groups-rings-fields

[proud bear]

i don't think so. Z/nZ is a cyclic Z-module

[coarse stag]

I assume where rx= rx mod n?

[coarse stag]

Where r in Z and x in Z/nZ

[proud bear]

yeah

[coarse stag]

So there just needs to be a homomorphism from R onto the underlying group then?

[proud bear]

if you consider R as an R-module over itself then there is always an R-module homomorphism f from R to any R-module M. basically you can define f(1)=x where x is any element of M and for any r in R, f(r)=f(r*1)=r*f(1). if you can find an element x so that this map is surjective, then M is a cyclic R-module

[lethal dune]

is this hint enough?

[dusty sapphire]

I is principal ideal then how to apply it on quotient ring?

[proud bear]

so yeah pretty much

[lethal dune]

I is principal so there is a+ib that generates it. Now every element of R/I can be achieved from euclidean div, so necessarily N(r) < N(a+ib)

[coarse stag]

I don’t understand why 1 is used?

[thorn delta]

maybe an easy way to think about it is if M = <m> is a cyclic R-module, then M is iso to R/I where I is the ideal {r in R : rm = 0}

[dusty sapphire]

ok then it will not allowed to be infinite. Thanks

[coarse stag]

That’s actually the exercise I was trying to prove

I was wondering if J was just the kernel of the homomorphism

[thorn delta]

right, the general procedure for finding an isomorphism between modules R/J --> M goes like finding a homomorphism R --> M whose kernel is J

[coarse stag]

And I assume in this context you just prove a homomorphism between the groups since they are both R modules

And you assume that the product operation plays nicely in M

[thorn delta]

well hmm, im not quite sure what you mean by that. an R-module homomorphism is not just a group homomorphism: it has the extra condition that f(rm) = rf(m) for every r in R and m in M

[coarse stag]

I see

I’m not quite sure about how that proof starts out

[thorn delta]

for a hint, take a look at the ideal i wrote down here. It is the kernel of some homomorphism. Can you see what it might be?

[coarse stag]

Oh wait the kernel is exactly the cyclic generator

[thorn delta]

wait hmm im not sure what you mean. Kernel of what homomorphism?

[coarse stag]

Nvm

[coarse stag]

I assume that the ideal is the kernel from R to M

But I’m not quite sure how f(rm)=rf(m) follows

[thorn delta]

yes, but kernel of what homomorphism from R to M?

[coarse stag]

Quotienting by cosets of I?

[thorn delta]

nah, so forget about I. M has some cyclic generator, call it m. We want to define a map f : R --> M
what should f(r) be? In other words, how can we get an element of M from an element of R?

[coarse stag]

Are cyclic R modules unitary in general?

I’m not quite sure how do define that without utilizing the kernel in some way

[thorn delta]

we don't know what the kernel is until we define a map. So it may help not to worry too much about the kernel thing yet. We just need to think of a natural mapping R --> M = <m>

[thorn delta]

also I don't think so, but idk i haven't really though much about modules over rings without identity. I'm not 100% sure why that assumption is present here

[coarse stag]

I guess let f(r) -> m define an function for <r> and work from there?

[thorn delta]

you're so close. f(r) = m is a constant function, so I don't think this is what you mean

also I think your ring need might need to have identity in order to have M = <m> = {rm : r is in R}

i think that's why that assumption is there

[coarse stag]

Because I want to write f(r+I)=m, but as you said we don’t have I until we have a homomorphism so this would be somewhat circular

[thorn delta]

the ideal we get from f in the end doesn't really matter. i.e. You don't have to force f to have the kernel we're expecting.

[coarse stag]

Also that’s what I was guessing, but I’m not sure if there could be rm=m such that r is not 1

[thorn delta]

you definitely can, and this would be some r such that (r-1)m = 0

well okay, this doesn't actually make it anymore obvious that r is not 1 lol

[coarse stag]

Yeah

I’m not sure if you can pull that 1 into (r-1) if ur not assuming unitary

[coarse stag]

So what, do we just assume f(r)=m iff r^|M| in preimg(0)?

[thorn delta]

You're overcomplicating this. We want a map f : R --> M. you can multiply elements of M by elements of R. Since M is cyclic, it has a very special element m such that M = <m> = {rm : r in R}. So we define f(r) = ||rm for all r in R.||

[coarse stag]

Oh wait

I’m dumb

I was too busy thinking of R and M as strictly groups

[thorn delta]

ah okay. I went ahead and spilled the beans because I thought I was saying too much and causing confusion.

[coarse stag]

Okay so now, i have to show f: M -> R/I such that rf(m)=f(rm)?

I assume this is where unitary is necessary

[thorn delta]

f is a map R --> M. We know that f(r) = rm. You have to prove that this is an R-module map (homomorphism of additive groups and f(rs) = rf(s))

[coarse stag]

Do you mind if I bug you about this tomorrow btw, I think I’m a bit too fried to think rn and I have to wake up pretty early tomorrow

Sry

[thorn delta]

sure, np

[coarse stag]

Thank you so much

[foggy merlin]

need to vent, just got back from commAlg oral exams and I'm pissed at myself for shuffling around the valuation Ring terminology

[dreamy fiber]

I am taking commAlg next semester and this is making me even more scared of the class

[foggy merlin]

it's an enjoyable subject, don't be intimidated by it, just don't be a disorganised moron like I am who needs to catch up the entire semester in 3 weeks, math just takes time to digest.

[dreamy fiber]

Thanks! Chances are I might quite like it, but we will see. I wish the best of luck to you, that sounds really rough.

Anything that has to do with condensing a longer period time of study into a very small period of time

[lethal dune]

[foggy merlin]

well that was luckily only the last third of the exam... the first part was decentish and there is a nonzero chance of passing so yeah,

[dreamy fiber]

You could always normalize the measure so that your chance is 1

That’s my very poor attempt at making a real analysis joke

[foggy merlin]

thanks, I hope you like it, I really got to love it during those weeks even under the pressure and uh... really regret not having started right from the semester beginning away

[dreamy fiber]

thanks! Did you like jump into the class the middle of the semester or?

[foggy merlin]

basically I had sort of a setback already because I was in a rush to finish my bachelor thesis for the first 2 weeks of the semester and some other exam I didn't take. so then I kind of had to catch up on subjects randomly, enrolled myself in too many classes without making a definite decision at the end. then I had a really work intense course on probabilistic methods with problems that needed really long to solve, so most of my time went into other subjects and commalg kinda drowned

probabilistic methods course was super cool as well, it just took way too much of my time lmao.

[dreamy fiber]

Wow that does sound like a lot of things to handle. Well hopefully the senester’s almost over and you seem relatively fine at the end? Maybe?

[foggy merlin]

I still got 2 exams ahead in 3 weeks, then enjoy my 2 f u l l d a y s of vacation until the next semester starts again

[dreamy fiber]

2 full days of vacation!?

Wow that’s like 3 months in mars time

Jkjk

[foggy merlin]

I know, what a luxury

[dreamy fiber]

And then next senester’s your last semester?

[foggy merlin]

but I'm looking forward to next semester since I won't have a setback this time

[foggy merlin]

no, I need at least two other ones

[dreamy fiber]

Ahh okay, well I am sure you will do great next semester then

You sound like you work really hard

[foggy merlin]

I hope so

https://cdn.discordapp.com/emojis/887072458803404890.webp?size=160&quality=lossless

[dreamy fiber]

Nothing a couple all nighters can’t solve

Jkjk

[foggy merlin]

I'm just ok at selling myself. if I had a least bit of average discipline and organisation this wouldn't sound as bad as I may make it to be. not saying I don't work hard. I just work hard very irregularly

[dreamy fiber]

I would suggest to prioritize your own health over academics, but I would be a hypocrite if I suggest that

But I am sure you have got this. Next semester’s gonna be great!

[foggy merlin]

oh trust me, I prioritise my pleasure a little too much if only you knew lmao. but no fr I'm doing pretty decent mentally, which is already an above average mental health for anyone studying mathematics

[dreamy fiber]

Ahh yea the dreaded 😂 experience every math major experiences

And another one bites the dust

But yes that is pretty good to hear

[foggy merlin]

well I mean I'm still passionate about the subjects I study and that's a great motivator, I hope you are as well

[dreamy fiber]

Thanks! I would say my passion is slightly above average

Considering I have decided to throw myself to into the pit of hell known as commAlg next semester

At my school it’s slightly notorious

[foggy merlin]

it's not a pit hell, just take your time for it, it's really beautiful imo

[foggy merlin]

ok a bad prof can ruin everything I agree

[dreamy fiber]

I totally agree, a professor can make or break a subject for you

That’s exactly what happened to topology with me

Everything was great in point set, but as soon as he introduced fundamental groups everything got super wack

[foggy merlin]

I was fortunate enough to have an amazing topology prof, at the end he went a little off the rails with all that deck transformation shit with covering maps at the end but otherwise an amazing course

[dreamy fiber]

Which I guess is slightly ironic since it has the word “fundamental” in its name

Lol my prof started talking about genus and polygons at the end

And our last pset was unironically the easiest one all semester

He just asked us to piece a polygon into a Genus-n, and then some other intuition stuff

[foggy merlin]

some proofs can get really nasty yeah, like the general van kampen thm. but the machinery itself is very comfortable to use

[dreamy fiber]

For real! I found Van Kampen a lot easier to work with when I learned what a free product with amalgamation is last semester

That and also fundamental Groupoids

[foggy merlin]

Genus-n always confused the heck out of me, luckily we only had it in exercise sheets

[dreamy fiber]

But I feel like my first intro to it was very unpleasant 😂 I was left rather confused

Like “wdym this loop generates things”

[dreamy fiber]

😂 exercises be exercises

[foggy merlin]

idk what a groupoid is lol, I assume it's sth from category theory, and yeah I should really study this by myself sometime, cuz no course fkn teaches it, but everyone uses some terminology of it

[dreamy fiber]

Undergrad topology classes that don’t take about what a free group is is absolutely blasphemous

[dreamy fiber]

Yeah a groupoid is just a generalization of the idea of groups 😂 in fact a group is a groupoid with 1 object

Which I know, doesn’t really help 😂

[foggy merlin]

we slightly scratched free groups in algebra 1. but not enough to the point where I do not get confused as to what happens when the kernel is nonzero using van kampen

[dreamy fiber]

Yeah I guess the best I have found is to just practice more and maybe try to convert them to some convenient group presentations

If you know what a category is, a groupoid is just a category where every morphism has an inverse

[foggy merlin]

well I think I might do so if I decide to ever pick up topology again, I've done topology and algebraic topology 1, both courses I enjoyed, but there are other areas I am more interested in

[dreamy fiber]

This my personal opinion on topology 😂 but

I found topology to be having the appearances of algebra, but in its execution there’s a lot of analysis

And that’s why I like algebra more 😂

[foggy merlin]

well isn't that kind of the point of the fundamental group and algebraic topology lmao?

[dreamy fiber]

Hey my algebraic topology education got butchered 😂 so bear with me here

[foggy merlin]

I learned more about short exact sequences and diagrams than topology in algebraic topology

[dreamy fiber]

Wat

I learned more about free groups and knot theory in my algebra class than topology

To be fair I have only taken 1 course in topology tho

And also my algebra professor was also a big fan of topology

But that’s besides the point

[foggy merlin]

you basically just prove some theorems and get your hands dirty every now and then to construct a formalism that allows you to say "fuck analysis, we don't need that anymore", (well you do but it's just covered up in the machinery so you don't have to deal with it)

[dreamy fiber]

😂 abstraction

Isn’t that the point of the word “abstract” in “abstract algebra”

[foggy merlin]

I can tell, because our algebra prof taught 0 things about knot theory

[dreamy fiber]

To be fair it was a tangent lol

Did you cover profintie groups in your alg class?

[foggy merlin]

after hearing about some of the stuff in category theory I feel like abstract algebra doesn't seem that abstract to me. seeing some of the texts and images makes me feel like I'm touching grass when working with a specific mathematical structure like a ring or a module

[dreamy fiber]

You are so right 😂 maybe we should change title to “concrete algebra”

[foggy merlin]

either idk about the terminology or no. I had algebra in german

[dreamy fiber]

You might also know it as Krull topology?

[foggy merlin]

nope

[dreamy fiber]

Germany must have a different direction in math education then

[foggy merlin]

switzerland to be exact

[dreamy fiber]

So at some point down in undergrad people might ask

Hey what if my Galois extension is infinite

It turns out there’s a really good part of topology that deals with this question

And it’s called Krull Topology

[foggy merlin]

makes sense, I guess there just wasn't time to cover some of that or our prof just focused on other topics

[dreamy fiber]

Yeah now I am curious what Switzerland’s algebra is like

I took my algebra in US btw

[foggy merlin]

or maybe we did look at it and I just forgot, my galois theory is basically nonexistent since that part I didn't study well, but no I don't remember seeing that

[dreamy fiber]

I totally get that, I feel this ambient void of nothingness when it comes to modules rn 😂 and that’s not good

[foggy merlin]

oh he did remark some of topological groups and infinite galois extensions at the end of the script I see

[dreamy fiber]

Yeah that usually comes last

[foggy merlin]

but that part was non examinable and he didn't go into detail so I just cut it

[dreamy fiber]

Ahh I see I see

If you like topology and algebra there’s a lot of interesting things that go on in topological groups

[foggy merlin]

yep, at least we got some good module practice and intuition from algebra 1. our prof really wanted to cover rings before groups because he really felt like showing the classification of finitely generated abelian groups as Z modules

[dreamy fiber]

So just the profinite groups (Krull topology) I talked about earlier, it turns out that if you have a group and it’s topology is compact, Hausdorff, and totally disconnected, then that group has to be a profintie group, and vice versa

[dreamy fiber]

😂 wow that’s a really creative approach, our prof just started groups and moved on to other things

[foggy merlin]

I also didn't learn what a profinite group is, I might be interested in that, but there are other areas for me to learn about still, I'm taking algebraic geometry next sem and I'm very excited about it

[dreamy fiber]

Wow you have a really intense next semester from@the sound of it 😂

I swear commAlg is gonna be the bane of my existence

[foggy merlin]

it's literally just ring and module theory in a different direction than in algebra 1

[dreamy fiber]

I haven’t done any algebraic geometry so I can’t comment

[foggy merlin]

I'm sure you'll make it, slow and steady wins the race

[dreamy fiber]

(Unless I have done it but my prof just never told us it is from alg geo)

Thanks you! Dr. j. stockfish

[foggy merlin]

our algebraic geometry is organised such that prequisites in commalg are recommended, so I think other unis might do it like that too?

[foggy merlin]

yeah I have an awkward nick on discord, friends just call me fish or doc

[dreamy fiber]

Oh yeah our alg geo is scheduled so that you can take it concurrently with alg 1

And then second semester concurrently with commAlg

That’s pretty interesting

[dreamy fiber]

Alright thank you Doc

[foggy merlin]

the doc is for a completely different unprofessional reason there btw, I don't want to appear pretentious

[dreamy fiber]

I mean I didn’t get that vibe from you lol

I ugh more thought about like

Doktor from MGR

[foggy merlin]

yeah, they also do it differently in some years, last year they just decided not to do a commalg course but algebraic geometry 1 and 2 instead. but I think you can do that because most people I ask say "algebraic geometry and commalg are essentially the same thing due to all the correspondances"

[foggy merlin]

haven't played metal gear lol

[dreamy fiber]

Now I know nothing of either 😂 so this will be prettt interesting I suppose

At some point down I have got to question the geo part of alg geo

Metal gear is a great game 😂

But it has never been the same since Kojima left

That’s my weeb side coming through

[foggy merlin]

I think the geo part of alg geo doesn't really have much to do with the usual inerpretation of geo, I don't know myself, but as far as I have heard it's about studying the set of zeros of polynomials in multiple variables over algebraically closed fields

[foggy merlin]

I don't watch anime, however I do play yugioh with friends on discord, so I am surrounded by a good amount of weebs

[dreamy fiber]

This is really embarrassing but I haven’t taken a proper geometry course at my university so thankfully it’s not as much zero as I would think 😂

Mostly bc the Professor who teaches the geo course at our university got a tenure in the 70s and hasn’t made a single publication since then

[dreamy fiber]

I totally understand that 👍

[foggy merlin]

well geometry is just some basic analysis mixed with basic algebra, at least here, easiest ace of my life lmao

[dreamy fiber]

I think the geometers have some bone to pick with you

[foggy merlin]

well that's more differential geometry, not geometry

and differential geometry was a really traumatic experience

[dreamy fiber]

I have not taken differential geoemtryyy to

I am sorry you had a bad experience in diff geo

[foggy merlin]

the prof who taught that course really has no feeling for appropriate work/ credit ratio

[dreamy fiber]

Ah

Lemme guess

20 hours/ week?

Excluding class time

[foggy merlin]

no, I gave up and did the same stupid thing like I did with commAlg, however looking back at it I probably could have worked for the entire semester nonstop and still have failed, because that was the experience of some of my peers, they said they had never studied that hard for a course and still failed

[dreamy fiber]

Your prof didn’t even curve?

Wow that’s brutal

[frail perch]

Diff geo seems to just generally have a really brutal reputation huh

[dreamy fiber]

A course I TAed in last semester, the curve was so bad everyone except for 1 person got an A

[frail perch]

Imagine having curves

[dreamy fiber]

Yeah grade inflation is 😂

A thing that exist

[frail perch]

The only course I've been with that had this sort of grade distribution was an undergrad combinatorics course with no prerequisitea that was actually a grad level combinatorics course in disguise, ended up getting the only grade above 80 in that class

[foggy merlin]

he just made a black bar over the grade statistics so no one could see lmao. I just found out from a friend who has connections to the interns how bad the statistics actually were, keep in mind this is just a 3rd year course, so there's no point in making people fail and kicking them out

[dreamy fiber]

Wat (￣▽￣)q

[frail perch]

In my diff geo course most people just ended up leaving lmao

[dreamy fiber]

The worst class I have been with was my quantum mechanics class where

The average for the class was a 55

[frail perch]

Damn

[foggy merlin]

smart decision

[dreamy fiber]

And getting a 65 on your midterm means you are doing fantastic

[foggy merlin]

wait how much do you need to pass/fail there? what's like the percentage of failed students

[dreamy fiber]

He made the failing grade to be 50% lol

It was initially 65

[frail perch]

Lmao,

[dreamy fiber]

And it’s that class that convinced me I shouldn’t be a physics major

So I jumped back in math

[foggy merlin]

oh you guys have this thing where to pass you need to be better than others, not have a fixed set of points?

[frail perch]

In my place it's a fixed set points

[dreamy fiber]

That’s part of the package that comes with curves 😂 I guess

I think it depends on the Professor tho

[foggy merlin]

the diffgeo prof changed the exam from oral to written and the pass rate dropped from like 90 to 40 percent

[dreamy fiber]

Bruh what happened on the written exam

[frail perch]

My diff geo was 100% exercises, some of them took more than 10 pages of writing but it was nice not having an exam to stress about

[dreamy fiber]

My diff geo is nonexistent bc the prof teaching it is really scuffed

[frail perch]

In what sense?

[foggy merlin]

he has this philosophy where he will never ask students to solve sth that isn't in the script already or in the exercise sheets, his course is basically just about memorization of proofs and the exercise solutions, 0 application or nothing where you would have to think for yourself

[frail perch]

Isnt that good though

[dreamy fiber]

He taught functional analysis a while ago and

[foggy merlin]

if it's too much to memorize then it isn't no, also it's hard to feel competent when nothing relies on thinking in new exercises, I perform a lot better in application oriented courses that don't just ask stuff you should know by heart

[dreamy fiber]

There were only 2 students who took it and they both got a C

[foggy merlin]

yo, that sounds like such an awkward course atmosphere

[frail perch]

Fair, diff geo has quite a bit to internalize (although I think it's one of those courses a good student should be able to recreate most proofs if they remember the central ideas of the proofs)

[weary terrace]

Sorry to interrupt with this tiny question: How do you refer verbaly to i,j,k of the quaternions? Is there a name for them? (Something like "imagenary elements")

[dreamy fiber]

Taking his class is like stepping in a landmine

I guess you just call them quarternions?

[frail perch]

I dont think they have a name, if you want to talk about the coefficent of i,j, or k in a number you could say something like "the j part of " or smth

[foggy merlin]

that is.. debatable, well it also depends on your definition of good lmao

[weary terrace]

Just trying to define them in my paper. Maybe "imagenary parts"?

Goddamn you Hamilton

[rustic crown]

i've heard someone calling them "vector part"

[dreamy fiber]

How about the non commutative part

[frail perch]

Well, what content specifically was covered in you diff geo course? The name "diff geo" refers to quite a lof of different stuff depending on university

[weary terrace]

Thanks guys

[foggy merlin]

oh yeah we're back at this part I'll send some screenshots

those were all the topics covered in the semester

well... all chapters

[dreamy fiber]

All 27 chapters??????

[foggy merlin]

yes

[dreamy fiber]

How many weeks did you have

[foggy merlin]

it's a semester course

[dreamy fiber]

Holy shit it’s like

[frail perch]

Ah, a lot more focused on diff top. My course atarted with the basics of diff top but the main focus was building up the theory of riemannian manfiolds (we got up to the sphere theorem!), so we had very differenr focused courses

[dreamy fiber]

2 chapters a week basically

[frail perch]

Yea diff geo courses gemerally tend to move at very fast rates

There's lots of content to cover

Anyway for diff top I can understand why a final exam like that sucks, memorizing topological proofs is one of my worst nightmares.

[foggy merlin]

after attempting to learn it I felt dumber in the end

I forgot what a derivative is, I forgot what a vector field is

all is very blurry

[frail perch]

Diff geo is a course you can only get good at by doing a lot of stuff by hand, I remember also starting the course with all the super abstract stuff and getting lost but once you get used to it all it flows very nicely

[dreamy fiber]

What textbook did you use?

[frail perch]

My course mainly used do carmo's riemannian geometry

[foggy merlin]

does it cover these topics, if yes I might consider getting it, the diffgeo from the other prof seems a lot more... doable

[frail perch]

Ok that pic is blurry excuse me I'll post a better one

[dreamy fiber]

So I should like

Take a course in real analysis before doing diff geo

Is that the vibe I am getting here

[foggy merlin]

....

yes

[dreamy fiber]

So funny story right

Our university’s official prerequisite for diff geo is linear algebra

[frail perch]

Depends, you should multivariable calc for sure, and linear algebra as well, but you dont need heavy analysis to start learning diff geo

[dreamy fiber]

And multi

[foggy merlin]

that is actually quite understandable, especially if you're going the more abstract route

[frail perch]

However for riemannian geometry you definetely do need soem analysis maturity (a lot of it analysing differential equations and such)

[dreamy fiber]

Okay so my Univeristy reallt low balled 😂 the pre requisite for diff geo

[frail perch]

Well, what is the sylabus for your uni's course? Again the word diff geo can mean quite a lot of things

[foggy merlin]

but ehm... at first glance I don't see how one would do diffgeo without some basics in real analysis like the implicit function theorem etc.

[frail perch]

I've even seen places where it was the name for the general relativity course

[dreamy fiber]

Yeah this is it’s description

Lemme see if I can get the syllabus

[frail perch]

Oh, classic diff geo, you should be fine with just calc 3 (multivariable differentiation) I'd say

[dreamy fiber]

Oh lmao 😂 very classical diff geo

[foggy merlin]

diffgeo just like the cavemen did in the good ol days

[wooden ember]

yeah undergrad diff geo isnt manifold diff geo most of the time

[wooden ember]

my diff geo course next year is like this too

[dreamy fiber]

Ahh okay okay

[frail perch]

In my uni it's called calc 4 and is a prereq for the grad level diff geo course

[wooden ember]

it's really just looks like a side note to calc 3 honestly

[dreamy fiber]

Imagine having a Calc 4

[wooden ember]

our calc 4 is weird

[dreamy fiber]

I thought that’s real analysis

[wooden ember]

it's lebesgue integration and fourier analysis

[frail perch]

Huh

[wooden ember]

which i thought would be two seperate courses

[foggy merlin]

is calc like a uni course for mathematicians? do you not just have real analysis instead

[frail perch]

I think fourier series were covered here in calc 3, whereas measure theory was it's own course

[wooden ember]

i call it calc but it's "advanced analysis I, advanced analysis II, analysis III, analysis IV"

[dreamy fiber]

So I guess in my university everyone’s required to have taken Calc

Yeah basically

And then read Folland for your grad class

[frail perch]

We dont have a distinction engineers and mathmaticians be taking the same

[dreamy fiber]

Which technically is like Calc 8

[foggy merlin]

[wooden ember]

manifold diff geo is just called introduction to smooth manifolds i think

it's a third year option

[chilly ocean]

In my uni there is no such thing as calc for math majors just 4 semesters of analysis

[dreamy fiber]

Like

You start with epsilon delta

[chilly ocean]

Yeah

[dreamy fiber]

And go all the way to manifolds?

[frail perch]

Actually wait no we dont have a classic diff geo course, we only cover the basics of manifold theory in calc 4

[chilly ocean]

Yeah

[dreamy fiber]

Bruhhhhhhhhhhhhhhhhhh

[chilly ocean]

In 2 years

[frail perch]

Stuff lile gauss bonnet and the theorema egregium we just proved as part of the grad diff geo course (usually as easy applications or exercises)

[wooden ember]

i dont think we introduce manifolds in analysis 4 but in topology in my course

[dreamy fiber]

In my university to get an ScB you are required to take one year of real analysis

[foggy merlin]

we have
analyis 1 and 2
fourier analysis
measure theory

those are the ones you have to take

[wooden ember]

basic measure theory is partially taught across probability and analysis 4

[chilly ocean]

Oh here we dont mention fourier pretty much anywhere lol

[wooden ember]

and for seeing it in depth it's a third year option

[dreamy fiber]

Lol first time i learned Fourier series was in my physics class

[frail perch]

Btw should we transfer this to #advanced-lounge ?

[chilly ocean]

No

[foggy merlin]

well, I never learned what a laplace transform is lmao

[wooden ember]

probably honestly

[chilly ocean]

Let the abstract algebra nerds suffer

[dreamy fiber]

Bruh okay

[chilly ocean]

Same lol

[wooden ember]

tbf i should get back to studying lmao

i still have two exams to go

[frail perch]

Gl

[dreamy fiber]

Good luck

Am I just on a timezone against everyone else but I am still in winter break rn

And it sounds like most people’s semesters aren’t over yet

[frail perch]

I'm also in winter break

[chilly ocean]

My Winter break hasnt started yet

[foggy merlin]

there is no such thing as a break in winter. the "break" is literally just study during your freetime or you'll fail

[tame fractal]

mine just ended a few hours ago

[chilly ocean]

Yo neko

[tame fractal]

hi

[dreamy fiber]

that is so trueeeeeee

[dreamy fiber]

Congratulations!

You must be so excited

[tame fractal]

excited enough to not be able to sleep yea

[frail perch]

I've got an alg geo course next semster and I can feel that lmao, I feel like I should step up my category theory during the break

[dreamy fiber]

My break ends in 2 days and

Suffice to say

I do not feel ready for commAlg

[chilly ocean]

Comm alg was the course that made me realize I shouldnt do algebra

[foggy merlin]

category theory seems unavoidable to learn at some point and no one cares to teach it, guess we gotta do it ourselves

[frail perch]

Comm alg is an interesting course...

[dreamy fiber]

Hey that’s the point of a DRP

[frail perch]

Yea basically lol

[foggy merlin]

yo... commalg was lit idk what u guys have lol

the exam was kinda bad but the subject itself, very enjoyable

[dreamy fiber]

i learned all my 🐱 theory from my DRP apparently it’s a tradition at this point

[frail perch]

My comm alg course was half comm alg and the the other half classical alg geo (the first part of hartshorne)

[dreamy fiber]

Bruh I can’t wait to start forming my own opinion on commAlg

[chilly ocean]

Not when youre covering most of AM in 2 months

I just couldnt keep up lol

[frail perch]

Am?

[dreamy fiber]

am I finally in the cool kids club after I take commAlg

[chilly ocean]

Atiyah macdonald

[frail perch]

Ah

[foggy merlin]

I had to do it in 3 weeks I can relate

[chilly ocean]

3 weeks lol not possible

[frail perch]

Yea I remember after the firsr week I told skmeone comm alg is like the professor has a machine gun of definitions to shoot at you

[foggy merlin]

our algebra course gave us a very solid basis in ring and module theory, so yes kind of, unless your commalg course was much larger

[frail perch]

Next semster I also have a course on noncommutative algenra (a.k.a rep theory)

[chilly ocean]

No way you were learbibg like 40 pages of AM a week man lmao

[dreamy fiber]

Yo when can we have a course on associative algebra

[frail perch]

Nonassociative algebras

[foggy merlin]

well we'll see how successful I was at that after I get the grade 😂 , got some bigger holes but having really liked ring theory previously that helped a LOT

[dreamy fiber]

And down the rabbit hole we go

[foggy merlin]

noncommutative ring theory

[dreamy fiber]

Yeah no 😂

Let’s see how much I can laugh about this two weeks later

[foggy merlin]

reminds me of our real analysis prof in the first week of uni saying.
"the shortest mathematician joke is let epsilon < 0"
silence
"yeah in one year you'll find that funny"

[dreamy fiber]

Hey it technically should work out

Just in the reverse order

[frail perch]

Yoo I remember once actually doing a proof with that, but I forgot where

[dreamy fiber]

My real analysis prof had a joke about him failing real analysis undergrad

And at the end of the joke he was like

“Yeah you will find that really funny when you fail my class@

[frail perch]

[dreamy fiber]

It was sarcastic

[foggy merlin]

he a freak

anyways I gtg, was nice talking to you guys, have a great day

[frail perch]

Goodbye

[coral shale]

What is the defn of a cyclic field?

Its multiplicative group cyclic??

[lethal dune]

for finite fields, they are always cyclic

and for infinite, they are never cyclic so not much of a classification

[wooden ember]

maybe they mean cyclic field extensions?

[grand cliff]

Hi, I'm having some trouble with this exercise from Dummit and Foote.
What I'm struggling with in general is how to prove two groups are isomorphic by their presentations (at least, that's how I'm guessing you're supposed to solve this problem). Maybe I'm a bit stoopid but I don't really know how to go about proving two groups are isomorphic via their presentations

,tex Let $ G $ be a finite group and let $ x $ and $ y $ be distinct elements of order $ 2 $ in $ G $ that generate $ G $. Prove that $ G \cong D_n $, where $ n = |xy| $. To remove notational ambiguity, $ | D_n | = 2n$.

[cloud walrus]

Cursor

[grand cliff]

My approach so far has been to write down the presentation of G and D_n, but I don't really know where to go from here. Would appreciate any guidance 😄

[lethal dune]

x and y both have order 2? you surely meant x has order n ig

[delicate orchid]

yeah I was thinking something was off

[lethal dune]

also |xy| = 2 always

like if it generates Dn

[wooden ember]

are you sure this is true?

im guessing finiteness of G forces this to be true but i have struggled to show it

[lethal dune]

yes it's true

[wooden ember]

how do you show it's true?

[lethal dune]

$D_n = \langle r, s \rangle$ with $r^n = s^2 = e$ and $sr = r^{-1}s$ so $(rs)^2 = rs\cdot rs = rs\cdot sr^{-1} = e$

[wooden ember]

oh if it generates D_n

yeah i agree there

i meant in general

[delicate orchid]

wait

[wooden ember]

i mean yes this should only generate D_2

[delicate orchid]

wait nvm I'm a moron s = s^-1, you're right

[lethal dune]

that's the general case?

[wooden ember]

but how do you show you only get the case |xy|=2 for a finite group

[wooden ember]

no but we want to show it generates D_n

[lethal dune]

r=r^-1?

oh

ninja edit

nvm

[delicate orchid]

I use r for reflection :troll:

[wooden ember]

how do we know that we cant ever get the case |xy| > 2

before knowing it generates D_n

is my point

finiteness must force it but i dont see how

[lethal dune]

🤔

like how do I prove the definition

[wooden ember]

no

how do you prove that if x and y are of order 2 and generate a finite group G, then |xy| = 2

that would conclude it

[lethal dune]

no that's not true

[wooden ember]

exactly

[lethal dune]

I never said that

[wooden ember]

but then the question is false

[lethal dune]

yes, it is

[lethal dune]

.

[delicate orchid]

_ _

[wooden ember]

alright i just misunderstood what you were saying originally

[delicate orchid]

I think I did as well tbh

[lethal dune]

no one understands me

[wooden ember]

im annoyed i cant find an example of a group with |xy|> 2 though

[lethal dune]

free group

lol

[wooden ember]

finite

is part of the restriction

[lethal dune]

hmm

I would have said matrices

[wooden ember]

ah maybe

[wooden ember]

so wait can you prove this

[lethal dune]

nah wait I also think finite ness forces |xy| = 2 or something

[wooden ember]

then the question would be correct yeah

i was trying to do stuff with the normal closure of {x^2, y^2, (xy)^n} with n>2 in F(x,y) but didnt manage

[lethal dune]

because see, if it's finite then x,y have some kind of relation, say xy=1 or xy=yx and these are the only possible one I think

[wooden ember]

should be able to show that quotienting by this gives an infinite group i imagine

[lethal dune]

both cases it turns out to be commutative so |xy| = ... = 2

[wooden ember]

yeah but i mean that "i think" isnt enough lol

[lethal dune]

what else can you have then

[wooden ember]

we need some kind of finite group where one group of order 2 doesnt normalize the other one

if that exists

or show nonexistence of such

[lethal dune]

any relation you come up with reduces to these 2 because order of x and y is 2

[wooden ember]

how though

[grand cliff]

woot, i get tea and the convo has started without me

[lethal dune]

like say xy=xy^k for some k, then it gives xy=xy (which is not possible because it gives an infinite set) and xy=x(which gives y=e also not possible)

[wooden ember]

lol easy counter example im dumb

take x = (1 2) and y = (2 3) in S_3

they generate S_3

[lethal dune]

ye they do

[wooden ember]

and |xy| = 3

[lethal dune]

hmm

[wooden ember]

but S_3 is isomorphic to D_3

[lethal dune]

ok so that was for nothing lol

[wooden ember]

so this isnt enough

[lethal dune]

why not?

[grand cliff]

What exactly are you trying to prove

[lethal dune]

idk really

[wooden ember]

well we havent proved or disproved the original question yet

okay it's true

embed G in S_n

wait no

[lethal dune]

[wooden ember]

was about to say something dumb

[lethal dune]

no x,y do not generate Dn

[wooden ember]

prove it

[lethal dune]

as mentioned in the question

[lethal dune]

[grand cliff]

The claim is that these two groups are iso

[wooden ember]

(1 2)(3 4) and (2 5)(4 6) ?

what does that generate in S_6

[lethal dune]

ok wait

I misunderstood

[grand cliff]

Wait, since the group is finite xy must have finite order

[lethal dune]

let x=s and y=rs then yes they do infact generate Dn

[wooden ember]

yes

[wooden ember]

yeah exactly

[grand cliff]

So I can remove the relater (xy)^n from the presentation of G

[lethal dune]

all the time I though x means r and y means s

[wooden ember]

oh lol

[wooden ember]

right that's why you gave this

which i was confused by lol

[lethal dune]

makes sense

ig the proof is now trivial?

[grand cliff]

I was trying stuff like this

But it feels so

[wooden ember]

is it?

[grand cliff]

algebra-shuffly

[lethal dune]

yes?

[grand cliff]

yes, you substitute the generators

[wooden ember]

oh fair enough lol

[grand cliff]

And check they satisfy the relations

[wooden ember]

just give that presentation of D_n

im an idiot

bruh

[lethal dune]

[grand cliff]

nono its fine lol

but I've been doing a lot of exercises

with these presentation isomorphism proofs

and

I always have no idea what to do

[wooden ember]

i mean this is early D&F right?

[lethal dune]

oh you were the original author, didn't notice

[wooden ember]

id recommend not spending too much time on those

[grand cliff]

oop

[wooden ember]

because you havent seen presentations rigorously then

wait till you get to the chapter on the free group

section 6.3

[grand cliff]

that's yonks away

[lethal dune]

you even remember the section numbers

[grand cliff]

something something universal property

[wooden ember]

yeh

[lethal dune]

cat theory

[grand cliff]

keep the cats away

[wooden ember]

the thing is most of the time you wont really be showing isomorphisms by presentations in the exercises

[lethal dune]

wait are you D&F in disguise ?

[grand cliff]

Could you rephrase this sentence

Like, in later chapters, do we not show isomorphisms between groups via their presentations

[wooden ember]

except for the exercises on presentations you wont be using them much to show isomorphisms

[grand cliff]

Right okay

[wooden ember]

it's less useful than you think

because finding a presentation isnt easy

in general

[grand cliff]

Huh

that makes sense

and answers a question that sorta popped into my head while learning them

[wooden ember]

it works well for finitely generated abelian groups and semi direct products of such but that's about it

[grand cliff]

I didn't really understand how you go from group to presentation

[lethal dune]

and they are super non trivial

[grand cliff]

like, is there some algorithm you can follow

[wooden ember]

not really no

[grand cliff]

to create a presentation from a group

oh weird

[wooden ember]

determining whether a given presentation is finite or not for example

is a very difficult problem

[lethal dune]

changing 2 to 3 may give you a infinite from a finite set

[grand cliff]

yeah, I remember this example

[delicate orchid]

presentation theory 😌

[lethal dune]

yeah and there's an algorithm I think to do calculate these

[wooden ember]

again for some groups it's easy though

namely finitely generated abelian groups and semi direct products of them

there is an "algorithm" in those cases

[lethal dune]

FTOFGAG

[wooden ember]

do you mean FTOFGZM

[grand cliff]

Okay, considering I don't really know much group theory. I'll just assume that: Finding presentations is generally a "?hard?" problem (what does hard mean in this context), but there are some groups where it's much easier to find their presentations.

[wooden ember]

yup

in the case of D_2n, it's a semi-direct product of two cyclic groups

so it's easy to get a presentation for it

[grand cliff]

I've heard the word semi-direct product, and someone tried to explain it to me

I did not fully comprehend

[wooden ember]

dw youll get there

it's a little farther down the line

[delicate orchid]

imagine the direct product but you shove a homomorphism in the middle

[grand cliff]

ya, I think it's something todo with like: take two groups, and combine them (in some way so that 1 is normal sg and other is just a sg)

[wooden ember]

it's like a direct product but as you move through the second factor it "twists" the first one

i like to think of it as a cylinder with the cylinder getting twisted as you move up

[lethal dune]

ZM? Z-modules?

[grand cliff]

I don't really have any intuition for what "twisting" is, but I'm happy to accept semi-direct product as a way to mix 2 smaller groups to make a bigger one lol

[wooden ember]

yeah lmao

[wooden ember]

yeah no just wait till you get there

[lethal dune]

FTOFGMOPID

[delicate orchid]

I could just state the definition if you want lol

[grand cliff]

That does not help with understanding lmao

[delicate orchid]

the "twisting" is conjugation usually

[wooden ember]

it's better for you to wait till you go through group actions in more detail

[grand cliff]

Yeah, that's the next chapter

or at least

[wooden ember]

youll get a better understanding of the semidirect product then

[delicate orchid]

actually yeah if you don't know group actions it won't make much sense

[grand cliff]

the next sub-chapter of ch1

[wooden ember]

i mean after ch4

[delicate orchid]

damn those are long chapters

[grand cliff]

yea, figured you meant this

[wooden ember]

semi direct products are ch5.5

which i think is at a perfect time

and it's the best part of the book imo with chap 6.2

you get to classify a bunch of finite groups it's fun

[grand cliff]

yaknow, quite a few people have advised me against doing all the exercises

but

it's kinda helpful

because when I get stuck i can have conversations like these

lol

[wooden ember]

ive done almost all of them and dont remember most

so take my word for it

it's not worth doing all of them

[grand cliff]

I don't think the point is to just remember them all tho?

[wooden ember]

just do a few simple ones, a few medium ones and a few difficult ones until you feel you understand the content

[grand cliff]

Maybe I'm wrong

[wooden ember]

yeah no it's not

but like

[lethal dune]

This man is the 'contents' of DF itself

[wooden ember]

im not much better at algebra than if i had done less of them

[lethal dune]

[delicate orchid]

law of diminishing returns

[wooden ember]

as my approach of your previous question shows

[grand cliff]

oki

LOl

so uh

wise sage

what do you suggest

[lethal dune]

btw refer me to some good video lessons on module theory

[grand cliff]

for me to get gud

[wooden ember]

just do what you feel is right

like

i say this

but i still try and do every exercise when im working through some book

even though i know it's not optimal

[grand cliff]

isn't it also satisfying though

[wooden ember]

idk

somewhat i guess

[grand cliff]

I've worked through textbooks cover-to-cover in the past

And it's pretty fulfilling

I do see your point though

[wooden ember]

you just need to make sure you do enough so that you dont need to flip back every two minutes later on in the book

[wooden ember]

ive yet to properly do a book cover to cover

i was planning to do that with d&f but got bored at ring theory lol

[grand cliff]

Yeah, I suspect I'll stop reading d&f at the end of group theory

[wooden ember]

the group theory stuff is gold though imo

[grand cliff]

Yeah, I want to learn what all the fuss is about

[wooden ember]

it would be nice if there was a chapter on infinite group theory though

at least for introductory techniques

cause all they talk about is the free group

[grand cliff]

Yeah, idk what the prereqs really are

[wooden ember]

more cat theory i think

i think a lot of the constructs are based on limits and colimits

[grand cliff]

By infinite group theory, is this the study of infinite groups?

[wooden ember]

i mean idk if it's an official field or anything but i just mean techniques in dealing with infinite groups yeah

cause all the combinatorial stuff in d&f just doesnt apply

[grand cliff]

I hear "lie this" and "lie that" all the time so I'm kinda interested to learn lie theory at some point

the continuum is based

I honestly don't really know what the field is about

but it sounds cool, since it combines analysis stuff with group theory (is my understanding)

[pliant raptor]

I need help proving that $\mathbb{Z}{(p)} / p \mathbb{Z}{(p)}$ is isomorphic to $\mathbb{F}{p}$, where $\mathbb{Z}{(p)}=\left{\frac{a}{b} \in \mathbb{Q}: p \nmid b\right}$ and $\mathbb{F}{p}$ is the field of $p$ elements. I thought about trying to create an homomorphism from $\mathbb{Z}{(p)}$ to $\mathbb{F}{p}$ whose kernel would be $p \mathbb{Z}{(p)}$, but I can't get any ideas.

[cloud walrus]

ImHackingXD

[hidden haven]

You can get a homomorphism with that property

The image of 1 has to be 1

No choices for the rest

[lethal dune]

like why not r -> r+pZ(p)

[hidden haven]

Are you defining a map from Z(p) to Z(p)/pZ(p)

[lethal dune]

well kinda, using this to decode the map should look like

[lethal dune]

but this is already enough

[hidden haven]

But the map needs to go to F_p

We are using

[lethal dune]

[hidden haven]

If you want the map that goes to F_p you use the universal property of localisation

[lethal dune]

idk what I was thinking lol

[hidden haven]

F

[frail perch]

Ahh why arent the stickers here not light mode friendly :(

[pliant raptor]

Wait, so the image of 1 as to be 1 is a consequence of being an homomorphism right? But how would I define one in this specific case?

[lethal dune]

because no one uses light mode

seriously

[hidden haven]

yes because it has to be a homomorphism

[frail perch]

I do, and many others do either for own comfort or for accessibility reasons

[hidden haven]

It is the only choice

[hidden haven]

if you are asking what the complete description of the homomorphism is then you can get that by first defining it for all natural numbers by writing them as 1+...+1 and then for all integers and then for all of those fractions

The part where you define it for all integers is a standard thing. For any ring R, there is a unique map from Z to R called the characteristic map defined this way

Here we are just using one more property of F_p to turn the characteristic map for F_p into a map from Z(p)

[frail perch]

(It's called the char map because the image of it depends on the characteristic of the ring)

[hidden haven]

Might be the other way around

[pliant raptor]

Another question: I've seen someone claim that there is a composition of maps $\mathbb{Z} \rightarrow \mathbb{Z}{(p)} \rightarrow \mathbb{Z}{(p)} / p \mathbb{Z}{(p)}$ that would be surjective and whose kernel would be $p\mathbb{Z}$. I can imagine the second map being the canonical epimorfism, but what would the one from $\mathbb{Z}$ to $\mathbb{Z}{(p)} $ be?

[cloud walrus]

ImHackingXD

[pliant raptor]

Thanks for the help guys

[hidden haven]

The characteristic is the non negative generator of the kernel of the characteristic map, so it feels like the map should come first

[frail perch]

Well depends how you define it, but most times when it's taught to 1st year students it's easiest to just tell them "the smallest ammount of times you add 1 to itself until you get 0"

[hidden haven]

The first map is x maps to x/1

but then characteristic 0 makes no sense

[frail perch]

Yea you just say that if you cant add 1 to itself to get 0 then char 0 by convention

Annoying but still the best for a first week linear algebra 1 definition

[pliant raptor]

I thought of that, but then how would I ensure that the composition is surjective? Since that first map is not surjective

[hidden haven]

damn you people look at characteristic in year 1

We had ring theory 2nd year

[frail perch]

I mean, we alsp have ring theory on year 2, but like for vector spaces and fields in linear alg

[hidden haven]

Put the isomorphism to F_p at the end of this chain

And prove that that is surjective

[hidden haven]

Oh we just defined a field and then didn't talk about the field at all apart from saying vector space over F in each theorem

[hidden haven]

By "that" I mean this whole big composition

[frail perch]

Huh

[pliant raptor]

Oh wait, the composition doesn't have to be surjective, nvm

Thank you for your help!

[hidden haven]

The composition is surjective though

[pliant raptor]

Yeah, but to apply the homomorphism theorem you only need to find an homomorphism whose kernel is the ideal, not anything else

Wait no

[hidden haven]

For this?

This only gives an isomorphism of the quotient onto the image

[pliant raptor]

It looks like what I'm saying but I've never seen that before

[hidden haven]

Oh damn

[pliant raptor]

Ye ye, you're right

I'll just check the surjectiveness on my own with your tip

Thanks

[modest prism]

Hey

In Representation theory, why is the isotypic component of the trivial rep,
$$ { v\in V | \forall g\in G, gv=v}
$$
?

[cloud walrus]

BenFire

[hidden haven]

What definition of isotypic component do you have?

[modest prism]

That if we decompose a rep to simple reps, the isoypic component with respect to rep z, is a direct sum of those isomorphic z

And I saw it doesn't depend on our decomposition to simple reps'

[hidden haven]

Right, so do you see what the decomposition into simple reps here is?

[modest prism]

What do you mean?

[hidden haven]

Can you decompose a trivial representation into simple representations?

[modest prism]

Isn't it simple itself?

[hidden haven]

By trivial do you mean 1-dimensional trivial?

[modest prism]

Yeah

That sends everything to 0

[hidden haven]

Right so you already have a decomposition

So there is only 1 simple rep to decompose into

[modest prism]

Yeah

[hidden haven]

Isotypic component corresponding to all other simples are 0

Isotypic component corresponding to the trivial simple is the whole space

And what you have written is just the whole space

[modest prism]

Wait

There is somthing basic here im not getting

Let's say we have a rep V = v_1 +...+ v_k

Where v_i are simple

The isotypic component of Trv is:
$$
U_{Trv} := \oplus_{v_i \cong Trv} v_i
$$

[cloud walrus]

BenFire

[modest prism]

Right?

[hidden haven]

ohh

So you are looking at some other rep V

[modest prism]

Yeah

[hidden haven]

and trying to find the isotypic component that corresponds to the trivial simple

[modest prism]

Exactly

[hidden haven]

So the set you have written

is a subrep

[modest prism]

Yes

[hidden haven]

and it contains all the simple subreps isomorphic to Trv

[modest prism]

Yes

[hidden haven]

And everything in it spans a copy of Trv

So it is exactly the sum of all subreps isomorphic to Trv

Oh wait you have to do it wrt a direct sum decomposition

So you decompose this rep by taking that set direct sum a complementary subrep

Which you can do assuming semi simplicity

[modest prism]

Yeah, assume semi-simplicity

[hidden haven]

Then decompose this set and its complement individually

All the simple reps you get outside, none of them can be isomorphic to Trv

[modest prism]

Yes

[hidden haven]

And everything in the decomposition of this set is isomorphic to Trv

So the direct sum of everything isomorphic to Trv

Is exactly this set

[modest prism]

This, you mean?

[hidden haven]

yes

Everywhere I say this set, read it as that

lol

[modest prism]

Sorry, but why? I still don't see it

[hidden haven]

So this gives a direct sum decomposition of V

Just by putting those 2 decompositions together

Right?

[modest prism]

Yeah

[hidden haven]

Then you have a decomposition of V as A_1 + ... + A_n + B_1 + ... + B_n where A_i form the decomposition of that set and B_i of its complement

Now you just have to identify, which of these A_i and B_i are isomorphic to Trv?

[modest prism]

Where we defined A_i to be a decomp of the set, right?

[hidden haven]

Yes

[modest prism]

Than yes

[hidden haven]

So do you see how to identify the ones isomorphic to Trv?

[modest prism]

Ummm

Ohhhh

I think i see it

It's because of the commutative nature of the trivial rep?

Sort of*

[hidden haven]

I don't think you need anything of that sort?

I mean we just have to see how the action of G restricts to each A_i and B_i

The restriction to A_i has to be trivial because by definition G acts trivially on all elements of A_i

So A_i has to be isomorphic to Trv

[modest prism]

If v_i is iso to Trv, it means that there is a T:v_i \rightarrow Trv, such that
T(gv) = g(Tv), right?

[hidden haven]

I guess I am using the fact that Trv is the only simple rep on which G acts trivially

[modest prism]

And if v_i is in our set, it satisfies that condition

Right?

[hidden haven]

Right

[modest prism]

Okay

I got it

You're a godsend, thanks so much!

[hidden haven]

[warm holly]

Yo i just have a question on multilinear forms and permutations. In my lin alg course if we are given a multilinear form $\Lambda \in Mult^{(n)}(V,K)$ with $V$ vector space, $K$ a field and $n=\dim V$ and $\tau,\sigma \in S_n$ (group of permutations over ${1,..,n}$. Then the claim is that $S_n$ acts on the left on Mult and so that $\sigma(\tau \Lambda)=(\sigma\tau)\Lambda$. But, the proof seems to me quite strange.. If we take $(v_1,..,v_n)\in V^n$ we have that

[cloud walrus]

𝔻аniil

[warm holly]

oups i didn't finish

.. we have that $(\sigma\tau)\Lambda(v_1,..,v_n)=\Lambda (v_{\sigma(\tau(1))},..v_{\sigma(\tau(n))} \neq \Lambda(v_{\tau(\sigma(1))},.., v_{\tau(\sigma(n)})=(\tau\Lambda)(v_{\sigma(1)},..,v_{\sigma(n)})$

[cloud walrus]

𝔻аniil

[warm holly]

So i don't really know if I misunderstand something or the claim is false

[hidden haven]

Yeah this should be a right action

Which means στ acts as τ after σ

[warm holly]

so we normally have that sigma.(tau.Lambda)=(tau sigma).Lambda?

[hidden haven]

Yes

[warm holly]

alright thank you

[hidden haven]

And the reason it's called right action

Is that this becomes just associativity

If you instead write it the other way around

As Λσ

Instead of σΛ

[rustic crown]

i feel it's correct

[hidden haven]

F

[rustic crown]

the problem lies on how you're computing that

we're permuting the places

not the indices

it shouldn't matter on how the vectors are labeled

[hidden haven]

[warm holly]

hm

[hidden haven]

Isn't that just a matter of definition

[warm holly]

but I searched a bit in the internet and in every course they propose it as the right action

[chilly ocean]

moldi and det what are prerequisites for quivers

[hidden haven]

Can you draw arrow

[chilly ocean]

no

[hidden haven]

Fail

[chilly ocean]

[hidden haven]

But if you mean rep theory of quivers then you should know algebras over rings and modules I think

And stuff like how direct sums work

And depending on the course, functors and possibility adjunctions

[rustic crown]

,tex

\begin{align*}
\tau\Lambda(v_{\sigma(1)}, v_{\sigma(2)} \ldots v_{\sigma(n)}) &= \tau\Lambda(w_1, w_2 \ldots w_n) \\
&= \Lambda(w_{\tau(1)}, w_{\tau(2)} \ldots w_{\tau(n)}) \\
&= \Lambda(v_{\sigma(\tau(1))}, v_{\sigma(\tau(2))} \ldots v_{\sigma(\tau(n))})
\end{align*}

[hidden haven]

Oh true

[cloud walrus]

det

[warm holly]

ye thats how it is proved in my course

[hidden haven]

F

[warm holly]

with variable exchange

[hidden haven]

Yes that is correct

[warm holly]

alright, thank you

[rustic crown]

you can't really define it with the labeling on the vectors, it won't be well defined. what if you were in a bad mood and denoted the vectors a_1, b_1, c_1, ... 🙈

[hidden haven]

Yeah I misinterpreted what you were saying

Also I'm never in a bad mood, I'm literally a beacon of joy

[rustic crown]

[chilly ocean]

same

[coral shale]

I am looking to refresh my Algebra I learnt 2 yrs ago as I'm starting Algebraic Number Theory/Galois Theory

Commutative Rings ⊃ Integral Domains ⊃ Unique Factorization Domains ⊃ Principal Ideal Domains ⊃ Fields

I basically need to brush up on this stuff (and a few things on polynomial rings/field extensions). I want to try get a feel what each of them is through some examples (I've re-read the definitions, but that alone isn't giving me much). Anyone has a good resource? 🙏

[frail perch]

Why not check some number fields as examples? Every field extension of Q has an associated ring of integers, which are always integral domains, sometimes those will be ufd, sometimes euclidean domains, sometimes PID's

For example, try showing the gaussian integers Z[i] is a euclidean domain

The other classic example for rings is polynomial rings, you should play around with those too for a bit

[hidden haven]

I can DM you a good and short lecture series

[coral shale]

Thanks for the replies. I've added @hidden haven

[prisma shuttle]

hey guys for this problem can someone explain the purple part

[tribal moss]

By definition u has the smallest norm possible outside tilde-R. Since the proof has (using Euclidean division) chosen r so it has smaller norm than u, r must be in tilde-R.

[chilly ocean]

How did my prof get the circled values?

[lavish nexus]

this is just multiplication

you get a bunch of cycles

[chilly ocean]

How did they get PIr

[lavish nexus]

left multiply 1 by r you get r

proceed

[chilly ocean]

But the other cycle?

For PIr

[lavish nexus]

left multiply s

by r

[chilly ocean]

Why does he split them?

[lavish nexus]

well can you left multiply anything in the first cycle by r to get something in the second one?

[chilly ocean]

So basically

He left multiplied everything in d4 by r

U can do in any order right?

[lavish nexus]

yes

[chilly ocean]

Got it got it

🙏

Ty

[lavish nexus]

I have two fields A and B. Take AB
how do I show the inverse of 1+ab is a finite sum of products

I know it's inverse is 1-ab+(ab)^2-....

like is there always a certain bunch of terms that sum to 0

[tribal moss]

Are A and B subfields of a larger ambient field?

[lavish nexus]

any fields

I heard the two definitions for AB are equivalent

that it is the set of finite sums of products

or it is the smallest field containing both A and B

so I wanted to show the inverse of anything ab+a_1b_1+...+a_nb_n is a finite sum
and to do that I figured I can show it works for 1+ab and then induction

[tribal moss]

Hmm, that sounds like it won't respect isomorphisms of the original fields. E.g. if A and B are both Q(X) then their product is Q(X) itself. But if A is Q(X) and B is Q(Y), then their product should be Q(X,Y) which has a different trancendence degree over its prime field than Q(X) has.

[chilly ocean]

Am i doing this right

[tribal moss]

Looks right to me.

[lavish nexus]

what is transcendence degree

[tribal moss]

https://en.wikipedia.org/wiki/Transcendence_degree -- not terribly important here, just the quickest argument I could think of that Q(X) and Q(X,Y) are not isomorphic.

[lavish nexus]

hmm

ok

ty

[chilly ocean]

i need help with this problem, i cant really visualise my own example for part a either so im kinda stuck

[lavish gale]

look at permutations (23) and (45) in S_5 for example

and see how permutations (23)(45) and (45)(23) act

[chilly ocean]

so let pi be (23)(45)

and sigma (45)(23) ?

[lavish gale]

sure

[chilly ocean]

but doesnt it say that they are disjoint

[lavish gale]

yes (23) and (45) are disjoint

so pi is (23) sigma is (45)

[chilly ocean]

ohh was i visualising this wrong

okay so

well can they get multiplied?

they cant even compose

for sigma

[lavish gale]

do u know what (23) means?

[chilly ocean]

2 goes to 3

3 goes to 2

[lavish gale]

and 3 goes to?

right

[chilly ocean]

yes

[lavish gale]

what about this (12)(23)

where does 3 go

[chilly ocean]

those are 2, 2 cycles

3 goes to 2

[lavish gale]

and?

2 goes to?

[chilly ocean]

1

[lavish gale]

right so (12)(23) says 3 goes to 1

[chilly ocean]

yes

so i do the other direction

but then generalize

So i would do (23)(12)

[lavish gale]

and see how permutations (23)(45) and (45)(23) act on set {1,2,3, 4, 5}

[chilly ocean]

Hmm

So if i go to (23)(45)

[lavish gale]

notice (23) deals with only 2 and 3, (45) deals with only 4 and 5

[chilly ocean]

Yeah so they cant really map to anything in f

[lavish gale]

ya, it does not map to anything in the permutation it follows since they are disjoint

so order does not really matter in this case

[chilly ocean]

So when two cycles are disjoint

They dont go to the other function

[lavish gale]

right

[chilly ocean]

Okay so

When we multiply (23)(45)

What would the answer be

[lavish gale]

(23)(45) here 4 goes to 5 as 5 is left unchanged by (23)
(45)(23) here 4 goes to 5 as 4 is left unchanged by (23)

[chilly ocean]

Yep

So how are they equal tho

Are they commutative

[lavish gale]

yes thats what you are trying to prove when they are disjoint

[chilly ocean]

So when you multiply 2 cycles

They are commutative

[lavish gale]

no it works for cycle of any lenght as long as they are disjoint

[chilly ocean]

Last question w

[lavish gale]

as for why they are equal
becuuse they act the same

[chilly ocean]

When i multiplied (23)(45) the answer will just be (23)(45) right

Now i gotta generalize all of this

And for the second part of question a

Lets say n was 4

[lavish gale]

work out the action of (23)(45) and (45)(23) on {1,2,3,4,5}

[chilly ocean]

How would that look like

[chilly ocean]

That would suffice the proof

For the first part of a

[lavish gale]

i gave and example to see whats really going on
it should give you a good idea on how to approach the problem

[chilly ocean]

For sure

Appreciate the help

[desert dome]

Hi, I am working on this problem. I only know that N_{K/k}(a) equals the product of all the roots of its minimal polynomial, but I am not sure how it's related to a_0. Is there some theorem about polynomial coefficients and its roots that I should know about?

[sly crescent]

Is the real rank of a simple Lie group the same as its rank as an algebraic group?

[proud bear]

if f(x) factors as (x-r_1)(x-r_2)...(x-r_d) when you expand that, the constant term is going to be (-1)^d*r_1*r_2*...*r_d

[desert dome]

I see. Thanks!

[frank gorge]

Any subfield of the reals must contain 1. Since a subfield is closed under addition, then the subfield contains all integers. (Ex: a = 1 + ... + 1) Since its a subfield, then it must contain all multiplicative inverses. So if the subfield contains some a, it also contains 1/a. Since the subfield is closed under multiplication (because it is also a subring) then it contains all a/b for integers a,b. Hence, all subfields of the reals contain the set of rationals.
Does this intuition make sense?

[hidden haven]

Yes

Negative integers come from additive inverses of the 1+...+1s

[frank gorge]

ahh the negatives right!

[next obsidian]

Your argument will extend to show any characteristic 0 field contains Q. That’s why Q along with F_p for p prime are called prime subfields, if K is a field then it’s char p iff it contains F_p, and char 0 iff it contains Q

[chilly ocean]

Hey guys are disjoint cycles commutative

[hidden haven]

They commute with each other

[chilly ocean]

So like

(12)(34) = (34)(12) ?

[hidden haven]

Yes

Proof by applying them to an element and seeing what happens

[cursive holly]

(probably) silly question: how can a' = a? isn't a' supposed to be the inverse such that a' * a = e, the identity element? how would we know that a * a = e?

are we just assuming since a * e = e doesn't make sense?

[sharp sonnet]

is this just a group with two elements?

[cursive holly]

ohhhh yeah my bad lol. since our group has just 2 elements we eliminate the other possibilities

thanks lol

[sharp sonnet]

[tough raven]

Call an ideal prime-radical iff it is the intersection of all prime ideals containing it. (Assuming the Axiom of Choice, this is equivalent to being a radical ideal.)
Call a prime-radical ideal I irreducible iff for all prime-radical ideals J1, J2; if J1 intersect J2 = I, then J1 = I or J2 = I.
Prove that if an ideal I is an irreducible prime-radical ideal, then it is a prime ideal. (The converse is easy.)