someone give me an explanation bcs I’m stuck

[next obsidian]

What part is unclear?

[mystic bolt]

Is someone able to explain the intuition behind what ideals represent, and what factor rings represent? Brushing up on my undergrad stuff and I never really understood the concept, just got bombarded with definitions and proofs

[void cosmos]

in your free time

grab paper and pen

and think of ideals in Z

think about divisiblity

prime ideals what are those

etc

you will have fun

and if you get to something just always remember that is just one way of looking at things

i think in algebraic geometry ideals represent points and etc..

[next obsidian]

Prime ideals do

[echo orbit]

AG is schizo

[next obsidian]

The idea behind a quotient is that you want to set certain things to 0. An ideal is just what ends up being spit out if you want what you end up with to still be a ring

I think it’s most instructive to look at things like quotients of polyomil rings

Like R[x]/(x^2 + 1) ≈ C

And what you’re doing here is adding a formal symbol x which at first doesn’t mean anything

But then by quotienting by x^2 + 1 you declare that x^2 +
1 = 0, or really, x^2 = -1

So now x acts like i, it’s a square root of -1, and that’s why you end up with C because the result is just R[i] which is what C is

[void cosmos]

yea

[next obsidian]

This is why the Gaussian integers is Z[x]/(x^2 + 1) as well

You can do lots of things like adjoining roots of unity to R, irrational numbers to Q, etc etc by this sort of construction

[mystic bolt]

That literally just made everything click, thank you

And now I understand the logic behind the fundamental theorem as well

[kind temple]

how many subgroups of Sn are isomorphic to Sn-1

me and my friend think that it should have n, each being V_k(k) where V_k(k) is the set of all permutations sending k to k

[delicate bloom]

edge case, S_2 has only 1, not 2 subgroups isomorphic to S_1

I suspect it might be possible for it to have more than n in general though, might be some weird ways of combining elements to make it happen or something, idk, probably an easy way to see that it's not the case by going through and throwing all the standard theorems at it though

[kind temple]

hmm.

[chilly radish]

@hidden haven I just remembered but did u end up thinking about the quotient of product thing we were talking about a while ago? It's still kinda bothering me (Ping me when u reply)

[next obsidian]

No

Sorry, I should’ve replied to the other one

But there’s (at the very least) an exotic copy of S_5 inside of S_6 not of that form

You use it to construct the outer automorphism of S_6

[prisma shuttle]

can someone explain why the red part is true

[next obsidian]

Try to compute the dimension of the d x d matrices of trace zero

You should get d^2 - 1, I think you can just do it grabbing the following d^2 - 1 linearly independent matrices

Start at the top left and put a 1 in the top left and a -1 in the bottom right

Or rather like

0 everywhere, and a 1 in exactly one of the spots other than the bottom right

But if you put a 1 in the diagonal, avtually put a -1 in the bottom right

[dreamy fiber]

Can someone explain to me, why, in very simple terms (pls or else my IQ can't handle it) that why is it that,

$$Ext_R^1(P, N) = 0 \text{ for all $R$-module M} \implies P is projective$$

Thank you!

[cloud walrus]

ScarletScorch

[next obsidian]

These all have trace 0, and give d^2-1 linearly independent vectors, and clearly [A,A] isn’t everything so it can’t be d^2 dimensional, then the dimension of the quotient is d^2 - (d^2 - 1) = 1

This is almost by definition

A module is defined to be projective if Hom(P,-) is exact, which because it’s left exact means that for any surjection kernel -> M -> N that the map Hom(P,M) -> Hom(P,N) is surjective

But due to Ext’s definition, you know that given such a surjevtion, the sequence
Hom(P,M) -> Hom(P,N) -> Ext^1_R(P,kernel) is exact

By assumption the last module is 0

But then Hom(P,M) -> Hom(P,N) -> 0 exact means the map is surjective

So P is projective

[dreamy fiber]

This is gonna sound really stupid but

[dreamy fiber]

Why does the definition of Ext tell me this?

[next obsidian]

Umm…

Well I guess it depends on how you defined Ext

I take it to be a derived functor of Hom in which case it’s like definitional

If you define it as the like n-th homology of blah blah resolutions, apply Hom…

That’s how you make derived functors, but in that case it’s a bit trickier

It isn’t actually that bad but

It requires you to develop some homological algebra

Like idk, what do you know about Ext? The most important thing is the long exact sequence you get from any short exact sequence which is what we applied here

[dreamy fiber]

Well so our definition followed from Aluffi's book which he just drew a bunch of arrows for a lack of better word

well so his definition is that

[next obsidian]

Sure but Aluffi then states that you get a long exact sequence for any short exact sequence

[kind temple]

if u could expand, that would be really cool. idk what outer automorphism is

[next obsidian]

The fact that this holds and well-definedness is done later

[dreamy fiber]

wow you are really well acquainted with the book

[next obsidian]

This is from the problem I did 2 years ago constructing the outer automorphism. I construct the faithful representation of S_5, and if you trace out what it does you’ll see it doesn’t fix any of the Sylow-5 subgroups so it doesn’t embed into S_6 as any of the distinguished S_5’s you listed

[next obsidian]

This is where I learned this material in particular from the first time hahahaha

So I’m familiar with this part of the book a lot more than the rest

[dreamy fiber]

Thanks so much I think I actually understood it

[next obsidian]

[prisma shuttle]

what does the carrot symbol mean here

[next obsidian]

Dual

It’s Hom(A,k)

[novel loom]

Let $\mathbf{M}$ be a $2^{N}\times2^{N}$ matrix with complex entries, and write $\mathbf{M}$ in block form as: $\mathbf{M}=\left[\begin{array}{cc}
\mathbf{A} & \mathbf{B}^{\dagger}\
\mathbf{B} & \mathbf{D}
\end{array}\right]$

[cloud walrus]

ComplexVariable

[novel loom]

where $\dagger$ denotes the conjugate transpose, and where all four blocks are $2^{N-1}\times2^{N-1}$. Assume that $\mathbf{M}, \mathbf{A}$, and $\mathbf{D}$ are all Hermitian. If both $\mathbf{A}$ and $\mathbf{D}$ are invertible, does it follow that $\mathbf{M}$ is invertible?

[cloud walrus]

ComplexVariable

[next obsidian]

How can a 2^N x 2^N matrix be a 2x2 matrix of 2^{N-1} x 2^{N-1} blocks?

[delicate orchid]

Split it into quarters?

[next obsidian]

Wait

I forgot how adding works

I thought 2^{N-1} = 2^N - 1 for a second

What is wrong with me

[delicate bloom]

when you do the fast fourier transform too fast

[next obsidian]

But also idk, it suffices to ask if AD - BB^dagger is invertible I think

But idk if this is true whenever A and D are invertible

But I’m also not totally sure if the adjugate matrix thing works over a noncommutative ring

[next obsidian]

Which is how much you’d know it’s equivalent to look at this

[next obsidian]

Okay I think I have a counterexample

I’m pretty sure a 2x2 matrix with 1’s in every coordinate satisfy this

And you can scale this up to any size by taking the 2^{N-1} by 2^{N-1} identity matrix for all four blocks

Or uhh… maybe if you go past 2x2 the entire matrix isn’t Hermitian anymore

No I think you’re fine still

[novel loom]

If it helps, I specifically need N≥2.

[next obsidian]

My counterexample works for all N

Just take each block to be I_N

[novel loom]

Yes, you are right.

Thanks!

[next obsidian]

[hidden haven]

Quotient of product?

[grizzled elm]

Can someone check if I messed something up here:\
Determine the splitting field and its degree over $\mathbb{Q}$ for $x^6-4$. \
$x^6-4=(x^3-2)(x^3+2)$, and both $x^3-2$ $x^3+2$ are degree $3$ irreducible polynomials in $\mathbb{Q}$. Hence, $\alpha=2^{1/3}$ is an element of our extension whose minimal polynomial is degree $3$. It also has complex roots of the form $\beta=\alpha\xi$ where $\xi=e^{i\pi/3}$. Notice that $\alpha$ is in the ideal generated by $\beta$ (as $\xi$ is the $6th$ root of unity in $\mathbb{C}$) and so $\mathbb{Q}(2^{1/3}e^{i\pi/3})$ is a splitting field over $\mathbb{Q}$ for $x^6-4$. $\beta$ has minimal polynomial $m_\beta(x)=x^3+2$ in $\mathbb{Q}$, and hence it is a degree $3$ extension.

[cloud walrus]

Dpao

[gritty sparrow]

alpha being in the ideal generated by beta is kinda meaningless since the ideal is always the whole field. You have to prove that all roots are in the subfield generated by beta, that is what will allow you to conclude that Q(beta) is splitting field. In fact it is not true that all the roots are generated by beta, I think.

[grizzled elm]

it definitley feels wrong that Beta would or that its degree 3

I think its prob best to go back and consider the polynomial (x^3-2) in $Q(\alpha)[x]$ which is $x^3-\alpha^3=-(\alpha-x)(\alpha^2+\alpha x +x^2)$

[cloud walrus]

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

[grizzled elm]

im overcomplicating things and it really is just Q(\alpha,\beta), alpha is not in the subring of beta

since it has 6 roots, should be degree>= 6 so just that alone shows my original is wrong

[chilly ocean]

is there an easy test to determine whether 2 elements of F_2 freely generate it

I'm trying to prove Out(F2)=GL2(Z) and I already have a surjective homomorphism of monoids End(F2)->(2×2 integer matrices), I just need to check that the intersection of Aut(F2) with the kernel of my map is exactly Inn(F2). this result ends up giving an easy test to determine whether a given endomorphism is an automorphism, but I was wondering if I could go the other direction with that

[next obsidian]

wat

F_2 has two things

if you don't contain 1

then you don't generate it

[chilly ocean]

I should have been more clear

not the field on 2 elements

the free group on 2 letters

sorry

[next obsidian]

oh

I think this is probably undecidable if I had to gguess

I think determining when things generate groups is a very hard problem or something

and also F_2 is fucked

¯_(ツ)_/¯

[chilly ocean]

well if the isomorphism holds (which I know it does) then I do have a test

provided you can look at an automorphism and determine whether it's inner or not

which you can

unless the map I already have is the wrong way to go about the isomorphism Out(F2)=GL2(Z)

but I doubt that since my function is pretty nice

[chilly ocean]

hahahaha

[patent crescent]

When talking about cosets, normal groups, factor groups, I keep seeing notation like an element multiplied by a group. This basically means an element multiplied by any element in the set since we are talking about equivalency classes right?

[novel loom]

Yes.

The idea is that gH represents the set of all elements of the form gh, where h is in H.

In additive notation, this would be g + H, the set of all elements of the form g + h.

You can think of this as the image of H under the map h —> gh

This map is called (left-)translation by g.

[patent crescent]

Thanks

[novel loom]

No problem. 🙂

[hidden haven]

oh ok I just remembered that as exactness of product lol

I did think about it

And it is true

Or at least I have been reading some weibel and it seems that some arguments also need it

So with that I can't see it not being true

And of course I didn't see any problems with the argument you gave

Ye I remember Weibel includes this as an axiom AB4 which says that product of epis is epi

And proves that R-mod always satisfies it

[next obsidian]

AB4 :3

[chilly ocean]

Is there a way to prove that any two distinct elements of order 2 of S3 form S3 besides brute force?

[next obsidian]

yeah

the subgroup they generate has to have at least 3 elements, but a group containing an element of order 2 has to have order divisible by 2

this means it's stuck at 4 or 6

by Lagrange, 4 is impossible as 4 doesn't divide 6

so you're stuck with the subgroup they generate being order 6, aka the whole thing

[chilly ocean]

nice

[chilly ocean]

Why do subgroups generated by two distinct elements of order two have an order >= 3?

Oh

Because of Lagrange's

[proud bear]

if x and y are the distinct elements of order 2, then the subgroup generated by x, y will at least contain 1, x and y

[chilly ocean]

I see

[open pilot]

Good proof

[chilly ocean]

Any hints for proving that if Z/Zm x Z/Zn is cyclic then gcd(m,n) = 1

[inland otter]

If gcd isn’t 1 show there cant be a generating element would be my first intuition

Or consider a generating element and try to get to bezout

[chilly ocean]

Well

I was thinking

Let the generating element be +1 for both m and n

You would still generate the entire group by carrying out the operation lcm(m,n) times

Unless I'm misinterpreting something

[next obsidian]

Compute the order of (x,y) in any product based on the order of x and y

You need an element of order nm for this group to be cyclic

From there you appeal to Bezout’s lemma

[rustic crown]

eevee

[coral shale]

Is there uh... a more 'obvious' example?

Z[sqrt-13] isnt clearly not a UFD to me D:

[rustic crown]

okie what about Z[2i]?

[coral shale]

Ok, its prolly cus i have to revise the meaning of UFD 😂

[rustic crown]

ah oops

[coral shale]

ill have a think and look again D:

[next obsidian]

I think the “canonical” choice is Z[sqrt(-5)]

This is the example I’ve always seen, but maybe because it’s in Dummit and Foote or something

[rustic crown]

here we physically removed the element i, so 2 and 2i aren't associates anymore! (as the only units are 1 and -1)

because of this 4 has two "different" factorizations
4 ~ 2 * 2 ~ (2i) * (2i)

[hidden haven]

How can you remove it physically when it doesn't physically exist?

[next obsidian]

Yeah it’s imaginary you dingus

[rustic crown]

C is nicer than R, C is more real

[hidden haven]

Listen to yourself

[coral shale]

soooo 2i becomes irreducible. And like you said we have 2i.2i = -2.2

I think I get this D:

[delicate orchid]

I’m team Z[sqrt(-5)] personally

[coral shale]

So whats so good about sqrt(-5) or -13

this looks totally nonobvious to me ><

[rustic crown]

yee, only factor of 2i are like 1 + i, and 1+i is no longer an element of Z[2i]

[hidden haven]

That's why it's good have fun trying to recall what the number in the root was when you write your next exam

[coral shale]

D:

[next obsidian]

6 = 2•3 = (1 + sqrt(-5))(1 - sqrt(-5))

[coral shale]

Is there some thm that says when Z[sqrt(-n)] is not UFD

n = 2, I recall is

[delicate orchid]

n >= 3

It’s not a UFD, I think

[next obsidian]

For 13, you can write 14 = 7•2 = (1 + sqrt(-3))(1 - sqrt(-3))

[rustic crown]

so like -3 and -4 are bad because they aren't integrally closed in their field of fractions... but -5 and -13 are even worse

[coral shale]

oh u do aa-bb = (a-b)(a+b)

[next obsidian]

Yeah

Why are -5 and -13 worse?

[rustic crown]

you can't fill in the "holes" to make it a UFD

[next obsidian]

Wut

[rustic crown]

but Z[2i] or Z[sqrt(-3)] are pretty close to UFD

[next obsidian]

How do you even quantify that

[rustic crown]

just need to add fill that missing i and (-1 + sqrt(-3))/2

[next obsidian]

Is this about the like Picard group or something

Or the ideal class group

[rustic crown]

[next obsidian]

In more classical language

[rustic crown]

yea this..

[next obsidian]

Like is it finite for -3 and -4

But infinite for -5 and -13

[rustic crown]

the class number for -3 and -4 are 1

[next obsidian]

Okay well the Picard group is the same thing as ideal class group

What’s the class number

[rustic crown]

size of the class group

[next obsidian]

Whaaa

Size 1 should mean UFD tho

[rustic crown]

class number for -5 is 2 i think

[coral shale]

I feel like I'm meant to know this stuff D:

[next obsidian]

Like the clas group is always non-empty and trivial iff UFD right?

No, this is algebraic number theory

[delicate orchid]

Ofc you’re not

[coral shale]

I am doing alg nt rn

[next obsidian]

Oh

[rustic crown]

like it's much more fundamental to have your ring be integrally closed in its fraction field to make it a UFD

[next obsidian]

Idk

Yeah

Wait but still, I thought the ideal class group being trivial iff UFD

[rustic crown]

which is why i used Z[2i] instead of Z[sqrt(-5)]

[rustic crown]

yea that's true

[next obsidian]

But if

[coral shale]

So does this always work out? what if a^2 - b^2 = (a+b)(a-b) gives you a prime 🤔

[next obsidian]

doesn’t this mean they’re a UFD

[rustic crown]

like class group of a number field K is the ideal group thingy of O_K

and O_K is automatically integrally closed

[hidden haven]

This stuff is all useless, I don't know any of it and yet I'm perfectly happy ignoring det whenever he talks about this

[rustic crown]

people don't care about non-integrally closed things ig

[coral shale]

they ask me if this blob is a UFD in the exam and im like 'wut'

D:

[next obsidian]

Was that only about integrally closed domains that the ideal class group says that

[weak oriole]

why
Prove it

[next obsidian]

Oh yeah I guess so, at least the definition I’m using

Has normal built in

[rustic crown]

it's pretty much same as the rational root theorem proof

[next obsidian]

I’m using scheme theory 🥸

[delicate orchid]

Idk the class group is kinda... swagtastic

[hidden haven]

🤓

[rustic crown]

AG hard

[next obsidian]

AG headass 🤓

I was looking at Utah’s old quals and one problem was to show C[x] is integrally closed

Lol

Like the first thing you learn when you learn what integrally closed means

Is that UFDs are integrally closed

Easy test headass 🤓

[weak oriole]

A is integrally closed implies A[x] is also integrally closed

iirc?

[next obsidian]

Yeah

Pretty sure

It’s at least true if it’s Noetherian

Probably even without

[weak oriole]

Neukirch discusses integral closure for ~2 paragraphs

so compactly

[next obsidian]

I think it’s probably easy to see using the R1 and S2 thing maybe

For Noetherian rings

[weak oriole]

what's R1 and S2 thingy

[next obsidian]

Uhh

Don’t worry about it, but if you want look up Serre’s criterion for normality

Actually it’s just

Easy

No need to bust out anything fancy lol

[weak oriole]

one more adjective for a ring

[next obsidian]

If R is normal, then consider an element of the fraction field of R[x] integral over R[x]

Then it’s also integral over K[x], and K[x] is a UFD so it’s normal

So the coefficients of its min poly live in K

But now you can like

There’s that thing about for a normal ring R, if you have an algebraic extension L of K = frac(R)

Oh holy crap nvm this might be hard

Okay I think this is hard

[coral shale]

🤔

[next obsidian]

I looked it up in the stacks project

At least as they present it, it’s very nontrivial

You have to do clever stuff to reduce to the Noetherian case and then use the notion of almost integral

[coral shale]

$\bZ[\sqrt{n}]$ for $n\in\bZ$

$$(a+\sqrt n)(a-\sqrt n) = a^2-n = a^2+(-n)$$

The difference between 2 consecutive squares is odd, so an $a$ can always be found to make this factorisable in 2 different ways.

I have to then think if $(a+\sqrt n) ,(a-\sqrt n)$ may be units

[cloud walrus]

Shuri2060

[coral shale]

Is this the right approach for the basic stuff we discussed earlier? 🤔

Also, I should probably make n non-square

[rustic crown]

you would also need to check for the irreducibility of the factors, which can be hard to generalize to weird values of n

[coral shale]

If the factors are reducible, that is fine right? It is only a problem if they are units

(I am trying to show I can find a number that can be split in 2 different ways)

[rustic crown]

yea it's fine... but you wouldn't know the two factorizations are "essentially different" unless you factorize them completely into irreducibles

[coral shale]

ahhh I see.

[rustic crown]

it coudl be something like
abcd = (ab)(cd) = (ac)(bd)

don't want this to happen

[coral shale]

and this is precisely what happens when n = -2, I suppose

[rustic crown]

yea it has to cuz Z[sqrt(-2)] is a UFD :p

(2 + a)(2 - a) = 6 = 2 * 3
where a = sqrt(-2)
both are just specializations of the factorization
a * a * (1 + a) * (1 - a)

[wooden ember]

If I quotient the free group on n generators by the product of those generators do I obtain the free group on n-1 generators?

I really need this to be true but I’m struggling to show it

I think it works right can’t I just take the first n-1 generators? They generate themselves and generate the inverse of the nth one so things should work right?

(Well this only show it’s a quotient of a free group on n-1 generators I suppose)

[terse crystal]

use some algebraic topology maybe?

Like covering space, those kind of things.

[rustic crown]

in think your case it's fine either ways...
we can easily define maps in both directions g : F_n/N --> F_{n-1} and f : F_{n-1} --> F_n/N
checking injectivity of f directly is hard, but you've already checked the surjectivity of f, which is enough to prove that the two compositions are identities on the respective groups.
gf = 1 is easy
and fgf = f follows, and since f is surjective fg = 1

[coral shale]

What is the process of quotienting here 👀
We have <a1, ..., an>
Is it <a1, ..., an | any permutation of the product a1...an > ?

[rustic crown]

i think we're quotienting by a fixed product, say a1...an

and by that we mean quotienting by the smallest normal subgroup containing a1...an

[coral shale]

Thanks, will explore this idea 👀

[rustic crown]

but in general if you wanna show something satisfies "no further relations" it's easier to act by the group on some nice object and show that any new relation won't act via identity so everything's good.

(although i've only seen one concrete example of this while showing PSL_2(Z) is isomorphic to C2 * C3)

[terse crystal]

I think it works, using subgroups of commutators in the picture, maybe you can check it for me

TG’/G’ is also the kernel because any Πa_i^s_i in the kernel, where a_i is the equivalence class of x_i, we must have that s_i=s_n for every 1<=i<=n-1 therefore all s_i are equal

[terse crystal]

What is n?

[lapis trail]

*of degree n

[terse crystal]

Also monic polynomials right?

[lapis trail]

No

I could adjust the formula if it were to be just monic

[terse crystal]

Please adjust it

[lapis trail]

Ok

[terse crystal]

Just by removing (p-1) in your picture?

[lapis trail]

Oh maybe I made a mistake

No it's fine

[terse crystal]

Then no, I don’t think it’s true. I calculated that number of monic irreducible polynomials of degree 3 over Z/pZ is p^3-p-C(p,3)-(p+2)C(p,2)

[lapis trail]

$p^n-{p \choose n}(p-1)$

[cloud walrus]

seth.delacroix

[lapis trail]

For monic

But let me think about what youbsaid

*said

It's an exercise in my book

[terse crystal]

So I have different result when n=3 then

$p^3-p-{p \choose 3}-(p+2){p \choose 2}$

[cloud walrus]

Cogwheels of the mind

[prisma shuttle]

what does the line over teh second term mean here

[terse crystal]

Conjugate

[prisma shuttle]

what do u mean conjugate in this context

oh wait

its a complex number

[terse crystal]

Conjugate of a+bi=a-bi

[prisma shuttle]

wait but like wut would happen if codomain wasn't C

then would there be a more general definition that could be used

[delicate orchid]

you'd use a different inner product

[terse crystal]

Don’t know, I only know that in C that bar means conjugate

[delicate orchid]

it's definitely complex conjugate in this context

lemme think

I think over R and Q you'd just use the fact that $\overline{a}=a$ and just write the inner product without the bar

[cloud walrus]

Wew Lads Tbh ✓

[delicate orchid]

for finite fields I have no clue

[terse crystal]

You can test it, like p=2, your result will be 8, mine is 2 which is the case: x^3+x^2+1 and x^3+x+1

[rustic crown]

There is a nice theorem which is pretty useful here.

[
x^{p^n} - x = \prod_{d | n} (\text{monic irreducibles of degree }d)
]
if you take degree of both sides, you get
[
p^n = \sum_{d | n} d \cdot N(d)
]
where $N(d)$ is the number of monic irreducibles of degree d.

use mobius inversion if you want a formula for it.

[cloud walrus]

det

[terse crystal]

Oh that’s brilliant… I thought I had to calculate for each n using induction or something 😂

[terse crystal]

Sorry made a mistake at the final line, G’ cup K =G’ cup T, not T cup K.

[coral shale]

with this kinda thingy from earlier, what goes wrong with this kinda thinking ---

$$\langle a, b\mid a^2, b^4, aba^{-1}b^{-1}\rangle \cong \langle a, b\rangle / \langle a^2, b^4, aba^{-1}b^{-1}\rangle$$

[cloud walrus]

Shuri2060

[coral shale]

Like this is not necessarily normal... right?

What's missing

[rustic crown]

right, you take the smallest normal subgroup containing it

not just the smallest subgroup

[coral shale]

But like uhhh

[rustic crown]

yea it's pretty eww to work with

[coral shale]

How do I put my thoughts into words...

[rustic crown]

unless there are some nice actions or finiteness conditions

[coral shale]

Like what's missing??? 😄

[coral shale]

I have to add the forward and backward conjugates by each of the generators to this to get the smallest normal subgroup?

[rustic crown]

probably conjugates by each element and not just each generator

[coral shale]

but each element is made from the generators

I think you generate the forward/backward conjugates of everything

https://images-ext-1.discordapp.net/external/VeXMpr0yd1Et7da7Ud0UgqaMDNsjDHIRugVQg5DgNz0/https/i.imgur.com/aLIrh0f.png

uhhh im looking at this result i saw on stack

[rustic crown]

yea but this lets you iterate the conjugation operation

a in H then gag' in H then hgag'h' in H

just a and gag' might not be enough is what i'm trying to say

[coral shale]

ah. uhhhh

ok ic ic

But as soon as you write this

https://i.imgur.com/mTQGBo2.png

[rustic crown]

only thing presentations are good to work with are constructing map from them :p

[coral shale]

It is implied

you quotient a, b

over the smallest normal subgroup containing the relations

[rustic crown]

yea that's the definition

[coral shale]

ok 👌 ty

Do we extend this above groups

rings, fields

[rustic crown]

there are free algebras just like free groups

and construction of tensor product is like a presentation

[coral shale]

does this go into category theory... or not that direction of thinking

[terse crystal]

presentation are different in things other than groups, like finitely presented module, it involves other stuffs like exact sequences.

[rustic crown]

at that level, category theory will be more language than any substantial theorem

and you can of course say the same in a different language

[coral shale]

ok . No clue what category is except a generalisation of algebra, somehow

[rustic crown]

so like a free commutative ring on n things is the polynomial ring Z[x1, ..., xn]

a presentation would be a quotient of this

[terse crystal]

It’s everywhere, like Jacobson basic algebra discusses category theory in volume 2 chapter 1

Limit and colimit aren’t defined generally though, but the rest is ok

[rustic crown]

i really like the description Aluffi gives

[terse crystal]

Gtm 211 Lang’s algebra also has category theory I think, but I remember it appears in the last few chapters

[rustic crown]

things, and things that go from things to things

it starts getting fun when you introduce more things that go between those things which were things going from things to things

[coral shale]

that's enlightening, thanks

[coral shale]

https://i.imgur.com/RexWqu7.png

For a general cayley graph

Is there an 'easy' way to visualize the right multiplication version

Like... uh...

idk how to describe it - but if you looked at just this, it's not 'obvious' what right multiplication does to an element

And so I don't find it obvious what conjugation does

[chilly ocean]

IS THAT A FUCKING YONEDA LEMMA

[hidden haven]

Did someone summon me?

[coral shale]

[hidden haven]

It's not yoneda lol

[coral shale]

wait wait i think I have an idea vaguely, can't put it in words much

[chilly ocean]

I see a square inside a square + arrows I think yoneda

[terse crystal]

So Whenever you see three objects six arrows then you think grothendieck…

[hidden haven]

If arrows are left multiplication, then right multiplication by x is all the arrows out of x

[coral shale]

So like I treat x as e

I basically consider x^-1(G)

I think

I think I kinda got it

[hidden haven]

Nice

[coral shale]

wasnt until recently i realised there are pictures for things u do in group theory

[terse crystal]

I don’t see the point. I draw a Cayley diagram when I want to use some algebraic topology on it not because I need to visualize anything…

[chilly ocean]

arrows

[coral shale]

😦

[coral shale]

idk, I feel like it's nice to visualize what these things are............

my way of seeing math

[hidden haven]

Nice

[terse crystal]

I see

[hidden haven]

They should make math 2 in which everything has a diagram calculus

[wise igloo]

Nice

[hidden haven]

[terse crystal]

Then he might enjoy diagrammatic algebra

[coral shale]

i probably would

the word diagram 👀

[chilly ocean]

dramatic algebra for short

[coral shale]

Me see words :

[wise igloo]

diaphragm algebra

[terse crystal]

We use Auto spelling… I can never spell diagrammatic myself

[hidden haven]

If anyone hasn't seen the proof of eckmann hilton by notation tricks I highly recommend doing that

[plush wasp]

diagrammatic

[hidden haven]

This has no prereqs btw

And is a very short proof on wikipedia 😌

[terse crystal]

In what book?

[hidden haven]

Wikipedia has it, titled as 2-dimensional proof

Actually how about I just send screenshots

[terse crystal]

Oh I see,thanks

[terse crystal]

Please

[rustic crown]

wait i don't get this

[lavish nexus]

wut

[coral shale]

you apply x^-1 on the graph

so x is now e

[hidden haven]

Statement and proof

[rustic crown]

on the right?

[coral shale]

wait i was thinking about conjugation

[hidden haven]

That second identity is called interchange identity

[hidden haven]

Idk maybe I don't remember what a cayley graph is lol

[hidden haven]

This is how you show that higher homotopy groups are abelian

And that 2 dimensional notation is extremely useful in category theory with some modification

[coral shale]

xa on the graph G
is where x is on a^-1(G)

is what I meant

is that what I mean 🤔

[rustic crown]

if the graph is corresponding to left multiplication you could only multiply the whole graph on right to preserve all the arrows and stuff

[coral shale]

What I mean by a^-1(G)

is you call where a was e now and redraw the graph

[rustic crown]

oh i kinda see now 🙈

[coral shale]

So this is what I meant by r^-1(G)

I hope.

hmmmmm

[rustic crown]

yea that's liek multiplying the whole graph on the right by r^-1

so it means if we multiply this new red graph on the right with r we get the old graph

[coral shale]

So if we want to know what (r^3s)r is................... (which should be r^2s)

We look at where r^3s is now (marked in red)

Which is what r^2s is on the old graph (the answer)

[rustic crown]

[coral shale]

So the red graph is more suitably called (G)r^-1

I see . heheh

Ok that makes total sense now 😄

The group/graph G acts on the left on r^-1 to make the red group/graph

[lapis trail]

If p(x) is irreducible then the ideal generated by p(x) is maximal. Is the best strat to show F[x] mod p(x) is a field? F is a field btw

[lethal dune]

F is field then F[x] is UFD so irreducible => prime => maximal ideal

probably easier

[lapis trail]

Thanks

[coral shale]

R[[X]] is defined to be the infinite series in X with coeffs in R

Can we make sense of this in any way with the original defn of
R[X]

[delicate orchid]

it's the completion of R[x] I believe

so it turns out we really did just let n -> infinity

[spice whale]

it's a superset of R[X]

[coral shale]

so what is this thing called

R[[X]]

[delicate orchid]

the ring of formal power series over X (with coefficients in R)

[coral shale]

for me googling

ok

[spice whale]

what happens if we do F2[[X]]/(1+X+X²+X³+...)

[delicate orchid]

I have no idea

I'm not 100% familiar with the actual ring operations I will be honest

I know addition is term-wise but multiplication?!??!

oh ok it is like polynomials

cool

[coral shale]

I think I prefer R^N

[delicate orchid]

oh for gods sake

[coral shale]

obvsly same thing

[delicate orchid]

R^N doesn't really have ring strucutre on it

and what if it's R[[X,Y]]?!?!

might get craaazzzyyy

[spice whale]

split it into R[[X]][[Y]] like usual

[coral shale]

it does

the functions from N to R

does have a ring structure?

You can add them times them

[delicate orchid]

point is R^N doesn't tell you how many terms it's in

[delicate orchid]

these are just the sequences in R

[coral shale]

🤔

well yes

[delicate orchid]

so.... yes.... cause R[[X]] is a ring

[coral shale]

and then how do we generalise this notion

to R[[X,Y]]

well isnt that the functions N^2 -> R

or no

[delicate orchid]

yeah that makes sense to me

I've never thought about it like that though

[coral shale]

if u have an infinite/maybe uncountable variables

well...