groups-rings-fields

[runic hemlock]

that's the definition of adjoining elements

[dusty sapphire]

If $a^3$ is algebraic over $\mathbb{Q}$ then $a$ is algebraic over $\mathbb{Q}$? Where $a \in \mathbb{R}$

[cloud walrus]

BLツ

[hidden haven]

You can construct a polynomial that it satisfies

[dusty sapphire]

Its true false question that’s why

If one can satisfy but if there is one a for that it is not true then it will be false so how to proceed

Okay😅

Is there any element from reals which is not algebraic over any subfield of Real number

[rustic crown]

any element alpha is algebraic over Q(alpha)

[dusty sapphire]

pi?

Element from real number

[rustic crown]

yea, Q(pi) is a transcendental extension (of Q), but still is a subfield of R

[dusty sapphire]

Ok or is there any element which is not algebraic over $\mathbb{Q}[i]$?

[cloud walrus]

BLツ

[rustic crown]

yee, pi

[dusty sapphire]

Haha

Okay thank you

$a \in \mathbb{R}$ could be algebraic over $\mathbb{Q}[\sqrt{2}] but may not be algebraic over \mathbb{Q}, how to find such element?

Lol

[cloud walrus]

det

[rustic crown]

that's never gonna happen

have you read about degree of an extension yet?

[dusty sapphire]

Yep

[rustic crown]

that's one of the simplest tool for understanding extensions

so if F/k is an extension, then alpha in F is algebraic over k if and only if [k(alpha):k] is finite

[dusty sapphire]

Its gone from my head

[rustic crown]

and this also behaves nicely with a tower of extensions

if k --> F --> E is a tower then [E:k] = [E:F][F:k]

if alpha was algebraic over Q(sqrt(2)) then [Q(sqrt(2), alpha):Q(sqrt(2))] is finite

since [Q(sqrt(2)):Q] is finite

[Q(alpha):Q] should also be!

[dusty sapphire]

Cool

[dusty sapphire]

How to find such poly for this?

$x^n-a^3$?

[cloud walrus]

BLツ

[dusty sapphire]

For some n

[rustic crown]

if a^3 satisfies a polynomial f(x), then we get that f(a^3) = 0, this means that a satisfies g(x) = f(x^3)

[rustic crown]

well, a^3 algebraic over Q, doesn't mean it actually lies in Q. that polynomial may not even have coefficients in Q because of that

[dusty sapphire]

Got it

[dusty sapphire]

Could you describe little more

f( a^3) =0 this is clear then?

[rustic crown]

um so let's take an example

a^3 algebraic over Q means it satisfies some polynomial f(x)
say it was x^5 + x + 1

so we must have that (a^3)^5 + (a^3) + 1 = 0

which is a^15 + a^3 + 1 = 0

so a satisfies the polynomial x^15 + x^3 + 1

[dusty sapphire]

Now we get polynomial in a^3 got it

[delicate orchid]

“The first problem sheet is just an easy refresher on group theory”
The problem sheet:

I am so fucked

[hidden haven]

omg

[delicate orchid]

First one I got is first iso

Fuck knows with the second

[hidden haven]

by second you mean (c)?

[delicate orchid]

Yeah

a) was just “what is the canonical homomorphism from G to G/N”

[hidden haven]

It's a standard theorem about universal properties it's a good thing to try and prove

[hidden haven]

The most important and often forgotten part of universal properties 😌

[delicate orchid]

it took an hour of intensive thought to understand a single diagram

Is it because the homomorphism defined by the universal property is unique? So if we have two quotients that result in the same group H they must be isomorphic?

[hidden haven]

Not sure what your argument is

[delicate orchid]

That makes two of us

[hidden haven]

Assume you have a group N and 2 universal pairs (Q,q) and (Q',q')

And you have to show an isomorphism between these

Don't use any elements

Only the universal property

[delicate orchid]

Yeah I gathered that doing it element wise is a fools errand

[hidden haven]

Well I mean you have to do some set theory, but don't do any beyond what is required in the universal property itself

[hidden haven]

But this should be a good hint about how to start

[delicate orchid]

I’m guessing it’s some combination of functions in both their diagrams that result in an isomorphism between the two

[hidden haven]

You have to apply to universal property to something, and there are 2 universal properties here (one for each pair) and very few things that you can apply this universal property to

[hidden haven]

Ye just draw both pairs in the same diagram

Instead of 2 separate diagrams

Then it should be clearer where you can start

[delicate orchid]

See that’s what I was thinking but there’s no arrows going out of H and none going into G so I don’t see how that helps me too much

[hidden haven]

H is any group

Don't draw H in your diagrams

The universal property works for any H

[delicate orchid]

Oh

Right yes of course

R is a group yes

Yeah I think I’ve got it now

[lethal dune]

where do I read about abelian categories and snake lemma

[hidden haven]

[lethal dune]

yes I'm into categories now

[hidden haven]

first chapter of homological algebra book

Lang has that at some point

[delicate orchid]

Wait shit the unique homomorphism doesn’t have to be injective my idea doesn’t work

[hidden haven]

Do not think of isomorphisms as bijective homomorphisms

Isomorphisms are homomorphisms with inverses

[delicate orchid]

Yeah I can see how $\overline{\phi}_q$ could be an inverse to $\overline{\phi}_r$ if they’re themselves bijective but I don’t see how it holds if they’re not

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

I wish I could latex the diagrams I have

[hidden haven]

Use the universal property to prove that their compositions are identity

[delicate orchid]

Alright I’ll go and think about it for a bit

Can’t believe I just used the words “trivial endomorphism” unironically... what have I become?

I’m pretty sure I’ve done it though

Ty for your help moldi I wouldn’t have even gotten started without you pointing out that H is an arbitrary group rather than

idk whatever I thought it was

I guess I thought it was kind of like a for all statement? Like Q is a quotient if it holds for all subgroups? So I couldn’t pick an arbitrary one? I see that’s wrong now though

[barren sierra]

I have said "trivially this is the trivial group" multiple times in some of my homeworks

Which just sounds so fucking dumb lol

[delicate orchid]

“Obviously this is the trivial group, you fool, you moron” is a good replacement imo

[barren sierra]

So true

I shall be putting that on my next HW

[gusty halo]

i think i've once read something along the lines of "even the most inexperienced reader can see that ..."

[elfin orbit]

given two proper subgroups G and H of a larger group, is there a term for their combination, i.e. the group composed of gh and hg for all g in G and h in H?

[lethal dune]

in general GH = {gh | g\in G and h\in H} may not be a subgroup, but if GH=HG then it's a subgroup

[unique berry]

I did that and got points taken off

My prof said: "don't use "obviously" or "trivially" when you don't even know what you're talking about"

[broken stirrup]

if one of them is a normal subgroup then GH=HG= G V H where last one denotes the join of G and H which is a group generated by the set G U H

[hidden haven]

G ∨ H being the general name 😌

[quick harness]

Can I ask something ?
Let's say you have a polynomial f.
What would be the properties of the polynomial so that f(2) = f(0) ?

[hidden haven]

x-2 divides f(x) - f(0)

[gusty halo]

tfw sniped

[quick harness]

Thanks

[hidden haven]

[lethal dune]

what is this product called?

[hidden haven]

Join not really a product

[lethal dune]

oh

subgroup containing both?

[hidden haven]

In a poset, a join of a set of elements is their sup and and the meet is their inf

Smallest subgroup containing both, that's what sup would mean

[lethal dune]

so subgroups are poset under inclusion

[hidden haven]

Yes

[lethal dune]

[pastel cliff]

when i say "this is an algebra" what do u guys interpret algebra to mean

moldi you already know this problem lol

[hidden haven]

[pastel cliff]

or maybe not

my prof defines an algebra as any set S closed under at least one k-ary operation

[gusty halo]

of course i think it means an object in the eilenberg moore category of some given monad

[pastel cliff]

https://tenor.com/view/eggman_slap-gif-24023679

[pastel cliff]

certain people have led me to believe this is missing something or just straight up no what is usually meant

[hidden haven]

There are many different kinds of algebras in math

[gusty halo]

sometimes they Lie to you

[pastel cliff]

i at least get that

couldnt tell you what it means but i know it's a thing

[hidden haven]

What you know is an algebraic structure. Then there are algebras over rings/fields. What Phil said is an algebra over a monad, and what the sticker talks about is the closely related algebra over an endofunctor

[south storm]

There’s also the notion of "an algebra" or "an algebraic structure" in universal algebra

Which just is a structure over a language without relation symbols

[pastel cliff]

when you say "over" something you mean like an added structure on something like Z\Z2

[hidden haven]

That's the thing nitezba knows

[south storm]

Is Z\Z2 a fancy way of saying 2Z+1?

[hidden haven]

Usually it's like the thing that it's over provides some way of acting on the algebra in some way

It's not a formal thing as far as I know

Like vector space over ℝ or over ℂ

Means that you are allowing scalar multiplication by ℝ or ℂ

But ℝ or ℂ need not sit inside the vector space for this to happen

[pastel cliff]

hmmm

ok somewhat get that

[south storm]

There‘s also an algebra of sets from measure theory

[pastel cliff]

[hidden haven]

I think you can treat that as a special case of an algebra over ℤ/2ℤ or something like that

[pastel cliff]

i think imma just roll with it for now

we spent the entire lecture today just playing with "aLgEbRaiC StRucTuReS"

[south storm]

Usually people don’t just say "Let X be an algebra" without any context, so I wouldn’t worry about it

[pastel cliff]

dude made us cut up a square and spin it 👶

[tight nova]

wait im like so confused. so there is a unique ring homomorphism from Z to any ring, but arent here multiple ring homomorphisms from Z to Z^2 given by projection?

[hidden haven]

Did you really throwaway 196882 accounts

[tight nova]

yes.... one for each question i get wrong 😦

[pastel cliff]

one for each of the hoes 💯

[south storm]

The projections are from Z^2 to Z

[hidden haven]

Well time for another one

[tight nova]

whoops you know what i mean

lmao

wait actually

[delicate orchid]

hello r/numberphile I have more diagram questions

[delicate orchid]

moldi you scoundrel is this a valid thing to do

can I just smash diagrams together like this

[pastel cliff]

who says you cant

[tight nova]

it really is time for another account

[delicate orchid]

moldi is about to

I can feel it in my bones

[lethal dune]

[hidden haven]

Wtf is this

What is H lol

What is phi

[delicate orchid]

any group you so please

[pastel cliff]

deez nuts?

[delicate orchid]

and it's just another universal property meme so phi is the product of the projection homomorph with the memer

[hidden haven]

Oh so you are saying that pi_1 is the quotient by G_2 map

[delicate orchid]

yes

did I get my G_1s and G_2s backwards

[hidden haven]

You don't get phi from phi_1 and phi_2 like that

[delicate orchid]

yeah I thought that might cause a problem

[hidden haven]

F

You can't even do that set theoretically lol

[delicate orchid]

oh well, I feel like I'm on the right track kinda anyway

[hidden haven]

Combining 2 maps out of a product group

[delicate orchid]

this whole universal product memery is so hard to use

actually no it's easy to use it's hard to prove that something satisfies it

[hidden haven]

Once you get used to it you will never look back 😌

Yeah proving that something satisfies a universal property can be annoying

That's how you realise that set theory is annoying

Lmao

[delicate orchid]

oh for gods sake I had the dashed arrow the wrong way around

[hidden haven]

Are you proving the univ prop of products

Lol

[delicate orchid]

it's fucking trivial with the arrow the other way around, since G1xG2 is homomorphic to a subgroup of H you can just map G1 (or G2) into whatever that homomorphic preimage is like you would if you were mapping it back into G1xG2

[hidden haven]

Yes 😌

[delicate orchid]

ffs I've spent half an hour on this

not sure if it's a or a

[hidden haven]

😌

[pastel cliff]

devastating kek

[delicate orchid]

I should go back and revise my answer to the first iso one because it is currently a complete mess

[lethal dune]

votes

[hidden haven]

The council will decide your emotions

[delicate orchid]

abelian diagrams

[pastel cliff]

ENOUGH with the abelian stuffs

[lethal dune]

no one asks about non abelian diagrams

[hidden haven]

We were being taught chain homotopies today

I was not paying attention

[lethal dune]

not paying attention is the norm for me

[hidden haven]

Everyone quick look at #cats

[strong valve]

why don't you use the tag bro

@\everyone

[hidden haven]

[pastel cliff]

yeah please moldi, ping everyone

[lethal dune]

https://tenor.com/view/just-do-it-shia-la-beouf-do-it-gif-4531935

[pastel cliff]

all 77000 of us

we're waiting

[hidden haven]

Guys honorable actually has @ everyone privileges

Is this a bug

[strong valve]

do it

[hidden haven]

[lethal dune]

let everyone know it's a bug

[pastel cliff]

#abstract-chill

[hidden haven]

[strong valve]

you mean let @\everyone know it's a bug

haha

[hidden haven]

We don't have the permission lol

[lethal dune]

prove it

type @everyone

[hidden haven]

uh oh what did you just do

[delicate orchid]

did you know I actually have admin perms

[hidden haven]

It went from 2000 online to 20000 online

Or did it

[lethal dune]

🙈

[strong valve]

yes i did know it

[pastel cliff]

give

[delicate orchid]

never

[next obsidian]

I have Chmonkey perms

[pastel cliff]

if i prove that the set of all positive integers P is the smallest subalgebraic structure of (R, +) that contains 1 and is closed under addition, can i say that the naturals (P u 0) must inherently have those properties as well

ik it's somewhat obvious i just dont wanna have to rewrite my justification so much

[rustic crown]

you just need to check closure of P u {0} right?
that can be done by taking a few cases

[pastel cliff]

i hope that makes sense actually

[rustic crown]

0 + 0 is there

0 + a = a is there for any a in P

a + b in P for a, b, in P by closure of P

[delicate orchid]

yeah you need to make sure the sets still closed, cause introducing some element could introduce some wackyness

[pastel cliff]

ok lemme type something unnecessarily verbose up rq

closure of P under addition is done before

idk if subseteq is the right symbol there or if i should just use epsilon for \in

this prof is dummy picky about proofs

"this also suffices to show that 0 is an identity element of the set"

[gusty halo]

looks good, but you didnt show that N is the smallest such set yet

[olive mirage]

this is a very unconventional definition of algebra

[pastel cliff]

yeah i know i know

my prof says algebra instead of algebraic structure

[pastel cliff]

probably pf by contradiction for this ig

[pastel cliff]

is this actually valid

[gusty halo]

no joke

[hidden haven]

I suggested that we should have more math stickers

[gusty halo]

nice

[hidden haven]

It was very quickly rejected

[gusty halo]

oof

[pastel cliff]

why

[hidden haven]

Shitty mods ngl

[pastel cliff]

mfw most mods arent honorable apparently

[hidden haven]

[gusty halo]

how many years do you need to spend here to obtain that title anyway

[hidden haven]

You don't need years you just need to get noticed apparently

[olive mirage]

just be loud enough ,it is what I did I suppose

[hidden haven]

Like we nominate someone and no one knows them because they hang out in just 1 channel and then they don't get in

[hidden haven]

Guess who I'm talking about

[pastel cliff]

im never gonna get it when i spend most of my time here shitposting in chill or asking dumb shit in advanced

[hidden haven]

Dw I'll nominate you someday

We'll overthrow the government together

[pastel cliff]

[rustic crown]

currently i'm hanging out in #groups-rings-fields#cats and #latex-help

[gusty halo]

i dont visit discussion and chill enough to have an accurate guess

i feel like most of the meta talk around the server happens there

[hidden haven]

Oh ye having impactful pfp helps

I have no clue wtf your pfp is Phil

[rustic crown]

like saketh

[gusty halo]

its a figure on the cover of the best analysis book

[hidden haven]

I don't either

Nominations are mostly out of advanced

[rustic crown]

(what's the book?)

[gusty halo]

amann & escher

[hidden haven]

I'll give you a hint, 4 letters, space, 1 letter

[olive mirage]

for undergrad analysis I really like the one I use which has a name that I will remember if I think about it hard enough

[hidden haven]

Unrecognisable pfp

[olive mirage]

Abbott there we go

[pastel cliff]

beans

[olive mirage]

but I think the problem almost all real analysis courses/texts I've learned from have is the Hartshorne problem. They in no way prepared me for any of the times/ways I would actually use analysis in my life.

[gusty halo]

thats the neat part

you just dont do any analysis later on

no but srs amann & escher is p good

[olive mirage]

I will give it a go next time I teach the course. But I think I'm teaching complex before I teach real again.

[gusty halo]

been a while since i used them though, since these days im doing nothing but category and topology and a bit of model theory

[hidden haven]

Homielogical algebruh when

[pastel cliff]

no homo

[gusty halo]

i wish i could find the time

probably soon though

hows weibel going?

[hidden haven]

Slow lol it's kinda boring but we also learning

Today I studied homie alg with saketh and clerk

On call

It was very nice 😌

[gusty halo]

oh wow

private tutoring from clerk

[hidden haven]

Yes

We were reading Beck's thesis

[gusty halo]

i was trying yesterday but at some point noticed i dont really know what to do with this stuff

[hidden haven]

We were just gonna read the proof of the monadicity theorem

[gusty halo]

and then i just looked at the examples

[hidden haven]

Clerk stopped us

Made us figure it out

While giving us some hints when we were being slow

Took 2 hours but I now understand this theorem really well

[gusty halo]

i saw the proof of monadicity already in riehl, but it was a different version than what was presented in the thesis

[hidden haven]

Yeah that's a more refined version

[gusty halo]

i skipped it in the thesis anyway

but no joke i was actually having trouble even reading the things

[hidden haven]

Oh yeah clerk doesn't use string diagrams for 2 categories

[gusty halo]

im a bit more comfortable now with string diagrams for monads at least, but still very shaky in basically everything else involving these

[hidden haven]

We were using it and he wasn't familiar with them, he's used them for monoidal categories only or something

Saketh and I cannot do cat theory without string diagrams

Clerk would say that the counit of the adjunction is UF → 1 and we'd have to draw a string diagram to make sure that's right

[gusty halo]

i feel like it would be pretty nice if you could write and read them fluently

[hidden haven]

I have become fluent

[gusty halo]

bruh

[gusty halo]

at what cost tho

[hidden haven]

I can draw them instantly pretty much

Lmao

I think it's a fair cost

They contain all the information of commutative diagrams and algebraic expressions anyway

[gusty halo]

that reminds me today i wanted to start the chapter in riehl on kan extensions after i finished the monads stuff 2 days ago

havent even started yet

[hidden haven]

Hell I proved a that a collection of maps is a chain homotopy using string diagrams a few days ago

[gusty halo]

how does that work

[hidden haven]

That involves adding maps

We just added 2 string diagrams lmao

And abused distributivity

Like take 2 string diagrams and just put a + in between

And now you can put this inside a larger string diagram

And distributivity memes

It works out

Somehow tho clerk could solve problems faster with commutative diagrams than we could with strings

[gusty halo]

ha

so maybe its not worth it after all

[hidden haven]

Lol it wasn't because of the notation

[gusty halo]

although i recall even clerk saying a while ago that string diagrams are useful

[hidden haven]

He was able to do cat theory faster

[gusty halo]

yeah ok, i guess he has a lot more experience

[hidden haven]

But he'd say an equation and we could instantly tell whether it was true or false 😌

Found a mistake in his equations because the string diagram broke 😌

And also was able to see what fixes the equation 😌

[gusty halo]

so did you just read marsdens note on string diagrams and then just tried using them until it clicks or how did you go about getting proficient

[hidden haven]

Is Marsden the arxiv paper

[gusty halo]

yeah

that one i recently complained about

[hidden haven]

I read that, struggled a lot, figured out my own formal system 😌

Which I have also explained to det and saketh

[gusty halo]

.

[hidden haven]

Det is not fluent though he hasn't used them much

[gusty halo]

i wish to learn the secret moldi techniques

[hidden haven]

Take the logic approach

Define an atomic string diagram

Define rules to put them together

[gusty halo]

but arent they domain specific?

[hidden haven]

Then prove theorems on manipulation from a couple basic manipulation rules

[gusty halo]

like the ones for monads work differently than the ones for adjunctions

at least it feels like it

[hidden haven]

Those are extra equations

Like axioms vs hypotheses in logic

[gusty halo]

i see

[hidden haven]

There are some manipulation rules which work regardless

[gusty halo]

but dont i need to know what objects im working with

[hidden haven]

Natural transformations

[gusty halo]

hmm

so objects and arrows you use functors from the one-point category?

[hidden haven]

Yes

[gusty halo]

[hidden haven]

Let me know if you want me to write something up

[gusty halo]

well its not like i'd refuse that

[hidden haven]

Lol based

[gusty halo]

even if i figure out my own system

[hidden haven]

Let me try to do it this week

[gusty halo]

nice

[hidden haven]

Maybe someone should give a talk on string diagrams

[gusty halo]

who else but you could?

i think ive only ever seen you talk about them

[hidden haven]

Lux

[gusty halo]

hmm

dont know them

[hidden haven]

He's the one who told me about them

Also Nobody, probably

And mniip knows about them too

Because he's asked before if there's a version for additive cats

[pastel cliff]

additive cats

cats are not closed under addition tho

[delicate orchid]

ok I think I'm right but I just want to check, this is just "A group homomorphism is uniquely determined by the image of the generators" right?

[hidden haven]

Yes

[delicate orchid]

no flippin way...

[sharp sonnet]

moldi i also need sanity check

Say we have finitely generated $k$-algebras $R_1 \subset R_1'$ and $R_2 \subset R_2'$ with $R_1[a_1, \dots, a_n] = R_1'$ and $R_2[b_1, \dots, b_m] = R_2'$.
Then $$R_1'\otimes_k R_2' \cong (R_1\otimes_k R_2)[a_1\otimes 1, \dots, a_n \otimes 1, 1 \otimes b_1, \dots, 1 \otimes b_m] \text{.}$$

[cloud walrus]

Lochverstärker

[hidden haven]

Don't you need some sort of flatness for this

Otherwise R_1 ⊗ R_2 might not inject into the larger tensor?

[sinful mirage]

on Wiki it is being stated that since the Lorentz group is semisimple,all its representations are completely reducible. can someone elaborate on this please? I thought only unitary reps are reducible

(sorry for interrupting)

[sharp sonnet]

this is sad

[hidden haven]

That's the definition of semisimple

[sharp sonnet]

but also probably correct

[hidden haven]

Completely reducible = direct sum of irreps and semi simple = all reps are completely reducible

[sinful mirage]

but why?

how to prove Lorentz is semi simple then?

I know on Lie algebra level

why is it true on Lie group level?

[hidden haven]

I don't even know what the Lorentz group is

[sinful mirage]

idk how this can be true

finite dim reps are not unitary

so how can they be completely reducible?

[hidden haven]

Might be by maschke's theorem if it's a finite group and you're looking at reps over ℂ or something similar

[sinful mirage]

yes but it's a lie group,so not finite

[hidden haven]

oh right F

Doesn't maschke have a converse that says that semisimple implies that the group is finite along other things

[sinful mirage]

I am not sure,idk

Lorentz is a Lie group

how can it be finite?

[hidden haven]

Ye

[sinful mirage]

I only know its algebra is semisimple and reps completely reducible

how to conclude group reps rae reducible?

[hidden haven]

Maybe it's 0 dimensional

[sinful mirage]

no

it's 6dim

[hidden haven]

Rip, I am out of ideas

[fallow plume]

damn now i know you're a hardcore (pure) mathematician

[next obsidian]

You wrote \otimes_k, in which case everything is flat

as long as k was a field here

[sharp sonnet]

so in this case it works?

[next obsidian]

Probably?

I'm not too sure, but tensor products over fields tend to work intuitively

especially if you embed things into big old overfield stuff

you can like do it all internally

[sharp sonnet]

well ok thanks

i will think about it again tmrw or something or just trust it works

[wind locust]

heya! I'm trying to prove/disprove that if $T$ is a semisimple operator on $V$, then so is $f(T)$ for any polynomial $f\in F[x]$, but I had no luck. Any ideas?

[cloud walrus]

iruneachteam

[cloud walrus]

please request a new nickname

[long obsidian]

What is x?

[dapper nebula]

The localization of the field at x

[lilac trench]

My book uses this notation that I've never seen before:
$\mathbb{F}_2[x]/I$

[cloud walrus]

please request a new nickname

[lilac trench]

What does the $\mathbb{F}_2$ mean here?

[delicate orchid]

yeah ok that one I recognise

that's the finite field with 2 elements

it's isomorphic to Z/2Z if you know what that is

[lilac trench]

$F_2$ or $F_2/I$?

[delicate orchid]

just F2

[lilac trench]

ok

[delicate orchid]

$\mathbb{F}_2[x]$ is then the polynomial ring with coefficients in F_2 as you'd expect

[cloud walrus]

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

[prisma shuttle]

can someone explain the purple part

why isn't that just an equality

i thought that was the definition of the product of two rings

[lethal dune]

you define $AB = \sum_{\text{finite}} a_ib_i$ here since $x_i+J \in I_1/J$ and $y_i+J\in I_2/J$ so it reduces to $(I_1/J)(I_2/J)$

[prisma shuttle]

wait so i am right in that it should be equality instead of a contains?

[lethal dune]

oh wait

no it should be contained

[prisma shuttle]

wait why

[lethal dune]

because that's an element not an set of elements of that form

[prisma shuttle]

oh shoot you're write

right

but if u take all such elements in that form and make a set it should be equality right

[lethal dune]

🙈

[lethal dune]

yes

10ⁿ ≡ 0 (mod 10)

[delicate bloom]

10-adics

[long obsidian]

Let a,b be algebraic in a field extension F/k . I'm trying to show that [k(a,b),k]=mn where m=[k(a),k] and n=[k(b),k] with m and n relatively prime.

By considering the tower of extensions k in k(a) in k(a,b) and k in k(b) in k(a,b) and as a,b are algebraic so their degrees are finite we know the degrees are multiplicative so
[K(a,b):k]=[k(a,b),k(a)][k(a),k]
And
[K(a,b):k]=[k(a,b),k(b)][k(b),k]
I feel like I basically need to conclude that [k(a,b),k(a)]=[k(b),k] and I think it comes from [k(b),k] and [k(a),k] being relatively prime

[next obsidian]

Because you can decompose [k(a,b),k] in the two ways you showed, it is a multiple of m and n, so it would be at the least their lcm, which as you noted, by being relatively prime is mn

So you just need to bound it by mn now

But you can do that by looking at it as [k(a,b):k(a)][k(a):k]

Or swap k(a) with k(b), doesn’t matter

[long obsidian]

Thank you

[next obsidian]

Whoops, I misspoke on the original one a bit, but I think you got the idea lol

I edited it so it’s right

[lavish nexus]

why is the second line true

[next obsidian]

You’d have too many roots to that

Look at what happens if you did

(a,1)^p

And (1,a)^p

Those will end up being the identity

Giving you > p solutions to that equation

[lavish nexus]

Oh ok 👍 ty

[hidden haven]

Where is this from

[lethal dune]

idk probably CGL

[rustic crown]

i really like the options 0 and 1

[lethal dune]

why tho

[lethal dune]

it's meth not math

[rustic crown]

🙈

[coral shale]

👻

[coral shale]

Automorphism between vector spaces --- both vector spaces have to have the same set V and same field F, right? (Not just V)

[wispy urchin]

a vector space is comprised of the set, field and two operations

so yea all of them are the same for autos

[coral shale]

Ah, we need the operations to be preserved as well

[wispy urchin]

in general existence of an arbitrarily large thing doesnt imply existence of an infinite thing

afaik finite groups have finite irreducible representations

[delicate orchid]

not irreducible ones

[sinful mirage]

let $\rho(g,g')=\rho_{c}(g) \rho_0(g')$. Assume $\rho_c,\rho_0$ are completely reducible. How can we conclude that $\rho$ is completely reducible?

[cloud walrus]

ProphetX

[sinful mirage]

for context:

[prisma shuttle]

can someone help me with this problem

for the nonmaximal ideal part

idk i considered adding something like (1, 0, 1, 0, \cdots) into the set (because it has a non-finite number 1s) but idt it forms an ideal anymore

[gusty halo]

what is D

what is M_n

[prisma shuttle]

D just means all the a_i are in Z/2z

M_n is the subset of D for which one of the a_i is equal to 0

[gusty halo]

how about you just post the full exercise

[gusty halo]

well if you add that sequence you would naturally consider the ideal generated by the previous ideal and the new sequence

then you definitely get an ideal

[prisma shuttle]

why is it an ideal though

i don't think its closed under addition

[gusty halo]

i said i am taking the ideal generated by this

so obviously its an ideal

wdym

its the smallest ideal that contains I and the element (1,0,1,0,...)

[prisma shuttle]

oh shoot whoops

oh i see

[gusty halo]

the question is then whether thats still a proper ideal

and im guessing it is because we do everything pointwise

[prisma shuttle]

it is becuase (0, 1, 0, 1,\cdots) is not in it

[gusty halo]

i dont think you can get (1,1,1,1,...) in it

[gusty halo]

or this yes

[long obsidian]

How can I show that the polynomial x^3+x^2+-2x+1 in Q[x] is irreducible?

Perhaps the rational root theorem?

[delicate orchid]

note that this polynomial is primitive, and then use gauss's lemma

that's how I'd do it

and then you can use the fact that the leading coefficient isn't 0 for any finite field Z/nZ and you can reduce it even further by looking it in Z/2Z[X] - I think
and then just check that the cases x = 0 and x = 1 aren't roots

[next obsidian]

It’s a cubic

If it isn’t irreducible it has a root

Apply rational root theorem

[long obsidian]

Thank you guys

[buoyant relic]

how does this follow from "every subgroup of a cyclic group is cyclic"

[delicate bloom]

they're listing every cyclic group generated by every element <n>

they don't have to list <-n> because it generates the same cylcic group as <n>

I sort of phrased that funny I am referring to n and -n as the generators of the group and <n> and <-n> as the groups generated by those elements, in case it wasn't clear

[buoyant relic]

wait they way u said, it doesnt even require the theorem to show Z's only subgroups are nZ

[delicate bloom]

well we know these are subgroups but hypothetically we could imagine there might be more complicated ones somehow

the theorem assures us this is not possible, this is it

[buoyant relic]

ooo ok ok makes sense

[delicate bloom]

cool

[buoyant relic]

thamks

[delicate bloom]

you're welcome 😎

[barren sierra]

is it inaccurate to say that every automorphism is an inner automorphism iff the center of the group G is trivial?

[gritty sparrow]

Yes this is inaccurate. Example: S6 has trivial centre but nontrivial outer automorphisms

[barren sierra]

ah

[patent girder]

this just says that /mu(G) is the smallest sum of trivial intersections of cores for the subgroups of a group right?

[long obsidian]

Supposed F/K is a finite field extension of characteristic p>0 such that a^p in K for all a in F. I want to show that there is a set of elements {a_i} of F such that {(a_{i}^{j_i}: 0\leq j_i \leq p-1} is a k-basis for F

I was given the hint to consider F=k(a) for a not in k. I know that a is a root of x^p- a^p and since F is characteristic p by the binomial formula x^p-a^p=(x-a)^p. Then it follows this is the minimal polynomial of a. ( The details of this are fuzzy but I believe it).

But idk how this leads into me getting a k-basis

[gritty sparrow]

The point is that if the powers 1, a, a^2,…,a^p-1 were linearly dependent over K then a satisfies a polynomial of degree <=p-1 which is impossible. To extend this to a basis just do the same trick but for F/K(a) to get {1,…b^p-1} that are linearly independent over K(a) using this you can show that the powers of a and b are linearly independent over K. Just keep repeating this until you get a basis

[dire summit]

i'm working through this problem about reflections in R^n, and i worked through part of the solution with a friend, but i don't know where to go from here. i'm just lost idk here's the problem:

[cloud walrus]

Snodlop

[dire summit]

here's my solution thus far:

[cloud walrus]

Snodlop

[dire summit]

there's slightly more dithering after this point, but nothing concrete or conclusive

[kindred jay]

first homework, kinda stuck on where to go:

went to office hours, my professor said something about how a, a^2, a^3, and a^4 work, and how you have to show that for any b, {a, a^2, a^3, a^4} and {b, b^2, b^3, b^4} are disjoint, but i honestly have no idea where he got those from, so any help would be appreciated

[dire summit]

if a^5=e, then (a^2)^5=(a^5)^2=e^2=e

similarly for the other powers of a

[kindred jay]

how do we know that a^2 is a nonidentity element?

[delicate orchid]

2 and 5 are coprime

[proud bear]

if a^2=e, then a^4=e^2=e, so a^5=a^4a=a

[kindred jay]

ty!

[dire summit]

i'd like to drop my ping on this one, i'm still quite lost <@&286206848099549185>

[subtle ivy]

you may solve this problem entirely without invoking the identity of the inner product involving cosine. in fact i would recommend you not choose a basis at all.

to get started, you may notice that inner product of x and v-hat can be distributed through the definition of v-hat to obtain a familiar formula (should you have seen the corresponding definition of a reflection). you may then easily verify that this definition of t(x) is indeed an orthogonal transformation (it is clearly linear) and then recall that a reflection is an orthogonal transformation which fixes some (non-trivial) subspace, and in fact is uniquely determined by this subspace.

much of this depends precisely on your definition of reflection/rotation, however that should be a nice start.

[tribal moss]

I'd suggest instead of choosing v itself as a basis vector, choose the normalizeation of v as the first basis vector, and extend it to an orthonormal basis. Since the basis is orthonormal, the inner product is the ordinary dot product on components, so you can write v=(a,0,...,0) for some a and start computing explicitly. For example, you get vhat=(2/a,0,...,0).

As Shortcut says, you definitely don't want to bring the cosine formulation into it, but simply choosing a good basis sounds like the right thing to do.

[hot hill]

I am confused about a step in the proof here : https://i.uguu.se/wbhuerPc.png . This is a preliminary lemma for proving Zorn's Lemma. I am confused about the second to last line: "or alpha <= beta for some beta in B, in which case c <= a." I have no idea how they got this.

[subtle ivy]

you've swapped it: their conclusion is if there is beta >= alpha, then alpha<=c

[hot hill]

Oh, I made a typo, sorry. I still don't see why it's true though.

Here is what I think is a counterexample: take B to be an interval of the real numbers. Take C to be an open interval of the real numbers contained in B. Take c = sup C. Take a to be in B, with c < a < beta. So a <= c is not true in this case.

[subtle ivy]

this follows immediately from transitivity of the ordering, no?

since c is defined to be the sup of B, and there is beta >=alpha, then beta<=c by definition, and your result follows.

[hot hill]

No, c is the sup of C, not B.

And C is just some subset of B.

So my understanding is that there can be beta's that are not in C. My counterexample used such a beta.

[subtle ivy]

i see

[hot hill]

Do you think this is a mistake? I found a semi-mistake a few pages earlier

[subtle ivy]

i am cautious. one moment

i suspect formulating a counterexample with this many hypotheses is a waste of time, however.

[hot hill]

Hmm you are probably right.

[subtle ivy]

well i am not quite sure about the logic presented, but noting that $C\subset B$, each $c'\in C$ is comparable to each $a\in A$. so if some $a$ were not an upper bound of $C$ (failing by $c'$), then we must have $a\leq c'\leq c$. and if it were, clearly $c\leq a$.

[cloud walrus]

shortcut

[subtle ivy]

does this seem plausible?

in short, c is comparable to any upper bound by definition, and comparable to any extant case by transitivity.

[hot hill]

Thinking 🙂 One moment

Right, c is comparable to any upper bound by the definition of least upper bound. For any a that isn't an upper bound of C, we have a <= c.

And if we have an a that is an upper bound of c, we have c <= a.

So indeed by your reasoning, c is in B.

So basically, this author is proving the right thing, but his argument in the book doesn't make sense to us, is that right?

Something similar happened a few pages ago, he had a nice diagram that showed something, but he left out a crucial fact that was necessary to make his argument work.

[subtle ivy]

yes i would think the author mangled things a little. in many cases like these, however, it is quite easy to get lost in the specifics while on the right path.

[hot hill]

I see. So I think moving forward in this book I should be prepared to create arguments to replace his, is that a good strategy?

[subtle ivy]

especially with such a widely used result like Zorn's lemma, it is most often more productive to spend time trying to prove the result the author intends, rather than debunking the argument given.

depending on your stomach and the text, it is frequently worth it to try to prove things yourself regardless.

[hot hill]

I see. I can't thank you enough for helping me with this

I'll keep moving forward 😀

[subtle ivy]

ofc!

[sinful mirage]

https://physics.stackexchange.com/questions/691989/non-unitary-finite-dimensional-representations-of-the-lorentz-group in case you're interested there is a mathematically precise answer

[chilly ocean]

why can't x+x=0 in a field with three elements F={0, 1, x}? i've forgotten

I think it has to do with the fact that 1+1=x

so (1+1)+x = 1+(1+x)=0

but 1+x=0 so that would mean 1=0?

[delicate bloom]

addition forms a group, try to write out the addition table

x+x=0 we can factor out x*(1+1)=0 and since x !=0 we can divide it out to get 1+1=0

that gives us problems trying to fill in x+1 since 0,1,x have already been used in the same row/column and that breaks our group

[devout crow]

yeah, that works

alternatively, every field F where 1+1=0 has the subfield {0,1}, which means 2 divides |F|

[barren sierra]

god Sagemath is so OP

[desert dome]

I want to show X^3 - X - t is irreducible over C(t). any suggestion?

[delicate bloom]

suppose it is reducible, then it has a linear factor, so it has a root r. Then r^3 = r+t. maybe we can do something with a third root of unity w^3 = 1 idk

thinking basically we can make the other 2 roots from one root or something like that, I feel like I've seen something similar before but for finite fields but they were like X^p-X-a over F_p but things cancelled a little nicer cause it's characteristic p

any root r of X^p-X-a will then make r+n for n in F_p also a root, feels closely related, maybe we can get the same thing to work in your case by taking r+w

[next obsidian]

We should just pin an answer on how to show a cubic is irreducible

I think it’s been asked 3 times in 2 days haha

I guess this one is kinda especially weird since it involves C(t)

I think you can use Gauss’s lemma and then rational root still?

[oblique leaf]

I have defined a map f from Z/n X Z/m to Z/gcd(m,n) in the natural way such that f(x,y)=xy in Z/gcd(m,n). How do i show this map is well-defined?

[next obsidian]

Write out what it means for (x,y) = (x’,y’)

Then show that this means they map to the same thing

[oblique leaf]

oooh of course

sorry for the dumb question lol

thanks!

[next obsidian]

No worries

[desert dome]

Thank you!

[dusty sapphire]

How to check normal subgroup for class equations?

Suppose we have class equation 1+3+4+4 then how we check that which subgroup is normal of some order?

[rustic crown]

use that normal subgroups are made up of conjugacy classes

[next obsidian]

I don't think you can use class equation stuff to really even prove that a subgroup is normal

just that it isn't because it couldn't be a union of conjugacy classes

[dusty sapphire]

So is there any way to check which order subgroup is normal by class equation?

[dusty sapphire]

I didn’t get

[rustic crown]

so if an element x is in N, then gxg' will all be in N, so it definitely reduces the number of options for the cardinality ofN

it has to divide |G| and also must be sums of these class equation things.

so like in your example the group has order 12
subgroups definitely contains the identity.
it's not hard to see that only possible non-trivial normal subgroup
could be because of 1 + 3

[dusty sapphire]

Okay

[rustic crown]

you can use this way to give a not very long brute force proof of A5 and A6 are simple

[dusty sapphire]

1+12+12+15+20 so there is no possibility for normal?

We need to add identity in conjugate class and then we need to check that number divided order of group or not?

[rustic crown]

yea something like that

you can also use stuff about A5

like non-trivial normal subgroups can't contain a 3 cycle

[dusty sapphire]

Yes got it

[rustic crown]

[dusty sapphire]

One more question, is there no other possibility number of element in order conjugate class?

[dusty sapphire]

Here 13, 16, 21 does not divide 60 so there is no chance of normal

[next obsidian]

Well

You have to do a few more

Like 1 + 12 + 12

[dusty sapphire]

Like all possible pair with 1

[next obsidian]

Not just pairs

[dusty sapphire]

Yeah got it

[next obsidian]

But it will end up none divide 60

So you’re right

[dusty sapphire]

Yep

[next obsidian]

[dusty sapphire]

[barren sierra]

So this is the question I'm working on and this is the part from the previous problem set that gives me a hint

So this in theory would give me that the action is well defined

IF

the order of GL(2, 3) was 24

since 24 = 2^3 * 3

and I could use Sylow's theorem

however couple issues

the order of GL(2, 3) is 48

and X doesn't have +- the identity (but I think I can get around that since for an action of G acting on X, X doesn't have to be a group so I don't need the identity)

however 48 = 2^3 * 6 and 2 | 6 so I can't use Sylow right?

so then I am stuck

but I feel this may be the right track

so how do I fix that idea?

because Q = {+-I, +-A, +-B, +-C} is the only group of order 8 in SL(2, 3)

and so it's normal by Sylow's Theorem

which gives me closure under conjugation if G' = SL(2, 3)

but G' = GL(2, 3) so this doesn't work

[frank lake]

quick question

if a^n = e, does that always mean a^n = a^0?

like the order of a is n

[barren sierra]

no

it means the order of a divides n

[frank lake]

oh wait let me comprehend that

yeah I'm still kinda clueless 😂

can you elaborate?

[barren sierra]

example. In Z_8 we have 2^6 = 0

[frank lake]

mhm

[barren sierra]

but the order of 2 multiplicatively in the ring is not 6

it's 3

2^3 = 0

[frank lake]

I haven't really covered rings that much, but is it a ring?

Isn't it just the group of integers mod 8?

closed under addition

idk

[barren sierra]

ok sure same point

then I'll write it additively

2*8 = 0 mod 8

but 8 is not the order of 2

4 is

[frank lake]

I see

[frank lake]

so how does that tie to this proof?

how is a^k = a^0?

[next obsidian]

We don’t have an “order” for rings

Because multiplication isn’t a group

[frank lake]

chmonkey pls help me out here 😭

ur my only hope 🥺

[next obsidian]

What’s theorem 4.1

[next obsidian]

Also a^0 is defined to be e

And a^k you assumed is e

[barren sierra]

oh yea 💀

[next obsidian]

Like idk what part you’re confused about, you started from a^k = e

And you also know e = a^0

[frank lake]

so is e always a^0?

[next obsidian]

So you just chain those together

Yeah

[frank lake]

is that applicable for all groups?

I see

[next obsidian]

It’s a•a^-1

[frank lake]

I got it 😄

[next obsidian]

It’s the same reason x^0 = 1 in R

[frank lake]

that's all I was confused about 😂

[next obsidian]

For any x

[frank lake]

tysm

[next obsidian]

Swag

[barren sierra]

ok wait can I get clarification about Sylow's theorem?

so say |G| = p^k * m

where p does not divide m

[next obsidian]

Ok

[barren sierra]

is the only Sylow Subgroup the one of order p^k?

that doesn't sound right

[next obsidian]

There isn’t “only”

[barren sierra]

ok

[next obsidian]

there can be multiple

[barren sierra]

it really sounded like my prof was talking about it like that

[next obsidian]

But the p-sylow is of order p^k

[barren sierra]

oh

[next obsidian]

Wel they’re all isomorphic

[barren sierra]

ok

[next obsidian]

Because they’re all conjugate

[barren sierra]

so yes but not really

hm

[next obsidian]

But you have one for every p

Well no

[barren sierra]

ok yea then I'm still stuck

wait wdym for every p

[next obsidian]

They are definitely not the same

Every prime

Has a p-sylow

[barren sierra]

oh

[next obsidian]

It’s just as much p as you can possibly have

Pushing P

I’m pushing P

Anyway

[barren sierra]

ok yea that doesn't help me but good to know

[next obsidian]

The fact there’s multiple matters

Or that there can be

[barren sierra]

ok so I can't use Sylow

but

the quaterion group is Q = {+-I, +-A, +-B, +-C}

and this is normal in GL(2, 3)

which is almost good enough

for showing this action by conjugation is well defined

but not quite since X =/= Q

[frank lake]

ok I'm back

so I'm trying to prove this here

idk if this would be valid but this is what I would do:

assume a = e, the order of e is 1 (since it's the identity itself), and the order of the trivial subgroup is also 1 since it only has 1 element

therefore |e| = |<e>|

😎

would this be a good proof?

[hidden haven]

That only proves this when a = e

You have to prove this for any a

[frank lake]

😢

that's where I'm kind of stuck

I've been looking at this problem for a while and I still don't have much of an idea, I tried looking at this theorem for insight but I'm still lost

any ideas?

[next obsidian]

Doesn’t…

Theorem 4.1 literally prove it?

|a| = n there

And |<a>| = n because they listed every element

[frank lake]

that's true

but is <a> always going to have every element?

[next obsidian]

The theorem says

<a> = {e,a,…,a^n-1}

That equals sign isn’t gonna start lying to you randomly

[frank lake]

why is that the case? Why can't <a> be {e, a, ...a^k}? Why does the order have to be the same finite number as the order of the element?

like I get what you're saying that every element is listed in the image, but what if the last element wasn't a^(n - 1)?

[next obsidian]

If k > n then the order is larger

If k < n then it’s smaller

Read the proof until you convince yourself of why it’s true and you’ll see what happens if it ended at k

[frank lake]

AGHHHHHHHHHHHHHH

I still don't understand

so wait

let me break this down, so since a^k is not a^n, k < n

why the hell do I struggle on such simple problems 💀

[sharp sonnet]

you basically have to convince yourself of two things: that there are not more than those n elements and that there are not less
the first follows from the order of a being n, since a^n = e and thus a^{n+k} = a^n*a^k = e*a^k = a^k, so you can reduce exponents to be smaller than n
for the second assumed that a^i = a^j for i, j different, say i < j, then you can multiply by (a^i)^{-1} and get a^{i-j} = e but i-j is a positive integer smaller than n, contradicting the order assumption

(this should be basically the proof in your source)

[frank lake]

yeah that's the proof for theorem 4.1

but

I'm supposed to use theorem 4.1 to help me with this problem

and that's where I'm stuck

[sharp sonnet]

as chmonkey said, the theorem does this by listing every element in <a>

[frank lake]

so to prove this all I need to do is say |a| = n, and <a> = {e, a, a^2 ..., a^(n - 1} and since the final element is n - 1 and the order of a itself is n, they are equal

right?

[sharp sonnet]

ye

e = a^0, so there are |{0, 1, ..., n-1}| = n elements listed

[frank lake]

I kind of feel unfulfilled with this, I just have a feeling like there's more to show

[sharp sonnet]

anything "more" would be a (partial) re-do of the proof of theorem 4.1

[frank lake]

🤦‍♂️ oh 💀

I read theorem 4.1 over and over again for 30 minutes but I still didn't figure this out

[sharp sonnet]

i'd say take a break

[frank lake]

yeah 😂

thanks a lot

😄

I'll read over theorem 4.1 tomorrow honestly but I really appreciate it Loch, Chmonkey, and Moldilocks 🙂

[dusty sapphire]

How to find?

I is not prime ideal so R/I is not integral domain, next how to find number of units?

[cloud walrus]

det

[rustic crown]

use this isomorphism to find the units. definitely multiples of x aren't units. what about the rest?

[lethal dune]

nice

I ended up finding all the elems of R that do not have (x-1) as a factor

lol

[delicate orchid]

I'm not 100% seeing the isomorphism between F_3[x]/(x-1)^2 and F_3[x]/x^2 - is it because they both just have 1 root?

[rustic crown]

nah, it's because of the shift of x and x-1

[delicate orchid]

oh yeah fucking duh

just map x-1 to x

[rustic crown]

yea

[dusty sapphire]

there are 9 element so how to get other except 1 and 2

sorry for my late

[rustic crown]

oh it's fine dw

one simple thing you can prove is that a unit + nilpotent is also a unit (in a commutative ring)

and because of this a+bx are all units when a is not 0.

[dusty sapphire]

okay

so total are 6

[rustic crown]

yep!

[dusty sapphire]

thank you so much

And it is isomorphic to Z_9?

[rustic crown]

nu

char is different

[dusty sapphire]

f map to f(2)mod9?

ok

[delicate orchid]

yeah it's definitely char 3 cause 3 is prime 🥴 I think

[dusty sapphire]

char of Z_9 is 9, wbu R/I

[rustic crown]

char of Z/9Z is 9

[delicate orchid]

like

by definition almost

[dusty sapphire]

yeap

[delicate orchid]

well R/I is the quotient of a char 3 field

[dusty sapphire]

not field

[delicate orchid]

sorry

polynomial ring over a char 3 field

i.e. a ring of char 3

same thing really

[dusty sapphire]

so here char of R/I is neither 3 nor 9?

[rustic crown]

it's 3 >.<

[delicate orchid]

it has to be 3 or 1

[rustic crown]

1+1+1 is 0 mod 3

[delicate orchid]

^

cause 3 is prime

[dusty sapphire]

by this🤔

[delicate orchid]

yes...?

so it's either char 3 or char 1 as the char of the quotient has to divide the character of the ring

I think

[dusty sapphire]

here R/I is not field because I is not maximal

[delicate orchid]

correct me if I'm wrong on that det

[delicate orchid]

it doesn't matter

[dusty sapphire]

what

[delicate orchid]

I said field instead of ring big deal no one cares 🤣

[dusty sapphire]

rip😂

[delicate orchid]

the character of ANY quotient of ANY ring has to divide the character of that ring

as I've said

three times now

[dusty sapphire]

now i get it

[delicate orchid]

3 is prime... so it's either char 1 or char 3, it's not char 1 cause then it'll be the trivial ring

so it has to be char 3

[dusty sapphire]

yes yes clear now

[delicate orchid]

nice

[chilly ocean]

or 2

[rustic crown]

you probably don't even need it to be a quotient lol

[delicate orchid]

alternatively just keep adding 1 to itself

[rustic crown]

R --> S is enough?

[dusty sapphire]

didn't get

[rustic crown]

Z --> R --> S
kernel of first map is (char R)Z and of the composition is (char S)Z
since kernel of composition is bigger, char S divides char R

[dusty sapphire]

okie

[delicate orchid]

something something homomorphism something something f(1+1+1) = f(0)

catfan is right actually I've thought about it for a solid 40 seconds

it could also be char 2 if it's a general quotient of a char 3 ring

[rustic crown]

wait really

no it can't >.<

if 1 + 1 + 1 = 0 and 1 + 1 = 0, then 1 = 0

[lethal dune]

wait what

lol

R=(0)

[rustic crown]

another proof could be this: if char R = 0, then nothing to prove, else
1 + 1 + ... +1 (char R times) = 0 then additive order of 1 in S has to divide char R

[delicate orchid]

I was thinking right

take R with char n and S an ideal of R, then in R/S we know that the sum n(1+S) = n+S = 0+S, so the characteristic of the quotient must be n or lower

but I couldn't see how to prove that it divides n

oh wait

yes I can

S is an additive subgroup of R so the order of the subgroup must divide the order of the group

[chilly ocean]

wait can it be 2?

I just wanted you to repeat the char divising thing for the fourth time

[delicate orchid]

oh

well you managed to convince me that I was completely wrong

[chilly ocean]

[delicate orchid]

should've stuck with my intuition ig

in other news chat I just unironically used cayley's theorem for the first time in my life

[hidden haven]

What's that about yoneda

[delicate orchid]

sorry moldi you uhh must've misclicked there friend! 😅 I believe you meant to go here: #category-theory 😁 thanks for being understanding 😅

[hidden haven]

ah of course 😅 I am very sorry for interrupting the conversation 😄

[rustic crown]

i want yoneda rant in #groups-rings-fields 🙈

[delicate orchid]

when the ARROW is... injective?!
hueh-whaaaatt??

[lethal dune]

#category-theory message ok that's a good explanation

[delicate orchid]

holy shit

[lethal dune]

but how does one use Yoneda to prove anything

[delicate orchid]

oh I kind of get it

[hidden haven]

How about G/N = (G/H)/(N/H)
Maps out of G/N are the same as maps out of G that have N in the kernel
which are the same as the maps out of G/H with N/H in their kernel
which are the same as the maps out of (G/H)/(N/H)

So Hom(G/N, -) is naturally isomorphic to Hom((G/H)/(N/H), -)

[delicate orchid]

are you sure about the third line moldi?

[hidden haven]

Yes

Assuming H is in N

[delicate orchid]

oh right

[hidden haven]

Same exact proof works for localisations of localisations

And many other similar constructions

Also this is not just for groups

[delicate orchid]

wait

localisation of localisations?

[hidden haven]

The above proof is for any quotient

yes

If you have T containing S

[delicate orchid]

I always thought localisation at J of a localisation of I is just a localisation of some combination of I and J

or is this the proof that there actually is an isomorphism

[hidden haven]

You localise at S then at the image of T, that is the same as the localisation directly at T

There you use the fact that maps out of S^-1 R are the same as maps out of R that map everything in S to a unit

And the same reasoning finishes the proof

[delicate orchid]

[lethal dune]

oh so from this we can say G/N and (G/H)/(N/H) isomorphic(different category)?

[hidden haven]

Yes, but idk what you mean by different category

Yoneda tells you that if Hom(A, -) and Hom(B, -) are isomorphic functors, then A and B are isomorphic objects

[lethal dune]

first is the category of (G, f) with ker f containing N and second is (G/H, f) s.t. ker contains N/H so the categories are different?

[hidden haven]

No both are objects of the category of groups

[lethal dune]

oh that's the cat we are taking

[hidden haven]

G/N and (G/H)(N/H)

And I am looking at maps to any other group

[lethal dune]

ohhhh

now I get it

[hidden haven]

But at the intermediate stages we are going through another category

[rustic crown]

lol i thought moldi going full cat and looking at group as groupoids on 1 object

[hidden haven]

Using the adjuction that we get from the universal property of quotients

[hidden haven]

[lethal dune]

nice, thanks