groups-rings-fields

[agile burrow]

https://math.stackexchange.com/questions/1781695/can-any-presentation-of-a-finitely-presented-group-be-reduced-to-a-finite-one

Oh but there are infinitely many generators here

Hmm this is a good question

I will think about it more

[coral shale]

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

[coral spindle]

Charles F. Miller III
wow

[rustic crown]

yea, these things feel very weird to me and i dunno whats the correct way to think about them

[next obsidian]

Woah this is wild! Because this is true for abelian groups

[agile burrow]

Group presentations are hard

[coral shale]

[coral spindle]

Free groups are wierd

[agile burrow]

Very hard

[next obsidian]

Because of “finitely presented => always finitely presented” for modules

[coral spindle]

A subgroup of a free group is free, but unfortunately even the free group with 2 generators contains a copy of the free group with infinitely many generators

[next obsidian]

Cursed fact

[coral spindle]

We just can't do the same things we would with Abelian groups in the general case 🤷

[next obsidian]

And the subgroup isn’t even that gross to write down

[coral shale]

nani

[rustic crown]

yea consider the subgroup generated by {x^nyx^-n} in F(x,y)

[next obsidian]

Free group on 2 elements being so fricked is one way to get at Banach-Tarski I believe

[coral shale]

@warm wyvern you'll love to read up later ^

[next obsidian]

I wanna consider the one generated by {y^nxy^-n} though

[rustic crown]

oopsie >.<

[rustic crown]

its also true for commutative algebras right

[next obsidian]

Yes

This one is kinda hard to prove tho

Like, it’s not as simple

I think it’s cuz you’re combining generation as a ring with generation of the ideal, module

[rustic crown]

yea >.<

how do you remember these lol

i always feel i might forget one of these cute sounding facts and they turn out to be massively wrong :p

[coral shale]

why is algebra:

1. look at proof of cool fact, understand it
2. forget proof, retain memory of cool fact

and sometimes skip 1

[next obsidian]

I have brain problems

[coral spindle]

That's like, all of math

Oooh there's too much out there to memorise ooooooh so spoooooky

[next obsidian]

It makes me forget important stuff and remember little nuggets of things

[coral shale]

But some of these are totally non-trivial to reconstruct proofs for

[coral spindle]

Yes

[coral shale]

but ig thats all of math when it gets hard

[next obsidian]

Also often I rmemebe these things because I went “hmm is X true?” Then I look it up and I go “oh wow!”

So I rmemebr it

[rustic crown]

my brain stops working if even slightest bit of atmosphere around me goes weird, like if my nose gets partially blocked i lose all thinking capabilities

.<

[coral spindle]

Ah so when you want to relax you can just plug your nose

and if you want to think again you can eat a curry

🧠

[rustic crown]

lol

[coral shale]

[coral spindle]

No

You should think of projective space as affine space + bits at infinity

the 'top-down' definition that has it as a quotient of lines in the larger space is misleading in this way

Also this is the wrong channel

[formal ermine]

oh sorry

[coral shale]

origin doesn't count as a line wut

[green shell]

hi do the rigid motions of a regular n-gon always leave at least one vertex fixed?

[coral shale]

is this linear alg?

man imagine deleting

[coral spindle]

This is geometry

[coral shale]

oh ic

[rustic crown]

no, (non-trivial) rotations fix nothing

[green shell]

ah yes I meant do the reflections always leave one fixed

[coral shale]

no?

do a line down the middle of a hexagon

that like doesnt touch any vertices

( what a great explanation)
https://i.imgur.com/HUQlWPD.png

[rustic crown]

make it a square

:3

hexagon too big

[green shell]

exactly so how can we possibly capture all reflections by labeling them s_k where k is the vertex that gets fixed

[coral spindle]

This works only in the odd case

[coral shale]

well regardless, there are only k reflections of a regular k-gon

[agile burrow]

nvm, I think you are correct det

[coral shale]

nani

[agile burrow]

that any finite generating set yields a finite presentation

The idea is that if you have a finite presentation, then you can consider the corresponding presentation complex

[coral shale]

wut about this

[agile burrow]

that has infinitely many generators

[coral shale]

ooo.

[agile burrow]

The idea is that if you have a finite presentation, the corresponding presentation complex has finite 1- and 2-skeleton

[coral shale]

skeletons. topology!?

[green shell]

weird yeah the way the proof in my book is written sounds like it's saying that we can enumerate all possible reflections based on which vertex it fixes

[agile burrow]

then the 2-skeleton of the universal cover is a partial free resolution of Z over ZG

and then it's an application of the same trick you use for finitely presented modules over a ring

[green shell]

but obviously not all reflections do that

thanks for the sanity check

[coral shale]

what is the best way to enumerate the reflections

I think s_n := s_1r_n

or similar

r_1 being the identity

ok slightly terrible notation

[agile burrow]

Right, because if you have some presentation with finitely many generators, then you can still consider the presentation complex. Again, the 2-skeleton of the universal cover yields a partial free resolution of Z over ZG, where the first term is finitely generated. But by Schanuel's lemma, the kernel of ZG^n -> Z is finitely generated

https://math.stackexchange.com/questions/1539598/finitely-presented-implies-always-finite-presented-for-algebraic-theories?rq=1

Here's something interesting that generalizes this to other algebraic theories

But I like my answer better

[past temple]

right

so given a subrepresentation U, we want to find a projector to U since the kernel of that projector will be a linear complement

but we want to make sure that this projector is a homomorphism of representations, as we want the kernel to be a subrepresentation

so the way we do that is, we start with any projector, then we apply the averaging idempotent to it

where "application" means the action of G on Hom(V,V)

being stabilized by this action is equivalent to being a homomorphism of representations

and then we know that the averaging idempotent applied to any vector in a representation of G is stabilized by G

therefore we have demonstrated a projector that is a homomorphism of representations, giving us a complement to our subrepresentation U, completing the proof

[formal ermine]

why are exact sequences? like I suppose they "appear naturally", but why?

what's so special about them? why not something else?

or is it just mere coincidence?

[void cosmos]

do u remember calc 3

a vector field is conservative iff its some F = del (f) for some function f?

yea turns out this shit nots always true

[lethal dune]

well failure of an complex being exact gives an invariant in some cases, so it's not like why exact sequence but when does it fail

[void cosmos]

u can have forms that are closed but not exact

a closed form is just a differential form which has derivative = 0

and an exact one is a differential form which is the derivative of some other form with less degree

[formal ermine]

are you talking to me?

[void cosmos]

yesa

yes*

[formal ermine]

uh I have no idea what you nor ryu are talking about, sorry

[void cosmos]

do u remember calc 3

[formal ermine]

I mean exact sequences as in 0 -> M -> M' -> M'' -> 0

[void cosmos]

yes

[formal ermine]

I never took calc

[elder wave]

moamen is trying to give an example

of de rham cohomology

[void cosmos]

okay

do you know

linear algebra

[formal ermine]

yes

[void cosmos]

okay

[elder wave]

since the failure of said statement is measured in the failure of a certain sequence to be exact

[void cosmos]

let V be a vector space and S be some subspace

do you know about the orthogonal complement

[formal ermine]

no

[void cosmos]

okay

first we need an inner product

do u know what that is

[formal ermine]

yes

[void cosmos]

okay so let S be a subspace

S^(dagger) is defined to be { x in V | x dot s = 0 for all s in S }

think of it similar to the annihilator (dot product-wise lmfao )

[formal ermine]

I see

[void cosmos]

now sometimes you can write the whole vec space as the direct sum of a subspace and its orthogonal complement

like with R^n

do you see that

[formal ermine]

yeah

[void cosmos]

thats not always true

exact sequences tell us when this is

when this is not.

thats it

relaly

[formal ermine]

what would the sequence look like?

[void cosmos]

think of it

x dot s = 0 ...

kernel? idk

try to experiment with is

with it*

now

u can think of these things but with modules now

u do not have inner products over modules but the quotient structure isnt trivial

so you can htink of this "completing the missing parts" like u did with orthogonal complements

as quotients and shit

so an exact sequence would be when this quotient is exactly whats missing .. something like that

or not this quoteint

this submodule

so basically exact sequences are things that capture a certain situation

cohomology is something that measures how much a certain occurance fails

ur doing the dry version

homological algebra which is this with no motivation or intuition

[green shell]

I just needed to read on the book wasn't claiming that any reflection s must fix a vertex - rather that we only need the reflections that fix at least one vertex to construct all elements of D_n

[void cosmos]

most of the time the motivation is geometric or topological ( i think? )

u would have learnt about those in AT or DG

but with a better understanding

[agile burrow]

Exact sequence

[void cosmos]

@rustic crown yo you free for a moment

remember when you proved that if a is in the intersection of all prime ideals then a is nilpotent using localizations

i read some stuff about them , can u rewrite it now

i proved (trivially ig) that S^-1 R collapses to 0 iff S has 0

is this enough

and i know how to localize wrt to an element and an ideal

[rustic crown]

iff S has a nilpotent

not 0

[void cosmos]

can you re do it now

[rustic crown]

oh wait

nvm

[void cosmos]

so i can do it on the exam if i see it XD

[rustic crown]

me dumb

[void cosmos]

no u smart

[void cosmos]

now that i know some langauge can u re do it

[rustic crown]

if it has a nilpotent then by closure it has 0

:3

[void cosmos]

and teach me 😦

[rustic crown]

okie

[void cosmos]

its enough to show that R_a collapses to 0

R_a is S^-1R with S = {a^n | n in N}

[rustic crown]

so first it's clear that a nilpotent element will be inside every prime right

[void cosmos]

using zorn?

[rustic crown]

nah, def of prime

[void cosmos]

oh yeah

yea yea

[rustic crown]

if a is nilpotent, then a^n = 0 in p, so a in p

[void cosmos]

yes

[rustic crown]

so we need to show that if s is not nilpotent, then it doesn't lie in some prime

[void cosmos]

yes

[rustic crown]

take S = {1, s, s^2, ...}

[void cosmos]

ok

[rustic crown]

so you have R --> R[s^-1]

[void cosmos]

which one

[rustic crown]

need to find an prime in R that doesn't contain s

and pullback of any prime of R[s^-1] can't contain s

[rustic crown]

oh me talking about the natural localization map

[void cosmos]

ok

XD what was it again

r --> r/1 ?

[rustic crown]

ye

[void cosmos]

ok ok

[void cosmos]

pl

ok*

there is a correspondonce

[rustic crown]

yep, so can we find a prime in R[s^-1]?

[void cosmos]

between prime ideals in S^-1R and ideals of R that do not intersect S

[rustic crown]

yep

[void cosmos]

(a)

?

[rustic crown]

hm?

[void cosmos]

1 moment 😦

okayy

[rustic crown]

when you can find prime ideals in a ring

[void cosmos]

did we assume a is nonnilpotent

[rustic crown]

(say maximals :p)

[void cosmos]

yea yea

[rustic crown]

where is a

[void cosmos]

a is in R

and we assumed a is not nilpotent

[rustic crown]

oh i think i wrote s, but okie

[void cosmos]

why cant we just preimage

nvm

ik this is local so there must exist atleast one

prime ideal

but idk which one but i dont really care ig

like we do not need to care right?

cuz this is zorn's lemma in disguise

?

ie every element is contained in some maximal (prime) ideal

right?

[rustic crown]

no

1 is not contained in any maximal

[void cosmos]

😠

okay but i think i got it

the whole point of localizing this is that a in R_a is now a unit

rihgt

[rustic crown]

yep

so a prime in R_a won't contain a

and hence it's preimage also won't contain a

[void cosmos]

yea contradicting that its in the intersection

cool ass

ty ty ty

[rustic crown]

yee so we only need to see why R_a has primes

and that's because it's not the zero ring

[void cosmos]

yes

isnt any localization local

like atleast has a unique maximal

?

[rustic crown]

whut do you mean?

[void cosmos]

local as in local ring

[rustic crown]

no no

[void cosmos]

XD ok

[rustic crown]

localizing at primes give you local rings

if R is a ring and p a prime, then take S = stuff outside p

so you invert everything that is not in p

and so this new p now beccomes a unique maximal

[void cosmos]

ohh

[rustic crown]

by new p, i mean p(S^-1R)

[void cosmos]

yea

the image of the canonical

[rustic crown]

this S^-1R is denoted R_p in this case to confuse everyone

[void cosmos]

yea

bad question

the localization is a field right?

oh

iff the S is the whole ring

-{0}

nvm

[agile burrow]

You'd need R to be integral domain for S to be closed in that case

[rustic crown]

yea

[void cosmos]

yea

cool guys and girls

tysm

[rustic crown]

[solar shore]

quick question

Indeed, D4, the dihedral group of order 8, has five
subgroups of order 2 and three of order 4.

this doesnt sound right to me

isnt it only 2 of order 4?

R_90 and R_270

is there a third im missing

[sonic coral]

which ones of order 4 do you have

[rustic crown]

you want subgroups of order 4, not elements of order 4

[cloud walrus]

MyMathYourMath

MyMathYourMath

[rustic crown]

what's that times cbrt2 - 1

[solar glacier]

id have to write it out i cant do that in my head lol hang on

but by times cbrt2-1 whered that even come from

[rustic crown]

x^3-1 = (x-1)(x^2+x+1)

lol

so what you wrote is just 1/(cbrt2 - 1)

[solar glacier]

i get -1

[rustic crown]

you mean +1?

but yea whatever

the point is Q(cbrt2) = Q(that thing)

[solar glacier]

that thing being the 1+cubrt2+cbrt4

[rustic crown]

yep

[solar glacier]

how do you see that

[rustic crown]

it's just a very specific thingy

we wanted to compute [Q(that thing):Q]

and it's clear that Q(that thing) is a subfield of Q(cbrt2)

so this degree [Q(that thing):Q] divides 3

now if you can ensure that thing is not rational, you're done

[solar glacier]

so it must be 1 or 3

but are the minimal polynomials identical for the two splitting fields

[rustic crown]

splitting fields where

we not making any polynomial split

[solar glacier]

or i mean Q adjoing

[rustic crown]

minimal poly are not the same

but the degree of the extensions are

cause like they're the same thing lol

(hi mniip )

[simple valley]

is it the case that if $N \lhd G$ then $<Ind>\nolimits^G_N V = (\text{permutation rep of $G/N$}) \otimes V$?

[cloud walrus]

mniip

[delicate orchid]

this is true when V is the identity representation, so it must be true in general!

[elder wave]

induction

[simple valley]

what I mean is normality of N "untangles" the induced rep

[delicate orchid]

I do indeed think this is true regardless

thinkin bout the character

[delicate orchid]

yeah, characters are the same - it's true

infact, I don't even think you need N to be normal

[simple valley]

wdym

[delicate orchid]

that iso holds even without N being normal if you just take G/N to be the set of cosets with natural G-action

[simple valley]

but then it's not \otimes V

[delicate orchid]

ah yes, duh

sorry

it does need to be normal

[main needle]

The prime ideals in $F[X]/I$ are the ideals the corresponding to the ideals in $F[X]$ generated by the prime factors of $f$ where $I = (f(x))$ ?

[cloud walrus]

ru0xffian

[simple valley]

are you sure? I'm no longer sure the question even makes sense

if V is an N-rep you can't quite make a G-rep out of it

[delicate orchid]

I'm pretty convinced both those reps have the same character when G is normal

[simple valley]

I was thinking something more along the lines of upgrading G/N-reps to G-reps

but G/N is not a subgroup of G unless direct product in which case trivial

[delicate orchid]

right that sounds more like lifting than inducing

[simple valley]

it's not induction if there isn't a coil of wire

[delicate orchid]

I'm checking le book to make sure

[simple valley]

ah I think what I really have here is a semidirect product

you can probably formulate something relating lift from (H \ltimes N)/N with induction from H

[solar glacier]

for this secon patt

part

do I write it as

$(1+\theta)(1+\theta+\theta^2)^{-1}$

[cloud walrus]

MyMathYourMath

[solar glacier]

then find that inverse

[delicate orchid]

cause I'm seeing we have the induced character for an arbitrary subgroup is [\text{Ind}N^G \chi(g) = \frac1N\sum{h \in G} \chi'(y^{-1}gy)] where [\chi'(g) = \begin{cases} \chi(g) & g \in N \ 0 & g \not \in N \end{cases}]
If this is equal to the product of the permutation character of $G/N$ with $\chi$ the corresponding $\bC G$-modules will be isomorphic.

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

the permutation character of G/N is obviously [\sigma(g) = \begin{cases} |G/N| & g \in N \ 0 & g \not \in N\end{cases}]

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

hmmm ok now I'm not quite seeing how they multiply to give the same thing

[simple valley]

ok I have an easy counterexample at hand

I just covered inducing S_5 reps from A_5 reps

[delicate orchid]

nothing involving tensor products and module isomorphisms is easy

[simple valley]

and it doesn't match up

[delicate orchid]

yeah my gut reaction was surprisingly.... wrong!!!!?!?!

[simple valley]

permutation rep of S_n/A_n is gonna be U+U' (trivial + sign)

[queen bluff]

Hello ! If M is a semi-simple R-module, why do we have that for every submodule N of M, M isomorphic to the direct sum N+M/N ? I don't know how to prove this

A semi-simple R-module is such that for every sub module N, there exist P such that it is the direct sur of N and P

[delicate orchid]

hint, each submodule and quotient of a semi-simple module is semi-simple

gonna take this opportunity to elaborate a bit on my hint, consider the natural surjection from M -> M/N as well

[queen bluff]

If I restrict this surjection to the "supplementary" of N I have to show that this is an isomorphism right ?

[delicate orchid]

you mean the submodule that sums with N to make M?

[queen bluff]

Yes

[delicate orchid]

yes, you would

[queen bluff]

But I don't know how to use your first hint sorry

[delicate orchid]

don't worry too much about it you seemed to know it already

it was to lead you to this realisation (if it was affective or not is up for debate)

[queen bluff]

I don't know how I am supposed to use that

[past temple]

im currently struggling on proving the following claim:

let G be a finite group and U, W representations of G, with dim U = 1. I want to show that W is an irreducible representation if and only if U (x) W is

[delicate orchid]

I'm telling you to just ignore it, lol

[south patrol]

Oh lol I misread it

[queen bluff]

Oh sorry !!

[south patrol]

I thought you meant (M + N)/N lol

[delicate orchid]

no worries boss, sorry for the confusion

[south patrol]

Oh lol this was hw for me

Basically to get you started, suppose you have a proper subrep of W. Can you concoct a corresponding subrep of U tensor W?

You can then do a similar thing in reverse

[delicate orchid]

this question is so much easier with characters why do they make you people work with modules...

[south patrol]

Yeah

Idk lol

We did it before chars introduced

[delicate orchid]

like I cannot see why ||U \cong CG as 1 dim => W (x) U \cong W|| - this is almost certainly wrong

[south patrol]

Kinda funny tho that they do general theory kinda painstakingly and then all applications we had were using C

What

Why does the first thing hold

Lol

[delicate orchid]

it's a dim 1 free module over CG? see this is what I mean

[south patrol]

Hm why would it be CG

[delicate orchid]

_ _

[south patrol]

Isn't that saying all 1 dim irreps r the same

[delicate orchid]

if I mean as group algebras yeah probably

I think I was thinking just vs-es

[south patrol]

ye which is not what we're interested in right lol

also like

CG isn't even a dim 1 representation right unless # G = 1

Bruh

[delicate orchid]

yeah this is why I just don't think of them as modules, ever

[south patrol]

Or am I smoking

Oh

Okay

[delicate orchid]

why on earth is the ring not dim 1 over itself

idiotic

[south patrol]

Fair nuff

Lol

I need to

Revise

the proof

Of p^alpha q^beta theorem

😩

[delicate orchid]

module theorists need pick some rules and stick to them

[queen bluff]

Does this works ? I have the short exact sequence : 0->N->M->M/N->0 and 0->N->N+P->(N+P)/N->0 , maybe I have to show that P/N is isomorphic to M/N ? I'm getting tired so maybe this is bullshit

And P/N is isomorphic to P right ?

Idk

This makes no sense

Sorry

[coral spindle]

With what action?

I mean tbh we already have a pretty great description of what happens in this case

I can't remember what theorem in Isaacs, maybe 6.16 is the one?

[south patrol]

You should take a complement to N and work with that

Say M = N (+) P

You have to show that N (+) P is isomorphic to what

[delicate orchid]

thank u stack exchange very cool

[queen bluff]

N + M/N

[south patrol]

Yup

So what isomorphism should we try to prove

[coral spindle]

I'm talking about Isaacs' character theory of finite groups

[delicate orchid]

yeah I've found it now dw

[south patrol]

To show N (+) P iso to N (+) M/N

[queen bluff]

P iso to M/N

[south patrol]

Yup

But really kinda easiest to just eliminate M entirely

so P iso to (N (+) P)/N

Have a go at that

[queen bluff]

Yes

Isn't it just a consequence of the first isomorphism theorem ?

[south patrol]

Exactly

[coral spindle]

6.16 was indeed it 😎

[delicate orchid]

shame I cannot decipher it

[south patrol]

So true

[coral spindle]

This describes the situation

So it is true that G/N is relevant

[queen bluff]

Oh nice ! And all the isomorphism I consider are isomorphism of modules @south patrol ?

[south patrol]

Ye

[delicate orchid]

the proof I can follow perfectly but the statement of the theorem might as well be Spanish

[queen bluff]

Nice thank you so much guys !

[coral spindle]

well maybe this special case is more comprehensible

Theorem 6.16 is quite general

[delicate orchid]

yeah ok I got that one

I should read this book actually good recommend

[coral spindle]

It is an incredible book

[pastel cliff]

what book is this

doesnt look like dummy and foot fetish

[white yoke]

Why does this "1 ≠ 0 and if x is any element of R, then x or 1 − x is a unit" imply that R is a local ring?

[south patrol]

Well if R is a local ring then you have a definite description of what the maximal ideal looks like ig

So try doing that and see what u get

Not sure how to say other than give it away

But maybe you want it given away lol

[white yoke]

All the non units right?

[south patrol]

Yes

[white yoke]

But then what

[rustic crown]

show that forms an ideal

[south patrol]

Ofc a boring way to do this is just to use known facts on the jacobson radical i think

So you're basically just reproving essentially that

In a special case

If you've seen that

[rustic crown]

what are the known facts about jacobson radicals lol

[white yoke]

Idk anything about Jacobson radicals so might be better to avoid that

[south patrol]

Ye dw

[rustic crown]

oh you mean that thing

[south patrol]

Yh

[rustic crown]

j in J if something something 1 - rj is a unit for every r

[south patrol]

Ye

Iff

[rustic crown]

i see

[rustic crown]

from there you would need to then show this is the maximal ideal of the local ring... i.e. every proper ideal is contained in this.

[white yoke]

If I let m be the set of all non units (i.e. elements of the form 1-x for x a unit) then as long as m is an ideal, it will be maximal (because it contains any other proper ideal since proper ideals cannot contain units)

[rustic crown]

right

but need to be a bit careful with what you wrote in the first parentheses

if 1-x is a unit, then that doesn't mean x is not a unit

[pastel cliff]

can somone help understand what's happening here pls

i know this is just linear algebra

but still

[rustic crown]

oh wait you've phrased it a little differently..

[pastel cliff]

we start with a matrix representing a submodule of a free module and row reduce it to arrive at a nice basis for that submodule?

[rustic crown]

what i mean is, x non-unit implies 1-x is a unit. but the other way is not true. x unit need not imply 1-x is a non-unit

[white yoke]

oh yea

good point thanks

[pastel cliff]

i promise this isnt a lot

[coral shale]

I see Smith:

try lalg channel too

[rustic crown]

yea it isnt'

okie lets do that in fields first

say you start with some weird matrix and then i ask you do to do row and column operations

if i just allowed you to do row operations, then you would be left with the reduced row echelon form

but if i allow you to do both, you can make that matrix look really nice

it will be a diagonal matrix with 1s and 0s and there are exact rank(M) many 1s

okie even before that... the question is why would you want to do this weird thing

you can think of a m x n matrix over R as a linear map R^n --> R^m right

if you allow yourself to change basis on both sides, can you make this linear map look really nice?

if R was a field, you can pick basis (x1, ..., xn) and (y1, ..., ym) such that the map is x_i --> y_i for i = 1, 2, ..., r and other x to 0

where r is the rank of your matrix

if R is not a field, then this becomes a little more interesting, since we can't easily kill of an entry of the matrix

over pids, you still can convert this matrix into a nice form using row and column operations

and they have illustrated this Z

so you started with a map Z^2 --> Z^3

the image is the subgroup generated by two weird vectors

when you do a row operation, it's actually multiplying your matrix on the left by an elementary matrix, which in particular is invertible

similarly, when you do a column operation, you're multiplying the matrix on the right with an elementary matrix

[pastel cliff]

[rustic crown]

and multiplying by invertible matrices is same as changing basis (on either the source or target)

[pastel cliff]

wdym change basis on both sides

[rustic crown]

like multiplying a matrix on the right is same as precomposing your map Z^2 --> Z^3 with a nice isomorphism Z^2 --> Z^2

[pastel cliff]

but we're thinking about this the opposite way right? like a submodule M of a free module R^n must(?) be the image of some linear transformation like this from a basis of the submodule to the whole module?

[rustic crown]

yea if you assume that the submodule is also free

[pastel cliff]

is it enough to assume it's finitely generated?

[rustic crown]

yep, then you get its image of some R^large enough thingy

and if you don't assume that you can just surject it by R^(size of your submodule)

anyway

the point is that if you wanna undersand stuff like (co)kernel or (co)image of this map, you can first convert this matrix to a nicer map and then be happy

[pastel cliff]

this is the motivation we get at the beginnning

[rustic crown]

say you started with the matrix A : R^n --> R^m and then did operations which ends you with B = P^-1 A Q.
where Q : R^n --> R^n and P : R^m --> R^m are nice invertible matrices.

[cloud walrus]

det

[rustic crown]

so with the isomorphism P, you can identify the image of A and image of B easily

and it also induces an isomorphism between the cokernels of A and B

only difference being that the matrix B looks wayyyyy nicer than A

[south patrol]

Upside down comm diagram

[rustic crown]

hoomans are so weird

when drawing graphs right and up are the natural choices

but everywhere else, right and down looks natural

[pastel cliff]

ok but back to the row reduction for a moment

we just row reduce to get a basis for a submodule

[rustic crown]

okie

yea in that example row reduction was enough to get you to a nice final matrix

but in general you'll also have to use column operations

[pastel cliff]

and that happens because the matrix encodes the image of Z^2 -> Z^3

which is itself the submodule we’re talking about

[rustic crown]

weirdly phrased, but yea something like that

[pastel cliff]

this is weird stuff

i know it’s just general lin alg

but still

[rustic crown]

you'll get used to it lmao

you never understand math

you just get used to it

[rustic crown]

anyway, so if M is a finitely presented module, i.e. you have finitely many generators, and the relations are also then finitely generated then this thing can help you understand M nicely

first use the generators to get a surjection R^m --> M

and then the kernel of this it eh relations among the generators

which we assume is finitely generated as well

so you surject that by R^n --> R^m --> M

so image of the first map is exactly the kernel of the second map

so finitely presented modules are exactly the cokernels of matrices R^n --> R^m

if you assume R is noetherian, then a finitely generated module is automatically finitely presented

and pid's are noetherian :3

so if you wanna classify finitely presented modules, a good start is to trouble m x n matrices

if you can classify how they look like up to equivalence, then you're really close to classifying finitely presented modules

[next obsidian]

I here’s my classification

[rustic crown]

[next obsidian]

{R^n/M | M is a fg submodule, and n is a natural number}

[chilly ocean]

you understand some math, and then you get more math that you don't understand
and it keeps on piling up

[rustic crown]

uwu

[white oxide]

bruh isn't this problem my professor posted for the practice midterm kinda like obvious for the <-- direction

[south patrol]

I mean

many iff things have one direction way easier than the other

So yeah shouldn't be too surprising overall

This is a typo actually lmao

Should be H subset K or K subset H, of course

[white oxide]

good man that potato

yeah makes sense

[south patrol]

Np

But yes with the modification one direction is indeed trivial

as u say

[past temple]

im confused by the application of the orthogonality of characters here to complete this character table

orthogonality of characters states that (X_U, X_V) = 1 if U and V are isomorphic, 0 otherwise, where U and V are irreducible

[south patrol]

That isn't orthogonality of characters though

That's column orthogonality

[delicate orchid]

This is row orthogonality, column orthogonality is a little different

[past temple]

how does column orthogonality work?

and can row orthogonality also be used to complete character tables?

[summer path]

both orthogonality can be used to complete character tables

[chilly ocean]

on this topic, does anyone know a good reference for developing some intuition on character theory

[past temple]

how could row-orthogonality be used to complete the above character table?

do we essentially get a system of linear equations?

[south patrol]

Yes

[past temple]

ok, that makes sense more or less

im a bit confused about the second-to-last line here

I understand why that identity is true; namely, the irreducible representations appear as direct summands in the regular representation FG, with multiplicities equal to their dimension

however, how does that identity actually help us to compute the character of a last unknown irreducible?

we would need to know what the character of the regular representation of FG is, but how do we even compute that?

[delicate orchid]

The regular representation is immaculately well behaved

It’s |G| on the identity and 0 everywhere else

Sorry, regular character

[south patrol]

My guy said immaculately

[delicate orchid]

You can see this as the regular rep is the permutation representation of G on itself, which only has any fixed points when the identity is acting on G, in which case it has |G| fixed points

[delicate orchid]

Yeah like Jesus

[south patrol]

That was a well-conceived joke

[delicate orchid]

Unlike the rest of the stuff I post

[past temple]

ahhhh right thats perfect

[south patrol]

Lol

[past temple]

the character of g is the number of points in G fixed by g

and e fixes every point, while every other element fixes no points

[delicate orchid]

Yus

[upper pivot]

right? nah left regular representation

[past temple]

ok so

knowing that makes the computation trivial

[delicate orchid]

Yur

[past temple]

so beautiful

another note

when computing the derived subgroup of a group

does it suffice to demonstrate some subset of that derived subgroup

then to show that the quotient by that sub-subgroup is abelian?

[toxic zephyr]

so i've seen \leq used to denote subspaces/subgroups/subrings. does it also work for subfields? just want to confirm

[tender wharf]

a field is a sort of commutative ring so sure

its just notation anyway

[formal ermine]

leq usually means "subset with structure"

so yeah

[chilly ocean]

How do we know that G_\alpha is a group?

[coral spindle]

Try proving it. This occurs because the lower groups form a chain of subgroups

[chilly ocean]

From transfinite induction

[chilly ocean]

they happen to contain the predecessors?

as a subgroup

The problem is I dont see the explicit construction, where is it?

Of what

oh wait its the next line

for non-limit ordinals

I get it now thanks

[dawn latch]

to find all the generators of a group (e.g. Z20), can i just list all the order of each element and conclude that those with order 20 (the same order as Z20), are the generators of the group?

[coral shale]

To be precise, you mean cyclic group

not just any group

That is the definition

[dawn latch]

oh yes, my bad for the confusion

[coral shale]

what u said

[dawn latch]

thank you

[coral shale]

element with order equal to order of group is a cyclic generator

[dawn latch]

how do i find all the distinct subgroups of a cyclic group?

in the case of Z20 (under + mod 20), i know |Z20| = 20. And the positive divisors of 20 are 1, 2, 4, 5, 10, 20.

thus,
<20/1> = <20> = {0}
<20/2> = <10> = {0, 10}
<20/4> = <5> = {0, 5, 10, 15}
<20/5> = <4> = {0, 4, 8, 12, 16}
<20/10> = <2> = {0, 2, 4, 6, 8, 10, 12, 14, 16, 18}
<20/20> = <1> = {0, 1, 2, ..., 17, 18, 19} = Z20

however, with the answer key that our professor provided us, {0, 5} and {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} are said to be subgroups as well. how is that the case?

[delicate orchid]

I disagree with your professor

[north sand]

they arent. probably typo

[dawn latch]

oh

thank you, i was scratching my head on how it got to there

[dawn latch]

then, if i could add, the ones I listed are all the distinct subgroups?

[north sand]

yes

[flint crater]

Just want to confirm something. To proof two modules M and N are Jordan-Holder isomorphic is it enough to proof that a chain in M and in N are isomorphic? Or do I need to proof that every possible chain is isomorphic?

[simple valley]

A way to see whether that's all of them is to see that <a> = <gcd(a, 20)>, and the values gcd(a,20) can take are exactly the divisors of 20

[dawn latch]

Is that only for Z_n? how about other cyclic groups, like U(18)?

[simple valley]

Z_n are the cyclic groups

not sure what you mean by U(18)

a cyclic group is one that is generated by a single element, so a cyclic subgroup of G is generated by a single element of G

[dawn latch]

oh

[dawn latch]

my bad, i mean this: U(n) is defined as the set of integers in the range [1, n-1] that are coprime to n, under the operation of multiplication modulo n

or, U(18) = {1,5,7,11,13,17}

[delicate orchid]

(Z/18Z)^\times

[simple valley]

that's not always cyclic is it

[delicate orchid]

It is in this case

But no not always

It’s only cyclic when n is 2, 4, some power of an odd prime, or two times some power of an odd prime

[dawn latch]

oh what

[simple valley]

right so in this case U(18) is isomorphic to C_6

[tribal moss]

Ok, U(18) is always cyclic, but U(n) in general may not be.

[dawn latch]

i'm still not in isomorphism, unfortunately. it's just that U(18) was under the list of cyclic groups in our module.

[dawn latch]

oh ok

[delicate orchid]

Ignored once more

[delicate orchid]

Two groups are isomorphic means you can think of them as being “the same”

[dawn latch]

oh apologies

[delicate orchid]

You’re just changing what the elements are called

[dawn latch]

i'm not sure i get this

[tribal moss]

It's true but didn't really need a response, did it?

[simple valley]

I'm irked by the use of an "always" quantifier on a statement without free variables

[delicate orchid]

Maybe just a thank u…

[dawn latch]

oh ok, so basically U(18) is cyclic since 2 multiplied by 8?

[dawn latch]

thank u ^ ^

[delicate orchid]

2*9 but yeah

[dawn latch]

i was absorbing all the information i am forgetting to reply lol

[tribal moss]

APologies, but it looked like Lowell was about to get himself confused by your "not always".

[dawn latch]

this is hard

[simple valley]

U(18) is cyclic because it is generated by 5

the theorem Wew cited might be a little too advanced here

[delicate orchid]

Yeah, you won’t be told to prove U(n) is cyclic unless it actually is lol

[delicate orchid]

And you prove things are cyclic by finding a generator

[tribal moss]

It looks like you're not being expected to get any deep insight into U(...) in general right now.
U(18) is just an example of a group where it is not obvious that it's cyclic when you look at its definition, but it turns out it is anyway.

[simple valley]

anyway my turn especially now that wew is here

[delicate orchid]

Sorry guys I have to go water my plants….

[simple valley]

why are all reps of an odd order group complex (except the trivial one)

this is in context of the formula for $\frac{1}{|G|} \sum_g \chi(g^2)$ for real, complex, quaternionic reps

[cloud walrus]

mniip

[delicate orchid]

I think it has something to do with the correspondence between conjugacy classes that square to themselves and real-valued characters

Like if there’s a conjugacy class in an odd order group that squares to itself then it has to just be {e}, but I can’t remember the details

If you want to use the frobby-schur indicator uhhh

ah I see, when the group is odd order $\sum_{g \in G} \chi(g^2) = \sum_{g \in G} \chi(g)$ as the map g -> g^2 is an automorphism of G I think

[cloud walrus]

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

[delicate orchid]

Whatever u get the point

[simple valley]

huh

[delicate orchid]

Maybe not an automorphism it’s definitely a bijection though which is the main thing

[agile burrow]

I'm thinking you can look at like \chi(g) = \chi(g^-1) and then use orthogonality to say something

[delicate orchid]

Then $\sum_{g \in G} \chi(g) = \langle 1, \chi \rangle$ which is 0 unless chi is trivial

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

And 0 implies the character isn’t real valued right?

[simple valley]

a group of odd order has no elements of order 2?

[delicate orchid]

Lagrange

[simple valley]

oh duh

then it's obvious

[delicate orchid]

How can they character be real when no odd roots of unity in R?!? Exactly….

[tribal moss]

You mean all irreducible reps here?

[delicate orchid]

I presume so, the regular rep is always integer valued

[simple valley]

yes ofc

[delicate orchid]

caring about non-irreducibles that's when the pain begins

[delicate orchid]

also u can just take any irreducible and add it to it's conjugate - what fun!!

tally ho!

[dawn latch]

If we let G = <a> and |a| = 24, how do i list all the generators for the subgroup of order 8?

i already listed down a^3
since <a^3> = {a^3, a^6, a^9, a^12, a^15, a^18, a^21, a^24} or |a^3| = 8

but how do i know the other ones? do i just manually check each element from a to a^24?

[delicate orchid]

you're missing the identity in <a^3>

[lusty marlin]

a^24 is the identity element

[delicate orchid]

oh yeah, duh

I quite literally cannot count

anyway - checking each one won't be too bad if you remember tricks like "a is a generator => a^-1 is a generator"

but after a few you might spot a pattern

[dawn latch]

oh

ok then, i'll just check each manually, hoping there's a pattern to it

thanks

[tender wharf]

look at the orders

[dawn latch]

from this

how was this implied? Because Z24 is a cyclic group of order 24 generated by 1, there is a unique subgroup of order 8, which is <‎ 3 · 1> = <3>

[delicate orchid]

what other subgroups of order 8 could there possibly be, really

trying to think of how to explain it without just sylow-theorem bashing it

assume there was another group of order 8, as it's a subgroup of a cyclic group it is itself cyclic - it must be generated by an element x of order 8, but all elements of order 8 are in <3> already, so <x> = <3>, thus there is only one subgroup of order 8

[dawn latch]

ohh, that makes sense.

this is the second part of this btw:

[delicate orchid]

alternatively C_24 isn't iso to C_8^3 so there's only one sylow 2-subgroup

[rotund aurora]

?? lol

[delicate orchid]

this is quite literally the same question

[rotund aurora]

you have even+even=-1, is that a typo or I'm missing something?

[dawn latch]

oh yes, i mean the first pic and the second pic are connected.

thanks btw

[delicate orchid]

bit of a lol, that looks like a typo to me

-1 is simply not in 2Z

[rotund aurora]

i guess I should replace Q3 for Q2 in the linear combination

?

fuck it that exercise is kinda dumb anyway

[delicate orchid]

yeah both coordinates are coprime so I can believe they span everything, if you can show it

[rotund aurora]

its supposed to create some insight on how you would obtain the points I think in the first place, but meh

[tender wharf]

notice anything about those ...

well I suppose since it says so

theyre coprime

if k generates a subgroup and n is coprime to k then n generates the same subgroup

[dawn latch]

oh, that's neat

[tender wharf]

There are many ways to check this

fundamental theorem of cyclic groups is probably the most accessible

You could also coset it

[dawn latch]

a separate (but maybe related) question, given 𝑈(18) = {1, 5, 7, 11, 13, 17} under mult. mod 18. how do i find the number of distinct subgroups of this group? is there a formula

1. {1}, the trivial subgroup consisting only of the identity element.
2. {1, 5, 7, 11, 13, 17}, the subgroup generated by 5, which has order 2.
3. {1, 7, 13}, the subgroup generated by 7, which has order 3.
4. {1, 11}, the subgroup generated by 11, which has order 6

this was what i came up with, but how do i know if there's more

[delicate orchid]

do we mean distinct up to isomorphism or literally every single one

[dawn latch]

i think the latter, since we haven't reached isomorphism yet

[delicate orchid]

you could always just google what an isomorphism

anyway there's one subgroup for each factor of 18

[dawn latch]

for 1, 2, 3, 6, 9?

[delicate orchid]

and 18

[tender wharf]

well

If you have classification of finite abelian groups

[delicate orchid]

they don't know what an isomorphism is of course they don't

[tender wharf]

Actually you don't need that

sigh

Do it by hand then

there's only 6 elements to check

[chilly ocean]

actually i though i understood the diagram but

if we select G' G and G''

as H G G/H respectively

obvioslu they are isomorphic

and the diagram doesnt tell nothing more than that the lower row is exact

the statement only says that there exists a diagram

and not that for all exact sequence G' G G"

.

[hot lake]

uh yeah ?

that works for any short exact sequence

if 0 -> G' -f-> G -g-> G'' -> 0 is exact then G' is isomorphic to ker g and G'' is isomorphic to G/ker g, and they give an isomorphism of short exact sequences

[chilly ocean]

yeah but diagram doesnt say for any such exact sequence but rather that there exists such groups

[hot lake]

a diagram is just a diagram

there is text accompanying it

[chilly ocean]

i mean the text starts from more precisely right?

and it doesnt say for all

[hot lake]

what's at the bottom of the previous page

[chilly ocean]

ah is the diagram statement starting from here?

[hot lake]

"for example to say that 0 -> G' -> G -> G'' -> 0 is exact ..."

they are considering any exact sequence

so whatever they say next works for all of them

[chilly ocean]

yeah okay now it makes better sense

cool thanks

someday ill be able to read diagrams like these

[cloud walrus]

DraK

[carmine fossil]

Well because if this is true and I understand correctly,you can describe all finite fields via multivariate polynomials in a sense

[hot lake]

yeah

it turns out every finite field F is of the form K(alpha1) for some alpha1 in F

[carmine fossil]

That's a significantly stronger statement than what I expected

[coral spindle]

This is a fine proof, but you can simply observe that F is now a K-vector space, hence has order p^n where n = dim_K F.

(this is me saying that your proof is very overcomplicated)

[carmine fossil]

Well I haven't really proved that

That's the problem

[coral spindle]

Well it is very easy to prove :)

In fact I might remind you that it is part of the definition of the degree F:K

[carmine fossil]

Is the order p^n because you can construct an isomorphism from a K-vector space of dim n to K^n?

And K^n will have p^n elements

[coral spindle]

Yes

[carmine fossil]

Ok,I definitely overthought this

[carmine fossil]

Is proving this easy?

[coral spindle]

It’s not too bad actually

You can prove it by showing that the group of units of a field is cyclic

That proof is a bit tricky but it’s, well, not too bad lol

[coral spindle]

N.b. this only applies to finite fields. The general statement is that any finite subgroup of a group of units of a field is cyclic.

[coral spindle]

When I first heard this saying, I didn't understand it, but now I'm used to it.

[white yoke]

In a group ring R[G], I don't understand what the trivial units are? Are they the units in R or are they the units in G?

[coral spindle]

The units in R

At least, this is the terminology I'm used to

You should double check with the text you're reading.

[white yoke]

Okay thanks

[coral spindle]

Context!!!

[white yoke]

It does say the units in R but something someone said to me confused me because they seemed to suggest it was G

Never mind then haha

So in Z[C_3] the trivial units are {1,-1} ?

[coral spindle]

If that's the definition then yes

I'm sure you can see why someone might be inclined to call the elements of G multiplied by the units of R "trivial units"

[white yoke]

Oh yes I've heard that as a definition too, that's what I meant

[odd abyss]

anyone know what that means?

tensor product to the exponent

[coral spindle]

The "tensor power"

[south patrol]

One defines it analogously to normal products

like $v^{\otimes n} = v \otimes \dots \otimes v$ n times

[cloud walrus]

potato

[odd abyss]

I mean seriously

[south patrol]

I mean that is the name lol

The [...]th/st/rd tensor power of the vector (or vector space or whatever according to context)

[odd abyss]

s $v^{\otimes n} = v \otimes \dots \otimes v$ n fois

[south patrol]

Oui

[cloud walrus]

Baba

[south patrol]

exactement

[formal ermine]

🥖

[south patrol]

Why are you translating one word to french

lol

[coral spindle]

I didn’t want to step on potato’s toes 😊

[south patrol]

I don't have toes

[obsidian sleet]

potatoes

[south patrol]

AYY

[coral spindle]

I didn’t want to step on any part of you

[south patrol]

thank

[odd abyss]

ok thank you

because im french

Just to be sure

If I have

$v^{\otimes 2}$

[cloud walrus]

Baba

[odd abyss]

$v^{\otimes 2} = v \otimes v$ ?

[cloud walrus]

Baba

[odd abyss]

or

$v^{\otimes 2} = v \otimes v \otimes v$ ?

[cloud walrus]

Baba

[delicate orchid]

why would we define it to be the second one

[odd abyss]

here

[south patrol]

n terms so the former

[odd abyss]

tensor product which must appear n times

[south patrol]

It's just like normal products

with multiplication given by tensor product

[delicate orchid]

yes but why in god's name would we define it to be this?? We'd keep it consistent with regular multiplication

[odd abyss]

OK so it's that

[delicate orchid]

yup

[odd abyss]

Ok thank you!

[rustic crown]

[karmic moat]

$v^{2\otimes} = v\otimes v \otimes v$ troll face

[cloud walrus]

anamono for anamono

[charred crescent]

dumb question, but if two conditions are "equivalent" that means one if and only if the other right?

[karmic moat]

yeah

[charred crescent]

okay thank you

[hollow mica]

what does this mean?

[formal ermine]

what part are you confused about

[hollow mica]

the isomorphism with C

[formal ermine]

you add an element that satisfies x^2 + 1 = 0

which is obviously our i

[hollow mica]

I don't get what you mean by "add an element"

[formal ermine]

modding by (x^2 + 1) is like setting x^2 + 1 = 0

forcing it

[hollow mica]

oh yeah that makes sense

because any term with (x^2 + 1) disappears

[formal ermine]

just like in R[x]/(x^2) we have x^2 = 0 and thus only linear polynomials

[formal ermine]

yes

now do minus one

and you get x^2 = -1

so every time x^2 appears, you can "replace" it with -1

which is equivalent to i, the imaginary unit

[hollow mica]

so the resulting polynomials look like:

f(x) = c
f(x) = ax + b

[formal ermine]

yeah

because anything with x^2 will turn into a linear term

thus by induction blabla only linear terms

[hollow mica]

this all checks out

but is there a general way to make this precise

like the notes I'm reading just dropped this on me

[formal ermine]

Yes

this was the ring theory explanation

I personally prefer the field theory one

it goes along the lines of

modding a field out by a polynomial is like taking its splitting field over that field

i.e. you add all roots of your polynomial to the field (minimal field that contains the base field and the roots of the polynomial)

the roots of x^2 + 1 in the algebraic closure of R (which is C) are precisely \pm i

so R[x]/(x^2 + 1) = R(i) = C

another way to see this is that [C : R] = 2, and C being algebraically closed. though this might mean nothing to you

R(i) means minimal field over R that contains i

[hollow mica]

hm idk any field theory probably a good time to learn some

I just picked up some comm alg notes so I could understand tensor product over modules

[formal ermine]

I'm gonna take a bath now, maybe someone else will be able to help you in the meantime while I'm gone

[hollow mica]

thanks for the help

[formal ermine]

np

[hollow mica]

dumb question but why does 0 = 1 imply a ring is trivial

[rustic crown]

because you can muliply both sides by any element of the ring

[hollow mica]

If 0 = 1 in a ring R, and x \in R is not equal to 0 (and thus or 1), then x = x * 1 = x * 0 = 0.

nice

[warm shoal]

I am trying to find the cosets of H

Aren't they just H+0, H+1..., H+4 ?

I know there are 5 because |G|/|H| = 5

Do I have to say anything about left and right cosets?

In this case it's all the same right?

[coral spindle]

In this case the left and right cosets are the same, yes

[warm shoal]

Should I state that 🙂 ?

[coral spindle]

They are. You should try and prove these are all of them, which is to say that you should try showing that they are all distinct.

[coral spindle]

If the question asked you just to 'find the cosets' it is likely that the asker was already aware that they are the same.

[warm shoal]

oh gotcha

how do I prove they are distinct?

I got all the elements of each

and they encompass all the elements of Z15

and I know H+0 = H+5 = H+10

[coral spindle]

So if you've done this, then you're done.

[warm shoal]

😄 wooo

thank you @coral spindle

[hollow mica]

It is true that there is a ring isomorphism between R[x]/(x^2 + 1) and C, but since multiplicative inverses exist in C, shouldn't they exist in the first ring too?

but like, while 3 + 4i has inverse (3 - 4i) / 25, the polynomial 3 + 4x in the first ring doesn't have an inverse

[hollow mica]

I guess this logic is faulty then

[coral spindle]

Multiplicative inverses do exist in R[x]/(x^2 + 1). Try describing them.

You've confused the element 3 + 4x of R[x] with the element 3 + 4x + (x^2 + 1) in R[x]/(x^2+1).

[hollow mica]

oh they have to multiply to x^2 + 1

[coral spindle]

No that's not true

[hollow mica]

ok yea they def exist

uhhhh

[coral spindle]

x^2 + 1 = 0 in the quotient ring

[hollow mica]

(x^2 + 1) = {f(x) * (x^2 + 1) where f(x) \in R[x]}

R[x]/(x^2 + 1) = {g(x) + (x^2 + 1) where g(x) \in R[x]}

in this quotient ring, the multiplicative identity is 1 + (x^2 + 1)

[coral shale]

yes

[coral spindle]

N.b. it is also x^2 + 2 + (x^2 + 1)

there are many ways to represent it

[hollow mica]

so to find the inverse of the coset (3x + 4) + (x^2 + 1), I need to find an f(x) \in R[x] so that
f(x)*(3x+4) + (x^2 + 1) = 1 + (x^2 + 1)

[coral shale]

set I = (x^2+1)

[coral shale]

seems so.

[coral spindle]

Hint: use the (extended) Euclidean algorithm

[formal ermine]

beat me to it

[hollow mica]

x^2 + 2 = (3x + 4) * (1/3x + 1/2) - 17/6x
(3x + 4) = (-17/6x)(-18/17) + 4
-17/6x = 4(-17/24x) + 0

am i doing something wrong

[coral spindle]

I suggest you think a bit more about why the hint is useful

[hollow mica]

yeah I'm suppose to get a 1 in the last two equations so then I can back substitute

ok scratch that those are just details, if R is isomorphic to S as a ring, and S also has a field structure, then R should also admit the same field structure and thus be isomorphic to S as a field?

[coral spindle]

Yes

[hollow mica]

well assuming the field structure S admits is identical to its ring structure + defining inverses, versus some completely different thing

well I guess R could also admit the "completely different" ring structure since they're already isomorphic as rings

[prime sundial]

is it possible to construct a group presentation of Z^3 with only 2 relations? so an abelian group of 3 generators, where the commutativity is given with only 2 relations?

[warm shoal]

Any insight would be greatly appreciated

[barren sierra]

what are the conditions to be a subgroup of G?

[warm shoal]

for all x, y in H, x*y must be in H (for H being a subgroup of G)

[barren sierra]

cool

[warm shoal]

and for all x in H, inverse x must be in H

[barren sierra]

ok so let x, y be in C_a

[warm shoal]

as I just learned 🙂

[barren sierra]

try to show x*y is in G

just start trying and see how far you get and where you get stuck

[warm shoal]

yes sir

my first stuckness is the *

what is our operation here?

[barren sierra]

whatever the operation in G is

you don't know more than that