groups-rings-fields

[novel parrot]

so 1 = linear combination of u1 and u2

[pastel cliff]

ok so what do you think c is

[vagrant zinc]

Thanks now I understand better! ❤️

[pastel cliff]

disclaimer that this material is still recent for me and i could say something wrong

[novel parrot]

so $c = x_1b_1 +x_2b_2$

[cloud walrus]

ActiveChapter

[novel parrot]

i dont understand why x_1 and x_2 are coprime

[pastel cliff]

go back to the definition of primes

[oblique river]

You were already told! Write b_1 and b_2 as multiples of c

Plug that in to your equation

Divide by c

[oblique river]

Or “cancel c from both sides” if you prefer

[novel parrot]

dont be angry with me please

im learning :2

trying atleast

[pastel cliff]

nobody gets angry at you here

[novel parrot]

we can write $b_1$ as multiple of because $b_1 \in (c)$ ?

[pastel cliff]

but you gotta put forward your own effort too

[cloud walrus]

ActiveChapter

[pastel cliff]

as helpful as people are here, there is no substitute for sitting down with a problem and tinkering with it yourself

talking helps but there are certain times when you just need to lock yourself in a room w/ pencil and paper

at least for me

[novel parrot]

this is correct?

[chilly ocean]

make sure it has good lighting and ventilation. water, too

[pastel cliff]

and preferably a standing desk

[rustic crown]

And have a cute plushie next to you.

Try explaining math to it. Cute and fun

[elder wave]

https://pbs.twimg.com/media/FlwCqybWQAAl5VP.jpg

[pastel cliff]

aint no way

this the kinda thing to show up on a "cursed set ups" video

[chilly ocean]

most normal moving setup

[elder wave]

yes it's from twitter

[elder wave]

lmfao

[rustic crown]

Don't see a cute plushie

[elder wave]

i sat on the ground and used the package of my mattress as desk for a bit

[chilly ocean]

interesting

I like to write in a notepad to explain stuff

and then delete it

[pastel cliff]

based chalkboards

[chilly ocean]

my strongest fighting game era was when i was moving and i played sitting on the ground with my monitor on a bunch of boxes. it's empowering

[pastel cliff]

i talk to empty classrooms (i fill the seats with the voices)

[elder wave]

goated setup

this is something i actually do quite often tho, either actually explaining the stuff to someone else or imagining to do so

i tend to find gaps or mistakes like that

[novel parrot]

if x1 and x2 are prime in PID why are they coprime?

(x1, x2) = (d) and we want to show d is a unit basically

but im stuck after that

anyone know?

[coral shale]

write down defns, try contradiction

thats what id do.

[novel parrot]

$x_1r_1 + x_2r_2 = d$

[cloud walrus]

ActiveChapter

[pliant forge]

the minimalist lifestyle 😍

[novel parrot]

i dont really get it

[formal ermine]

maybe try bezout

[novel parrot]

i am in a PID

idk bezout

[formal ermine]

look it up then

idk if it works tho

but this situation kinda looks like bezout idk

[novel parrot]

is it bezout identity

the one ur talking baout

[formal ermine]

yes

[novel parrot]

so yeah already i know that $(x_1, x_2) = (d)$

[cloud walrus]

ActiveChapter

[novel parrot]

that means that d is the gcd of x_1 and x_2

but x_1 and x_2 are primes / irreducibles

[molten silo]

Can anyone help be with part a?

Im not sure how to prove its maximal

[rustic crown]

R[x] is a pid, so compute the GCD to write it as an ideal generated by a single polynomial

[molten silo]

ok thank you

[chilly ocean]

I think you could write this like this even in a general ring since the polynomials are monic?

[obsidian loom]

Hi, I need some help with this question

Is it correct to conclude that the operation for this group is x * y = x^-1y?

Also, I think I'm supposed to show that $x^{-1}y \in H \implies xy \in H$ but I can't figure it out

[cloud walrus]

trayan_b

[coral spindle]

No.

The operation for H is the operation for G, restricted to H.

This is what it means to be a subgroup.

[obsidian loom]

what do you mean by restricted to H

[coral spindle]

I mean x * y = xy.

[obsidian loom]

oh

but in this case

for G, x * y = xy but for H, x * y = x^-1y, so it's not closed under the same operation as G

so by this logic it's not a subgroup

[coral spindle]

I don't know how you're getting this idea that x * y = x^-1 y

I said already that this isn't right

[rustic crown]

That's only for the first division though. The reminders might not be Monic >.<

[chilly ocean]

ah right

But (p(x), q(x)) is principal when p(x) and q(x) are monic, right

[rustic crown]

Not necessarily. Like what about (x, x+2) in Z[x]

[chilly ocean]

I see

[obsidian loom]

We are given that for H, xy = x^-1y, no?

For it to be a subgroup of G, it must be that for any x,y in G, xy =x^-1y also

[coral shale]

no.

you're given x^-1 y is in H

for x, y in H

[coral spindle]

No.

[obsidian loom]

So the "closed under the same operation as G" property is already given?

@coral shale

[coral spindle]

We are given that if x and y are in H, then x^-1 y is in H too

[coral shale]

All youre given is whats in the Q

[coral spindle]

Sorry mu I kinda trod on your toes there

[coral shale]

u started helping dw

• G is a group, H is a non-empty subset
• forall x, y in H: x^{-1}y is in H

These are your 2 premises

The first step is to show the closure property; ie. xy in H (for all x, y in H)

[obsidian loom]

Yeah

[tribal moss]

The first step is proably to show that H contains the identity!

[coral shale]

forgot

[obsidian loom]

Is that not one of the things you have to check?

Oh right

[tribal moss]

Yes.

[coral shale]

Indeed, and it should be the first step

[obsidian loom]

Thats what you said

I misread

I dont see how I would do that

[tribal moss]

Hint: This is the place where you need the assumption that H is not empty.

[chilly ocean]

what's with all the questions about the subgroup test recently?

i've seen so many over the past week

[tribal moss]

Perhaps we're in the part of the semester where such-and-such natural organization of an intro to algebra course reaches subgroups? Dunno.

[coral spindle]

This is why we should have a elementary group theory channel in early uni >:(

[chilly ocean]

actually agree

[coral shale]

if ur still stuck, take Tropo's hint. There is only one way to combine this with the other premise: x^{-1}y in H

[formal ermine]

yeah in the pre uni category

just like #elementary-number-theory

[thorn delta]

i understand the sentiment, but the lines between "elementary" group theory and group theory seem much more blurred than with NT idk

[chilly ocean]

what are the arrows of that category

[coral shale]

theres elementary group theory, then the rest of group theory and algebra

[thorn delta]

idk exactly what you mean by elementary though

anything in a first course could be considered elementary it seems

[obsidian loom]

I did it but I used: xy = x^-1y

and i think earlier you said that was wrong?

[coral shale]

well yes its wrong

where exactly are getting this from

How do you 'use' it???

[obsidian loom]

I'll tell you what i wrote

[coral shale]

Let our group G be integers under addition. Let H be the even numbers.
2, 4 are in H. Inverse of 4 is -4.

You are claiming 2 + 4 = 2 + -4

[obsidian loom]

We know that H is non-empty. Suppose that H has exactly one element, say x.
Then since for all x,y in H, x^-1y in H, it must be that x^-1x is in H and x^-1x = e, so the identity is in H.

[coral shale]

Thats good

Ok, so identity is in there

next i think u will want to show inverses

[coral shale]

its great - tiny nitpick: you dont need to suppose H has exactly one element

nor do u want to

You just want to say 'Let x be in H' and this is justified by it being non empty

[vagrant zinc]

xde

[obsidian loom]

yeah makes sense, i see why its unnecessary now

[obsidian loom]

I did

[coral shale]

ok then closure and you're done right?

[obsidian loom]

true

it seems closure is the hardest one to show

out of all the properties

[coral shale]

The reason we got u to do it in this order

is because its necessary

[obsidian loom]

oh so theres an order you need to show things

good to know

[coral shale]

in this particular case the proof needed it yes

(which is why u were stuck at the beginning)

In general, there isn't necessarily an order, sometimes there is.

[vagrant zinc]

Guys, a question if I have Z_12 and it tells me that <5> is a generator of Z_12

[obsidian loom]

maybe its just me but group theory is much harder than I expected it to be

[coral shale]

early on it comes down to writing down or pulling up the definitions relevant to the question

and applying them in the only way possible

later on there will be more thinking / more things u could apply as more theorems are developed

[obsidian loom]

anyways I finally completed the question

you were very helpful thank you

[coral shale]

theres no question; elaborate

[vagrant zinc]

exactly 5 is a generator of Z_12 but the question is to know if it really is a generator, I did that check

[summer path]

So given that you did that check, do you think 5 is a generator of Z/12Z?

[vagrant zinc]

yepi

[summer path]

Ok so there is no question

[coral shale]

There is a shortcut

You want to show 5n is not a multiple of 12 for n an integer from 1 to 11

5n = 12m

5 and 12 are coprime, ie gcd(5, 12) = 1

[coral shale]

See if you can finish the rest of the proof

[vagrant zinc]

I understand, it's good to know first where things come out in order to understand what you're saying

[solar shore]

does anyone have any good, free additional resources to study up on some abstract algebra?

[summer path]

dummit and foote is nice i guess?

[solar shore]

ive just been looking at my old exams and homework assignments and then redoing them to brush up on my next part of the class

hmm ok noteed

[prisma ibex]

Keith Conrad's notes are good for reviewing various topics

[solar shore]

okay thank you

appreciate it

[quaint raptor]

Hey, is this boxed fact true? I don't think it is tbh -- my understanding is:

"if all elements are their own inverse, the group is abelian" is true, but I don't believe you can say it is true the other way around".

This is for a finite abelian group

[formal ermine]

what is the context

[quaint raptor]

just proving that performing your operation on all elements of an abelian group^2 equals the identity element

[coral spindle]

Z/3Z is an Abelian group, and none of its elements are of order 2

Hint for a correct version of this proof: try pairing elements up with their inverses. What if a^-1 = a?

[quaint raptor]

awesome - ty. Yeah that makes sense and thats pretty much what I am thinking but just wanted to make sure I wasn't crazy

[south patrol]

Also if the boxed fact were true, then you could immediately conclude that the first term is e anyway

[barren sierra]

Proofs by diagram chasing are fun and feel like cheating lol

[coral shale]

profound argument. no cheat.

[barren sierra]

well yea

[chilly ocean]

why do you only need to prove that ab^-1 is in the subgroup to prove that it is indeed a subgroup?

and nonempty!!!

it needs to be nonempty!!!!!!!

yeah and that

[chilly radish]

what about the empty group

[chilly ocean]

i meant only as in how does it imply closure and inverse closure

[chilly ocean]

aint a subgroup

three lines:

if it contains some element z, then it contains zz^-1 = e. if it contains x, then it contains ex^-1 = x^-1. if it contains x and y, then it containts x(y^-1)^-1 = xy

[chilly radish]

Sounds sus

[chilly ocean]

ah u could do zz^-1

[chilly radish]

(I am joking if it wasn't clear)

[chilly ocean]

oh come on

fuck you lol

[elder wave]

i'm reaching, i was just waiting for a chance to post it again

[chilly radish]

Ok but nonempty is actually important

[elder wave]

[chilly ocean]

so many subgroup test questions

all of the group theory courses in the world are doing the same thing right now

but im different

im self studing so it aint count

[chilly radish]

It's such a superfluous thing

Maybe as an exercise

[elder wave]

i'm starting to see the point of an extra channel in early university

[chilly radish]

It's usually easier to check closure properties individually anyways

[chilly ocean]

i personally think it's absolutely useless

[chilly radish]

I think letting students prove the equivalence of the condition as an exercise is nice

But not more than that

[chilly ocean]

if someone is struggling with proving something is a subgroup by the definition and someone suggests the subgroup test, i don't think they're going to get any further with that than they did trying to use the definition

[barren sierra]

what extra channel?

[barren sierra]

same

great exercise

[chilly ocean]

i see it happen too much here

it aint even a exercies its like two words

[chilly radish]

Oh absolutely, I think understanding why it's true is good for getting the noggin jogging if nothing more than to practice abstract symbol pushing

[chilly radish]

That's silly

[chilly radish]

Another aa channel in early university

[chilly ocean]

yeah i was wondering why it wasnt up there

[barren sierra]

makes sense but then would #real-complex-analysis also get moved there?

[chilly radish]

This channel is kind of a catch-all at the moment. Another algebra channel was at least momentarily discussed

That was another point of discussion

[barren sierra]

cause we do have #advanced-analysis

so I guess analogously #adv-algebra makes sense

[chilly ocean]

i think we should just get rid of all of the channels

[chilly radish]

We have a button for that, it's called the "leave server" button

[barren sierra]

just one channel, #math

[coral shale]

im waiting for my advanced-anime channel

[chilly ocean]

you have NO idea

[barren sierra]

#advanced-chill

[solar shore]

i am incredibly stuck on this one problem 😭

Suppose the group $R*$, find elements $a$ and $b$ such that $|a| = \infty, |b| = \infty$ and $|ab| = 2$

[cloud walrus]

blanket

[coral shale]

ok for starters, what infinite groups do u know

[solar shore]

well

real numbesr

[coral shale]

oh wait sorry i misread

wait im confused, is that the full question

what is R*

[solar shore]

yeah

positive reals

[coral shale]

uhhhhhh

|ab| = 2 ...?

[chilly ocean]

$\bR^*$? that typically means the non-zero reals (not just non-negative) equipped with multiplication...

[cloud walrus]

ТТерра

[coral shale]

yeah i dont think positive reals

should be reals without 0, multiplication

[chilly radish]

The positive reals with multiplication don't have order 2 elements, you should try and see why this is true

To add to what was said above

[chilly ocean]

I have found that the group elements are $e, \sigma, \sigma^2, \sigma^3, \tau, \tau \sigma, \tau \sigma^2, \tau \sigma^3$ but how would i prove them? do i just say since they are finite its trivial?

[cloud walrus]

Jester

[chilly ocean]

or maybe try proving first that $\sigma^n \neq \tau \sigma^m$ for all m, n integers

[cloud walrus]

Jester

[chilly ocean]

oh nvm its trivial really

[sweet sundial]

are the groups Z4 x Z18 x Z15 and Z3 x Z36 x Z10 isomorphic

I think they are cause they have the same order

is that enough or do I need to show more?

[delicate orchid]

what order are each of the groups?

[sweet sundial]

360

i mean 1080

sorry

[delicate orchid]

are you sure the first group is order 1080?

[sweet sundial]

yes

4185

4 times 18 times 5

wait

wait wait

15 im sorry

[delicate orchid]

ok now I believe you that they're the same order

see if you can construct an isomorphism

[sweet sundial]

construct an isomorphism

like a bijective function between the two?

[delicate orchid]

no, a bijective homomorphism

[sweet sundial]

i see

what would be an example of that

[delicate orchid]

or use some other method of determining if two products of cyclic groups are isomorphic, if you know any

[sweet sundial]

okay

i will take this into account

[tender wharf]

this is almost always insufficient

for example S_n and Z_n! have the same order and are most definitely not isomorphic for quite a lot of values

infinitely many of them

[coral shale]

Z2 x Z2 and Z4 have the same order; non-isomorphic

[patent girder]

what is V_t, is it just a part of the representation?

[south patrol]

Basically if groups of the same size were isomorphic then finite group theory would just become counting

[chilly ocean]

finite group theory is just counting with a little bit of spice

[coral shale]

missed my cue for

Finite group theory is counting

[coral shale]

better resolution ?

[patent girder]

dont have

its a scanned thing from many decades ago

[coral shale]

well im sure wew can make it out when he sees it

[barren sierra]

I don't really have many thoughts on how to start.

[cloud walrus]

Spamakin🎷

Spamakin🎷

[barren sierra]

but otherwise I got nothing really

[kind temple]

H = Hom(Z,A)/im(m*) = coker(m*)
say q is the quotient map.
im(m*) = ker q so it’s exact at Hom(Z,A) (the one furthest to the right)
trivially exact at H since q is surjective
@barren sierra

[barren sierra]

Oh that's much more simple than what I was thinking

yea make the image "zero" which solves everything

[kind temple]

im not really sure how i would explain how i got this answer

that’s a good point.
i was just thinking about natural maps out of Hom(Z,A)

it kind of reflects the symmetry of the original SES

where we were mapping Z —> coker(m) = Z/mZ

[barren sierra]

yea

I saw that in the original one but didn't make the connection

I literally went "Z / mZ is Z quotiented with the image of m"

and did not make the very small next leap

cokernals are strange

I do not have intuition for them outside of some concrete examples

although that probably stems from my lack of intuition about certain category theoretical objects

[kind temple]

i don’t have a lot of intuition for them either yet

[barren sierra]

cokernals, equalizers / coequalizers, pullbacks, not much intuition yet

[kind temple]

category theory i feel like takes a lot of time to develop intuition for

[barren sierra]

just cause I'm only now learning them

I think

[kind temple]

same

[barren sierra]

yea it sure feels like it

[kind temple]

slowly

[barren sierra]

probably need more exposure

[kind temple]

yea, more examples and such

[barren sierra]

It would be nice if my alg geo class was any good

so that I could maybe look at examples relating to that area

but my prof is absolutely awful and I've kinda given up on the class

[kind temple]

profs can really make or break a class

[barren sierra]

it's totally broken this class lol

[kind temple]

dropped my alg top class because my prof really just didn’t want to be there

[barren sierra]

my algebra prof this sem tho is amazing

[kind temple]

same

[barren sierra]

really really enjoying that course

[kind temple]

what school do u go to?

[barren sierra]

See I'd do that but then I'd only be doing 1 math class and 2 CS classes

UIUC

[kind temple]

was gonna say lol. i’m at umich

[barren sierra]

Ooooo wait

do you know a prof, Ronnie Chen

does logic type stuff

[kind temple]

i do not

[barren sierra]

damn

he was one of my math profs my first semester here, now he's at UMich

he's super cool

[kind temple]

idk if i’m doing anything in logic but i’ll def keep an eye out if he’s teaching any interesting courses

[long nebula]

So, projective modules are summands of free modules; what's the corresponding statement for injective modules?

[barren sierra]

Every module is a submodule of some injective module

[long nebula]

oh alright

hm, how is that a dual notion though?

I don't see how that's related to projective modules being summands of free modules

[barren sierra]

Oh I guess it isn't really a dual notion

at least not totally

[long nebula]

are there cofree modules or smth lol

[next obsidian]

There is none

Welcome to the horror of injective modules

[long nebula]

I thought injective modules were just projective modules with the arrows flipped though

Why can't you just flip all the arrows on that statement

[sweet sundial]

I think these are actually isomorphic

[long nebula]

They are not

Think about the order of each element in Z2 × Z2 and the order of each element in Z4

[barren sierra]

bijection as sets does not imply isomorphism

both have 4 elements but yea, order

[lusty marlin]

ℤ_m × ℤ_n is isomorphic to ℤ_mn iff gcd(m,n)=1

[vagrant zinc]

Introduction to group theory and abstract algebra (from Judson)

[astral stream]

Question about Algebra 1. How many proof/theory questions should I do (there are roughly 20-25 in each exercise) and how many computation questions should I do (there are roughly 20-25 questions). I'm reading Fraglieh.

[summer path]

i think in general when self-studying and doing exercises, i'd say to look to do a couple computation problems and maybe a few proof problems until you feel comfortable and then move on to the next section and then come back and do more if you feel like you needed more practice

but that's just me personally; also fraleigh was pretty nice to read

[lusty marlin]

How does one show that the center of a group of order 105 cannot have order 7?

[astral stream]

thanks, is it ok to continue if I can't do the last 3-4 questions?

[agile burrow]

If Z(G) has order 7, then G/Z(G) has order 15 which implies it's cyclic. This implies that G is abelian

[lusty marlin]

The conclusion about order 15 groups, does it come from a theorem stating that a group whose order is the product of two primes is cyclic?

[lusty marlin]

I'm not saying this is actually a theorem, I am yet to come across such a statement.

[summer path]

i think that's really just up to you tbh; i'd just be comfortable enough with the material in each section to continue
if you want to do all the problems, you can do that too; if you feel like you need some hints, there's probably people who would be willing to help out here (and elsewhere as well)

[obtuse bear]

you need additional hypotheses; S_3 has order 6 = 2x3 and is not cyclic

[astral stream]

Ok thanks for the help

[summer path]

or at least that's kinda how i approach self-studying; i do know other people who like to work through all the exercises in every section when they read a textbook

[lusty marlin]

Right, that example was in mind

[tribal moss]

https://groupprops.subwiki.org/wiki/Classification_of_groups_of_order_a_product_of_two_distinct_primes

[lusty marlin]

Nice, so since 3 does not divide 5-1, any group of order 15 is cyclic

And from this, the solution to my problem follows immediately.

Thanks guys

[solar shore]

Give an example to show that if H and K are subgroups of a group G the set HK = {hk | h ∈ H, k ∈ K} need not be a subgroup of G.

im blankin on this

[tribal moss]

It will be a subgroup if G is abelian, so you'll need to look to non-abelian groups for examples.

[solar shore]

do you have any good ideas for non abelian groups?

any suggestions ive gotten were with $S_n$ which we havent covered yet

[cloud walrus]

blanket

[solar shore]

all i really got are $D_n, \mathbb Z_n$

[cloud walrus]

blanket

[tribal moss]

Pity, that's the simplest non-abelian groups. How about matrix groups?

[solar shore]

oh

i know $GL(n,\mathbb R)$

[cloud walrus]

blanket

[solar shore]

and $SL(n, \mathbb R)$

[cloud walrus]

blanket

[next obsidian]

D_n works

[tribal moss]

Hm yes it does, and D_{2·3} is even the same as S_3.

[solar shore]

ok ill take a look at D_2 and 3

tysm!

[tribal moss]

(I meant the dihedral group with 6 elements).

[solar shore]

oh

so d3?

[summer path]

some people write the dihedral group of order n as D_{2n} and some write it as D_{n}

[solar shore]

ohh i see

[summer path]

though i believe D_{2n} is used more commonly in algebra settings (i may be wrong on this)

[astral stream]

Why is the yellow part true?

[formal ermine]

what's phi_x

[formal ermine]

you can construct an explicit isomorphism by sending one generator to another

[tribal moss]

(a) By Rice's theorem. (b) By Rice's theorem. (c) By Rice's theorem. (d) By Rice's theorem. (e) by Rice's theorem. (f) By Rice's theorem. (g) By Rice's theorem.

Computability is mostly #foundations, tough we also get some questions in #discrete-math.

[astral stream]

what does sending one generator to anther mean?

[astral stream]

nvm I think I undertand now

[tribal moss]

That's more of an ad-hoc diagonalization proof -- which is perfectly fine and probably the general style you're supposed to attack the other ones in.
Rice's theorem (google it!) gives a generic construction that happens to settle all these cases in one fell swoop.

It's too late in the night for me to check details with any confidence, sorry.

[prisma shuttle]

why does the relation [A,BC] = B[A,C]+[A,B]C hold for eleemtns of sl(2)

where sl(2) is a vector space $V$ with a triple of operators $E,F,H$ such that HE-EH=2E, HF-FH=-2F, EF-FE=H$

[cloud walrus]

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

[prisma shuttle]

I don't see how taht relationship comes directly from these defining properties (for those of u who have worked with it before, this is just the special linear lie algebra on 2 elemetns)

[rustic crown]

that's not sl(2). that's a representation of the lie algebra sl(2).

sl(2) is the sub-lie-algebra of 2x2 matrices over C with trace 0

and the lie bracket is [A, B] = AB-BA

so you just have to plug in this definition and verify this works [A,BC] = B[A,C]+[A,B]C

[smoky cypress]

Why does the statement bE subset A hold?

I don’t see how it’s connected to bB’=0 or any of the previous statement

I think I’m supposed to use exactness in the middle somehow but I don’t see it

[rustic crown]

i didn't read the whole thing, but that should simply follow from that fact that image of bE under E --> B is zero

so by exactness, it should lie in the image of A --> E

[smoky cypress]

Why is the image of bE 0?

[rustic crown]

because image is bB

[smoky cypress]

?

I think b is an integer here

[rustic crown]

yep

[smoky cypress]

Oh duh

[rustic crown]

say that map E --> B is called p, then p(b * e) = b * p(e) in bB = 0

[smoky cypress]

Got it

Thanks

[rustic crown]

hehe np

[prisma shuttle]

ah i see thx u

ima bit stuck on this problem part (h) involving the casimir operator does anyone get how i can finish this

btw the V_{lambda} are defined as this:

I managed to show that the action of C on V is an operator with one eigenvalue, and that this eigenvalue is the scalar that its act on some irreducible subrepresentation of V

are all irreducible subrepresentations V_{lambda} for some lambda or smth? cuz then the problem would immediately follow

but ima not sure how i can show that

[tribal moss]

Do you have any information about what E, F, and H are?

[prisma shuttle]

oh yea

its a representation of the lie algebra sl(2)

[rustic crown]

yea, take v to be the highest weight vector.
if V is some f.d. irreducible representation, then take v to be the an eigenvector of H with largest (real part of) eigenvalue. since you're working over alg closed field this should be nice. then the submodule generated by this v looks like V_lambda

was this not one of the problems above?

like seeing that this lambda which maybe apriori a weird complex number turns out to be a nice integer

[prisma shuttle]

huh

wdym one fo the problems aboe

the notion is first introduced from problem f) so idt its part of a prev problem?

[rustic crown]

oh yea this

[hot lake]

in f) you showed that any finite dimensional irreducible representation was a V lambda

[prisma shuttle]

isnt a representation though

not a subrepresentation

am i missing smth obvious

[hot lake]

subrepresentations are representations too ?

[rustic crown]

iirc the idea here was that E(F^kv) = (some nice polynomial in k and lambda) * F^{k-1}v. and you use this for the smallest k such that F^kv = 0 which gives you that dim V = lambda + 1

so every f.d. irred rep does look like V_lambda

[prisma shuttle]

wait rlly

i thoguth subrepresentatins were just subspaces closed under the action of a representation

[hot lake]

just like how a subgroup of a group, ... is a group too

[tribal moss]

That makes them (or rather the representation restricted to them) representations too, doesn't it?

[prisma shuttle]

Omg wut was I thinking

Thx so much

[prisma ibex]

another way to think about this if you know about modules, representations are just modules over the group ring (e.g. if G is a finite group then k-linear representations of G are the same as modules over the ring k[G], or if g is a Lie algebra then k-linear representations of g are the same as modules over the universal enveloping algebra U(g)...)

submodules of modules are still modules

[bleak abyss]

My turn

So to recap basic construction

Chamber system C over an index set I, means each element i in I forms a partition of C

So then, for i in I, we say two thing are i-adjecent if they're in same part of the partition corresponding to i

And if J is a subset of I, then J-connectivity means you find a path (or formally a "gallery") where any two elements are connected by some element of J (the element can differ within the path)

Residues are connected components

Morphism C->D for two chamber systems over the same index set is a function such that if c is i-adjacent to c', then phi(c) is i-adjacent to phi(c')

Actually I'll just link these notes

https://people.maths.ox.ac.uk/demartino/LectureNotes.pdf

Key point is, if R is a type J residue and S is a type K residue

Then we say S is a face of R if S contains R and K contains J

Now, let's say R is a residue of cotype J (meaning type I\J)

For each K subset J, R has a unique face of cotype K. This is important, if you imagine a simplex with vertices {v_1,...,v_n}, well excluding a vertex defines a unique face, excluding something in that a unique codim 2 simplex, etc

So if we have a chamber system

How do we get its geometric realization?

Well, residues of cotype {i} are vertices

Now, given a residue of cotype {i,j}, it should have a unique face of cotype {i} and a unique face of cotype {j}. So that's two vertices, glue along an edge

Given a residue of cotype {i,j,k}, has a unique face of cotype {i,j}, cotype {i,k}, and {j,k}. 3 edges, glue a triangle in. Keep going

gg

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

And yeah an r-flag has as its codim 1 faces just "exclude one of the r elements of the flag"

I'm getting the idea a bit

Embarrassingly slowly but still

[ripe glade]

may I ask what course might cover what you're talking about rn?

[bleak abyss]

I linked the notes earlier. For me it's not a class I'm "taking" exactly but

I'm learning the material off a course

About Bruhat-Tits buildings

[cloud walrus]

Sloth King Daminark

[ripe glade]

thanks, seems way out of my scope of understanding but interesting

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Okay why am I blanking now?

I need to wake up

So

Vertices of cotype {2}

Well, we take one non-zero vector, then another that's not in its subspace

Divided by q^2

err no

[pastel cliff]

what does an element of a direct sum look like

in my notes it says that it can expressed as a sum?

[prisma shuttle]

its just a linear combination of the elements in each of the parts of the sum

[pastel cliff]

ive been thinking about this completely wrong then

i thought it was like

[prisma shuttle]

so an element of $V_1\oplus V_2$ is just an element $k_1v_1+k_2v_2$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

where, $k_1, k_2$ are scalars in the field

[cloud walrus]

JustKeepRunning

[prisma shuttle]

yea i thought about this wrong too before

[pastel cliff]

a product with all but finitely many zeros

this changes everything

[prisma shuttle]

yea be careful

direct product would be written as

$V_1\times V_2$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

and then another type of product to be careful not to get confused with is the tensor product

which sorta looks like a "tilted" direct sum sign

tensor product of $V_1$ and $V_2$ is denoted as $V_1\otimes V_2$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

and note that all the $V_i$ here are vector spaces

[cloud walrus]

JustKeepRunning

[pastel cliff]

not there yet but soon

so what is the relationship between direct product and sum

i know the categorical defn of like flipping the arrows

but that doesnt do much for me

[elder wave]

this is what a direct sum is

[pastel cliff]

[prisma shuttle]

wait are u talking about a direct sum of vector spaces

or a direct sum of rings

[pastel cliff]

oh i shouldve been specific ig

R-modules

[elder wave]

i'm talking modules

[pastel cliff]

but shouldnt they also be analogues

[elder wave]

the point is that every element in your direct sum can be expressed as sum of elements of the components embedded into your product

[prisma shuttle]

idk i think my def is right for vector spaces?

tahts wut my mentor told me so like...

[elder wave]

the finitely many non zero condition makes the sum finite

[prisma shuttle]

idk correct me if ima wrong ig

[elder wave]

so that it makes sense

[prisma shuttle]

wait was my def wrong

fro vector spaces

[pastel cliff]

so it's just a way of representing it?

[elder wave]

your definition is related to two subspaces

of the same vector space

because if you take arbitrary families of R-Modules (which K-vectorspaces are a special case of) you can't make sense of addition like that

[elder wave]

yes, you can't add elements from arbitrary families of modules

[prisma shuttle]

oooh that makes sense

so like

my def was a subset of the real def?

[elder wave]

your definition is something slightly different

[pastel cliff]

probably too general but does this mean we can say like, (..., 3, 4 ,5 , ....) is represented as 3 + 4 + 5?

[elder wave]

there is the external direct sum and the internal direct sum

[pastel cliff]

that cant be right but i think i need an example if you have one

[elder wave]

the external direct sum can be constructed for arbitrary families and the internal one only makes sense for subspaces of the same module

[prisma shuttle]

so wut i was doing was internal direct sum?

[pastel cliff]

external

i think, since you had a scalar operation on it

[elder wave]

no internal

[prisma shuttle]

i thought u said mine only made sense if they were subspaces of teh same module

yea

its internal

[pastel cliff]

nvm

[elder wave]

external is what we are doing rn sebb

[prisma shuttle]

scalar is from the same field

[pastel cliff]

[prisma shuttle]

like k_1, k_2\in K

[elder wave]

you can even leave the scalars out of it

since k*v1 is in your subspace again anyway

okay where were we

[pastel cliff]

this

[elder wave]

sure, take R^n

[prisma ibex]

for R-modules, \oplus is both a product and a coproduct

[pastel cliff]

only in the finite case no?

[prisma ibex]

yeah

in the infinite case there's an annoying difference between the product and direct product

[elder wave]

this is the n-fold direct sum of R

so think R being embedded in R^n as (R,0,0,0,0...)

and that for every component

now you can think of every element in R^n as a sum of Elements in R after embedding them like that

and the condition that finitely many entries are non zero ensures that this makes sense even for infinite index sets

such that your sums are still finite

but unless it is infinite it is indeed just a normal product

and nG will tell you the rest/correct me because i gotta go eat something

[prisma ibex]

iirc in general the direct sum is a coproduct, the direct product is a product; the direct sum is smaller because of this condition of finite support, and you have a canonical morphism from the direct sum to the direct product that is an isomorphism in the finite case

[elder wave]

[pastel cliff]

i can see this, what confuses me is the sum

also since when is timo honorable

[prisma ibex]

since a while lol

[pastel cliff]

[elder wave]

it's called a sum because of what i said

you can't do this for a product

take Z^N for example

[pastel cliff]

i guess i should be explicit then, for the R^n example, does that mean we have some kind of map from \oplus R_i -> R

[elder wave]

yes the projection

[pastel cliff]

lemme just post my notes hold on

ignoring the actual proof

the sum part in the first bit

it's saying what you're saying obv

but doesnt that give us multiple elements that might map to the same sum?

[solar totem]

if N a normal subgroup of G, for any a in G, show that (Na)^(gcd(ord(a)),ord(Na))=N
any hints?

[elder wave]

i'm not sure what you mean here

[pastel cliff]

(..., 1 , 2, ....) = (...., 2, 1, ...) = 2+1 no?

even though those would be two diff elt's

[elder wave]

no those are different components

you embed your m_i first and then take the sum

and 2+1 does not make sense in your setting

you're still working with tuples

[pastel cliff]

and this is part of the categorical defn right

i think im close but not there yet

sorry to keep running circles around this

[elder wave]

i'm working explicitly rn not in categorical slang

take R^3 would you say that (1,2,3)=1+2+3

no right

let's stick to R^3 for now

and think of it as 3 fold direct sum of R

[pastel cliff]

that's what im confused about! like that's what sounds correct but doesnt seem like it is

that didnt make sense but still

[elder wave]

now we have \iota_1(r) = (r,0,0)

we embed the first component into our direct sum

same for the other two

[pastel cliff]

oH

it's almost like a basis representation

[elder wave]

and now (a,b,c) is equal to (a,0,0)+(0,b,0)+(0,0,c)

by just embedding a from the first copy of the real numbers, b from the second copy and c from the third copy

you're probably overthinking this

[pastel cliff]

i most certainly am

[elder wave]

you're still in R^3

how would 3+2+1 make sense in R^3

[pastel cliff]

in the infinite case it's just a sum of (0, ..., R, ... 0) finitely many times

[elder wave]

yeah

because finitely many terms of the element you're trying to represent are non zero

[pastel cliff]

i got it

[primal tusk]

Are ideals different in algebraic geometry and abstract algebra?

[pastel cliff]

so it also becomes obvious why prod and direct sum are the same in the finite case

if we have a finite product of things of course we can just express it as a sum

[elder wave]

right?

[pastel cliff]

but the finiteness of the indices in the direct sum guarantees we can do it anytime

thank you timo

[elder wave]

you're welcome

[bleak abyss]

Finally

Okay so

[elder wave]

your turn again dami

enjoy

[bleak abyss]

Interrupting fucks lol

Residues of cotype {2} are 2-dimensional subspaces

[pastel cliff]

it's me im fucks

sorry tho

[bleak abyss]

GL(3,q) acts on the space of 2-dim subspaces of F_q^3

[barren sierra]

no, ideals specifically refer to the substructure of rings

definitions are the same

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Top left 2x2 bit should be some invertible 2x2 matrix

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

How many chambers are there?

Well for each 2-dim subspace, we take a line

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[primal tusk]

@bleak abyss is this related to what this guy studies? https://lsa.umich.edu/math/people/faculty/smdbackr.html

[bleak abyss]

Yeah

[primal tusk]

thats awesome

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

And I'm pausing to eat so

[chilly ocean]

inb4 dami gets interrupted by a subgroup test question

[prisma shuttle]

wait so just to make sure

any subrepresentation of a vector space is ALSO a representation of that same vector space?

is that wut u guys were saying

like basically wut this is saying?

[delicate orchid]

why would we call it a subrepresentation if it wasn't a representation

[rustic crown]

subbasis moment

[bleak abyss]

JustKeepRunning careful

"sub" means subpsace

Not subgroup

So you have a representation of G on V, well is W invariant? If so then you can get a representation of G on W. That's a subrep

[wet zodiac]

artin mentions bezouts lemma

[sweet sundial]

how would i go about showing that the two groups z4 x z18 x z15 and z3 x z36 x z10 are isomorphic

i know i have to find an isomorphism between the two somehow or show that there cannot be an isomorphism

could i maybe just do a counterexample?

[south patrol]

How would there be a counterexample

But yeah generally hint: Chinese remainder theorem

for checking these things

[sweet sundial]

chinese remainder theorem

interesting

okay

ty

[quaint raptor]

Maybe stupid question, but want to clarify something about functions being vectors.

For the most part this makes sense. The thing that is tripping me up is my understanding is that 1) functions do not always have to have their range be the real numbers and 2) it is a requirement that vectors map onto the real numbers.

Wouldn't this mean that you can only say functions are vectors if you know your function maps onto the reals. Asking because I am using (f+g)(x) = f(x) + g(x) for some subgroup problems

[chilly ocean]

if S is a set and V is any vector space, then the set of functions from S to V becomes a vector space with the operations you expect

1. functions do not always have to have their range be the real numbers
   then you can clarify "real-valued functions" or "functions into R" or whatever as you write

2. it is a requirement that vectors map onto the real numbers.
   this phrasing makes me think you're saying all vectors are in some way functions to the reals, which is just false

[quaint raptor]

I think for part 2 I misinterpreted what I read as it said "A vector maps {1,2,3} onto R"

but thats only for that example / you can map some set of n integers to what I conventionally think of with a vector of that size

ty

[rotund aurora]

Is Daminark clinically insane?

[rustic crown]

we all are

[rotund aurora]

What's this

[chilly ocean]

notation

[rotund aurora]

?

[S : R] is a number

what is that

R is a subring, so you cant quotient

[rustic crown]

If R and S are fields

[rotund aurora]

lmao

I didnt read all of that xdd

[rustic crown]

ik

:p

[rotund aurora]

thanks

[rustic crown]

[formal ermine]

isn't this just like the degree of a field extension

[rustic crown]

yes

it is exactly that

[formal ermine]

ic

[chilly radish]

Perhaps

[lapis latch]

inconceivable

[chilly ocean]

i can see that both xH and Hx are subsets of f^-1(f(x)) but how are they equal?

also its saying rewrite like its so very obvious but isnt H a set how are you able to just use them like normal group elements is it notation abuse or are the group elements with subgroups form a group or sth like that?

like i wouldnt know if (xH)x^-1 = x(Hx^-1) from the first sight

if y is in the preimage, then f(y) = f(x), which implies that x^{-1}y is in H or y = x(x^{-1}y) is in xH

[chilly ocean]

oh yeah that makes sense

xHx^-1 dont even feel stable

shouldnt |xHx^-1| >= |H|

and for appropriate x, >

?

conjugation is a bijection

conjugation?

g -> xgx^-1

well yeah but thats not the same as xHx^-1

since its accumulating

xHx^-1 is the image of H under this

like H = {e} then xHx^-1 would be {e, x, x^-1}

no

when x!=e

that is not xHx^-1

its not?

ohh its not

yeah it aint accumulating the cardinality keeps same trhought

yeah okay since |H| = |xH| = |xHx^-1| ofc there is a bijection

Thanks

[tulip shuttle]

Is there an Engineering channel here or no?

[next obsidian]

No, this is a math server

[agile pine]

wrong channel

[tulip shuttle]

oh okay

[coral shale]

why isnt 'engineering' censored in here

[chilly ocean]

engin**r

[summer path]

Wrong server*

[tribal moss]

Mentions of it don't usually lead to toxic discussion. ¯_(ツ)_/¯

[chilly ocean]

check out the servers in #old-network , cheers

just a quick check my verification that $N_S$ is a subgroup:
$$\forall x, y: xy^{-1} S (xy^{-1})^{-1} = xy^{-1} S y{x^-1} = x S x^{-1} = S$$
right?

[cloud walrus]

Jester

[chilly ocean]

so N_S is a subgroup

its wrong?

[tender wharf]

i see you're trying to do the subgroup test

[chilly ocean]

yeah

sure its correct

although y^-1Sy is not super straightforward but it comes from ySy^-1 = S of course

x, y are supposed to be in N btw

yeah

[tender wharf]

if so that works lol

probably mention where x, y come from next time

[chilly ocean]

yeah i just remembered it as Sy = yS

just picking on details

[pastel cliff]

what's the point of using a direct sum in this question

like isnt it the same as a product

[chilly ocean]

if you wrote a product, i could ask you what the point of using the product as opposed to the direct sum is

[pastel cliff]

true

why am i even here typing away silly little symbols at my silly little mac with my silly little brain

[south patrol]

Direct sum better

[pastel cliff]

why

[south patrol]

Also nice mac flex

Uhhh direct sum is stronger

And cooler symbol

[pastel cliff]

the only good argument

[south patrol]

Lol

[pastel cliff]

w regards to the actual question, we can say M and N are isomorphic right

going by this def'n

[agile burrow]

no, because we don't know that sigma o pi = 1_M

[pastel cliff]

honorable walter

true tho yeah i realized that at the same time i sent it

well deserved honorable

now to get a walter emoji

,av walter

[cloud walrus]

walter#1555's Avatar

Click here to view the image.

[pastel cliff]

i love this dog

what might the iso look like then

im trying to just define it explicitly

if m \in ker(\pi) map it to (m, 0)

but i dont think saying if m isn't in the kernel just map it to pi(m) is enough

[agile burrow]

right

[pastel cliff]

so it's not enough right?

i figure i need to use that identity map somehow

[agile burrow]

I'm trying to think of how I want to explain it

I guess my intuition comes from vector spaces

The way I'm thinking of it is that pi is projecting M onto N and then sigma is a section of pi so that sigma(n) is in pi^-1(n)

And this condition on sigma comes from the fact that pi o sigma = 1_N

So now if I have an element m in M, I can project it to N via pi and then look at what representative it maps to under sigma, namely sigma o pi(m)

I'm explaining this terribly because I'm hungry

[prisma ibex]

it's better to write \oplus instead of \times if you really have a direct sum, since it's good to remember that you have universal arrows going in both directions

although for instance if you have an argument which ONLY uses the universal property of being a product, then it's fine to write \times to emphasize this

but it's definitely abusive to write \times and then be like "oh btw this is also a coproduct haha"

[agile burrow]

Right, but now the idea is that both m and sigma o pi(m) will map to pi(m). And this is easy to check since we have that pi o sigma o pi(m) = pi(m)

And in particular, this means that m - sigma o pi(m) is in the kernel of pi

so this is how you sort of break an element of M down into a part that lies in the kernel of pi and a part that lies in N

[pastel cliff]

damn i need a chalkboardd

i need a min to read this over but im starting to see it

[agile burrow]

I probably could've texed it too but again, i'm hungry

[pastel cliff]

all good, i still appreciate

[tender wharf]

steal one

[bleak abyss]

It's my turn again

Dam walter you are a loyal fan

Thanks fam

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Right I forgot to write lol

So we represent this by a diagram

[cloud walrus]

Sloth King Daminark

[bleak abyss]

(See page 7 of notes for visuals)

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Okay we back

So

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Okay this proof is long

It basically gets into root system shenanigans

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Back

[bleak abyss]

$s_i(e_j) = e_j + 2\cos(\theta)e_i = (2\cos(\theta)\sin(\theta), 1- 2\cos^2(\theta)) = (\sin(2\theta),-\cos(2\theta))$

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Now, $(1,0) = a e_i + b e_j = (a\sin(\theta), b-a\cos(\theta))$, so $a = \csc(\theta)$ and $b = \cot(\theta)$

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Jesus

Eh frick

Forgot to put \ in front of a theta

nbd

holazach why'd you make me do all that work smh

[lapis latch]

I think you’ve went insane

[bleak abyss]

no u

[upper pivot]

'went'

[bleak abyss]

I'm mostly being a bitch lol this is ez

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[solar shore]

im having a bit of trouble understanding this notation

i understand that <a> is the cyclic subgroup, but what does <a, b> mean? does it mean the numbers produced by ab?

[summer path]

<a,b> is the set of elements generated by a and b

[solar shore]

oh

so ${a^n, b^m|n, m\in\mathbb Z}$?

[cloud walrus]

blanket

[summer path]

not quite

[solar shore]

or rather

[summer path]

you can also have stuff like

aaabbabababaaba

[solar shore]

is it the union of <a> and <b>?

hmm

[summer path]

it's basically the set of all combinations of "multiplying" a's and b's

[solar shore]

ohh

i see

is there any notation

to describe it succinctly?

i think i get it instinctively but it helps if i can attach notation to it in my head

if not its okay

oh

[summer path]

i don't think so? im not sure though

[solar shore]

this actually helped out a lot

its not general but it helps it out a bit more

[summer path]

yeah that's nice in the case where your binary operation commutes :)

[solar shore]

hehe yep

but so say for matrices, then its different to have, say matrices A,B and the subgroup <A, B>

that would be like, A, B, AB, BA, ABB, AAB,...

not necessarily commutative?

[summer path]

yeah

[solar shore]

ok tysm

thank you

im writing notes up rn and i just wanted to like, clarify that for myself later on lolol

[summer path]

[solar shore]

🙏 i appreciate it

[tender wharf]

I think you meant a^n b^m?

[solar shore]

yeah sorry

also

i dont get why this is true

[tender wharf]

oh gallian

[solar shore]

yeah

[tender wharf]

there's a couple of typos in gallian

TODO: Release gallian-typos.pdf

[solar shore]

is this a typo?

[tender wharf]

nope

[solar shore]

bcs im not understanding why the order of an element is equal to the order of the cyclic subgroup

[tender wharf]

woah you just got active

[solar shore]

oh

shit

LOL

[tender wharf]

I mean

think about it for a moment

[solar shore]

wait

[tender wharf]

how could it be higher or lower

that would be pretty absurd

[solar shore]

as soon as i typed that it literally just clicked

💀

my bad LMFAO

we chillin

[tender wharf]

all good

haha

[solar shore]

yk those brainfarts

[tender wharf]

I feel you

[solar shore]

@tender wharf

lovin the typos

[summer path]

at least it wasn't a content typo lol

unless there are those too...

[tender wharf]

UH

you say that

there's a severe content typo in the chapter on normal subgroups

[tender wharf]

you have a bad pdf lol

[summer path]

oh no

[solar shore]

lovin it

[tender wharf]

open your dms

[solar shore]

o

[tender wharf]

i have a better copy

[solar shore]

omg thank you

ive been seeing so many typos and its bad

i usually correct it when i speak to my professor but damn

[tender wharf]

yeah there are quite a lot

I'll probably post an errata list sometime

[solar shore]

wait holy shit

the quality on this pdf is so much better than the one i was using

im crying now

i feel cheated.

[tender wharf]

lol

it's almost like

I got it from an access code

the 9th edition might have fewer typos

feel free to like dm or @ me if you find something that you think is a typo

[summer path]

honestly, sometimes it's nice to have a couple typos in proofs so that i can check my understanding

but if there's typos in definitions and theorems, then it's pretty yikes

[pliant forge]

jesus

[trail stump]

is it true that, if there exist a G-set for a group G, such that G is not faithful on the G-set, and not every element of G act as an identity on the G-set, then G is not simple

[coral shale]

exists a homomorphism from G with non-trivial/proper kernel <=> G has non trivial/proper normal subgroup <=> G not simple

What you describe results in a non-trivial/proper kernel, I believe.

[trail stump]

yeah idk, it seems like a direct application of the theorem

but i don't think i understand the theorem enough to know if it is the "right" application of it

[void cosmos]

do you need zorn's lemma to show that if a is in the intersection of all prime ideals then there exists n in N such that a^n = 0 ?

over a commutative ring

[rustic crown]

yee

the question is equivalent to asking whether there is a prime ideal in the ring A[a^-1]