groups-rings-fields

[void cosmos]

so u just like

like consider these "additive factors" of 1 but only the finite indices of them

and then any element is the sum of those

does this work

[rocky cloak]

Yurr

[void cosmos]

lmfao math is literally made up i sweare

okay so

wait so thats it

its artnian

cuz simple rings are artinian

[rocky cloak]

Yup

[void cosmos]

trivially

and the direct sum is ( i wont prove it )

cool

[rocky cloak]

simple modules are Artinian, simple rings need not be.

(Simple rings are not related to semisimple rings)

[void cosmos]

what

[rocky cloak]

In case you wanted to go on a needles tangent

[void cosmos]

how

do simple rings need not be

they literally have no 2-sided ideals

buyt they can left ones tho

have*

[rocky cloak]

Yeah, they can still have loads of left ones

[void cosmos]

yea

and they can be garabge

they cna be full of garbage literally

full of league players

who knows

okay..

but i mean

i have it as the direct sum of semisimple modules which each should be isomophric to a submodule ( by the direct sum property )

right?

[rocky cloak]

Yes

A direct summand is a submodule

[void cosmos]

okay so time to show it has 0 radical

any hints? or is it easy on my own?

[white oxide]

can somebody explain to me how this hypothetical operation would have been defined? I'm a little confused about that part

like I get that the action of A on End(M) is given by (af)(x) = af(x) but what f(x) would we choose, or how would it be well-defined

oh right

wait

[void cosmos]

f_a(r) = ar

[white oxide]

a(m1 + m2)

= am1 + am2

or something

and a is the homomorphism

[void cosmos]

so the assignment from the element a to this f_a

is a homo

[white oxide]

wait no i'm stupid

lmfaop

[white oxide]

oh

[void cosmos]

cuz am is in M

[chilly ocean]

actually what i wrote and deleted was correct. i just misread myself

[white oxide]

$a \mapsto f_a$?

[cloud walrus]

okeyokay

[void cosmos]

yea

[rocky cloak]

A hint for a quick solution is Chinese remainder theorem.

[void cosmos]

yo eren jeager

what does it mean for a monomorphism to split

an epimorphism*

that is there exists a right inverse?

as in splitting lemma or what

[crystal turtle]

:o

[south patrol]

idk the point of the terminology but maybe i'm not algebra pilled enough

[rocky cloak]

If A ->> B splits, then A = B(+)C splits into the sum of two summands

[south patrol]

I mean yes agreed

[void cosmos]

yea its as what i said i think

ty

[south patrol]

but perhaps it is worth it lol i just find it weird given it makes perfect sense in any cat

please don't call me daddy

[void cosmos]

yeah

[dull hornet]

That's not how that works

[void cosmos]

it literally is

[summer path]

kirby in algebra channel

[dull hornet]

rare occurrence

[rocky cloak]

Surely, nobody cares about categories that aren't additive

[solar inlet]

hey can we maybe not engage in creep behavior and make other users uncomfortable?

[void cosmos]

sure why not

[white oxide]

Just to be clear, $V is a k[X]$-module where the action of $p(x) \in K[x]$ is given by $p(x)v = $\psi(p(x))v$ where $\psi$ is the association map?

[cloud walrus]

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

[white oxide]

bruh

ykwim

[void cosmos]

i did not make anyone uncomfy th

tho

[solar inlet]

that's not for you to decide

[south patrol]

i mean it does make me a bit uncomfy lol like it is a bit weird to be called daddy by somebody i don't know on the internet lol

[void cosmos]

now yeah cuz now he/she just said that

didnt say it last time but its okay i have no intentions on making anybody uncomfy obv

[rocky cloak]

The map is called \rho here, but yeah

[void cosmos]

cool king didnt mean to

jk jk

[prisma ibex]

you have such a horrible attitude good lord

[void cosmos]

wdym

what did i do now lmfao

[white oxide]

oh right thanks

[coral shale]

bruh.

[dull hornet]

saying they is only 4 letters instead of typing 6 for he/she 😭

[void cosmos]

yeah see

im a nice gy

guy

i put in effort uknow

lmfao

[dull hornet]

what, their pronouns are they/them

[void cosmos]

oh really

yea mb

[dull hornet]

Yeah there are pronoun roles to check, that's what the squares mean

[void cosmos]

yea i know about them but i just forgot or idk

[dull hornet]

yellow is they/them, blue is he/him, red is she/her, grey is any or ask

[void cosmos]

i thought in no way can someone comment on me saying he/her but still u managed to win

[maiden ocean]

Yeah i know about your pronouns i just cant be assed to use the right one

sorry haha

[void cosmos]

i literally said he/she

[toxic zephyr]

jesus christ

[void cosmos]

some lol

[coral shale]

[chilly ocean]

...

[formal ermine]

wtf is going on here

[toxic zephyr]

i just default to they/them to be safe. some people don't like he or she.

[void cosmos]

yea mb from now on i will be calling potet they/them

[formal ermine]

how does a non binary samurai kill people? ||they/them||

[void cosmos]

now the world can relax

even though i called he/her like 5 times

[open sluice]

oh boy

[void cosmos]

but its okay

[coral shale]

new bladewood pfp

[void cosmos]

why is he not banned for this

hahahahhaa

[coral shale]

oh fuck i get it

[south patrol]

oh okay i meant like

The idea makes sense in any category so feels funny to give a name for it that only makes sense in additive cats

[summer path]

i thought gray is any, slightly tinted green is ask

[south patrol]

Or whatever the correct level of generality is

[long nebula]

he/she is not a substitute for they

[tiny pagoda]

this is not an ok attitude and repeatedly ignoring pronouns is against our rules

[long nebula]

There are people who do not go by either he or she

[rocky cloak]

I mean theyre also called sections and retractions (which only makes sense in the context of fibrations)

[south patrol]

Okay fair this may be my bias lol

[summer path]

ya know there's definitely a point where you just admit to being wrong and move on

how is this still going on

[void cosmos]

okay say i call you by a wrong pronoun ( which i allegedly now did against potet ) wouldnt you just go like " hey can u please call me she" thank you?

[long nebula]

If you can't read the room and stop your behavior when you make other people uncomfortable, you're not welcome here

[tiny pagoda]

if you "can't be assed" to follow people's pronouns then we can't be assed to have you around

[void cosmos]

thats not what potet do tho

[white oxide]

yo sorry to interrupt can i ask an aa question or is the third world war still going on

[void cosmos]

why are you quoting someone else for me Hhahahaha

[long nebula]

ask away!

[south patrol]

I view section and retractions as neutral words but I suppose they do only really "make sense" in more geometric contexts etc but fair lol

so that is a good point, thanks

[tiny pagoda]

you didn't just make a mistake, you were corrected and then said you can't be assed to use the correct ones

[tiny pagoda]

that is what you said

[void cosmos]

corrected by not even the same person

[white oxide]

what is the purpose of studying representations? is it that we can gain more information about the vector space by studying the algebraic structures which "act" on that vector space or more or less the other way around?

[tiny pagoda]

the person you misgendered didn't want to make a fuss out of it

[void cosmos]

i called potet he/she mutliple times and THEY did not have any problems

[tiny pagoda]

just because someone else had to say it for them doesn't mean they're totally fine with it

[void cosmos]

its just some other they/them commented on it for no reason idr who

[long nebula]

moamen, honestly, either apologize and own up to your mistake, or leave

[tiny pagoda]

"for no reason" you're really close to a line

[void cosmos]

what

wdym

[tiny pagoda]

you misgendered someone

there is a reason you were corrected

[void cosmos]

who could i have known if i am misgendering you if you arent even bothered enough to comment on it

like whats the point

but again

[long nebula]

there's literally roles for this

[void cosmos]

if potet felt uncomfy for me calling them he/her

[tiny pagoda]

because there's a little symbol right next to the name?

[void cosmos]

i apologize for sure for potet

[tiny pagoda]

which tells you which pronouns to use

[void cosmos]

they are very nice to me

and i never mean any uncomfy

[south patrol]

okay i mean i don't massively care but others will/do care

[void cosmos]

ikr

i literally know

[tiny pagoda]

you just called potato him again

💀

[void cosmos]

?

what

are u

saying

[void cosmos]

hello?

[tiny pagoda]

for call him he/her

[void cosmos]

potet themselves dont care why is there some other they/them just ocmmenting for no reason hahaha

but again it's okay

mb potet

[void cosmos]

yeah mb typo

[tiny pagoda]

why do you say they don't care?

[void cosmos]

cuz they never

commented

??

[tiny pagoda]

they said they don't massively care

that doesn't mean they don't

[void cosmos]

it does for anyone

lmfao

[tiny pagoda]

they cared enough to set that as their pronoun role

[void cosmos]

it literally does queen

[long nebula]

Okay enough

[tiny pagoda]

don't call me queen and don't tell me whether people care about you misgendering them

[long nebula]

Yeah a big part of algebra is gaining information about something by studying its actions

[dull hornet]

I'm not some other they/them, it was more just me commenting on it because you've referred to people as he/her before, including me, and I did not comment on it in the past
I made a slight joke about it because he/she is 6 characters so you're putting in more effort than just saying they as a default pronoun
Anyway carry on

[long nebula]

For example you can study groups by looking at how they act on other objects

The point of the representations is to gain info about the group

[coral shale]

why lin spaces gud

in particular

[topaz solar]

Because LA nice

[coral shale]

wg

dunt lie tuh me

[summer path]

more like LA is well understood, and other stuff is not so well understood, so ooga booga make other stuff into linalg problems

[long nebula]

Yeah, we already understand a lot about vector spaces

So if we can turn groups into actions on vector spaces, then we gain information about the group

[coral shale]

tell me more about R over Q

but ok yh

[topaz solar]

Choice says it has a basis

[coral shale]

is rep theory done on modules too?

or can

[dull hornet]

Vector spaces are fun spaces because they behave nicely most of the time

[coral shale]

[topaz solar]

I don’t know enough reps, but I don’t see why it couldnt but it would suck relatively

[solar vessel]

can

[coral shale]

maybe u get monoids instead of groups or something? rando guess

[rocky cloak]

So representations are the way algebraic structures appear "in the wild". A symmetry group acts on something, and if you want to understand that action you want to know how a representation breaks down into simpler pieces and what those pieces can be.

Just think about groups that appear in physics. They're all matrix groups acting on some space. The action is the important thing, not necessarily the group or even the space

[south patrol]

Well depends what you mean by rep theory lol

[coral shale]

ur missing field inverse but what does that mean

[topaz solar]

If you have a quiver and like functor it into k- finite Vect, you can say some cool stuff too

[south patrol]

In fact, iirc, the first reps course I saw was phrasedd in terms of reps of modules

[topaz solar]

But it’s hard to just point towards basis dimensions and grassmanians on modules in general

If you have particularly nice ones ig sure but idk why you’d do this then

[coral shale]

ok so is there a rep theory for monoids?

not sure what id have to do

as opposed to automorphisms

take endo...

[topaz solar]

Can you make em into a quiver?

Then gg

[coral shale]

idk what that is opencry

[topaz solar]

Uhhh directed multi graphs

I think they like conditions like no A->A and no A->B->A too for alg combo

[rocky cloak]

Sure, a monoid is a category and there is representation theory of categories

[topaz solar]

Just uhhh

F:C -> k-Vect gg

[coral shale]

just skip to rep of cat then...

[coral spindle]

I happen to know something about this

[topaz solar]

Rep theory of syntactic category of the theory of monoids

[rocky cloak]

representation theory of groups/rings/algebras/quivers are all just representation theory of categories

[coral spindle]

Yes, there is a monoid-specific rep theory and a fair bit of active research is happening on modular monoid reps atm

[topaz solar]

Interesting

[south patrol]

how useful is representation theory of category?

[coral spindle]

There are nice results regarding the semisimplicity of monoid rings, for example as one might expect, (von Neumann) inverse monoids produce semisimple complex monoid rings

[rocky cloak]

Is the charactersitic 0 case well understood?

[coral spindle]

Well yes sorta but also it fucking sucks

Like I think there's very little in general we can say

[topaz solar]

I’ve seen quiver-y reps connected with cluster algebras and triangulations of (oriented) Riemannian surfaces w/ boundary

[coral spindle]

but we do have individual nice results

[spice whale]

yes afaik there is rep theory on everything

[coral shale]

statistics

[rocky cloak]

depends what you mean by useful

[south patrol]

Inch resting

Oh I mean like

Idk i suppose I mean how fruitful does it seem as a field of study more generally

even if it's just interesting in its own right

[crystal turtle]

it's categories, of course it's not

[south patrol]

but that isn't really smth that can be quantified well usually lol

[coral spindle]

One example where it is well-understood is this: there is a paper on these things called R-trivial monoids, where they argue their representation theory is the most general way to analyse monoid actions on Markov chains. I'll find the paper.

https://arxiv.org/abs/1401.4250

The semigroup group did a reading group on this paper at my institution, which I participated in

[coral shale]

Rep theory on categories is still 'viewing things' as acting on linear spaces or not any more?

[warm wyvern]

every time a mathematician is asked whether something that's dead useless was useful:

[coral spindle]

You always appear in these conversations

It's so strange

[warm wyvern]

yea, I dunno why either tbh lmfao

[coral spindle]

Every time someone asks what something is used for, DarQ comes along

[coral shale]

i wonder if he found a use for measure theory

[warm wyvern]

probability theory

[coral spindle]

Anyway

[summer path]

what's probability theory used for

[warm wyvern]

literally everything

[coral spindle]

I was gonna mention that there are some nice results known (perhaps Stuart Margolis told me about this? I cannot remember) about this but I can't bloody remember

[rocky cloak]

So for a finite category its representation theory is the same as the rep theory of its path algebra, which is finite dimensional. And rep theory of fintie dimensonal algebras is a big field. Any abelian category with enough projectives is the module category on its category of projectives, there is some active research in generalizing things from rep theory of algebras to exact categories.

But just pure categories in general, idk...

[summer path]

super sully

[topaz solar]

Ultimate in toxicity

[south patrol]

noice

[coral shale]

please sphere this cube

[coral spindle]

If it were anywhere, it would be in this godforsaken channel

[summer path]

i wonder what the most super reacted message is on this server

[coral shale]

Thanks for the insights. Time to bump rep theory up on the list todo

[summer path]

rep theory is already on my todo list

[coral shale]

yes but it gets a higher seat now

[south patrol]

rep theory is fun

[topaz solar]

Chill, discussy, & alg chill are bad too

[coral shale]

im not sure what my highest is

[frosty pecan]

Representation theory has been on my to do list for too long.

[coral shale]

this is alg chill

[coral spindle]

Ben Steinberg has a book on complex representation theory of finite monoids btw 👀 it's a little heavy on the category theory but it's good

[summer path]

why would you split aa into 3 channels smh

[south patrol]

i think you made a typo -

should be "so" instead of "but"

[coral spindle]

Well ok that was an unfair thing for me to say

[south patrol]

i expected to be sullied for that lol

[coral spindle]

The thing is they jump straight into the however many adjunctions of all the restriction-induction stuff without talking about the background very much

Like it's fairly clear if you know about Frobenius reciprocity, but if you don't I saw it to be quite confusing

It's a lot to deal with

Woah category theory is a helpful tool if you know enough but if you don't then it can be confusing and impenetrable? :O colour me surprised

I'll say more about the rep theory of monoids because I can't help myself...

Unfortunately the irreducible representations are fundamentally not very interesting. They boil down (in a mildly subtle way) to the representations of the maximal subgroups of the monoid. As it turns out these maximal subgroups can be embedded in an interesting way in the monoid, but in any case at some point you will just be doing group theory.

The non-irreps are, as one might expect, extremely difficult to get and frankly I don't know if we understand them even in relatively nice cases. Perhaps the book talks about that

[dim widget]

tox usually hangs out in combinatorial structures

[next obsidian]

Tox tends to hang out in the back of my brain calling me mean names

[coral spindle]

Steinberg wrote a paper fairly recently in which he put a bit more into the modular representation theory of monoids. For those uninitiated, this is in a field of positive characteristic rather than C. He formulated some conjectures (which I confess I have not looked into) but recently Eisele I believe found a counterexample to one (or perhaps two?) by computer-aided search

I have nowt more to say about monoid reps

[coral spindle]

I can now say: this is funny

[coral shale]

Is there a type of ring quotient where you take things to 1 rather than 0? or just cant make sense?

[crystal turtle]

addition sounds like it would be hard with that

[coral spindle]

Well if something, say a, is sent to 1, then a - 1 is sent to 0 :)

[coral shale]

im just thinking here about trying to quotient with fields

[coral shale]

hmm but yh. So these things are kinda indivisible

[south patrol]

also like

what's nice about 0 is that 0 + 0 = 0 and 0 * anything = 0 lol

so that you can quotient out by subgroups nicely

or whatever

[coral spindle]

The fundamental stupid trick of algebra: if the difference between two things is zero, they're equal.

[south patrol]

Idk how you'd have a meaningful object otherwise

[solar inlet]

what the hell are you talking about

I'm a #foundations user

no one uses combinatorial stuctures

[dim widget]

Isn’t counting the real foundation of math?

Food for thought

[solar inlet]

oh god oh fuck, but I can't count

[white oxide]

wow, yea this makes a lot of sense to me, i'm not familiar with that much physics lol altho i do remember edward frenkel talking about the group representation or special unitary group or some shit on a podcast

[white oxide]

can somebody please give me a hint for this problem? I'm trying to define an isomorphism from M_r to Hom_R(R/(r), M). I know that it can't involve multiplication by m (at least for the examples I tried) because then everything would map to the 0 map and hence not be injective. i've also tried some other maps involving m in M_r, but they all ended up being the 0 map or not being well-defined.

or not even to M lol

i suppose I could go the other way and try to define a function from the Hom group to M_r, maybe something to do with the kernel

but I don't know how to generalize homomorphisms in Hom_R(R/(r), M), or rather what form every homomorphism has if that makes sense

I feel like the canonical way to define a homomorphism $\varphi: R/(r) \to M$ would be $s + (r) \mapsto sm$ but I'm pretty sure this is not well-defined

idk

[cloud walrus]

okeyokay

[white oxide]

moreover even if it is canonical other homomorphisms may come in different forms

so i feel like defining a function from Mr to the Hom group is better

but anyways yea if i could get a hint that'd be great

but multiplication is the only way we can transform s + (r) into some element of M right? since M is an R-module

[south patrol]

Well

[white oxide]

please let it be please let it be....

[south patrol]

You seem to be just defining a map without thinking about what you've been given or viewing this map as given

Like

There are two ways to do this problem but let's continue with your way

Suppose you are given a map R/(r) -> M

[white oxide]

mhm

[south patrol]

How are you going to associate with it and element of M_r

[white oxide]

yea now i'm thinking about going the other way....

like

[south patrol]

Actually it is probably best to think in terms of "data"

[white oxide]

like

i feel like it's staring me right in the fucking face lol

[south patrol]

What is a map R/(r) -> M

[white oxide]

with the quotient being (r)

[south patrol]

Like what is an equivalent amount of data

Hint: it is a theorem

Sorta lol

[white oxide]

uh does it have to do with torsion stuff

and modules over PIDs

[balmy belfry]

woah what is this channel

[south patrol]

No it is way simpler

[white oxide]

ok thank god

[white oxide]

groups rings fields!

uh hm whenever i see like

G/H --> Y or some shit i immediately think first iso

[balmy belfry]

yeah but wheres abstract algebra or modern algebra lol i just find this weird

[white oxide]

but like

[balmy belfry]

like #linear-algebra could be "Linear operators-Vector Spaces"

[white oxide]

this is algebra lol

oh

[balmy belfry]

ya

[white oxide]

i see ur point

[balmy belfry]

non of the other channels are like this

ig its cool idk

watcha guys talking bout

[white oxide]

yea it's a fun problem

[south patrol]

Yes

How do you define maps out of R/(r) ?

[white oxide]

this some meta-homo-homo type shit ya feel me

[south patrol]

no

Well lol

[south patrol]

think about this

This is probably the key point to underline imo

[open sluice]

it used to be called abstract algebra but we ended up splitting off the very big bren stuff into #advanced-algebra

[white oxide]

yo if this problem involves first iso that would be so cool

[white oxide]

ok

[south patrol]

Well

[balmy belfry]

i see

[south patrol]

it is kinda related to it lol

[balmy belfry]

what big brain stuff?

galois?

why not seperate it into "modern algebra" and "abstract algebra"

[south patrol]

This is more like a first course in algebra

[balmy belfry]

thats how universities do it

[south patrol]

Uh wdym

That seems a weird distinctino lol

[balmy belfry]

modern algebra

[south patrol]

what would modern algebra be

[balmy belfry]

groups-rings-fields

exactly this

like i mean in universities

[open sluice]

modern and abstract algebra are synonymous in the catalogues I've seen

[south patrol]

That's an odd disction imo like

[balmy belfry]

its called modern algebra, and then there is abs alg

[south patrol]

Both are modern in some sense and both are abstract ig

this is imo preferable as it is descriptive

[balmy belfry]

well i mean

yeah but you do calculations in arithmetic

[south patrol]

Which isn't algebra

[balmy belfry]

so you could say calculus is a weird name

[south patrol]

Sure

[open sluice]

do you know where the word "calculus" comes from

[balmy belfry]

not really, im sure my example doesnt fit but

see through my literal words

[south patrol]

I mean to say like we needn't be imprisoned by weird naming schemes of unis lol

when we can just describe the topics

[balmy belfry]

yeah but not many of the other channels do this

[open sluice]

it is related to arithmetic but historically calculus meant the arithmetic of pebbles--very, very small things
there's a lot of math terminology I hate tbh but we're stuck with it

[balmy belfry]

idk

XD

im just talking for the sake of talking and making conversation

this channel just stands out is all I mean

[white oxide]

oh wait, s + (r) |-> phi(s) where phi is some homomorphism from R to M or something

i just had to look up the first iso theorem proof again lol

at least i think

idk

[south patrol]

Well it is analogous to the split between #real-complex-analysis and #advanced-analysis too

But ye

[south patrol]

Sure so I mean

Basically what I'm getting at is that

[white oxide]

ok wait

wait

[south patrol]

oh lol

[white oxide]

WAIT WAIT WAIT

ok lol

i'll just like

not look at that last sentence you sent

my fault

thaank u

like it's literally staring me in the face rn uhhhhh

[balmy belfry]

whats adv analysis for?

functional analysis probably

whats between complex analysis and functional analysis

[open sluice]

well it's in the channel description

[balmy belfry]

genius

oh measure theory

[white oxide]

and by that homomorphism i'm assuming that has to do with representations or some shit

like the action of r on m

wait let me think about this a bit

[south patrol]

still a bit overcomplicated

[white oxide]

yeah that often happens when i try to prove things for some reason

i'm not really sure, s + (r) --> phi(s) which is in M

like

i think the trouble that i'm having is identifying it with a single element of M_r and not all of M_r if that makes sense

well

$M_r \subseteq $\bigcap_{i \in I} \text{Ker }\varphi_{r_i}$ where $\varphi_{r_i}$ is left multiplication by $r_i$

does that have anything to do with it?

i think

and that's for all ri in R sorry

[cloud walrus]

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

[white oxide]

$s + (r) \mapsto $\varphi_s$ maybe

[cloud walrus]

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

[white oxide]

but then that's the whole entire kernel associated with each homomorphism lmfao, don't know how to associate a particular m in M_r

[topaz solar]

If we have R/(r) -> M, then we have R -> M where (r) is in the kernel ye?

[white oxide]

oh i didn't know that, but i guess that makes sense

hm ok i'll think about that thanks

but (r) is not the kernel unless it's an isomorphism

[topaz solar]

Well I mean, you have R -> R/(r)?

[white oxide]

ya

i give up

i'll return to this problem later prolly

i mean i have to since it's due next tues lol

[celest furnace]

What is G’?

[coral shale]

Let M be a monoid. Fix x in M. Let y, z in M.
Then yx = xz = 1 gives
yxz = z
y = z

So left-inverses must equal right-inverses. And both existing means you uniquely have one double-sided "inverse".

But then if say left-inverses don't exist, you can have multiple right-inverses?

[rocky cloak]

Thats right, for exmaple the monoid of functions from N to N should give some examples

[coral shale]

hmmm ill have a see

i kept thinking about matrix ring

but ig theres more going on there

[rocky cloak]

A square matrix has left inverse iff it has right inverse iff it is invertible iff it has nonzero determinant, so that might be hard

[coral shale]

ill have a think whether 0 in ring stops it

well yh i was thinking why

[coral shale]

ie. is that specific to just this ring

aaa i dont know many non comm rings

ring of functions ig. . .

[rocky cloak]

its a finiteness condition. If x has a left inverse, then multiplication by x is injective. If the set is finite (or finite dimensional with some linearity condition) then injective = surjective, so it also has a right inverse

So you need something infinite/unbounded for this to work

[coral shale]

interesting

much food for thought

[rocky cloak]

If you want it to be a ring, you can take the endomorphism ring of an infinite dimensional vector space. Which is pretty close to the matrix example I guess

[coral shale]

do we have a name for a ring structure but its monoid for addition as well?

say natural functions u could do this

ig ur probably looking at endomorphisms of monoids then

[rocky cloak]

I think semiring or rig is the word

[coral shale]

ah there's also a near ring

rig

[coral shale]

hmm yh, thinking in terms of functions was v helpful

definitely proved these about functions once, but didnt appreciate it back then

ig its because "left" and "right" arent very meaningful to me. vs "before" and "after"

pre-inverse, post-inverse

[mossy lintel]

i'm stuck at using phi to construct an automorphism for G

[coral spindle]

The easiest way to do this is to use the classification of finitely generated Abelian groups imo

If you want to construct an automorphism I would suggest rather using a very large subgroup H of G (perhaps of order |G|/2) and extending the automorphism on that. You will need to think about how the structure of H and G/H interact.

[mossy lintel]

well, I'm not familiar with the concept of classification, can u explain more of it

[coral spindle]

I would suggest looking up the classification of finitely generated Abelian groups in your own time. Alternatively you can look at my second suggestion.

[mossy lintel]

I'll look into it later while thinking about the second hint

[mossy lintel]

Assuming there exists a H of order |G|/2, i can indeed construct an automorphism for G with an automorphism of H

And it seems i can prove the existence with induction, thanks for reaching out

[indigo plinth]

Where do you get that G is abelian?

[delicate orchid]

Every exponent 2 group is abelian

[indigo plinth]

Ah ok

[delicate orchid]

Yeah I’d do something similar to what “Schur thing boss” says from that point

[indigo plinth]

Yeah the classification makes this pretty simple then

[delicate orchid]

You then know that it has to be some number of copies of C_2 so then you can just permute them around to get automorphisms

But the classification seems over kill for this

Perhaps you can do it inductively

[indigo plinth]

(proof for myself: ab a^-1 b^-1 = abab = (ab)(ab)^-1 = 1)

[coral spindle]

You can, I believe, using the same strategy: find a subgroup H of order |G|/2, and show that H has a complement. The sketch proof I had in mind for showing there exists an automorphism essentially requires this

This is fairly similar to proving that every finite Boolean algebra has an F_2-basis

[open sluice]

for any a, b in G, ab = (ab)^-1 = b^-1 a^-1 = ba

[indigo plinth]

neat

[rocky cloak]

Or just notice that's it a vector space over F2, then use linear algebra

Don't even need the group to be finite

[delicate orchid]

Ok mr smarty pants

[hybrid pike]

I'm trying to solve this problem and I genuinely have no clue where to even start: Suppose G is a group where each element has a finite order. Suppose $f: G \rightarrow GL_n(\mathbb{C})$ is a group morphism. Prove that $f(g)$ is diagonalizable for each $g \in G$ by looking at the minimal polynomial of $f(g)$. I understand that I'm supposed to prove this minimal polynomial is a product of polynomials of degree 1, but how do I even go about finding this?

[cloud walrus]

NotAPenguin

[coral spindle]

You don't need to find the minimal polynomial. Instead, simply find some polynomial that f(g) satisfies, which the minimal polynomial therefore divides. You may recall the fact that if the minimal polynomial has no repeated roots, the matrix is diagonalisable.

[hybrid pike]

hmm okay, I'll give this a shot

Thanks for the hint

[next obsidian]

What is even the point of f lol?

[rocky cloak]

Just there to ensure f(g) has finite order I suppose

[next obsidian]

Yeah but like

[rocky cloak]

The point is just to tie the exercise into representation theory later I suppose

[coral spindle]

This is in reality proving that the simple reps of abelian groups are of dimension 1

[hybrid pike]

hmm no, its just a question on one of the previous exams of this course

[rocky cloak]

This is maybe a motivation for schurs lemma I guess

[next obsidian]

The problem immediately becomes “show a finite order element of GL_n(C) is diagonalizable”

Like if you needed to find a g so that gfg^-1 was diagonal then i would get why you want f lol

[coral spindle]

I don't see what you mean by that conjugation

g is in G, f is a map, so what exactly does that mean?

[next obsidian]

g was a bad notation

Lmfao

Call it h

Another map

Or like, a matrix A

Like if you needed to make the diagonalizations coherent among all f(g)

[rocky cloak]

I'm having a hard time thinking of a solution that involves the minimal polynomial actually. Wouldn't it make more sense to use the Jordan form? Or what solution is the exercise hinting at

[hybrid pike]

Okay so what I currently have is
Suppose $g \in G$ with $g^n = 1$. Then take a polynomial $P_g = X^n - I$. Then $P_g(f(g)) = (f(g))^n - I = f(g^n) - I = f(1) - I = I - I = 0$.

[cloud walrus]

NotAPenguin

[hybrid pike]

So I found some polynomial that f(g) satisfies

And now I show that this means the minimal polynomial has no repeated roots?

[rocky cloak]

Ahhh, I see. That's what they were getting at

[hybrid pike]

im still a bit confused on how this follows though

[next obsidian]

Do you know about the derivative test

[hybrid pike]

Oooh

Yeah I remember that

[next obsidian]

That proves over a char 0 field any irreducible polynomial has no repeated root

You can compactly say this as “char 0 fields are perfect”

Where perfect is a technical term, not just me complimenting the fields

[rocky cloak]

This polynomial is very much not irreducible though.

But it has n distinct roots, and you know exactly what they are

[hybrid pike]

what I remember about a derivative test is that f has a repeated root a iff a is a root of f'

[hybrid pike]

Oh wait theyre just the nth roots of unity of course

[next obsidian]

Just take the minimal polynomial. Once you know it satisfies a polynomial a minimal one exists

I mean you can also do what you said I guess lol

[rocky cloak]

The minimal polynomial of what?

[next obsidian]

f(g) yeah?

[rocky cloak]

You can't expect that to be irreducible

[next obsidian]

Or does this fail for matrices(

What? A minimal polynomial is by definition irreducible

[rocky cloak]

You're mixing up to different meanings here

[next obsidian]

Maybe it does fail for matrices

[hybrid pike]

What I wrote down now is that $P_g$ has a repeated root in x iff x is a root of $P_g'$ iff $nX^{n-1} = 0 \iff X = 0$, but 0 is not a root of $P_g$ so $P_g$ has no repeated roots

[rocky cloak]

In a field if 0 = f(x) = g(x)h(x), then either g or h is 0. But this is not true in arbitrary rings

[next obsidian]

Kekw sooooo trueeeeeeee

I was thinking harder about it and it was making less and less sense

[cloud walrus]

NotAPenguin

[next obsidian]

Yeah that would work, but jagr2808’s observation was that you can just enumerate every root by multiplying f(g) by roots of unity

[hybrid pike]

Yeah that's true, probably even easier

[coral spindle]

It does fail for matrices – for example take the 2x2 matrix M with 1 in every entry not below the diagonal (you might write it M = [1 1 \ 0 1]). This matrix satisfies x^2 - 2x + 1 yet it is not the identity.

[dim widget]

a matrix is semisimple iff its minimal polynomial is separable, then it follows immediately from x^n - 1 = 0

[coral spindle]

This is what I had in mind

[quiet pelican]

How do you motivate these constructions
Sure I kinda get lying the cube roots of unity over an even permutation group
But then why are these covers exceptional?
And why do we add (2, 0, 0, 0, 0, 0) to get 3.A_7

I can kinda guess that for the stabiliser of the partition to surject onto A_n we might need n = 6 but I don't intuitively see that

[fossil vapor]

Does someone have a reference/proof of the first product formula given on the Wikipedia page for the Legendre symbol? The reference simply states this as an exercise.

[mighty kiln]

Aluffi chapter 0

Regarding the proof: How is B' guaranteed to be maximally linearly independent? Doesn't one run into problems of the following sort if v is not a successor:
{v1, v2, v3, ...}
{v1+v2, v2, v3, ...}
{v1+v2, v2+v3, v3, ...}
...
{v1+v2, v2+v3, v3+v4, ...} ← does not span V
Also are there any other references to the stronger result that is proved here? Namely, that if S is linearly independent and B is a basis then one can replace |S| elements of B with elements of S to get another basis

[long obsidian]

Hey stupid question. In the natural numbers m|n implies m<=n right? It seems obvious but I can't think of a short proof

[coral spindle]

That is indeed true

[ashen heron]

1 <= n for all n in Z+ so if s|t and t = sk then 1<= k implies s <= sk = t

[white oxide]

i'm confused, what's the difference between the family and the single element x? since every element in that family is equal to x

do they just mean as in like

x + x + x + x = 0 vs x = 0

ok thzx

thx

[rocky cloak]

x - x = 0

[white oxide]

ye or x + x + x + x = 0 whence 4x = 0

or like nx = 0

wait

does that make sense

[rocky cloak]

Are you working over Z/4...?

[white oxide]

wait

every module is a Z-module right

wait

am i trippin

[rocky cloak]

If nx = 0, and n isn't 0, then {x} isn't linearly independent

[white oxide]

right

wait why can't i just write x + x + x + x = 0 and take the scalars to be 1

[rocky cloak]

Because 4x isn't 0, or why would it be 0?

[white oxide]

wait i'm so stupid lmfao

i forgot the definition of linear independence

[white oxide]

yea lmao

[white oxide]

what does this blue underlined phrase mean? w = b_1x_1 + \dots + b_rx_r + a_{r + 1}x_{r + 1} or....

[next obsidian]

Yes

But like uh

The other b_i are existential

They just exist

Cuz of the definition of a guarantees such a w exist

[long obsidian]

If you have two free modules with bases the same size, are the two modules isomorphic as modules?

I think the isomorphism is just mapping the basis elements to each other.

[white oxide]

ah i see what you mean

thanks

[next obsidian]

Yes

Precisely

[toxic zephyr]

what about a free module that can have a basis of any size? is that module isomorphic to all free modules? something like this

[white oxide]

yeah this makes sense, then (r) is clearly in the kernel but i'm having trouble seeing the other direction/when R \neq M

[topaz solar]

For every f in Hom(R/(r), M) we do have an f’ in Hom(R, M)

And, if (r) is in the kernel, it goes the other way ofc

In particular, r in R has \phi_r(m) = rm as a map

[elder wave]

this can only happen for rings that don't have IBN

[toxic zephyr]

what's IBN?

[elder wave]

Invariant Basis Number

[topaz solar]

Invariant basis number

[toxic zephyr]

ah

[elder wave]

Otherwise what chmonkey said isn't true

[topaz solar]

Basically: the property that definitionally kills it

[elder wave]

but the question seemed to imply that the size of basis is unique

[topaz solar]

But commutative rings with identity have it

[elder wave]

so i would've said the same as chmonkey

[toxic zephyr]

so if it doesnt have IBN, then it isn't isomorphic to another module that does have IBN?

[topaz solar]

Idk what nasty properties you need to get -IBN

[elder wave]

it's not a module property

it's about the ring the module is over

[topaz solar]

IBN is a ring property

[elder wave]

https://en.wikipedia.org/wiki/Invariant_basis_number

here if you want to read about this

[topaz solar]

Since it says all the relevant modules have this nice basis size

[toxic zephyr]

oh that makes sense. sorta like an apples and oranges type thing wrt the ring

[topaz solar]

Your ring has to suck as a ring for this to fail tho

[elder wave]

yeah it's a weak assumption

[toxic zephyr]

that's the technical term i assume

[elder wave]

a lot of stuff implies it

[white oxide]

correct me if I'm wrong; $\varphi: R/(r) \to M$ induces a homomorphism $\psi: R \to M$. in particular, $\psi \in \text{Hom}(R, M)$ where $r \mapsto rm$. If $\psi$ is the zero map then $\psi \in \text{Ker }\varphi$. but this also implies that $m \in M_r$. hence each element in the kernel of $\varphi$ can be associated with an element of $M_r$?

[topaz solar]

How bad does it have to be

[cloud walrus]

okeyokay

[white oxide]

maybe there are more implications to show but I feel like that's the general idea

only problem is I didn't use (r)

[topaz solar]

\psi \in ker \phi is nonsensical

So I’m not reading further

[white oxide]

oh right oops

[elder wave]

uuh like

[topaz solar]

I think it has to be non commutative?

[elder wave]

non commutative, not (left) noetherian

not the thingy that i forgot the name of

AB = 1 implies BA=1

[topaz solar]

Inverses are 2 sided

[elder wave]

not ring elements here

[next obsidian]

Even if you don’t have IBN it’s true

[elder wave]

matrices with entries in the ring

[topaz solar]

Ah

[next obsidian]

If you have a basis of cardinality kappa

Any other module with a basis of cardinality kappa is isomorphic

It’s just that kappa isn’t unique

[topaz solar]

R^n \cong R^m

[toxic zephyr]

damn thats crazy

[elder wave]

oh

right

[next obsidian]

The map is just “send basis to basis”

[topaz solar]

(This R^n ~ R^m thing is mentioned in the wiki page even)

[elder wave]

yeah i'm smoking

sorry for doubting you chmonkey

[topaz solar]

[toxic zephyr]

so without IBN, a 1d R module M1 is isomorphic to a 2d R module M2? since M2 also has a 1d basis?

[elder wave]

I drew this

just throwing that out there

[topaz solar]

Irony

“0 days”

[coral spindle]

An example of a ring without IBN is the countable product of copies of Z. Then the free module on one generator is also the free module on two generators, et cetera. this is wrong sorry uwu

[topaz solar]

Well, if they have a basis of the same cardinality

Idk if you always get bases of all sizes

[toxic zephyr]

modules are wack lol thats cool

[coral spindle]

Well rings without IBN are generally pretty nasty I think! Certainly a ring in which R^1 is iso to R^2 can't be Noetherian

[topaz solar]

Basically you simply do not care about them

[topaz solar]

Isn’t this commutative

[coral spindle]

Yeah I was about to say, I'm full of shit aren't I

[topaz solar]

Yeah

[coral spindle]

The module structure I wanted to try out on it wasn't working in my head

[toxic zephyr]

you call them nasty, i call them sexy chaotic 😆

[toxic zephyr]

gotta read up more on modules sometime

[white oxide]

idk. maybe all of the maps in Hom_R(R/(r) --> M) are zero or some shit and that we can then associate each one with an element of M_r idfk

lmao

[topaz solar]

Moreover, any commutative ring (except the zero ring) satisfies IBN,[1] as does any left-Noetherian ring and any semilocal ring.

[coral spindle]

Yeah indeed

Apparently End(k (+) ... ) over any field, where we take countably many copies, does not have IBN

[elder wave]

yes

[coral spindle]

So let's say V = k^(+)N, countably many copies

[elder wave]

the right shift there is also the standard counterexample for stable finiteness iirc

[coral spindle]

right there you go

[topaz solar]

This looks similar to the given example

[elder wave]

god i hates this part of my comm alg class

[topaz solar]

L(F^N) or wtv

[elder wave]

takes comm alg class
rings are not commutative

[topaz solar]

Repost the original question rq

[coral spindle]

They meant by r, not by R

[white oxide]

so I followed potato's hint, an element phi of Hom_R(R/(r), M) can be given by phi(s + (r)) = psi(s) where psi is an induced homomorphism from R to M

[white oxide]

oh

bruh

well hopefully that makes things easier lol

[coral spindle]

Well it is somewhat easy to determine all elements annihilated by the unit in the ring, tbh

Unfortunately the fact written would be generally false

[white oxide]

as in the problem?

[coral spindle]

If M_r denoted the set of elements of M annihilated by the entirety of R? Yes.

[white oxide]

ah okay

[coral spindle]

Maybe you can think about what the set of elements annihilated by the entirety of R, and therefore in particular 1, is.

[white oxide]

ok will do

this problem's murdering me

maybe it'll be easier now to go back to trying to define an isomorphism from M_r to Hom_R(R/(r), M)

[cloud walrus]

okeyokay

[white oxide]

wait

maybe I'm on to something

pls pls pls

[white oxide]

$m \in M_r \mapsto \varphi_m \in \text{Hom}_R(R/(r), M)$ where $\varphi_m(s + (r)) = \psi_m(s)$ given by $\psi_m(s) = sm$ lol`

[cloud walrus]

okeyokay

[white oxide]

i hope this is well-defined, i have no clue how to show so for the moment

[topaz solar]

if tm = 0 for t in (r)

Then if a-b is in (r)

What’s going on with am, bm

[white oxide]

ah there we go

x + (r) = y + (r) => x - y \in (r) => x - y = rk => (x - y)m = 0 => xm = ym I believe

[topaz solar]

Rather than => I’d just consider that the basic notion of R/I kinda tbh

[white oxide]

too late

[topaz solar]

Also yes

[white oxide]

but yea

you're probably right

i just like to bring myself pai

n

[topaz solar]

am - bm = (a-b)m = 0

[white oxide]

yea

holy this problem is so fun

probably because i made it three times as hard for myself

but whatever

[topaz solar]

Can you characterize hom(R, M)

[white oxide]

wait why didn't I just define $\varphi_m$ as $s + (r) \mapsto sm$

[cloud walrus]

okeyokay

[white oxide]

💀

[white oxide]

uhh yea let me try to think about this after writing the proof

[topaz solar]

Well I think proving this will tell it to ya since

Ya know

R/(0)

[white oxide]

Ya know

[topaz solar]

[white oxide]

here's my issue; $\varphi_m(x + (r)) = 0$ for all $x + (r) \in R/(r)$, or for all $x \notin (r)$ satisfying $xm = 0$, but at the same time $0 \in (r)$

[cloud walrus]

okeyokay

[white oxide]

(assuming m is in the kernel of the map from M_r to the Hom group)

wait

but then it has to be true for all x

yea

oh no

now i have to show that every single homomorphism in Hom(R/(r), M) has the form I defined

fuckkkkk

wait sorry

what's like the best way to show this

that the only way to define a homomorphism from R/(r) to M is multiplication by m

[white oxide]

i've tried experimenting with universal properties of homomorphisms

all I've written down is now, any homomorphism R/(r) --> M naturally involves multiplying representatives of R/(r) with fixed elements of M 💀

like idk how to prove this

and then after that I have to show that it only involves multiplying elements of r with fixed elements of Mr and not just any arbitrary element of M

[topaz solar]

Try looking at Hom(R, M)

[white oxide]

ok

uh

was i supposed to do this exercise before hand lol

[crystal turtle]

probably not needed, given that the order of M and R is switched

[white oxide]

o

right

.

oh

M is finitely-generated

i think that helps

[crystal turtle]

probably

[void cosmos]

M = R^n + Tor(M) for some n --> Hom_R(R^n + Tor(M) , R) = Hom_R(R^n,R) + Hom_R(Tor(M),R)

if you can show that Hom_R(Tor(M),R) is 0 ur done

[white oxide]

oh boy

the next three problems involve torsion submodules and shit

[white oxide]

i'll understand this soon hopefully

[void cosmos]

in general this is how finitely generated modules over a pid are

read the hungerford section if ur lost

[topaz solar]

It’s before it, so yes, but I mean R->M

[white oxide]

prof assigned us lang reading for htat 😄

[topaz solar]

Moamen just throwing an answer in even

Smh don’t just give answers to homework

[void cosmos]

yea i think lang must have it too ig :d

[hushed hamlet]

wait people use other books than atiyah macdonald for commutative algebra

[void cosmos]

sadly hungerford is not commutative algebra textbook

[crystal turtle]

they shouldn't

[white oxide]

yea, time to postpone this problem to tomorrow yet again lol

my alg topology class is also calling 😄

i've got 5 hours on this problem, sharp's got 1 minute

[void cosmos]

try to do it

it's not hard

[white oxide]

same with moamen

yea i'll do it later

i need to save time for my other classes

[void cosmos]

no i think these problems are hard

if they werent why would they be given you for hw

dont be discourage king

d

[white oxide]

thanks g

if anything these problems are a great learning experience lol

[void cosmos]

gl and hf

[fast stratus]

I don’t understand, what it means to be a non-negative complex number…

[chilly ocean]

it's a nonnegative real number

real numbers are also complex numbers

while R admits a nice ordering, C doesn't, so when we refer to a complex number as less than / greater than another we have reals in mind

by "nice ordering" i mean one which respects the field structure in the ways you'd expect

[fast stratus]

Yes… that is why I don’t understand this

[open sluice]

nonnegative complex number = number of the form x + iy where x >= 0 and y = 0

[chilly ocean]

go prove it or look it up

[open sluice]

ok how about this phrasing
"a complex number z is equal to a nonnegative number iff ..."

[chilly ocean]

this is the relevant fact

[fast stratus]

I see… authors definition only admits these as non negative numbers

[fast stratus]

Yeah… i understand this… any total order you define on C, will not obey ordered field axioms

[topaz solar]

So helium, lemme ask this

Can you define a partial order on C?

Which is compatible with field operations

[fast stratus]

Field operations don’t care about order

Ordered field by definition requires total order

[fast stratus]

So no?

[topaz solar]

I mean, can you define $\leq$ such that $a\leq b$ means $a+x\leq b+x$ for all x, and where $0\leq y$ and $a\leq b$ implies $ay\leq by$

[cloud walrus]

Dragonslayer Sharp

[fast stratus]

Lemme think

Yes

[fast stratus]

$\leq = {(a,a) | a \in C}$

[cloud walrus]

helium

[crystal turtle]

[open sluice]

genius

[crystal turtle]

now come up with a nontrivial one

[mossy lintel]

I need some help with this problem

[topaz solar]

Good

[mossy lintel]

the mapping i fond is Na-> N(aT)

i have proved it's a homomorphism and onto mapping of G/N

but i have no clue about how to show it's 1-1

[void cosmos]

What does (N)T mean

does it mean T(N)

[mossy lintel]

yep

[void cosmos]

and what mapping did you find?

[mossy lintel]

Na->NT(a)

[void cosmos]

aN --> T(a)N ?

[void cosmos]

like this?

[mossy lintel]

N is normal, so it's equivalent to what i wrote

[void cosmos]

okay so suppose there exists xN , yN in G/N such that T(x)N=T(y)N

then what can you do

[mossy lintel]

T(y^{-1}x) \in N

[void cosmos]

remember that T is a bijective homomorphism

bijective

that is its both 1-1 and onto

so we are basically going to look to contradict this somehow right?

that is if xN is not equal to yN

so what happens if we suppose for the sake of contradiction that xN is not equal to yN

[mossy lintel]

y^{-1}x \notin N?

[topaz solar]

Yes but why would you write it in that order

[mossy lintel]

that's how it's done in the textbook, lol

[void cosmos]

yea

sorry i went brb

anyways

if you restrict the automorphism T to the ssubgroup N u still get an automorphism on N but u would have y^-1(x) not in N with T(y^-1(x)) being in N

[mossy lintel]

so Ty is equivalent to Tx while x is not equivalent to y

[void cosmos]

what

i dont follow u

wdym by "so"

[mossy lintel]

the cosets of N defined an equivalence class

[void cosmos]

yea

what about that

are u following or am i stupid or what

[mossy lintel]

ehh, since T(y^{-1}x) = [T(y)]^{-1} T(x) \not in N, so Ty, Tx are not equivalent

[void cosmos]

no

okay let's recap

we first supposed that T(xN) = T(yN)

[hollow mica]

do you know why (in the link) the proof of theorem 2 doesn’t readily work for C?

[void cosmos]

we claim that xN = yN correct?

doesnt this what it means for ur map to be injective on G/N?

or would u rather show it has trivial kernel

or what

[mossy lintel]

injective homomorphism means trivial kernel right

[void cosmos]

yea they are equivalent

so u following?

i gtg elgato i have boring internship at 1:pm and its 6:30 am now

soo hopefully someone else will be able to help you better

gl

[mossy lintel]

thanks for trying bruh, i'll try to work this out:)

[void cosmos]

yea just remember that T is a bijection

it has an inverse that is a bijection aswell

[open sluice]

what

have you been awake all night

[void cosmos]

yea i have bad sleeping schedule

[open sluice]

oh my god

[void cosmos]

preworkouts + diet coke they have shit ton of caffeine

its so sad

like i am living the worst time of my life rn

no joke

i cant sleep

[hollow mica]

and to top it all off you’re doing algebra

gl man

[void cosmos]

yea i was studying actually too for algebra

haha

anyways please help king/queen/they/them elgato

bye everyone

[chilly ocean]

it does work for C. i guess they phrased it the way they did because they wanted to clarify that only the properties of Z[i] were needed

i didn't actually read the link when i posted it

[hollow mica]

Ah that would make sense

[rocky cloak]

This isn't true actually. Unless T(N) = N, you can only guarantee a homomorphism from G/N to itself.

For example if $G = \mathbb R^{\mathbb Z}$, $N = \mathbb R^{\mathbb N}$ and T just shifts the indexing up a degree, then the induced map on G/N will not be injective.

[cloud walrus]

jagr2808

[void cosmos]

[mossy lintel]

What is R^Z? R^N means a collection of sequence of reals right?

[rocky cloak]

Yeah, sequences indexed but integers/natural numbers respectively

[mossy lintel]

I see, this is a cool example

[void cosmos]

yea this guy is the real deal

[mossy lintel]

so is this channel(server)!

[void cosmos]

nah not really its only me

jk

[white oxide]

ugh just returned to this problem - i'm assuming that it has to do something with the fact that (r) is a subset of the kernel from R to M

idek

i mean i know that M is finitely generated

so phi(s) = r1m1 + r2m2 + \dots + rnmn sure

should i be familiar with "modules over principal rings" in lang

i feel like there's a theorem i'm missing

[agile burrow]

There is a classification of finitely generated modules over a PID

[white oxide]

do I need this to show it

[agile burrow]

That's the statement of the classification

Or at least one form of it

[topaz solar]

(r) being a subset of the kernel is clearly important, since M_r are objects such that rm=0 ye?

[white oxide]

yeah that's true

somebody told me that i should do problem 2 and that it's solution might work for 3 with some tweaks

or rather it'll be instructional

so i'll probably sleep and figure it out tomorrow hopefully

[lethal dune]

hint: ||finitely generated torsion free modules are free ||

[hybrid pike]

I was solving some more exercises on extension fields and just want to check if I did this one correctly. Let $\alpha = \sqrt{1 + \sqrt{3}}$ and $K = \mathbb{Q}(\alpha)$. First I found the minimal polynomial $f$ of $\alpha$ to be $(X^2 - 1)^2 - 3$ and used that to determine that $[K : \mathbb{Q}] = 4$. Then I had to find the factorisation field of $\alpha$ over $\mathbb{Q}$. I found the two linearly independent roots of $f$ to be $u_1 = \sqrt{\sqrt{3} + 1}$ and $u_2 = i\sqrt{\sqrt{3} - 1}$. From that I concluded that $E = \mathbb{Q}(u_1, u_2)$ and that $[E : \mathbb{Q}] = 16$, since there are 16 distinct elements in the form of $u_1^xu_2^y$ with $x,y \leq 3$. Is this conceptually correct? I'm mostly not entirely sure about the last step really, but it seems to at least be below the upper bound of $4!$.

[cloud walrus]

NotAPenguin

[rocky cloak]

So the roots of f should be

±sqrt(1 ± sqrt(3))

It can be a good idea to try to break the extension up into smaller pieces. For example you know that the extension contains sqrt(3), so it might be easier to determine the degree over Q(sqrt(3))

[hybrid pike]

Yeah that makes sense, the polynomial that I found has roots ±sqrt(1 + sqrt(3)) and ±i*sqrt(1 - sqrt(3)) (which lines up with another subquestion asking to prove that this polynomial has complex roots)

so it might be easier to determine the degree over Q(sqrt(3))
Would this not lead to the same answer, since that should be 2, and then since the degree of Q(sqrt(3)) over Q is also 2 the result is 4

[rocky cloak]

Are you taking about a different polynomial than the one you wrote?

[hybrid pike]

I think not no

🤔

[rocky cloak]

(x^2 - 1)^2 - 3 has roots

±sqrt(1 ± sqrt(3)), right?

[hybrid pike]

ah, yes it does

the imaginary roots then come from the case of ±sqrt(1 - sqrt(3))

[rocky cloak]

Yeah, I see I just misread your sign I think

Anyway, these roots have degree 2 over Q(sqrt(3)) so the degree must be either 4 or 8

Not 16

[hybrid pike]

hmm I see

I guess its not as simple as grabbing all those x^i*y^j like I did

[rocky cloak]

Yeah, you would have to check whether they are linearly independent to do that

[hybrid pike]

Right, which is probably too much work here

Well I feel like [E : Q] can't be 4 because [Q(a) : Q] is already 4 and is missing the imaginary roots

[rocky cloak]

Exactly

[hybrid pike]

That just leaves 8 then

https://cdn.discordapp.com/emojis/904779355757113366.gif?size=48&name=party&quality=lossless

Thanks

[rotund aurora]

epilepsia moment

[rotund aurora]

the reason why divisible groups are not a Q-vector space is because scalar multiplication is not necessarilly well defined?

so like torsion-free divisible groups are Q-vector spaces right

so if for all g in G and for all n in Z there is exactly one h in G such that g=nh then G is a Q-vector space

[formal ermine]

how did you pull this off so perfectly

[rotund aurora]

I have no idea of what you mean

[rocky cloak]

That's right. In a divisible group x/n isn't necessarily unique, that happens iff the group is torsion free

[oblique thorn]

Anyone know the extended Euclidean algorithm?

I’m in a cryptography class nd we are learning about Ferrara little theorem and EEA I need someone to explain to me bc my teacher is just a TA and is always in a panic

[shy fossil]

sure, what do you want to know about it

[oblique thorn]

I need to use it to find a multiplicative inverse of a prime number

Like over a field

[shy fossil]

right yes, that's what it is used for

but what is your issue? Do you want to understand it?

[oblique thorn]

So a = 51 and Fp = 2^17 - 1

[oblique thorn]

Yea

I know my prof is not making sense

They are trying to make it seem impossible but they are giving us like all the answers and not explaining anything

2^17 - 1 is a Mersenne prime btw

I’m not a math major I’m a computer science major taking a cryptography class

[oblique thorn]

Would you be able to show me?

[shy fossil]

i do have some hand written notes I can share