groups-rings-fields

[delicate orchid]

Ah so I am completely blind

Good to know

[next fulcrum]

Here's the solution

[delicate orchid]

Good golly miss molly

[heavy dagger]

Is it possible to construct an infinite order cyclic group <a> with identity as the limit point of the sequence {a^r} ?

[delicate orchid]

every infinite cyclic group is isomorphic to Z so I doubt it

[subtle ivy]

Z with the trivial topology satisfies.

the question is a lil funny. you have to narrow what spaces you're interested in.

[heavy dagger]

But like, with an actual metric

[heavy dagger]

Hmm I get the intuition a little but topological properties are not preserved in an isomorphism, are they?

[delicate orchid]

uhhhhhhh

you mean the isomorphism is a homeomorphism?

[heavy dagger]

I meant like even though the elements of the group correspond, how close they are w.r.t to their metrics will differ

So limit points don't have to correspond

[subtle ivy]

the answer is obviously no, again by just imposing a non metrizable topology on the group.

[subtle ivy]

take the metric to be the inverse of the standard metric on Z, i think.

[heavy dagger]

inverse as in?

[subtle ivy]

reciprocal

maybe actually uh

[heavy dagger]

Sorry what was the standard metric on Z?

[subtle ivy]

absolute value

i actually am not confident that is right lol

[heavy dagger]

oh okk

[delicate orchid]

my main problem is I still have no idea what {a^r} is, lol

are we taking a to infinity?

r?

[subtle ivy]

what

bruh it is the infinite cyclic group generated by a

[delicate orchid]

wait we're not taking a limit of groups

[subtle ivy]

.<

[delicate orchid]

#point-set-topology leave

[heavy dagger]

yeah lim a^r

[delicate orchid]

ok so r is going to infinity got it

ok this is actually a lot more approachable than I thought

[heavy dagger]

noice

[subtle ivy]

ok so good news what i suggested is actually a metric

so i think we are done.

erg wait

[heavy dagger]

1/|x-y| ?

[delicate orchid]

discrete topology on a countable set is metrizable right

I know it is on finite

[subtle ivy]

uh yes it is on all sets

[delicate orchid]

I lost my train of thought

[subtle ivy]

yeah so this is totally wrong lol

[heavy dagger]

Okie tag if anyone has some idea

[heavy dagger]

A person from my uni answered it, consider (Z,+) with the metric |1/x - 1/y| and d(x,0) = d(0,x) = |1/x| and d(0,0) = 0

[lethal dune]

i doubt that's a metric

[heavy dagger]

It is, isnt ?

[lethal dune]

lemmi see

[delicate orchid]

axiom checking time

[heavy dagger]

Satisfies the triange boi

[lethal dune]

ig it is then

[delicate orchid]

is that not just the standard Euclidean metric on [-1, 1] lol

or, equivalent to

yeah

[heavy dagger]

yems

[lethal dune]

[subtle ivy]

that is weird

[lethal dune]

you mean after composing with 1/x?

[delicate orchid]

way I'm viewing it is that |x-y| is a metric for all real numbers so if you just write x = 1/k, y = 1/n then |1/k-1/n| is also a metric (after considering what happens if they're 0 obvs)

[heavy dagger]

samesies

[full panther]

my prof notes says for any semisimple lie algebra g, [g,g] = g

i thought this was only true for simple lie algebras

i know that semisimple ones are isomorphic to direct sum of simple

[coral shale]

What's the motivation for studying Finite Fields - anything in particular? (Galois theory context, Frobenius Map, (In)separability)

[next obsidian]

Number theory

[plucky torrent]

I'm doing Galois theory, and am showing that the Galois group of x^p - 1 is the cyclic group of order p-1 for p a prime. What I don't understand is, why do we get a group of order p-1 if the degree of the extension we are condsidering ( Q adjoin omega, a primitive pth root of unity / Q ) is surely p (we have basis for the vector space of our extension that is 1, omega, omega^2, ... , omega^p-1). Isn't the degree of the extension supposed to equal the order of the Galois group? Have I computed one of the two incorrectly? Any help would be appreciated.

[lethal dune]

if an extension is galois then yes

[plucky torrent]

what makes this extension not Galois?

[latent anvil]

it is galois, because x^p - 1 splits over Q(omega). I'm not sure how to reconcile this with what you said

thinking

oh wait x^p - 1 factors

[plucky torrent]

ah

[latent anvil]

as (x-1)(1 + x + x^2 + ... + x^{p-1})

right?

[plucky torrent]

yes

[latent anvil]

I think you're counting 1 an extra time or something, thinking once again

right yeah the set you wrote is linearly dependent

because $1 + \omega + \omega^2 + \ldots + \omega^{p-1} = 0$

[cloud walrus]

Enmaidened Shamrock

[latent anvil]

does that make sense?

[plucky torrent]

yes thanks

so the basis only needs p-1 elements, not p

[latent anvil]

yep!

[plucky torrent]

thanks

[limber vale]

After having finished dummit and foote, how far is the road to actual research level algebra? I heard Langs book is the standard for grad school, is there still more algebra after this?

[sharp sonnet]

short answer: yes

long answer: yeeeeeeeeeeeeeeeeeeeeeeeeeees

[limber vale]

haha

[sharp sonnet]

you can look at like arxiv articles tagged algebra

[limber vale]

I am almost done with undergrad, I think research sounds like a great future but the problem is i have no idea what is out there

[sharp sonnet]

see how many words you recognize

[urban moth]

what does p +q \leq 1 means in the context of C*-algebra?

[thorn delta]

wild and totally not certain guess, but maybe it could mean r(p + q) <= 1, where r is the spectral radius?

[sharp turret]

what does this symbol mean

[thorn delta]

"proper subset of"

[sharp turret]

ok thats what i thought i was used to just the top curve with no line to mean that

thanks

[thorn delta]

np

[thorn delta]

and this notational idea would carry over for p(p+q) <= p: if, r(p + q) <= 1, then since multiplication is analytic, sigma(p(p+q)) = p * sigma(p + q)
so r(p(p+q)) = p*r(p+q) <= p, but again im just spitballing ideas here.

Hm, no nvm, the stuff I wrote down doesn’t quite make sense

[patent crescent]

How can the order of a semidirect product be the product of the order of the two groups if the semidirect product is isomorphic to NH?

Because the order of NH is not necessarily the product of the order of N and H right?

[south patrol]

|NH| = |N||H|/|N intersect H| in general and we require N intersect H to be trivial for an internal semidirect product so we get the result you state, |NH| = |N| |H|

[pastel cliff]

quotient groups are inherently subgroups righ

[south patrol]

No and what is 'inherently' meant to mean there

[pastel cliff]

yeah nvm

was thinking incorrectly

[south patrol]

Sure dw

[delicate orchid]

for finitely generated abelian groups this holds

[south patrol]

In what sense?

I mean (Z,+) has no subgroup isomorphic to (Z/2Z,+) for example unless I'm being thick

[pastel cliff]

no that's right

bc (Z,+) has no cyclic subgroups

but he specified finitely generated

[south patrol]

1. all subgroups of (Z,+) are cyclic
2. Z is finitely generated

[pastel cliff]

unless that means something besides what im assuming

[delicate orchid]

sorry, finite finitely generated abelian groups

[south patrol]

Oh sure

[delicate orchid]

you're right

once more I forget infinite groups exist

[pastel cliff]

double

im on a roll

[south patrol]

Sorry lol I shouldn't've put it that way

[pastel cliff]

you're fine

[south patrol]

But yeah remember cyclic groups can be infinite, but then they're always isomorphic to Z

Kind of cute as it means when you work with cyclic groups you can just consider quotients of Z

[pastel cliff]

studying algebra feels like there's so many just like

random facts i should have memorized

[delicate orchid]

not really

this all follows from the structure theorem

[pastel cliff]

the waht

[south patrol]

I'd say yes algebra can feel very discretised into random facts but eventually when the structures begin to click a lot more and you learn the bigger picture stuff like that it becomes easier

[pastel cliff]

ahh

im assuming that just takes time

cuz i can see how everything fits nicely together sometimes

[south patrol]

There's a theorem that (roughly) says every finitely generated Abelian group is isomorphic to a direct product of Z/nZs and Zs

[pastel cliff]

but i still also feel like it takes me a good deal of effort to put things together/remember stuff

[south patrol]

But for example, for a finite abelian groups G say of order N, then we have G isomorphic to a direct product of some Z/nZ where n is a power of a prime and the product of the ns is N

e.g. abelian groups of order 4 are isomorphic to Z/4Z or Z/2Z x Z/2Z

abelian group sorder 6 are all Z/6Z or Z/2Z x Z/3Z, etc etc

[delicate orchid]

"Z/6Z or Z/6Z"

[south patrol]

good point lol that was poitnless and went against what I said above, too sleepy ig

[delicate orchid]

the numbers need to divide the previous numbers

so like

C_2 x C_4 and C_2 x C_2 x C_2 work for order 8

[south patrol]

Eh for the invariant form they do but this is the primary form i was talking about ig

[coral shale]

https://i.imgur.com/ubhye65.png

Is part (a) doable by avoiding Fermat's little (or similar reasoning)? 🤔

I'm guessing not

[south patrol]

it essentially implies Fermat's little theorem so ye you should probably use it

[coral shale]

I was thinking like an inductive approach

but I might end up just proving Fermat

idk, is it the same?

It feels like that avoids it

[south patrol]

is there a reason you want to avoid it, it's a fairly basic fact

[coral shale]

ehhhhhhhhhhhhhh

just running away from number theory

$$f(\alpha + 1) = (\alpha+1)^p-\alpha - 1 + \lambda$$
$$=\alpha^p-\alpha + \lambda+1^p - 1 = 0$$

[coral shale]

yh this works right? But am I just proving fermat

[south patrol]

seems like just proving fermat ye

[coral shale]

never seen a proof like this 🤔

[south patrol]

I mean Fermat is just lagrange's theorem on F_p right so eh

[coral shale]

Fermat by finite induction

[coral shale]

oh it is

[chilly radish]

On F_p^x actually

[south patrol]

Yes lol oops

[coral shale]

https://images-ext-2.discordapp.net/external/wch3y3vHqT_mdMcnCutIWH6Fbz6GodqubWzVS-LZ2vI/https/i.imgur.com/ubhye65.png

For part (b), that ^d should mean composition d times?

apply phi to alpha d times

$\varphi^d(\alpha)=\alpha^{p^d}$

right?

[coral shale]

If so, I seem to have some error. . .

[void cosmos]

yes

[coral shale]

,align \varphi(\alpha)&=\alpha^p=\alpha-\lambda\
\varphi^2(\alpha)&=\varphi(\alpha)-\varphi(\lambda)=\alpha-\lambda-\varphi(\lambda)\
\varphi^3(\alpha)&=\varphi(\alpha)-\varphi(\lambda)-\varphi^2(\lambda)=\alpha-\lambda-\varphi(\lambda)-\varphi^2(\lambda)\
\cdots\
\varphi^d(\alpha)&=\alpha-\sum^{d-1}_{i=0}\varphi^i(\lambda)\

[coral shale]

I reckon this

,align f(\varphi^d(\alpha)) &= \phi\left(\alpha-\sum^{d-1}{i=0}\varphi^i(\lambda)\right)-\left(\alpha-\sum^{d-1}{i=0}\varphi^i(\lambda)\right)+\lambda\
&=\alpha^p+(-1)^p\sum^{d}{i=1}\varphi^i(\lambda)-\alpha+\sum^{d-1}{i=0}\varphi^i(\lambda)+\lambda\
&=-\sum^{d}{i=1}\varphi^i(\lambda)+\sum^{d-1}{i=0}\varphi^i(\lambda)\
&=-\varphi^d(\lambda)+\lambda\
&=-\lambda^{p^d}+\lambda\
&=-\lambda^d+\lambda\

[coral shale]

But that last thing isn't necessarily 0

So must have error somewhere

oh im dumb, that last line is the error

ahhhhhh ive got it

😅

[void cosmos]

good

do you know about modules

specifically

are u good with computing cerrtain stuff

like Hom

im tryuing to ask

something

Hom_R(C,C) where R is matrices of the form (a,b,0,c)

and C is complex numbers

[coral shale]

I'm not, no

,,\mat{a&b\0&c}

[coral shale]

I could have a think, but doubt I would get far.

Can you define Hom_R(C, C) for me?

[coral shale]

All the linear transformations represented by matrices of this form with complex entries?

[void cosmos]

yes

no

C module(vec space)

R-module

think of C as this R-module

and i want the homomorphisms

from C to C

[coral shale]

So we are seeing C as this R-module

,,M=\qty{\mat{a&b\0&c}z:z\in\bC}

[coral shale]

or wait i wrote this dumb

uhhhh so are we viewing C as R^2 space

[void cosmos]

Yes

[coral shale]

So when you multiply by this matrix

z in R^2 really

Sorry if I misunderstood

I'm just not 100% on what this scalar multiplication is

Is it matrix multiplication on the cartesian form of the complex number

,,\mat{a&b\0&c}\mat{x\y}

[coral shale]

tower law

I will write this Q(a, b) in the same order

[Q(a) : Q] is a degree 4 extension

[Q(a, b) : Q(a)] is a degree 2 extension... this is harder to see

You need to figure out what the minimal polynomial of b in Q(a)[t] is

2

[coral shale]

figure this out and then you will get the answer

As for the basis vectors...

It will be products like so

,, a^mb^n

[coral shale]

m = 0, 1, 2, 3
n = 0, 1

@void cosmos sorry so did we have this?

And a, b, c are real entries, or are they complex?

and x, y represents the complex number (real entries)?

confused 😵

no no, surely they have to be real entries

unless I misunderstand this completely

Deciding where the basis vectors map to determine your homomorphism if we can find a basis 🤔. If the scalar multiplication is as above, there's no basis for this one

[tall jay]

If $G$ is an abelian group, how can I argue that the set$$ K = { a \in G: \abs{a} \leq 2 } \leq G? $$

[cloud walrus]

beeswax

[tall jay]

|a|= order of a

[valid basin]

use a subgroup test

e.g. first show the identity exists in K

[lethal dune]

note (ab)²=a²b²

[tall jay]

Since e is in G, then e is in K bc it has order 1

[valid basin]

yeah

also, for any a,b in K, show that the product ab is also in K

and also show that for any a in K, the inverse of a is also in K

[tall jay]

Right, but what I'm a little shaky on is that why does ab have order less or equal to 2?

[valid basin]

you can use this fact

since in an Abelian group, we have that (ab)^2=a^2b^2 for any a,b

this is useful because, since a and b are from the set K, they have order 2 or less

which means a^2=b^2=identity

[tall jay]

meaning a²b²=e?

[valid basin]

it implies that (ab)²=a²b²=e

which means the order of ab must be less than or equal to 2

and from the definition of K, we must have that ab is in K.

[tall jay]

Okay, let me try to repeat to see if I understood

So, if we assume that a,b is in K, then the order of those 2 elements are each less than or equal to 2.

Since G is an abelian, then for a,b in G in K, (ab)²=a²b²=e. This implies ab has order less or equal to 2.

[valid basin]

yeah nice

one small thing to point out - in the second line, we are considering a,b in K (not a,b in G)

[tall jay]

Ah

Okay, I understand it better, but now I am not 100% on

(ab)²=a²b²=e ---> ab has order less or equal to 2

[valid basin]

yup

since when you square it, ab returns to the identity

so ab must have an order of 2 or less

[tall jay]

OH OHHH

Wow literally just the definition of order

[valid basin]

after showing that the product ab is in K, the final step is to show that if we have some a in K, then a^-1 is also in K

[tall jay]

Well if a is in G and has order 2, then a inverse will also be in G argue why a inverse has order 2 then done.

[valid basin]

sounds good to me

[prisma shuttle]

hey guys

can someone explain why we need the power of $n$ in the expression $(\mathbb{Z}/p^i\mathbb{Z})^n$?

[cloud walrus]

JustKeepRunning

[prisma shuttle]

cuz if its just plugging the values into a univariate polynomial why do u need to plug an ordered pair in

[hidden haven]

f isn't univariate though

[prisma shuttle]

oh shoot i didn't realize that

ok thx

[barren sierra]

Is the only difference between direct product and direct sums of modules is that direct ones have strictly finitely many non -zero entries?

Why is one called "product" and the other called "sum"? Just curious about the naming lol

[oblique leaf]

Is it true that $$K^p\subseteq K$$ when K is infinite and of characteristic p? Or is this fact only true for finite fields

[cloud walrus]

alyosha

[lethal dune]

they are same in case of finite sums/products

the difference comes in case of infinite prods and sums

it's called product as it comes from categorical product in R-mod

direct sums are co-product in category, idk why they are called direct sums

[lethal dune]

take Zp(x) and see what happens

[oblique leaf]

sorry what is Zp(x)?

[lethal dune]

well u only cate about subset right?

[oblique leaf]

yeah i want to know if K^p is going to be contained in K

[lethal dune]

field of fractions on Z_p[x]

[oblique leaf]

okok

[lethal dune]

why not, K is a field so closed under multiplication

[oblique leaf]

I don't know, I am trying to figure out why the Frobenius automorphism is not an automorphism for K being infinite

it seems that it is only surjective in the finite case

[lethal dune]

yes

[oblique leaf]

why is that?

[lethal dune]

.

do you know what this is?

[oblique leaf]

not really ..., but essentially you are saying that we can have K^p contained in K AND the map being injective and still not have surjectivity?

[lethal dune]

no that's not what I said at all

I am asking, can you tell me an example of an infinite field of char p

[oblique leaf]

ok wait thats all wrong sorry

[delicate orchid]

pretty sure I've had a discussion about this in here

uhhh

[lethal dune]

[oblique leaf]

ok but K^p is always contained in K regardless of finiteness correct?

since fields should be closed under multiplication

[lethal dune]

like yeah

because K is a field so

obviously

[oblique leaf]

and the map is injective

[delicate orchid]

yeah just taking the infinite direct product of F_p should work

[lethal dune]

any hom from field is injective or 0

[oblique leaf]

true

[lethal dune]

product in fields

[delicate orchid]

like some kinda completion type beat

[oblique leaf]

but so then somehow, this injective map going from K into K^p (that is at most as big as K) is NOT surjective

[lethal dune]

yes

I was trying to give you a counter ex

but looks like you don't know what Z_p(x) js

[oblique leaf]

yeah, i actually only have to prove why it is surjective in the finite case

[lethal dune]

do you know polynomial rings over a field?

[oblique leaf]

but i realized that my proof would work for the infintie case

so it is wrong

[oblique leaf]

yes

[lethal dune]

well since it's finite...

[lethal dune]

do you know field of fractions over an integral domain?

[oblique leaf]

no

[lethal dune]

u divide by non zero elements and create a bigger set

which turns out to be a field

like Z is an ID, divide by non zero integers and you get Q

[oblique leaf]

hm ok

[lethal dune]

same thing u do with Z_p[x] so your elements are kf the form p(x)/q(x) where q ≠0

which gives us a field

of char p (which u can check)

[oblique leaf]

hm ok i see

yeah that makes more sense

[lethal dune]

yea

now can u show that the frobenius map in this field is not surjective?

use irreducibility

[oblique leaf]

x^p-a is irreducible? where a is not in K^p

[lethal dune]

why not just take x

[oblique leaf]

ok lol

[lethal dune]

do you see now?

[oblique leaf]

oh right

x is not a pth power

of anything

correct?

[lethal dune]

yes

so is it surjective?

[oblique leaf]

no

[lethal dune]

[oblique leaf]

lol thank you!

[wooden ember]

What am I doing wrong here: working in $F_2[t]/(t^3)$ i get $(t+t^2)(t+t^2)(t)=(t+t^2)(t^2)=t$ on one hand but on the other hand it is $(t+t^2)(t)=t^2$ since $t+t^2$ is idempotent??

[cloud walrus]

𝓛ittle ℕarwhal ✓

[south patrol]

(t+t^2)^2 = t^2 right

[wooden ember]

t^2 + t^3 + t^3 + t^4 -> t+ t^2 no?

[south patrol]

Where did that t term come from

if t^3 = 0 then t^4 = 0

[wooden ember]

god im a massive fool you're right

[south patrol]

In fact the first term should just be 0 too so yeah

[wooden ember]

ugh i have zero motivation to do this all over again ree

[potent briar]

book I'm reading says

More precisely, a scalar-valued function w on W is a bilinear form if
      w(a1x1 + a2x2, y) = a1w(x1, y) + a2w(x2, y)
and
      w(x, a1y1 + a2y2) = a1w(x, y1) + a2w(x, y2)
identically in the vectors and scalars involved.

my first question is what does "identically in the vectors and scalars involved" mean??

[south patrol]

it just means 'this holds for all possible scalars and vectors' vik

[wooden ember]

ig at least i learned that cyclic multiplication aint good for rings then

[south patrol]

wdym by cyclic multiplication?

[wooden ember]

in the sense that i was basically cycling through the orders like i would in a cyclic group

[south patrol]

oh ok

[wooden ember]

i dont mean anything rigorous by it since i had nilpotent elements anyways

[south patrol]

Sure

[pastel cliff]

any suggestions on online resources to learn about lie algebras? besides reading the wiki

no access to a book rn but if anyone's got suggestions about that i can find pdf's as well

actually i'll ask some better questions

def reads "A Lie algebra is a vector space g over a field F"

why do we say say "over" another thing

[maiden ocean]

I mean that's just the standard term?

[delicate orchid]

cause it's... over the field? lol?

[south patrol]

unless you mean why do we use that terminology for vector spaces in the first place lol

[delicate orchid]

you're applying a set of vectors + scalar multiplication to a field so it's "over" the field I guess?

[south patrol]

but i guess the idea is that the field in some sense forms the basis of the structure lol

[pastel cliff]

yeah more along this line, i get that it's standard i was just asking like what's the idea of using that term specifically

[pastel cliff]

like this

so it's just a statement about what the elements of the vector space look like right

[maiden ocean]

Possessing a lie algebra structure?

[pastel cliff]

being over a field

i know it's a super basic thing to question

[maiden ocean]

It's not really a property so much as it is just that being a vector space isnt defined without reference to a given field?

[delicate orchid]

yeah, it just means the vector space has that field as its set of scalars

[pastel cliff]

[delicate orchid]

I guess it's kind of a short hand for writing a vector space "that acts on via a group action on" a field

[chilly radish]

Humphreys

Lie Algebras and representation theory

is there a reason that Gauss' Lemma is usually stated for UFDs? I'm pretty sure the proof really only uses the fact that irreducibles are prime, so wouldn't it work over a GCD domain?

[tranquil parcel]

It seems it does generalize to a GCD domain, TIL

[pastel cliff]

prof defined SL like this but isn't it usually matrices s.t. det(A) = 1??

and afaik trace = 0 does not imply det = 1

[upper pivot]

so this is isnt SL

this is sl

the lie algebra of SL

[pastel cliff]

[tranquil parcel]

yikes SL \neq sl

that's confusing lol

[upper pivot]

you can show that differentiating the det=1 condition shows trace=0

not really, the point is little sl (usually written in some fancy font like frak) is the lie algebra of SL, so you want to be able to relate these

[pastel cliff]

his handwriting is so ass that i just assumed he overlooked capitalizing lol

duly noted tho

[pastel cliff]

does this imply they still have det(A) = 1 tho?

[upper pivot]

nope

[tranquil parcel]

No. Consider the zero matrix

[upper pivot]

so while SL is a group

sl is a lie algebra, so in particular a vector space

[pastel cliff]

then why do we say "lie algebra of SL"

or am i thinking too hard about that detail

[upper pivot]

right so how much do you know about lie groups/lie algebras

and if not how about like, manifolds

[pastel cliff]

day 1 of trying to learn it lol

[pastel cliff]

the basics but probably/hopefully enough?

[upper pivot]

right so do you know like, what tangent spaces and vector fields are?

[pastel cliff]

i started looking into that recently but got confused with the basis of tangent spaces being the partials

i overthink things easily though

[upper pivot]

i see

ok so the easiest way to say it for now would be that to each lie group G you can associate functorially a lie algebra g

this turns out to be simply T_1 (G) or the space of left-invariant vector field, but dw too much about this for now ig your book will define this

[pastel cliff]

i know im punching a bit above my weight but how is this related to manifolds tho? im doing this for some independent study with a prof and i did think it odd that he introduced the idea of manifolds and then (seemingly) randomly switched to algebra

[upper pivot]

well lie groups are manifolds with group structure right

[pastel cliff]

sure yeah

[upper pivot]

its a class of manifold where a lot of things end up working out nicely

[tranquil parcel]

for example, the circle

[upper pivot]

so ithink its like, a good way to ease into like, diff geo or w/e

[pastel cliff]

:O

[upper pivot]

altho i will say it is probably more useful to see this after you know a bit about tangent spaces and vector fields imo

cause otherwise things like this sl is a bit random

[pastel cliff]

yeah he just gave me this as an exercise and promised more details later

[upper pivot]

i see, yeah thats just confirming the details

dont forget to show that the lie bracket is indeed defined, i.e [A,B] also has trace 0

[pastel cliff]

oop

[prisma ibex]

(it's easy to show this is well defined for 2x2 matrices by a direct computation; what if you're looking at sl_n more generally? what property of the trace can you use to show this is well-defined?)

[tranquil parcel]

I (think) I want to map $F(a,b) \to G_2$ and prove the kernel is the normal closure of $\langle abab^{-1}\rangle$ abusing the first homomorphism theorem. I can determine the map by sending a,b and then just extending it to the unique homomorphism (universal property of the free group). The issue is, I don't know where to send a nor b. My thought is to make the image of $abab^{-1}$ be $c^2d^2$, but I don't think this works.

[cloud walrus]

Michael Harp

[delicate orchid]

funny substitution test

hmm

[pastel cliff]

silly algebra man

im curious about this

silly algebra man = wew lads

[tranquil parcel]

me too man

[prisma ibex]

you use that Tr(AB)=Tr(BA), even if A and B do not commute

[delicate orchid]

I know fuck all about lie groups

[pastel cliff]

"fuck all"

[delicate orchid]

something something sums of eigenvalues

no seriously

I've never worked with them

[prisma ibex]

so then Tr(AB-BA)=Tr(AB)-Tr(BA)=Tr(AB)-Tr(AB)=0

[delicate orchid]

anyway back to seeing if the substitution test works

[prisma ibex]

(btw this does not imply that trace is invariant under permutations, only cyclic permutations)

[delicate orchid]

tfw

[tranquil parcel]

i even assumed abab^{-1} = a^2b^2 and tried to see if i could get a work-backwards solution and left empty handed

[delicate orchid]

ok, so abab^-1 implies that a and b are kinda anti commutative right
like you can swap ab = ba^-1
could do something with this

unless I'm tripping hard

[tranquil parcel]

no that's totally right

a is kinda a conjugate of its inverse

a b-conjugate specifically

[delicate orchid]

yeah yeah good

[delicate orchid]

I'm trying to do this, btw

kind of

[tranquil parcel]

ok wait

[delicate orchid]

oml I've managed to get abab^{-1} = b^{-2}a^{-2} they're actually trolling me
and now I've reduced that to a^{-2}b^2

[tranquil parcel]

Is that is true, then a^4 = 1

if*

[delicate orchid]

what's your reasoning

cause that would be fantastic

[tranquil parcel]

nope never mind

i made an error haha

i just derived the same thing you did yikes

[delicate orchid]

zoinks

lemme try again

I had to do something very similar to this for THREE relators a couple of days ago

pure agony

[tranquil parcel]

yeah that sounds bad

[barren sierra]

I do not understand tensors

[tranquil parcel]

would you have guessed this is a topology class

[delicate orchid]

it was the same conjugate inverse type deal as well

no lmfao what kinda topology class is this

[tranquil parcel]

It's labelled topology 2 but we basically cover algebraic topology fundamentals. like fundamental groups, connected sums and wedge products, the weird unions of two open sets theorem

forgot its name

but it gives you the fundamental group of the union of two spaces under certain conditions

[pastel cliff]

dumb quesetion

matrices with trace 0 summed together still have trace 0 right

[barren sierra]

what is trace again?

[pastel cliff]

yes

[delicate orchid]

yup

[barren sierra]

is that sum of diagonals?

[tranquil parcel]

The trace of an nxn matrix is the sum of the diagonal entries

[delicate orchid]

dIaGoNaL SuM

[barren sierra]

ya then yes

[delicate orchid]

but who uses that lol

[barren sierra]

^

[delicate orchid]

it's the sum of the eigenvalues

[tranquil parcel]

it is?

[barren sierra]

again tho what is the use

[pastel cliff]

i want this to be less annoying

[delicate orchid]

95% of character theory

actually, no

[barren sierra]

ye it's kind of nice

[delicate orchid]

100% of character theory

all of it

[barren sierra]

similar to how determinant = product of eigenvalues

idk what character theory is

[tranquil parcel]

characters? as in the homorphisms from groups into the units of a particular field F characters?

[delicate orchid]

you can completely reconstruct a (complex) representation of a group just from knowing its trace

[barren sierra]

interesting

wait what that's actually really cool

[delicate orchid]

those are... 1 dimensional characters I thinkkk

yeah

linear characters

same thing just in a restricted scope

[tranquil parcel]

Ah, so yo consider affine spaces over fields too

[delicate orchid]

ok linear

is a bad name for them (it's not really)

[barren sierra]

also does anyone have any idea with the tensor question I sent above 💀

[delicate orchid]

these are all derived from linear transformations, it just that the trace a matrix is only a linear function when the matrix is one dimensional

hence, linear characters

[delicate orchid]

wtf is a quotient field

[barren sierra]

field of fractions

[tranquil parcel]

The field of units that can complete an integral domain

[barren sierra]

idk why they call it that either

[delicate orchid]

got it

oh ok these are module quotients

how annoying

[barren sierra]

it's just like the definition of a tensor product is very contrived as far as I can tell

and i have absolutely no idea what elements of this product look like

[tranquil parcel]

You just wanna turn bilinear maps into linear maps

[delicate orchid]

first things I need to understand the structure of Q/R as an R-module

then it'll probably be easy to see

[barren sierra]

so I mean each element in Q/R takes the form a/b + R but after that

¯_(ツ)_/¯

[tranquil parcel]

An example would be R^2 tensor R^3, which would be formed by a standard basis of 2x3 matrices

[delicate orchid]

R^6

[delicate orchid]

can we represent each one of those equivalence classes with just 1/b+R for some b?

[barren sierra]

I don't think so?

wouldn't you need multiplicative inverses

[tranquil parcel]

$\mathbb{Q}/\mathbb{Z}$ is a funky group to think about. It's an infinite order torsion group, with unbounded torsion,

[cloud walrus]

Michael Harp

[delicate orchid]

if it was just a ring you would

but this is a module

[barren sierra]

hm

[tranquil parcel]

oops i see

[delicate orchid]

but thinking about it more I believe you

there must be something we can shove in that R or the quotient would just be Q

[proud bear]

I think you can say like [\frac ab\otimes\frac cd=\frac{ad}{bd}\otimes\frac cd=\frac a{bd}\otimes\frac{c}1=\frac a{bd}\otimes0=0]

[barren sierra]

wait what how

first to second one I get

I think my issue is more I do not understand what tensors are or how they work

and reading over the notes I have is not helping

[delicate orchid]

ok, so we can reduce elements of $Q \otimes_R Q$ from $a/b \otimes_R c/d = ac(1/b \otimes_R 1/d)$ ok coollll
so if we quotient this by R now we get that elements of $(Q \otimes_R Q)/R$ are (1/b \otimes_R 1/d), can we then say that $(Q \otimes_R Q)/R \cong (Q/R \otimes_R Q/R)$?

[cloud walrus]

Wew "Quaternonic" Tbh (201🧊) ✓
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[delicate orchid]

cause if we could that would be rather nice

[delicate orchid]

oh good god that's clever

I like it

I feel like I was trying to do basically that but with the whole space at once

[barren sierra]

ngl I do not get how any of those manipulations work

[delicate orchid]

ok so, a/b ~ ad/bd by definition of field of fractions
move the factor of d on the top outside the tensor (cause da O b = d(a O b))
and then move it back in on the other side (d(a O b) = a O db)

cancel

boom

c/1 is in R

double boom

[barren sierra]

ohhhhhhhhhhhh

[delicate orchid]

kablammo

[barren sierra]

it uses the bilinear map

damn

that's slick

[delicate orchid]

I tried to see something like that but I was missing the ad/bd step

[delicate orchid]

as you can see here

[barren sierra]

jeez yea I have no intuition for tensors

[next obsidian]

This is a much easier computation than what you all did

[barren sierra]

like I kinda understand parts of this but manipulating it is all

[next obsidian]

(a/b (x) c/d) = (ad/bd (x) c/d) = (a/bd (x) dc/d) = (a/bd (x) c) = (a/bd (x) 0) = 0

[barren sierra]

yea that's what @proud bear did

[next obsidian]

Ah

So you don’t even need two Q/R’s

In general, this computation shows that if M is torsion, then Q (x)_R M = 0

And Q/R is all torsion

[delicate orchid]

I just remember 1, 2, 3 tbh

[barren sierra]

how does this use the fact that Q/R is torsion

[delicate orchid]

and of course the funny quotient def

[barren sierra]

I do not understand how 1, 2, 3 are used

in like defining elements

[delicate orchid]

think of a (x) b as just

(a, b) but with some funny properties

[next obsidian]

You used the fact that on the right, c/d is killed if I can multiply it by a d

So what I did was on the left, multiply by a d/d, then moved the top d across the tensor

[barren sierra]

yea

that step used the bilinarity

[prisma shuttle]

can someone explain wut the notation used here means

[chilly ocean]

the integral might be defined elsewhere

but its saying that you are integrating over the ideal pZp to the n’th power where Zp is are p-adic integers and p is a prime

[chilly ocean]

what book is this from

[chilly ocean]

I was reading involutive algebra or *-algebra. Just for confirmation, in that structure, we have a ring R being commutative, where an associative algebra A is present over R. I don't understand the over R part. In associative algebra, we have associative Multiplication amd addition operations forming a Ring; and addition and scalar Multiplication operations forming a Vector space over a field K. Is this over K being replaced by over Ring R in involutive algebra? Or is it something different?

Any hints for proving that if the index of a center of a group is n, the most elements any conjugacy class of that group can have is n elements?

[prisma shuttle]

its from a paper on the igusa zeta function

https://perswww.kuleuven.be/~u0005725/articles/2020/3/Santander.pdf

[oblique leaf]

I have to use field theory to prove that $$\phi_n(x) $$ and $$\phi_{2n}(x)$$ have the same degree. We can only use cyclotomic polynomials, but we cannot use the standard argument that $$\phi(nm)=\phi(n)\phi(m)$$ when $$gcd(n,m)=1$$

[cloud walrus]

alyosha

[steady axle]

IF we have a short exact sequence 0 -> A -> B -> C -> 0 such that A is direct summand of B non canonically, does it imply that B is direct sum of A and C

[hidden haven]

No
0 → 2ℤ → ℤ → ℤ/2ℤ → 0
is not split, even though 2ℤ ≅ ℤ is a summand of ℤ

[lethal dune]

why don't epimorphisms in Ring don't mean onto ring homomorphism?

[hidden haven]

ℤ → ℚ is epi

[lethal dune]

[next obsidian]

There’s a remarkable fact about this summand thingy

If you have any SES of the form
0 -> A -> A (+) B -> B -> 0 where A,B are finitely generated modules over a commutative Noetherian ring, this diagram is split

It doesn’t have to be the canonical maps

[wooden ember]

where am I going wrong here? If $I$ and $J$ are principal prime ideals (not necessarily coprime) then $IJ = I \cap J$. To see this take $I=(x), J=(y), IJ=(xy)$. We want to show $I\cap J \subset IJ$, ie that if $x | a$ and $y | a$, then $xy | a$. Write $a=px=qy$ to see that $y|p$ since $(y)$ is prime, and hence $px=kxy=a$ so that $xy|a$ as required

[cloud walrus]

𝓛ittle ℕarwhal ✓

[hidden haven]

y could divide x instead of p

[wooden ember]

oh right i should have also added that x and y are irreducible

though i guess then only one of them need to be

[hidden haven]

y=x still is a counterexample

[wooden ember]

fair enough

[hidden haven]

Irreducibility of x and y follows from their ideals being prime

[wooden ember]

thanks, i was mainly asking for the specific case of x and y in F[x,y] for which the argument is true then

[hidden haven]

oh wait

we are not in PID

[wooden ember]

yeah not necessarily

[hidden haven]

F

[wooden ember]

so does this coputation work then? I want to compute the nilradical of F[x,y]/(x^2y^3) for F a field

i just compute that rad(x^2y^3) is (xy)

by noting that if x^2y^3 divides P then x and y divides P so xy divides P

and the opposite inclusion is clear

and so the nilradical is (xy) + (x^2y^3)?

[hidden haven]

Sounds legit

[wooden ember]

coolio thanks

[wooden ember]

what does the notation $\mathbb{F}_p[\mathbb{Z}/p\mathbb{Z}]$ mean?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

i have to do stuff with it in an exercise series but i cant find any place where the prof defined it

unless he means a group ring 🤔

yeah nvm im a fool this is a group ring

[devout crow]

what proportion of finite groups are commutative? when listed by order or something

[wooden ember]

id expect it grows far slower than the number of finite groups

for p groups for example you already have a lot more non-abelian than abelian groups

from what i can tell there's a lot of research on the asymptotics of these numbers though

so we can say stuff about which ideals to quotient by to get no zero divisors (prime ideals) or no nilpotency (nilradical), but what about ideals such that the quotient isnt a product?

a related question is what ideals to quotient by such that there's no idempotency, but can we get tighter than this?

[chilly ocean]

Why does this hold..

[chilly radish]

Since M is a proper ideal and R has unity

M being proper means 1 is not in M

[chilly ocean]

Why 1 is not in M@chilly radish

[chilly radish]

Because M is a proper ideal

Maximal ideals are proper

If 1 was in M it would be the whole ring

[devout crow]

1 in M => r=r.1 in M for all r in R, so M=R

[devout crow]

huh i couldn't find anything on first search, what do i need to look for?

[frail perch]

Suppose F(s),F(t) are transcendental extensions of a field F, how can I describe the primes of the tensor product F(s) x F(t) over F?

[chilly ocean]

Pls help

[wooden ember]

this could be a good starting point https://mathoverflow.net/questions/21265/how-many-groups-of-size-at-most-n-are-there-what-is-the-asymptotic-growth-rate

the asymptotics for abelian groups should be a lot more precise given we have the fundamental theorem of finitely generated abelian groups to help us count them

[chilly radish]

For abelian groups this just boils down to the number of nontrivial factorisations of a number

[celest mantle]

If A is a cohomogically trivial G-module, and N is a torsion free G-module, how would i go about proving that Hom(N,A) is also cohomogically trivial pls ? I know it's a torsion free G-module, maybe finding a short exact sequence where it fits with 2 others cohomogically trivial G-modules would be good perhaps, idk

[lavish nexus]

Oh never mind I see why 5s after taking this picture

[chilly radish]

I need help understanding what they even mean by c being an identity, considering c is just an element of the algebra of nxn generic matrices, not a polynomial in noncommuting variables. I assume they mean the associated central polynomial, in which case I don't understand why it's so easy to see that this is an identity for M_n-1. So basically, we have a polynomial in noncommuting variables, such that for every evaluation in nxn matrices, you get a scalar matrix, and this is supposed to imply that for every evaluation in n-1xn-1 matrices, you get 0. Any ideas?

[terse crystal]

What book is it from?

[chilly radish]

It's a paper, quasi-identities on matrices and the Cayley-Hamilton polynomial

[terse crystal]

Thank you

[chilly radish]

Np

[tribal moss]

Is it implicit here that the polynomials have no constant term? Otherwise, say, c=2 would seem to be a counterexample.

[chilly radish]

if you mean in the theorem, it is explicit in the statement of theorem 2.2

it

it's still weird to me that they refer to elements of Z_n as if they were polynomials themselves

I think something is going on here that i'm missing, maybe this identification is known to be bijective so it's okay to make

[tribal moss]

Oh, right.

[echo orbit]

I have a free group on three generators a, b, c.

I am trying to take the quotient of this group by the normalizer of the cyclic subgroup <a^2bcb^-1c^-1>.

I need to find the presentation of this quotient group. I'm having a little trouble with this.

[tribal moss]

Evaluate the polynomial on a set of block matrices of the form $\begin{bmatrix}X_i\&0_{1\times 1}\end{bmatrix}$ where $X_i$ are arbitrary $(n-1)\times(n-1)$ matrices. This produces $\begin{bmatrix}c(X_1,\ldots,X_k)\&0_{1\times 1}\end{bmatrix}$, but the only way for that to be a scalar is if it vanishes completely.

[cloud walrus]

Troposphere

[chilly radish]

oh

clever

thanks!

[chilly ocean]

Suppose A is a Dedekind domain and B is a domain containing A that is finitely generated as a module over A.
I need to prove that B is also a Dedekind domain, so far I've been able to prove that it's noetherian and that every non-zero prime ideal is maximal, but I can't seem to be able to use the fact that A is integrally closed to prove that B is.
Can anybody give me a hint?

use that B is finitely generated over A

[next obsidian]

It isn’t true @chilly ocean

[tribal moss]

Is that normalizer even a normal subgroup?

[next obsidian]

Look at Z[sqrt(5)]

[chilly ocean]

oh, thank you. It was in my professor's notes so I didn't think about finding a counterexample.

[next obsidian]

So the like

Using Krull-Akizuki you can say something which I assume is what they wanted to say

Which is that if L is a finite extension of Frac(A)

Then the integral closure of A in L is a Dedekind domain

You show that the integral closure is a dim 1 Noetherian ring using Krull-Akizuki

And then being integrally closed is free because the thing is the integral closure of A

[echo orbit]

the normalizer of it is.

[tribal moss]

How do you know? Normalizers are not automatically normal.

In fact, I think the normalizer in the case is the same as the subgroup itself, which is definitely not normal.

Perhaps the question should have said "normal closure" instead of "normalizer"?

[tranquil parcel]

I have tried so far to find a map $F(a,b) \to G_2$ with kernel equal to the normal closure of the first presentation's relation. I don't even know where to send $a,b$. I do know this uniquely determines a homomorphism, which we can then use the first isomorphism theorem to give us $G_1 \cong F(a,b) / \ker(\varphi) \cong G_2$.

[cloud walrus]

Michael Harp

[tranquil parcel]

In reference to this image.

[indigo ravine]

You can try a bunch of combination of c and d for phi(a) and phi(b) until you find one such that phi(a)*phi(b)*phi(a)*phi(b)^(-1) = ccdd .
I think || a -> cd b -> dc || works ?

[barren sierra]

normalizer of <(1 2 3 4)> in S_4 is itself right?

or no elements like (1 2) are in the normalizer right? as is (1 2 3)?

wait nvm no it doesn't

so it's just itself

[tranquil parcel]

Yeah any two 4-cycles are conjugate

specifically they are conjugated by the permutation of the entries in the cycle

[barren sierra]

yea

[chilly ocean]

How can I figure out if the quotient group formed by the center of a group is cyclic or not?

I'm not sure how I to use cosets to figure out whether I can generate all cosets from a single coset

[woven delta]

I wasn't able to show the one you posted worked, but I was able to show that the map ||(a,b)\mapsto (cd,d)|| works

[lavish nexus]

I've shown R((x)) = S^(-1)R[[x]], S = {x^n|n in N}
how do I show R((x)) = K^(-1)R[[x]], K = {power series with no constant terms}?

or maybe there's another way to show if R is a field then R((x)) is a field

[woven delta]

Can you state this as formal Laurent series with a finite tail (eventually 0 in negative coefficients)

Anyway you can give an algorithm for division in the ring of formal finite tailed Laurent series. Let $f=a_0x^n+...+a_kx^k+...$ be a formal finite tailed Laurent series with $a_0\neq 0$. We want to show that there is some corresponding g with $g=b_0+...+b_kx^k+...$ with $fg=1$

[cloud walrus]

Emma

[woven delta]

This gives us an infinite sequence of equations we can solve inductively to find g

@lavish nexus

[cloud walrus]

Emma

[woven delta]

The equations we have are given by the fact that $fg=1$ so the 0 degree coefficients of $fg$ is 1, and all of the other degree coefficients are 0

[cloud walrus]

Emma

[woven delta]

And then you can recursively find the coefficients of g by starting from the bottom and solving each equation individually, then plugging your solutions into the next equation

[woven delta]

(I edited above to assume that f is a power series with a_0 not 0, since we can divide or multiply by x^n to get every other element)

So this is just applying the fact that power series with nonzero constant term are units

If you already know that you can skip the above

[lavish nexus]

Yeah I know that algorithm for finite termed Laurent series but not for something with infinite negative power terms

[barren sierra]

I'm pretty stuck on (b)

So my understanding is that S is this subring of polynomials such that the terms without any y's also have no x's (i.e. are just constants from the underlying field F)

from that I'm not sure how to leverage that into proving (b)

[woven delta]

That ring is finite termed Laurent series

That's what adding in x^-n for each n does

Oof I should have asked before I gave exposition

Ah well

[carmine fossil]

So f(x,y) being in the ideal means each f_i is a combination of {1,x,x^2,....x^{k-1}} except for f_0 which has to be 0

So first cancel f_n(x)

You know {1,y,y^2...y^n} are all elements in S

[barren sierra]

how can you cancel f_n(x)?

[woven delta]

Anyway you can't even have a ring of formal power series with infinitely many negative power terms, multiplication isn't well defined

[carmine fossil]

So using this you consider {y^n,xy^n,x^2y^n...}

Now some linear combination of this has to cancel f_n otherwise you can't reduce f_n to 0(and hence the whole thing to 0)

[woven delta]

Not without a notion of convergence, and even then you need more then that I think

[carmine fossil]

Because only x^k y^m terms can cancel x^k y^m terms by subtraction

[barren sierra]

oh you just mean cancel by subtracting

[carmine fossil]

Yea

[barren sierra]

I thought you mean multiplicatively and I was so lost

ok

[woven delta]

The reason why products of formal power series work is there is only a finite sum of coefficients for each degree, in the infinitely many in the negative case you have an infinite (indexed by Z) sum for each coefficient

[regal carbon]

Hi, I'm very much afraid of abstract algebra. I've completed learning proof of contradiction, Contra positive and direct proofs but I haven't completed proof of induction. Still I am very very much afraid I don't know from where to start. I am very much of just about prerequisites such as mathematical induction, functions sets and relations I think I am overlearning. I am really struggling with the functions theorems. I am unable to understand really feeling discouraged. Currently I am reading "mathematical proofs: a transition to advanced math" textbook, it is very big that I am unable to complete it. I think I have completed 4 chapters from it.

Again it is very much discouraging for me for not understanding functions theorems.

Experience nearly 3 months I have completed the proof techniques. I completely forgot I think I have to recap it now! Should I directly jump into the actual subject ?

[tranquil parcel]

Well, to learn abstract algebra, you would want some basic set theory and proof techniques (precisely all the ones you listed above - including induction). But after that you should be totally ready. Honestly, the only thing you need to know about functions before you start abstract algebra is that they have inputs from a domain and outputs in a codomain, cardinality, what surjectivity and injectivity are (or what onto and one-to-one are, respectively), bijectivity (onto and one-to-one), and pre-images. But even these are all usually covered in an introductory algebra text.

[regal carbon]

Oh really?

[tranquil parcel]

You sound mostly ready though if you've been working on proofs in another subject. The main emphasis is that you are comfortable with proof structure.

[regal carbon]

This is the book I've read

I've practiced sir

[barren sierra]

I'm lost as to why we are looking at the n-th term

[carmine fossil]

Well, There's actually no reason actually when I think about it

You can just cancel f_1(x) directly

[barren sierra]

?

[carmine fossil]

Consider {y,xy,x^2y,...x^{k-1}y}

Some linear combination of this will have to cancel f_1(x)y

[barren sierra]

oh basically show that if deg f_1 >= k then there is no additive inverse in S?

[carmine fossil]

Well,yea

[sly crescent]

Are HNN extensions commutative?

[hidden haven]

If you start with a non abelian group, it will embed into its extension so the extension can't be abelian

[broken stirrup]

Hi, how do you prove A_1 and A_2 are isomorphic?

[sly crescent]

Not what I meant

If you reverse the order of the associated subgroups, do you always get an isomorphic group?

[next obsidian]

You can break up both of them as Z_p^2 (+) Z/pZ in different ways

[broken stirrup]

Can i map the generators to generators?

Let's say there is a mapping which sends (p,0,0) to (p,0,1) and (0,0,1) to (0,p,0)

would they be isomorphic?

[broken stirrup]

this another one that i got stuck

I know that it's universal property and I'm free to choose A whatever I want but got nothing so far

[terse crystal]

Suppose there are a_j such that Σa_j l(x_j)=0 .Let f:X—>R^n. Mapping x_j to (0,…,0,1,0,…,0) (the j-th element is 1), mapping X \ {x_1,..,x_n} to zero

[broken stirrup]

oh so (r_1,...,r_n)=(0,...,0) thus r_i=0

brilliant

thanks

I took A as a free R-module but didn't consider it as direct sum of copies

[trim grove]

Someone Can give an example of Infinite Field With Finite Characteristic (not zero).

[delicate bloom]

rational polynomials with coefficients in a field of finite characteristic, F(x)

[trim grove]

What is they way of representing this?

Any special notation?

[tribal moss]

Usually just the same notation as rational functions over the reals, e.g, (x³+1)/(x²+x+1)

[trim grove]

$\mathbb{F}_{p}(x)$ ?

[cloud walrus]

Solution

[tribal moss]

Ah yes, the name for the entire field. Yes, that's it.

[heavy dagger]

Are the following statements true-

1. When |alpha| = 1 and alpha has an argument that doesn't divide 2pi, Z[alpha] is dense in a wierd subset of C whereas if the argument divides 2pi it's dense only in Z[alpha]
2. When alpha is an algebraic complex number, Z[alpha] is bijective to the set of equivalence classes (of polynomials with integer coeffs) based on the value they take at alpha
   ?

[tribal moss]

(1) No -- for example if alpha is a primitive 5th root of unity, then Z[alpha] is still dense in C.

[heavy dagger]

Any hints on how I can go about showing that?

[tribal moss]

Still for the 5th root of unity, alpha + alpha^4 is a real that is smaller than 1; therefore Z[alpha] contains arbitrarily small positive reals, and so its intersection with R is dense in R.

[heavy dagger]

Wait 2cos(pi/5) is greater than 1 right?

[tribal moss]

In fact assuming |alpha|=1, the only way for Z[alpha] not to be dense in C is if alpha^4=1 or alpha^6=1 (in which case it will be a lattice). Any other divisor will lead to a way to make numbers with modulus strictly between 0 and 1, and then you inevitably get a dense set.

The sum is 2cos(2pi/5) = 0.618.

[heavy dagger]

Oh right

[tribal moss]

(2) looks right, though.

[heavy dagger]

Why is this statement true?

As in the any other divisor statement

[tribal moss]

Divisors 1,2,3,4,6 give nice lattices according to my claim.

5 we've just handled.

And if the divisor is greater than 6, then |alpha-1|<1.

[heavy dagger]

Oh right got it

Thanks mate, this stuff is really cool

[tribal moss]

~~In the case where the argument of alpha is not a rational multiple of pi, then alpha is transcendental, and the evaluation map Z[X]->Z[alpha] is an isomophism. This doesn't seem to be immediately obvious (someone prove me wrong, please!) but follows from the https://en.wikipedia.org/wiki/Gelfond–Schneider_theorem~~
Oops, I missed one of the assumptions of the theorem.

[trim grove]

Upto isomorphism how may algebrically closed finite fields are there? please explain.

[tribal moss]

Zero.

[trim grove]

can you please help me with this, how can i show this?

[tribal moss]

If the elements of the field are a1,a2,...,an, then consider the polynomial (x-a1)(x-a2)···(x-an)+1.

[trim grove]

how this will help me to solve , can you please elaborate ?

[tribal moss]

It will show you that a finite field cannot be algebraically closed.

[trim grove]

Thankyou

[coral shale]

did not know this

[delicate bloom]

it's clever cause you immediately think "ahh I could have thought of that cause I know euclid's proof of infinitely many primes!"

[chilly ocean]

Mero

[broken stirrup]

so I already showed that {f_1,f_2} generates R, but how are they linearly independent?

[tribal moss]

The multiplication i R is function composition?

[chilly radish]

Anyone know what the reduced trace might refer to?

[tribal moss]

Assume f1r+ f2s = 0 for r,s in R, and prove r(en) and s(en) must both be 0.

(This assumes we're viewing R as a right R-module. The claim doesn't seem to be true for the corresponding left module).

[broken stirrup]

yes sir

[broken stirrup]

oh yes i was doing multiplication from the left and got nothing

and from right, it's easy to see that they are independent

maybe there's a problem in the statement of the problem

[tribal moss]

I'd say it does. The left module cannot possibly have two independent elements.

(At least not when K is a field. Module endomorphisms are too tricky for me to commit to the claim in the general case...)

Wait, that's not even true in the field case. It's just the particular f1 and f2 that don't work.

[broken stirrup]

i just assumed it's right R-module

thank you

[pastel cliff]

is it accurate to say that a normal subgroup of G is just one that is like

closed under conjugation

ik the def of a normal subgroup but that wording isnt used in my book i just wanna make sure it doesnt miss anything/is wrong

[coral shale]

forall g, gN = Ng

forall g, gNg^-1 = N

yeah sounds ok to me.

You can say conjugation by any element in G acts on a normal subgroup I think?

Just remember N has to be a subgroup in addition to this condition

[pastel cliff]

yeah i mean it's w regard to normal subgroups so

is there any point to talking about normal groups that arent subgroups

chmonkey said sumn about this the other day

[coral shale]

Another way of phrasing it is

A subgroup N is a normal iff all inner automorphisms of G act on it
I think? If I haven't remembered my terminology wrong

[pastel cliff]

that "normalness" is a statement about how H lives inside G

[coral shale]

And we have a special type of normal subgroup

characteristic subgroups

if all automorphisms of G

act on it

N subgroup G is normal iff Inn(G) acts on N

N subgroup G is characteristic iff Aut(G) acts on N

Think we have this

[broken stirrup]

is direct sum of two finitely generated modules finitely generated?

[terse crystal]

yeah

By definition

[pastel cliff]

moldilocks

[hidden haven]

Try proving that if N is a submodule of M, then
M is noetherian iff N and M/N are both noetherian

The result you posted then follows from induction

[broken stirrup]

I believe I'm supposed to prove the fact by following instructions and we haven't covered M is noetherian iff N and M/N are both noetherian

[pastel cliff]

moldi why should i care about normal subgroups

i know what they are and that "i can quotient them" but like

why

[hidden haven]

They don't really give much lol those are the usual definitions. The reason you'd come up with the lemma I stated is that induction is the natural thing to try here, and if you try using induction then you should see why you need what I stated

[hidden haven]

They the kernels

Why should you care about subgroups

[pastel cliff]

why do i care about any of this

[hidden haven]

Just one more week bro

Quotienting is important because you can study maps that "behave the same way" on some part of the group

If you wanna study collections of maps that behave the same way, you start by studying collections of maps that behave the trivial way on a part of the group

That's what quotienting allows you to do

Normal subgroups are just parts of the group on which maps can act trivially

[pastel cliff]

hmmm

but this refers to maps from the group to itself right?

also the fact that cosets of a normal subgroup forma another group seems important?

ive talked about this before but algebra feels like so many just

[hidden haven]

In the sense that if you ask for maps that act trivially on a subset S of the group, that's the same as asking for maps that act trivially on the normal subgroup generated by S

[pastel cliff]

facts

[hidden haven]

I don't get it

My messages are delayed I think

[pastel cliff]

they def are

[hidden haven]

[hidden haven]

Yes

Did you read the first iso thread

[pastel cliff]

not in it's entirety

[hidden haven]

[pastel cliff]

[hidden haven]

T!cat

[pastel cliff]

honestly same

[pastel cliff]

oh wait a group cant have two different normal subgroups

them shits are isomorphic

interesting

[next obsidian]

What do you mean

[pastel cliff]

uhhhh

normal subgroups of thte same group are isomorphic

or am i mistaken about that

[next obsidian]

No

G itself is always normal

And {e} is always normal

As an example

[terse crystal]

R is a noetherian ring therefore a noetherian module . so R^n is a noetherian module .And any submodule of a noetherian module is finitely generated (any submodule M of a noetherian module T, let A be the set of all finitely generated submodules of M, since T is noetherian there exists a maximal one in A, that maximum must be M, otherwise you can construct a bigger one

[pastel cliff]

oh wait nvm, conjugate subgroups are isomorphic

wait but

confusion

normal subgroups are conjugate subgroups arent they

[pastel cliff]

also the kernel changes depending on the map?

[next obsidian]

No

Normal means that it is conjugate only to itself

[pastel cliff]

im getting mixed up then my b

[cursive temple]

butting in, gowers has a nice blogpost about normal subgroups with some nice motivation https://gowers.wordpress.com/2011/11/20/normal-subgroups-and-quotient-groups/

[lavish nexus]

Do we get 6 dimensions?

oh wait actually no

seems to be 4

2 from W + 0 -> 0 + W and 0 + W -> W + 0

another 2 from E_k + E_k' = W + W -> W + W

[next obsidian]

lol

xD

[hidden haven]

That sounds a lot like "prove that A and B are isomorphic" "A is isomorphic, and B is isomorphic, therefore both A and B are isomorphic"

[vestal snow]

Sanity check, the kernel of a map can be expressed as the colimit indexed by the empty category right?

[chilly ocean]

Let $(G,\cdot)$ be a group and $H$ be a subset of $G$. Does anyone know if there is a name for the smallest nonnegative ineger $n$ (if it exists) such that ${a_1\cdot a_2\cdot \cdots \cdot a_n: a_1,a_2,\ldots, a_n\in H}=G$?

[coral shale]

Can we define \ominus by quotienting to reverse \oplus?

A\oplus B = C
C \ominus B = A

=========

For groups, vector spaces, etc.

[woven delta]

Not if you want nice properties under isomorphism

But also we just use the quotient symbol for that

I don't see why using a different symbol would matter

[dreamy fiber]

For a more I guess philosophical standpoint, subtractions are addition of the additive inverses, you can’t really find a nice additive inverse with respect to direct sum

[coral shale]

I feel the same about oplus though

dont we use that in place of cartesian product for the same reason

[woven delta]

No

oplus is more than a Cartesian product

A Cartesian product is only defined for sets

[coral shale]

i mean direct product

[woven delta]

Oh

[coral shale]

but i suppose theyre not the same

[woven delta]

You haven't seen infinite direct products

[coral shale]

only finite combinations with oplus

But for the finite case they are synonymous right

[woven delta]

Yes

With infinite they can even have different cardinality

[coral shale]

which makes me feel... idk \ominus would make sense

[woven delta]

It's not too hard to understand

Oh so the problem with Ominus

[coral shale]

direct product vs quotient
direct sum vs 'direct minus'

[woven delta]

So oplus is defined for all isomorphism classes of groups

Your ominus is heavily dependent on that B is a subgroup (substructure) of A

So you don't really have an invariant definition of ominus for isomorphism classes of groups

[coral shale]

no, but the same could be said of the quotient operation on groups

[woven delta]

Yes

But oplus is defined on isomorphism classes of groups

And we would want ominus to be as well I think

[coral shale]

Just because how \ooperation is usually used?

[woven delta]

Yes

[coral shale]

👌

[woven delta]

I guess it's not just that though

This is also only defined when you have that B is a normal subgroup of A, and so the domain of this operation would be Groups in the first coordinate, and normal subgroup of the first coordinate in the second coordinate

and also the quotient analogy for the operation makes a lot of sense

Much more than the minus analogy

For minus I would expect set minus

[coral shale]

This quotient is defined for more general structures than groups right? ik it works for groups, vector spaces, havent thought further

yh nvm it is

since it is the same as direct product for finite case

Oh wait, ig infinite case matters too

I can 'see' the quotient should naturally exist I think

[woven delta]

For for quotient the basic idea is of collapsing equivalence classes to points

[coral shale]

You map the coordinates you want to quotient over to 0

and the others are fixed

[woven delta]

Which is what division is

[coral shale]

If a module has a basis, then does a submodule also necessarily have one?

[woven delta]

You're asking if submodules of free modules are free?

[coral shale]

I think so

I haven't studied them much beyond the definition

[woven delta]

Well then you can do the ring Z/(6) and the ideal (2)/(6) (={0,2,4}) inside Z/(6)

So the ideal generated by 2

[coral shale]

huh but neither are free right?

[woven delta]

Well Z/(6) is free over Z/(6)

Which is where I'm working

[coral shale]

oh I see

[woven delta]

So yeah in the finite case any proper ideal is not going to be free

And being free is a very rare property

(among modules)

In the vector space case everything is free

[coral shale]

ah ok, found this

[woven delta]

Yes

That's a good definition of PID

[coral shale]

PID is quite a strong requirement

[woven delta]

Lol

It's equivalent

Yeah

Which makes sense