groups-rings-fields

[rustic crown]

lol

[desert dome]

I am reading this note. Does anyone know why "its minimal polynomial over K(β) divides it minimal polynomial over K"? I don't quite get it..

[rustic crown]

if f in K[x] is the minimal polynomial of alpha over K, then it's also a polynomial in K(beta)[x].
and minimal polynomial over K(beta) has to divide any polynomial in K(beta)[x] which alpha satisfies!

[desert dome]

right! I see. thank you!

[lethal dune]

i used to run my old laptop with no battery inside

[rustic crown]

it's 100% now

[upbeat juniper]

there's an example I'm having some trouble with

so recall a field extension $L/K$ is separable iff $\overline{K}\otimes_K L$ has no nilpotent elements. in this case, I'll set $K=\mathbb{F}_2(t)$ and $L=\mathbb{F}_2(\sqrt{t})$ where $t$ is transcendental

[cloud walrus]

wy

[upbeat juniper]

so $\overline{K}\otimes_K L$ is isomorphic to $\overline{\mathbb{F}_2(t)}[x]/(x-t)^2$, which has the nilpotent element $\overline{x-t}$

[cloud walrus]

wy

[upbeat juniper]

I wanted to figure out how this element looked like in $\overline{K}\otimes_K L$

[cloud walrus]

wy

[upbeat juniper]

but passing $\overline{x-t}$ through the isomorphism, I get $1\otimes\sqrt{t},+, t\otimes1 = 1\otimes(\sqrt{t}+t)$ which doesn't seem to be nilpotent. so I'm confused

[cloud walrus]

wy

[hidden haven]

The minimal polynomial of sqrt t over F_2[t] would be x^2 - t, not (x-t)^2

It will factor as (x-sqrt t)^2

And then it is clearly nilpotent

in the tensor

[upbeat juniper]

right, that was a dumb mistake 😳

[upbeat juniper]

hmm so the nilpotent element is $\overline{x-\sqrt{t}}$, and passing through the isomorphism now gives $1\otimes\sqrt{t},+,\sqrt{t}\otimes 1$

[cloud walrus]

wy

[upbeat juniper]

that makes a lot more sense now

[hidden haven]

minus

I think

[upbeat juniper]

it's char 2 so isn't it the same

[hidden haven]

oh right

[upbeat juniper]

because the tensor is over F_2(t), you can't bring over one of the sqrt(t)'s to the other side

[whole basalt]

So this is probably not the usual question that gets asked in here, but would anybody be willing to share their class's syllabus? 😛 Trying to look at syllabi to figure out how to pace my own course I'm teaching starting in February

This is for undergrad intro to abstract algebra. We're going to use Judson's book

[upbeat juniper]

I tried a similar idea for $\mathbb{Q}(\sqrt{2})\otimes_\mathbb{Q} \mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{2})\times\mathbb{Q}(\sqrt{2})$, and found that the non-unit $(0,1)$ is mapped to $-\frac{1}{2\sqrt{2}}\otimes\sqrt{2} + \frac12\otimes 1 \thonk \catthink$

[upbeat juniper]

$\thonk$

[cloud walrus]

wy

[upbeat juniper]

cool

[cloud walrus]

wy

[upbeat juniper]

there

[hidden haven]

Magic

[broken stirrup]

any tip guys?

I don't even know where to start

[hot lake]

pick any nonzero matrix A and show that 1 is in RAR the ideal generated by A ?

[broken stirrup]

oh so I is the whole ring itself

[hot lake]

well you have to show that the only ideals are {0} and R

so you show that if I is not {0} then it is R

that's the same as showing that for any nonzero A, 1 is in RAR the ideal generated by A

[broken stirrup]

thanks its clear

[hot lake]

maybe show that 1 0 // 0 0 and 0 0 // 0 1 are in RAR separately

actually RAR is kinda a bad notation

maybe I should have just written <A>

because strictly speaking, if RAR = {r1 A r2 | r1,r2 in R} then that may not be closed by addition

but the ideal generated by A is

[next obsidian]

You know, under Bourbaki’s definitions this would be a field I think

Since they explicitly distinguish between a field and a commutative field

[hot lake]

[hidden haven]

You'll thank him when you don't have to say "A vector spaces over a field or a division ring always has a basis"

[rustic crown]

why stop at division rings then, go up to semisimplicity

if a module is sum of simples, then it's direct sum of simples

[hidden haven]

Semisimple rings are fields

[next obsidian]

[broken stirrup]

I got stuck a bit. What if an ideal contains an element 0 0 // 0 a for a non-zero a, how do you get that it contains 1 0 // 00 by matrix operations?

got it

[rose axle]

I despise ring theory

[next obsidian]

:(

[chilly ocean]

:(

[next obsidian]

I revise my stance to be >:( as well

[wicked zephyr]

Let $I$ be a two-sided non zero ideal of $M_n(R)$ (where $R$ is any ring with unity). Then prove that a matrix $A$ is in $I$ if and only if the matrices obtained by placing any entry of $A$ in any position, and 0 elsewhere, belong to $J$. Using this prove that if $J$ is a non-zero two sided ideal of the ring $M_n(R)$ (again, $R$ is any ring with unity) then the set $K$ of (1,1) entries of matrices in $J$ is a two-sided ideal of $R$.

If $D$ is simple, you can use the last statement to prove that $M_n(D)$ is simple.

[cloud walrus]

Potitov06

[pastel cliff]

help me understand why i should care about rings

probably an anger inducing question but i have yet to really understand the use of restricting operations and saying this is a ring

shit, why do i care about any math who am i

[hidden haven]

Cry more

[pastel cliff]

getting abstract in #groups-rings-fields

[hidden haven]

Det gave some reasons right after that message lol

But also a very realistic reason for caring about rings is that you will need them later

[pastel cliff]

thank you generic professor

ok then about that example they gave tho

[viscid pewter]

it's a surprise tool

the use of restricting operations
lots of things might have an addition and a multiplication but not everything else

[pastel cliff]

ok well first is Z[i] the integers mod i ? i feel like ive seen that notation mean something else but i dont recall

also integers mod i feels... weird

[hidden haven]

It is the lattice of integer points in ℂ

All the grid points

It is the smallest subring of ℂ that contains ℤ and i

I mean in the sense that most of the time if you see some elementary application of the theory you could feel like yeah I could have just used integers here why do I need to develop ring theory just to do some integer stuff. But then there are a lot more places which will really tell you how powerful the general theory is but then you can't get there without doing ring theory first

[hidden haven]

Was continuing from here

I felt this way about modules for a long time and whenever someone would give me some bs I'll eat it but still feel hungry

Until I did actual stuff with modules

[pastel cliff]

that's exactly how im feeling rn

like i feel like i just memorized the definition of a ring and can do nothing with it

[hidden haven]

And now if someone asks me why they are useful I can confidently say that they are useful in developing homological algebra

[pastel cliff]

ik that'll change later w my actual class in algebra but me impatient

[hidden haven]

If they ask me why that's useful I'll eventually loop back to saying to develop module theory so it becomes a self sustaining loop 😌

[hidden haven]

Number theory and polynomial factorisation problems etc is the simplest motivation I suppose

[pastel cliff]

that's what im trying rn

[hidden haven]

Factorise x² - 4, but you are not allowed to use any ring theory

[wicked zephyr]

Yeah Fermat's Last Problem is a motivation.

[hidden haven]

Not so easy now is it

[pastel cliff]

you mean like none of the operations/properties associate with a ring?

oh wait that was probably a joke

[hidden haven]

Ye lmao

[pastel cliff]

woooooooooosh

[hidden haven]

You need an optimistic person like det to answer a question like this why did you even ask me 🙈

[pastel cliff]

im stuck in a loop of try to do problems -> get bored -> get upset at myself for being lazy and not persisting -> try to do problems

with random "ask math discord for motivation" sprinkled in at all parts

[hidden haven]

Oh yeah if you have just seen the definition then you're at the most boring part because you'll be solving the shittiest problems

It will improve significantly in a week or two

[wicked zephyr]

Maybe the answer is not that great. For example, why would you care about studying group theory? Well, the real answer is why not? If they are natural and appear in maths, why should we resist in developing group theory? The same happens with ring, modules and etc. Usually they come up when mathematicians tried to solve other problems. For example, as Moldilocks said, many of algebraic structures are used in number theory and algebraic geometry.

[pastel cliff]

these are the problems in my book

[hidden haven]

🤮

Necessary grind

[pastel cliff]

[hidden haven]

It's like GTA before they introduce guns

Just give it a couple weeks

[pastel cliff]

im one of those people who drives normally in GTA every once in a while just cuz

[hidden haven]

🤝

[pastel cliff]

but yeah back to school in 2 weeks thankfully

but it's online so

[wicked zephyr]

What book are you using?

[pastel cliff]

Algebra, by Theodore Shifrin

only bc i cant stand pdf's and this is what i was able to steal from my math department

[hidden haven]

Never even heard that name

[pastel cliff]

ik it's random but it's what i have

i was talking about this before but it's apparently a weird book

starts w rings

[wicked zephyr]

I think Aluffi's, Algebra: Chapter 0 is the best book for learning algebra. He writes beautifully.

[pastel cliff]

groups are chapter 6

moldilocks = namington

[wicked zephyr]

It is common to think that group theory comes before ring theory. Lately, many profesors I know are inclined in teaching ring theory first because it is more "natural".

[hidden haven]

Please don't compare me to a discord m*d

[pastel cliff]

is this valid

[wicked zephyr]

Yeah, it is valid, but long.

[pastel cliff]

im making my solutions purposefully verbose so when my monkey brain looks back at them it gets the help it needs

[wicked zephyr]

ahh ok

Then it is perfect.

[pastel cliff]

yeet

is there a way to do it without Bezout

i think think det said there was but me no remember

[chilly ocean]

also the implications can all basically be reversed so it's kinda obvious but did you do the reverse direction too?

[wicked zephyr]

I don't think so

[chilly ocean]

the task looks like it was to prove if and only if

[pastel cliff]

yeah just need to type it up

[chilly ocean]

nice

[coarse forge]

also idk whether using the "there exist" symbol is good practice

Usually I'd prefer to write "there exists some b \in Z_m" instead

[pastel cliff]

oh nah im always gonna use $\exists$

[cloud walrus]

nitezba

[coarse forge]

Hmm as long as your professor is happy with it. Our professors always told us to avoid it for clarity I guess

[pastel cliff]

im just typing it up for myself

that's definitely a professor specific thing tho

[coarse forge]

Yeah. As long as yours is okay with it 🙂

[chilly ocean]

yeah it's standard for math literature that everything, even equations etc are part of sentences and symbols should be replaced by English (or whatever language) whenever possible so I always did that writing up homeworks

but tossing symbols down is great for scrawling in your notes cause it's fast af

[pastel cliff]

i just like shortening my already long stuff as much as possible

fuck the reader 💯

[chilly ocean]

lul

[coarse forge]

yeah my professor's lecture scribbles always use those symbols and they're like "ok but in a proof, don't do that pls"

[chilly ocean]

hahaha

[long obsidian]

Is it accurate to say that a subset H of a group G is a subgroup iff each element in H is left/ right congruent modulo H with each other element in H?

[uncut ridge]

could you define left/right congruent modulo H?

[chilly ocean]

I would guess a = bh1 and a = (h2)b respectively, h1, h2, in H

Actually either gives you subgroup criterion right? And the other direction is obvious from closure

[acoustic pine]

I'm a bit confused on something from Eisenbud. In chapter one, they vaguely introduce the notion of the Hilbert function, and then say simply that it agrees with some polynomial for large s, but doesn't really explain how to find it. In the exercises, they ask you to compute, explicitly, the hilbert polynomial for k[x,y,z,w]/(x,y)\cap(z,w). I know the dimension of each M_s is 2(i+1), but is that the hilbert polynomial? or how would I go about computing that

[wild sapphire]

Can anyone here help me prove $x^6-6x^3+4$ is irreducible over $\mathbb{Q}$?

I've thrown everything I know at it. I've used the Rational Root theorem to prove it has no linear factors, but other possible factors still exist. I've tried Gauss's Lemma, Eisenstein's Criterion, Manual factoring into arbitrary quartic/quadratic pairs and cubic pairs. I'm not sure if I'm missing something obvious here.

[cloud walrus]

darkninja175

[next obsidian]

Find all the roots in C, then arbitrarily bash together all of the possible linear factors (in C) and show they don’t result in rational coefficients

This is going to be unbelievably tedious but it would work

You can set y = x^3 then use the quadratic formula to find roots for y

Then taking cube roots gives you all the roots as complex numbers

[wild sapphire]

But I feel like directly dealing with roots only implies information on linear factors, how does it prove it can't be reduced into two cubics for example

[next obsidian]

You have to group up all linear factors

In every possible way

[wild sapphire]

oh

[next obsidian]

You have Gauss’s lemma which says it would have to even have integer coefficients

[wild sapphire]

yeah I see why its tedious now

yike

[next obsidian]

So you probably can just show certain combinations won’t work

Yeah, this maybe isn’t the best way

But it’s guaranteed to work

I mean it either factors like

4-2

3-3

Or 2-2-2

So this reduces the stuff you have to check

Maybe handle every pair

Or something idk

[wild sapphire]

would it not be like 6! ones to check or something

[next obsidian]

No

It’s far less

Look at how I broke it up

For the first one, it’s just a choice of the two

So 6 choose 2

The other one is 6 choose 3

Then the last one is a a bit harder to count

But it’s picking 3 pairs from 6 things

And like I said, you might be able to argue more abstractly certain things can’t lead to integer coefficients

Since Gauss’s lemma is also here to help you

There’s probably better ways

This is just one way

[delicate bloom]

I think there's some trick using the fact that $\omega^3=1$ gives a symmetry $f(x)=f(\omega x)$

[cloud walrus]

Merosity

[wild sapphire]

for context, this polynomial is constructed based on showing $\sqrt{3+\sqrt{5}}$ is algebraic. So that's like the core (real) root it revolves around.

after that I'm looking to simply show this polynomial is minimal, so irreducible.

[cloud walrus]

darkninja175

[wild sapphire]

I tried that trick that irreducibility of f(x) holds iff $f(x+a)

but my "searching" yielded nothing

[delicate bloom]

like there are 2 distinct roots $r,s$ such that all the other roots can be written as $r, \omega r, \omega^2 r, s, \omega s, \omega^2 s$

[cloud walrus]

Merosity

[delicate bloom]

if it's irreducible that is, otherwise wlog we could say r=s and are both on the real axis I think

then I think we just have to look at gcd(f,f')

gonna go for a bit, curious to see what happens

[wild sapphire]

I mean WolframAlpha says it's irreducible, and I've tried to look into the algorithm they use for that but i couldn't find anything

Wolfram doesn't give context or information though

[delicate bloom]

well are you picking up what I'm laying down? 🤔

[wild sapphire]

I... think so

[frank gorge]

my book kinda mentions that polynomial rings are commutative but im not seeing it

any help?

[iron vessel]

Multiplication between two elements is the same regardless of the order

(X+1)(X-1)=(X-1)(X+1)

[gritty sparrow]

Fist show that x^n and x^m commute for all m and n. Then show that if a and b commute then any linear combination ra+sb and ta+lb commute. Now notice that a polynomial ring is made up of linear combinations of powers of x and elements of the base ring, all of which commute with each other

[iron vessel]

You just have to check that multiplication commutes

[frank gorge]

ahh ok

[long obsidian]

Suppose I had a cycle $\tau$ and a permutation $\sigma $ both in $S_n$.

Is the cycle type of $\tau \sigma \tau^{-1}$ related to the cycle type of $\tau$ or $\sigma$?

[cloud walrus]

fajitas

[south patrol]

conjugate cycles have the same cycle type

i.e. same type as sigma

[long obsidian]

Thank you!

[long obsidian]

Is it true that $\tau \sigma \tau^{-1}=\sigma$ if $\tau$ and $\sigma$ are 2-cycles?

[cloud walrus]

fajitas

[south patrol]

not necessarily

try a few examples e.g. in S3

[long obsidian]

That's actually where I ran into this. If $N={e,(12)(34),(13)(24),(14)(23)}$ then this is all the 2-cycles in $S_4$. Im actually interested in showing $S_4/N\cong S_3$

If $\tau\in S_4$ I know $\sigma\in N $ and $\tau \sigma \tau^{-1}$ have the same cycle type so implicitly $\tau \sigma \tau^{-1}:(N\setminus e) \to (N\setminus e)$ is a permutation. Then since $|N\setminus e| =3$ so $\tau \sigma \tau^{-1}\in S_3$ is an induces permutation of 3 elements where $\sigma\mapsto \tau \sigma \tau^{-1}$. I figured that if $\tau \sigma \tau^{-1} =\sigma$ for $\sigma\in (N\setminus e)$ then $ker(\tau \sigma \tau^{-1})=N$ and the isomorphism is proved by the first isomorphism theorem. Is there some key idea I'm missing?

I know that N is normal

[cloud walrus]

fajitas

[south patrol]

(To me, none of those are 2-cycles, i mean things like (12) for example)

[long obsidian]

I must be using a wrong definition ...

Is it sensible to say the element of N are the elements which can be represented by 2 disjoint transpositions?

[south patrol]

I think this is a bit unclear though e.g. when you say ker(tau sigma tau^-1) i assume you mean the ernel of the map sigma |-> tau sigma tau^-1 or smth

[long obsidian]

Yeah that map exactly

[south patrol]

ye (plus e)

[long obsidian]

I think what I need follows from normality of N

[south patrol]

I mean also your map is an isomorphism so the kernel will just be trivial

I'm not really sure what you're trying to do with that map if it's just conjugation

My lazy method for this would just be to say at there are only two groups order 6, and S4/N can't be Z_6 (think why)

[long obsidian]

Yeah that. I didn't like it cause what if I forgot the groups of order 6 lmao

[south patrol]

Yeah sure, fair enough

I had this as a homework question last year and the guidance in the question was to consider the subgroup H of S4 of elements that fix 4

[Sym{1,2,3} if you like]

Show that that map sending x to xH is injective and then you can use that to deduce the isomorphism

[long obsidian]

Hmmmmmmmmm okay thank you I'll try that

[south patrol]

The idea is just that H is a 'copy' of S3 within S4

[rustic minnow]

If I have a group like G= (Z,+) I cant take a subgroup to be a modulo group right?

[uncut ridge]

subgroups of Z are nZ

elements of Z/nZ are cosets of the form m+nZ which are not elements of Z, so Z/nZ is not a subset of Z let alone a subgroup

[rustic minnow]

Is n here an integer?

[uncut ridge]

yeah

[rustic minnow]

right, so even numbers form a subgroup

[uncut ridge]

yup

you can prove there's no NON-TRIVIAL finite subgroups of Z by considering that if there was such a finite subgroup, then it would have a largest element M but also has to include M+M = 2M as it's closed under addition. Which is a contradiction as 2M > M

[rustic minnow]

oh, right thats smart

odd numbers wouldnt work either under closure

[uncut ridge]

odd numbers fail epic style

they don't have the identity (0) either

[rustic minnow]

Why did you mention the condition of cosets. Does it refer to the fact that groups must have cosets which are distinct and contain new elements not found in the group?

[uncut ridge]

groups don't have to have cosets...?

I mentioned them because they're elements of Z/nZ

[rustic minnow]

ok

I thought the idea of cosets was to construct new sets, although in these case with integers we dont get any new sets.

[uncut ridge]

cosets are formed when you quotient a group by a normal subgroup - just so happens they also form a new group

they're not for arbitrary "new sets" they're specifically x+N where x is a member of the group and N is a normal subgroup

[rustic minnow]

Yes, I understood that they are not arbitrary "new sets" and for group operator multiplication we would get xN right?

[uncut ridge]

yup!

I'm still confused by what you mean with "in the integers we don't get any new sets" though

[rustic minnow]

I just mean that when we use x + N we dont obtain any elements outside of Z

[uncut ridge]

x+N is always a subset of whatever group you're working in so you should never get any elements in a coset that's not in the group

sorry I'm just confused

[rustic minnow]

no worries. Im still confused as to why its called to "quotient a group"

[uncut ridge]

ok there's a cool way to explain this

a quotient group G/N is basically whatever group you get if you look at N and say... "hmm... what if this was ZERO?!"

it turns out if you do this via cosets it's actually a valid thing to do

this relates very closely to the first isomorphism theorem if you know about that

[rustic minnow]

oh, that is cool. Does the first isomorphism thm. relate the condtions that must be satisfied for isomorphism because I do know that or is it something else?

[uncut ridge]

it relates the "kernel" of a homomorphism to the "image"
the kernel is basically the set of all elements that go to 0 under the homomorphism, and the image is every possible value the isomorphism can take.
For example, take the map x -> x mod 2, the "image" would be {0, 1} and the "kernel" would be all the even numbers, cause every even number is 0 mod 2
the first isomorphism theorem basically says you can take the group and, just like with a normal subgroup, "pretend" that the elements that map to 0 (elements in the kernel) ARE zero
and it turns out this G/ker(f) is actually both a group itself AND is isomorphic to the image of f

the actual statement of the theorem is
for all homomorphisms $\phi$ we have $G/ker(\phi) \cong img(\phi)$ but that's boring and doesn't tell you what it's actually saying

[cloud walrus]

AbelianRings

[next obsidian]

TFW {0} isn’t a subgroup

[uncut ridge]

correct

[next obsidian]

TFW M + M > M for M in Z

[uncut ridge]

there has to be a positive element due to the existence of additive inverses THERE

my lord this is why I hate posting in here lmfao

[next obsidian]

:^)

[rustic minnow]

@uncut ridge you have been super helpful

[uncut ridge]

ty

[rustic minnow]

So do we treat kernels in linear algebra and abstract algebra as different topics?

[uncut ridge]

very similar

[rustic minnow]

yeah, cause Ax = 0

[uncut ridge]

yup

[rustic minnow]

there are also kernels in machine learning

[uncut ridge]

it's all the elements that are zero under a linear map (which are actually vector space homomorphisms)

[rustic minnow]

yes

[uncut ridge]

machine learning is uhhh different I think

[rustic minnow]

yep

[uncut ridge]

oh btw @rustic minnow since you mentioned linear maps, the first iso theorem holds for vector spaces as well - just replace the group homomorphism with a linear map

it also holds for rings

it really says something about the structure of structure imo which is why it's my favourite result of all time™️

[rustic minnow]

cool. When proving homomorphisms exist is it based on the operator of the group, i.e addition , multiplication whether to use f(a*b) = f(a) * f(b) or f(a+b) = f(a) + f(b) ?

This was confusing when I was doing exercises

[uncut ridge]

groups only have one operation - use that one

say you have a group $G$ with operation $$ and another group $H$ with an operation $#$, then $\phi$ is a homomorphism between $G$ and $H$ if and only if $\phi(gh) = \phi(g)#\phi(h)$

[cloud walrus]

AbelianRings

[rustic minnow]

thanks.

Also wondering how we can use thm. of Lagrange when we dont have finite groups.

[rose axle]

you can always use that cosets partition

[rustic minnow]

you mean the fact that every coset of a subgroup H of a group G has the same number of elements as H?

[uncut ridge]

that is true, but also that cosets are disjoint

meaning if something is in one coset, it isn't in any of the other ones

[rustic minnow]

But how can the order of H be a divisor of order G if both are infinite

for example if we take all real invertible matrices and choose the subgroup as the real invertible matrices with det= 1 under mtx multiplication

are not both groups infinite?

[uncut ridge]

yes

keep in mind that even for finite groups, n divides n

G is a subgroup of itself

[rustic minnow]

infinity divides infinity

[uncut ridge]

yurrrr

well, for our purposes yurrrr

[rustic minnow]

Oh would H not be a proper subgroup in this case then?

as G is a subgroup of itself

[uncut ridge]

no H is absolutely a proper subgroup

it's a proper subset

it's just an infinite subset

[rustic minnow]

We just need H to not equal to G then in order for it to be a proper subgroup

So even numbers under the integers would be a proper subgroup as well right?

[uncut ridge]

yes

[rustic minnow]

What did u mean with G is a proper subgroup of itself?

[uncut ridge]

I didn't say that

I said G is a subgroup of itself

[rustic minnow]

oh, my mistake

thats like saying the whole group is also a subgroup.

[long obsidian]

A group action induces a permutation right?

[rustic crown]

not just one permutation, you get a permutation for each element of the group,

more precisely a group action on a set X can be thought of as either a function G * X --> X satisfying some nice enough properties or as a group homomorphism G --> Perm(X)

[long obsidian]

Is there a way to determine the kernel of this homomorphism?

I've seen some criteria like conjugacy

[lethal dune]

kernel will be the elements that fix all element of X, i.e. stabX itself

[gusty cave]

I was reading on Sylow's Theorem 1 and in the proof here there was a few parts I didn't understand, I was hoping if someone could clarify them for me.

THe proof is as follows, Orb means orbit and Stab means stabiliser

The first thing I don't get, is why m has to equal p^ru. Desnt such a form for m mean that the proof does not work for all multiples of p^n?

Secondly, it says "there must be an omega element of S such that |Orb(0mega)|=p^st. Why is it that the size HAS to be in the form p^st?

Lastly, I don't understand the last sentence which says "FOr this choice of .... is thus a subgroup of p^n". That part just kinda doesn't make sense to me

pls help

[rustic crown]

any natural number k can be written as (p^a) * b where p and b are coprime

so writing m = (p^r) * u doesn't restrict any values of m, and proof works for all multiples of p^n

same for the second question, every number can be written like that. the real question is why can we require s to be smaller than or equal to r. And that's because if each orbit had size divisible by p^(r+1), then |S| would be divisible by p^(r+1) which it is not.

[lethal dune]

[rustic crown]

in the first page of the proof, we showed that for any omega, it's stabilizer is at most of size p^n
in the second part, we find a specific omega for which size would be bigger than (or equal to) p^n
both these inequalities show it has to be exactly p^n

[gusty cave]

OHHH I see now that makes a lot of sense thanks

the part of |stab(omega)|<=p^n completely went over my head lol

[untold sapphire]

Can anyone help with part b)? I'm pretty stumped

[next obsidian]

Honestly, I’m not too sure myself

One thing I noted was, if f is in R_0, then consider g(x,y) = f(x) - f(y) for x,y indeterminates

Then g(a,b) = 0, so that g is in the kernel of the evaluation map k[x,y] -> k sending (x,y) to (a,b) so it’s a routine result that g is in the ideal (x-a,y-b) of k[x,y]

Perhaps you can try to reinterpret that in terms of f??

Honestly I don’t really have any good ideas for this one

[coarse storm]

The given ring is certainly a subring of R_0. It is not clear to me that they are actually the same.

Is the given ring the same as (x-a)(x-b) k[x]?

Uh... I did not phrase that right. Hm..

[next obsidian]

It is not

x^2(x-a)(x-b) doesn’t exist in the purported R_0

[hidden haven]

Right side in the left side is obvious. Now take some f in R_0, divide it by (x-a)(x-b) and show that the remainder is constant. You will get some quotient q(x). Divide it by the same thing and this time you can't argue that the linear thing is 0 but that linear thing multiplies with the first divisor to get you x(x-a)(x-b) so you are fine. Induct.

@untold sapphire

[hidden haven]

bruh moment

It is there I think

Write x^2 as (x-a)(x-b) + linear

then distribute

[next obsidian]

Eh?

[hidden haven]

argument by mostly saketh lmao he was just too lazy to type it out

[coarse storm]

I feel like I have seen this same argument for a question ... two weeks ago?

[hidden haven]

x^2 = (x-a)(x-b) + (a+b)x - ab

[next obsidian]

Nothing in the purported R_0 is linear

[hidden haven]

Now substitute and distribute

[next obsidian]

Eh?

[hidden haven]

[next obsidian]

How can you add in (a+b)x tho

[hidden haven]

multiply with (x-a)(x-b) on both sides

[next obsidian]

[hidden haven]

[next obsidian]

[gritty sparrow]

[next obsidian]

No algebra at 2 am challenge

[untold sapphire]

@hidden haven God damn it thanks

this made me feel really stupid for hours

[lethal dune]

what book would you guys recommend reading for fields and extensions other than Dummit Foote?

[sharp sonnet]

i like field and galois theory by morandi

[rustic crown]

i like aluffi

[next obsidian]

I’ve mostly just read the stacks project

[hidden haven]

Milne goob

[lethal dune]

4

[rustic crown]

catpill everyone

[untold sapphire]

yeah aluffi is what bridged the gap to CT for me

[rustic crown]

for me it was aluffi and moldi

[delicate orchid]

how wholesome

[hidden haven]

[next obsidian]

For me it was trying to do AG

[untold sapphire]

hah

[lethal dune]

I read cat because not knowing it made me feel dumb

[untold sapphire]

I still can't do modern AG tbh but thankfully I think I'll be able to avoid it for a bit

[lethal dune]

(abstract nonsense aside)

[rustic crown]

i wanted to understand this

[delicate orchid]

I don't read cat because not knowing it makes me feel smart ( )

[lethal dune]

I still don't understand this

[untold sapphire]

I picked my research topic so I wouldn't have to understand scheme theory 🙈

[rustic crown]

(i think it's not supposed to make sense, unless someone enlightens me)

[lethal dune]

moldi

[hidden haven]

It is not supposed to make sense

[lethal dune]

lol

[hidden haven]

This one is tho

[delicate orchid]

that's just what induction is

[hidden haven]

You prove that some claim is true on a subalgebra of N and therefore must be true on N 😌

We need more nonsense stickers

[lethal dune]

makes zero sense

[hidden haven]

You can visit #category-theory when every fifth day the whole gang comes together to explain Yoneda to some noob 🤓

[rustic crown]

Oh woah

invite me next time

[lethal dune]

[hidden haven]

lol just search the channel for Yoneda

And you will find good past explanations

[next obsidian]

The meme makes sense

Because the person saying the category theory nonsense is making no sense

And it’s a complex analysis class

Perfectly capturing the behavior of the UGCT

[hidden haven]

Apply zero functor to chmonkey

[next obsidian]

[hidden haven]

oh ok I am not ugct 😌

det is though

[next obsidian]

Applying Chmonkey functor to moldi

Chmoldilocks

[rustic crown]

[hidden haven]

oh god

Everything is

[next obsidian]

Wait a second

You det and Saketh all go to the same uni?

[lethal dune]

choldilocks

[hidden haven]

Go is a strong term

[next obsidian]

Hurb

[hidden haven]

But we attend the same uni sitting 1000km apart

Actually

[next obsidian]

[untold sapphire]

hooray for zoom classes

[hidden haven]

Saketh is literally on the other side of the world for the last year

[next obsidian]

Why

[hidden haven]

He lives in the US

[next obsidian]

What’s on the other side of the earth from India anyway

Why tf does he go to an Indian university?

[hidden haven]

Used to live in India when enrolled

[next obsidian]

[hidden haven]

Born in the US lived in India for few years then moved back

Now he doesn't want to come back, they gave us an option to return this sem which eventually got cancelled but he said he doesn't want to come back so he will continue to attend remotely until I peer pressured him into saying yes in the form in a 15 minute call by calling him a bunch of names

[next obsidian]

[lethal dune]

I'm staying at the campus and attending online classes

[rustic crown]

[next obsidian]

I haven’t slept

Good night

[hidden haven]

4am algebra

[next obsidian]

,ti

[cloud walrus]

The current time for Chmonkey 2.0 is 07:01 AM (PST) on Fri, 14/01/2022.

[hidden haven]

oh my god

7am algebra

lets go

[dusky sparrow]

Who is Saketh?

[rustic crown]

lol

[hidden haven]

oh Saketh is in the same time zone

WA

uh oh doxxed

[next obsidian]

👉😁👈

Let’s gooooo

[lethal dune]

wait so saketh attend classes at midnight

lol

[hidden haven]

Yes

[lethal dune]

[dusky sparrow]

Your class or just a meme?

[hidden haven]

random guy on server

don't look him up if under 18

[dusky sparrow]

Okay.

[rustic crown]

just a meme

[lethal dune]

,av @gritty sparrow

[cloud walrus]

saketh#0733's Avatar

Click here to view the image.

[lethal dune]

oof

[dusky sparrow]

Clearly 13+ or maximum 16+.

[hidden haven]

You don't get to decide, major conflict of interest

[dusky sparrow]

Okay .

[rustic crown]

chmonkey says i'm lewder

,av

[cloud walrus]

det#7067's Avatar

Click here to view the image.

[lethal dune]

lul

[dusky sparrow]

Yes you are.

18+.

[delicate orchid]

check #groups-rings-fields
ecchi

[next obsidian]

[rustic crown]

[dusky sparrow]

Chmonkey det ryu and moldi made abstract-algebra a better #discussion.

I came here cause when I ctrl-escaped this channel was the first to have a message.

[delicate orchid]

chat if you don't mind I have an actual abstract algebra question

ik a bit of a moment

does anyone know of any nice degree 3 representations of Q_8

[rustic crown]

the irreds either 1 or 2 dim?

[delicate orchid]

doesn't have to be irreducible

I've tried constructioning a few from smashing together irreducibles but they don't act on a nice geometric object

it's just for a presentation about representation theory from a really introductory level I have to do

[hidden haven]

trivial direct summed with itself thrice

ok I will be serious

sign direct summed with itself thrice

is that the same

in this case

[rustic crown]

what do you mean by sign?

[delicate orchid]

I think that's the same

[hidden haven]

embed into S_n

then sign

[rustic crown]

S_8

right all are even

[delicate orchid]

yeah I thought they would be

wait no surely they should all be odd

[hidden haven]

Maybe this embeds into S_4 and then it's not in A_4 😠

[rustic crown]

idt it embeds into S_4

[lethal dune]

https://tenor.com/view/matrix-math-grafics-gif-20047924

wth is this

[hidden haven]

Sigmaoldilocks

[rustic crown]

like sylow subgroup of S_4 will be unique and stuff and so it should be D_4

[delicate orchid]

I'll just use the regular representation of Q_8 lol

[hidden haven]

Maybe just start by finding number of irreps then direct sum then make char table then direct sum them as you want for interestingness

[delicate orchid]

oh I've done all that

[hidden haven]

oh wow

Then what's the issue

[hidden haven]

ok what about S_5

ha

owned

[rustic crown]

idk >.<

[hidden haven]

[delicate orchid]

the idea for my presentation is that you can actually see the representations acting on a geometric object which isn't fantastic if they're complex valued

maybe I should just bitch out and use S_4 acting on a cube

that's probably eaiser

anyway ty for the advice

[rigid depot]

Can someone explain to me why it follows that A is noetherian? I guess it's obvious but I don't see it right now

[gusty halo]

well any chain of ideals in A is a subchain of the written one with the m_k, but in that one, no matter where you start, you have only finitely many steps until you reach m, the largest one

its basically the same argument why the natural numbers are well ordered i guess

[median pawn]

Hey

[cloud walrus]

Hausdorff

[median pawn]

Note that subrings may not contain the identity

Does anyone have any ideas? I've already come up with examples where S,R have different identities or one of them does not have an identity element

[gusty halo]

if you think being able to invert elements is algebraic enough, then how about Z in Q?

otherwise you could maybe look at things like integral closure

[median pawn]

Z in Q sounds nice

[median pawn]

What's that?

[gusty halo]

its like algebraic closure but for rings instead of fields

[median pawn]

I'll read about it!

[gusty halo]

seeh e.g. atiyah macdonald for more details

[median pawn]

I'm currently trying to produce S,R where S is an integral domain/field/division ring but R is not

[gusty halo]

then Q in Q[x]?

i mean Q[x] is still integral but not a field

you could also just adjoin some nilpotent element maybe to get something that isnt an integral domain anymore

[median pawn]

Q[x] is not a field because certain non-zero elements don't have inverses, right? For example, x+1? (x+1)p(x) = 1 means p(x) = 1/x+1 which is not in Q[x]

[gusty halo]

yes

[median pawn]

I suppose Q[[x]] is a field though

[gusty halo]

dont know actually

[median pawn]

Ah, I'll check

[gusty halo]

but division ring is like a field without commutativity right?

[median pawn]

Correct

[gusty halo]

if so then maybe take invertible matrices included in all matrices

lower thing should be division ring, but the ring of all matrices has zero divisors

[median pawn]

That's a nice example, yeah

Nah that's problematic

Invertible matrices don't contain the zero matrix

[gusty halo]

oh right

[median pawn]

So that set is not a subgroup, hence not a subring

[gusty halo]

also not closed under addition now that i think about it

[proud bear]

the reals are contained in the quaternions for field contained in division ring

[delicate orchid]

I like this one

[proud bear]

wait nvm

you already have field contained in not field

[delicate orchid]

C is algebraically closed R isn’t

Another clanger

[median pawn]

Yup, did that!

[proud bear]

ah yeah you can embed the quaternions into M_2(C), so you have division ring contained in not division ring

[delicate orchid]

How many do you need lol

[delicate orchid]

Omg degree 2 representation my most loved

[median pawn]

I guess this is good enough lol

I was thinking of something with Boolean rings

Z/2Z is a Boolean ring, can we view it as a subring of a non-Boolean ring?

[delicate orchid]

Is Z/4Z Boolean?

[median pawn]

Right, it's not, since 2^2 = 0

[delicate orchid]

Excellent my random shot in the dark well educated guess payed off

[median pawn]

Actually it didn't 😦

Z/2Z is not a subgroup of Z/4Z

[delicate orchid]

is 0+4Z, 2+4Z not a subring

[median pawn]

because 1 + 1 = 2, which is not 0 or 1

[delicate orchid]

I mean true, Z/2Z is only isomorphic to a subgroup of Z/4Z

[median pawn]

Yep yep

[delicate orchid]

I assumed that would be good enough though

[median pawn]

We need S to be a subring of R, not just isomorphic to one 😦

[delicate orchid]

in that case take {0+4Z, 2+4Z} as S

[median pawn]

Yes, but that's not a Boolean ring (a^2 = a for all a) so what's the point LOL

[delicate orchid]

[proud bear]

take R=(Z/2Z)[x]?

[median pawn]

That's brilliant!

[median pawn]

I need three more examples

[hidden haven]

Being non zero 🤡

[median pawn]

Idk what even is the point of this stupid exercise

[hidden haven]

Having a non trivial automorphism group 🙈

[delicate orchid]

wtaf do rings even have that many properties

[hidden haven]

Being initial in the category of rings

[median pawn]

Well I'm trying to find S such that S is a PID/PIR but R isn't

[hidden haven]

Field and polynomials over it

[delicate orchid]

^

[hidden haven]

Wait

Non field PID and polynomials over it

Not field

[delicate orchid]

I mean field is also a PID

you weren't wrong

[hidden haven]

But then polynomial ring is PID

[delicate orchid]

[hidden haven]

Embeddable into ℤ 🙈

Field of fractions of some other ring

[delicate orchid]

F[x1, ..., xn] is a noetherian subring of F[x1, ...] which is not noetherian

[hidden haven]

Polynomial ring of some other ring etc

Orderable

[delicate orchid]

problem is we don't really know how advanced our properties can be

like I'd say noetherian is a bit of a stretch given that this seems to be a question showing that not all properties are preseved in substructures

[median pawn]

Z[i] is a PID which is not a field

[hidden haven]

Yes

Endomorphism ring of some abelian group

Idk if this one works lol

Countable

[delicate orchid]

uhh

yeah I'm kinda out of ideas lol

[median pawn]

Interesting, what's this property?

[hidden haven]

ℚ is an ordered ring, no way to make ℂ into an ordered ring

[median pawn]

Ahh okay

[delicate orchid]

I still believe

[median pawn]

Here we are using the fact that R[X] is a PID iff R is a field, right?

[hidden haven]

No, ℤ[i] isn't a polynomial ring

It's a quotient of 1

[delicate orchid]

it's the image of Z[x] through some homomorphism

same thing (first iso my beloved)

[median pawn]

Put R = Z[i]

[hidden haven]

Oh lol

[delicate orchid]

oh you want Z[i][X]

[median pawn]

Yep

[hidden haven]

You can find some example directly too

[delicate orchid]

hmmm

[hidden haven]

Example of ideal

(2,x)

Which also works for ℤ [x]

[delicate orchid]

Z[i][X] is going to be isomorphic to something wacky and I want to find it

[hidden haven]

And also you could do k[x,y] over PID k[x]

[delicate orchid]

i = y, then y^4 = 1 so take Z[X, Y]/(Y^4) yessssss

[hidden haven]

It's isomorphic to ℤ [i, X]

[median pawn]

Is the second part an exercise?

[hidden haven]

Are you sure about that

[delicate orchid]

absolutely not

[hidden haven]

Ye you just use axioms of ordered rings to find a contradiction

[delicate orchid]

but I don't see where it fails

[hidden haven]

You've proven that the quotient ring you have surjcets onto the ring you want

Because of

[delicate orchid]

ah yeah it might not be injective

[hidden haven]

Yep

[hidden haven]

Oh this isn't the general statement

[delicate orchid]

yeah cause y^2 = -1 so they'd both map to the same thing

I worded that poorly

[hidden haven]

Yep

But then quotienting by (y²+1) will work

[delicate orchid]

ah yes

we have found the funny thing

[hidden haven]

[delicate orchid]

I'll credit you in the paper

[hidden haven]

Yo I thought I was getting that coauthor clout

[delicate orchid]

you know what... I've had enough of you category theorists and your "co" nonsense

what's next? copapers?

[hidden haven]

A coauther is just a reader

[delicate orchid]

ah yes cause the arrow from the writer to the paper has been reversed

the paper is "writing" your brain if you will

quite intuitive...

[hidden haven]

Yes, I have coauthored quite a few books at this point

[delicate orchid]

HAHAHA

I'm going to start saying that and when people call me out I'll just blurt a load of category theory-adjacent nonsense

[median pawn]

This is probably the funniest thing I saw today

[median pawn]

Did you have an example in mind for this, or?

[delicate orchid]

Z is a good one

[hidden haven]

ℤ and ℂ lul

ℤ, ℚ, ℝ, all have trivial aut groups

[delicate orchid]

fields are so boring honestly

[hidden haven]

Oh wait

Non unital rings

[delicate orchid]

wait we're supposed to be working with non-unital rings?

[hidden haven]

That's what the hausdorff say

[cloud walrus]

Hausdorff

[delicate orchid]

I need bleakdevastation

[hidden haven]

Oh but even for non unital this is true

[median pawn]

yes

[hidden haven]

Negation is not an automorphism

[delicate orchid]

I'm actually going to cry

[hidden haven]

That's what I thought for a moment

[delicate orchid]

waitttt a minuite

[hidden haven]

But -1 and 1 can be distinguished by squaring

[median pawn]

Wait, I'm confused - so the Z and C example does not work right?

[hidden haven]

It do

[delicate orchid]

the automorphism group of Z is not trivial moldi just had a moment

[hidden haven]

ℤ has trivial aut group even when non unital

[delicate orchid]

wait are you sure

[hidden haven]

Because of this

Yes

Negation ain't an automorphism

[delicate orchid]

I need to get out of "group mode"

[median pawn]

Oh shit wait, Aut of rings is not same as Aut of groups

[delicate orchid]

I was thinking Z is homomorphic to {-1, 1}

yeah

[hidden haven]

[delicate orchid]

I'm being an actual moron

sorry moldi

wait so, sub-rings in non-unital rings are just ideals aren't they?

[hidden haven]

No they are weaker

Ideals absorb multiplication

[delicate orchid]

oh yes true

[hidden haven]

Subrings are just closed under it

[delicate orchid]

anyway how many more examples do you need @median pawn

before my brain dribbles out from my ears

[median pawn]

that's it haha, these are good enough

[cloud walrus]

Hausdorff

[median pawn]

Right? I'm trying to show that Aut Z is trivial

[hidden haven]

yep

[delicate orchid]

that's true for groups so something something category theory yes the image of the generators uniquely determines the image

[hidden haven]

This is the category definition of generators 😌

[cloud walrus]

Hausdorff

[hidden haven]

subset such that agreement on it implies agreement everywhere

Not sure if it always lines up but yeah

[delicate orchid]

more over you can just see that f has to be of the form f(n) = mn and invertible, the only such m are -1 and 1 and you can then show those two cases are equivalent

that's how I did it on my 2nd year algebra exam

[median pawn]

wait why is f(1) = 0 not valid

oh ok nvm that's not an auto

f(1) = 1 gives f = identity, wow yes

[delicate orchid]

yeah f(1) = 0 is definitely not injective lol

[cloud walrus]

Hausdorff

[delicate orchid]

in fact I'd go as far to say it is the exact oppositve of injective. Outjective if you will

[hidden haven]

very big

[median pawn]

C is the complex plane lmao (as a ring)

[hidden haven]

as they say in the textbook I coauthored, absolute chomker

[delicate orchid]

[hidden haven]

But a non trivial automorphism is conjugation

Other automorphisms may not be constructed because very choosey

[cloud walrus]

Hausdorff

[delicate orchid]

I was thinking about groups again that's why I said it was trivial

[hidden haven]

That does sound good to me

As long as you give the example of the non trivial automorphism

[median pawn]

Yep that suffices

Yay, amazing

[delicate orchid]

i -> -i my most beloved

[hidden haven]

Interesting thing is that the reals have no non trivial automorphisms

[delicate orchid]

hmmm

does the ol f_m(n) = mn not work?

it's definitely bijective

[hidden haven]

No because np maps to mnp not mnmp

[delicate orchid]

quotient it out

I am getting some kinda tensor producty vibes from that actually

f(ma, mb) = m^2f(a, b) hmmmm

[hidden haven]

Yes multiplication is always bilinear

That is exactly what distributivity says

Some books define an algebra as vector space A with a multiplication A ⊗ A → A

[delicate orchid]

ahhh there we go

there's the tensor time

I knew there was one hiding there

[rustic minnow]

If I want to show closure for the following set I can multiply two arbitrary elements and show that the result belongs to the field of complex numbers, but do I need to write the elements in a + bi form first for the set $\varepsilon_n = { z \in \mathbb{C} | z^n = 1}$?

[cloud walrus]

Fredrikpiano

[delicate orchid]

I'd write them in polar form for that set

i.e. z = re^(theta*i)

[hidden haven]

You don't need to do that actually

a^n = 1 = b^n, just prove that (ab)^n =1

Chmmutativity

[rustic minnow]

dont I get ab = ab^2n ?

[hidden haven]

(ab)^n = a^n b^n

when things chmmute

[rustic minnow]

oh right, now I got it, since a^n and b^n are equal

[delicate orchid]

so you know the complex numbers are an abelian group under multiplication because they're a field, take a and b inside that set, then a^n = b^n = 1, then (ab)^n = ababababa...b = aaa.aab.bbb. = a^nb^n = 1*1 = 1

so you know that ab is also inside that set

[rustic minnow]

yes field not ring

but I understood thanks

[hidden haven]

non zero complex numbers

Also you could just say ring and then take multiplicative monoid lul

works just as well

[delicate orchid]

https://tenor.com/view/no-i-dont-think-i-will-captain-america-old-capt-gif-17162888

[rigid cave]

moldi be like

[hidden haven]

Who made this lol

Did you just now

[rigid cave]

this is real lmao

[hidden haven]

[rustic crown]

[hidden haven]

bruhhh

[delicate orchid]

moldilocks in real loife

[rustic minnow]

[hidden haven]

2019, my name is older 😌

[rustic crown]

what does moldilocks mean

[hidden haven]

Not gonna say

[rigid cave]

you can buy moldi here https://www.amazon.com/Moldilocks-Three-Scares-Zombie-Tale/dp/1454930616

[rustic crown]

[hidden haven]

Just think of it as a misspelling of my real name

[regal arrow]

Now consider the ring of polynomials in the variable X with real coefficients, R[X], and the ideal I = (X2 + 1) consisting of all multiples of the polynomial X2 + 1. The quotient ring R[X] / (X2 + 1) is naturally isomorphic to the field of complex numbers C, with the class [X] playing the role of the imaginary unit i. The reason is that we "forced" X2 + 1 = 0, i.e. X2 = −1, which is the defining property of i.

Can I get some help understanding why we force x^2+1=0 for this equation

like

why do we think of R/I as sending everything in I to zero

I get that R/I comes in with a universal property which associates it with a homomorphism where I is its kernel

but is still doesn't make sense to me

[rustic crown]

the zero element in R/I is the coset containing 0, which is nothing but I

[hidden haven]

I know that this isn't particularly helpful but I really feel the need to do this

[rustic crown]

[regal arrow]

It's just that when I see something like R[X]/(X^2+1) my brain only sees equivalence classes, so when I consider a specific element, I think of some p(X)+X^2+1 since the equivalence classes still contain polynomials, so how does that make sense if X^2+1 goes to 0

[hidden haven]

It's not p(X) + X² + 1, but p(X) + (X² + 1)

Brackets denote ideal generated by

(X² + 1) is the zero element so you might wonder why we even write it here

That is because the + between p(X) and the ideal is not the + of the ring

p(X) + (X² + 1) is a single object

p(X) alone isn't an element of the quotient ring

[rustic crown]

I'm assuming you've verified that the operation on these cosets is well defined, that is we can take the representatives, do stuff with them and then look at the coset containing that to find what we wanted.

in ℝ[X], we want to force the relation X^2 + 1 = 0, which is why we quotient by the polynomial. This works out in the quotient ℝ[X]/I where I = (X^2 + 1), the coset [X] actually satisfies the equality, [X]^2 + [1] = [0]

because the left side is precisely [X^2 + 1] by the well-definedness and the coset containg X^2+1 is just I itself = coset containg 0 = [0].

two elements are equivalent if and only if the equivalence classes containing them are equal

[regal arrow]

So here's how I see things. We've taken a polynomial ring, in our case R[X], chosen a polynomial, X^2+1, considered the ideal it generates, (X^2+1), and then created a new ring of equivalence classes R[X]/(X^2+1) which is understood by declaring that X^2+1 must go to zero. What I want to get at is how we go from "choose the polynomial X^2+1 in R[X]" to

we want to force the relation X^2 + 1 = 0

Why are we forcing any relation

why do we care about sending it to 0

what does =0 do

[rustic crown]

it gives you a way to abstractly add roots of a polynomial

for the ring ℝ, the polynomial X^2 + 1 has no roots. That gives you a recipe to find a bigger field in which that polynomial does have a root

[hidden haven]

One thing that might be confusing is the equation X² + 1 = 0 is only true in the quotient, even though it's written as if it becomes true in the orginal ring somehow

[regal arrow]

From my perspective I'm not exactly interested in talking about its roots. I just want to know why "forcing the ideal to be zero" paints a picture of the quotient ring.

[hidden haven]

In the quotient, X² + 1 + (X² + 1) = 0 + (X² + 1)

And the right side is the zero of the quotient ring

You can say this about any other element of the ideal (X² + 1) as well

So in the quotient ring, they all become 0

And nothing else becomes 0, because it won't satisfy that relation

[regal arrow]

And what happens to an arbitrary polynomial, say X+1

Does it just belong to [X+1+X^2+1]

[hidden haven]

Nope

It's not in that ideal

[regal arrow]

I mean in the quotient ring

what happens to it

[hidden haven]

In the quotient ring its image is X + 1 + [X² + 1]

Using square brackets now to stick with your notation

[regal arrow]

() for ideal, [] for equiv class right

[hidden haven]

Oh

Then it belongs to the element [X+1] of the quotient

Which is the element X + 1 + (X² + 1)

[regal arrow]

And could one say that [X+1]=[X+1+X^2+1]=[X^2+X+2]

[hidden haven]

Yes

[regal arrow]

woo gaming

I'll have to reread this 1000 times but thanks bros

[frank lake]

would Q be{..., -3(1/2), -2(1/2), -1(1/2), 0, 1/2 2(1/2), 3(1/2), ... }

[next obsidian]

Yeh yeh yeh yeh yeh

[frank lake]

thanks

[next obsidian]

[frank lake]

just to make sure

is this the group of rational numbers under multiplication?

{...1/4, 1/3, 1/2, 1, 2, 3, 4...}

[next obsidian]

Not quite

It’s just everything of the form m/n for m,n integers

Except m,n ≠ 0

3/7

-27/4

4/2 = 2

[frank lake]

[next obsidian]

Under multiplication doesn’t mean anything when listing all the elements

[frank lake]

then what set would this be

[next obsidian]

Idk

It’s like N U N^-1 or something

Naturals and their inverses

[frank lake]

damn

[next obsidian]

And then you rip out 0

[frank lake]

I'm trying to get a random group and use it to prove that for any group the order of the element is the same as the order of it's inverse

just like a test or something

[next obsidian]

I mean it’s true

And it’s not too hard to prove

Just show either |g| <= |g^-1|

[frank lake]

OH

WAIT

[next obsidian]

Or |g|>= |g^-1|

[frank lake]

LEt me figure this out

[next obsidian]

You only need one inequality

[frank lake]

oh that's simple

💀

[next obsidian]

Cuz (g^-1)^-1 = g

[frank lake]

just prove that one is a subgroup of another

[next obsidian]

Wut

[frank lake]

?

[next obsidian]

You’re dealing with elements

Or do you mean like

<g>

[frank lake]

what does >= and <= mean then

[next obsidian]

And <g^-1>

That’s a less than or equal my guy

[frank lake]

oh

[next obsidian]

|g| meant the order

Maybe you’ve seen it as o(g)

[frank lake]

in my textbook it's also used to represent that one is a subgroup of another

math notation :/

[next obsidian]

Yeah it is, but those aren’t sets

|g| is a number

[frank lake]

hm

let me try this out

[frank lake]

I mean how would you prove |g| <= |g^-1| for a general case

like if you were given a set this wouldn't be so difficult

[next obsidian]

Here’s a way to show the other one, >=

Note that (g^n)^-1 = (g^-1)^n

[frank lake]

oooh

[next obsidian]

If g^n = e, then (g^n)^-1 = e^-1 = e = (g^-1)^n

This shows if g^n = e, (g^-1)^n = e

So |g^-1| <= |g|

[frank lake]

bro my braincells

so if g^n is equal to the identity

then the inverse of g^n is equal to the inverse of the identity which is just the identitiy

[next obsidian]

Yeh

[frank lake]

and the identity would be equivalent to the order of the inverse of g

[next obsidian]

Inverse of g to the n-th power

[frank lake]

oh yeah

so if g^n is equal to the identity then the inverse of g to the nth power is the identity

[next obsidian]

Yeah

[frank lake]

that makes sense, but it would be pretty difficult for me to think this off the bat

😂

[next obsidian]

¯_(ツ)_/¯

[frank lake]

how do you think of how to prove things like these so quickly

[next obsidian]

I’ve been doing algebra for like 3 years

[frank lake]

damn

[next obsidian]

Anyway

You have to now show |g| <= |g^-1|

But my claim was that you get this for free just from |g^-1| <= |g|

Do you see why?

[frank lake]

why <= instead of just =?

[next obsidian]

Because all you’ve shown is that if g^n = e then (g^-1)^n = e

This is only enough to get you an inequality

[frank lake]

so to show |g| <= |g^-1| I just show if (g^-1)^n = e then g^n = e?

[next obsidian]

Sure, but you can be more clever

[next obsidian]

We know this for arbitrary g

If you “plug in” g^-1 for g

[frank lake]

OOO

[next obsidian]

You get that |(g^-1)^-1| <= |g^-1|

But this just says |g| <= |g^-1|

[frank lake]

so since |g| <= |g^-1| and |g^-1| <= |g| that means |g| = |g^-1|?

[next obsidian]

Yes

[frank lake]

dang

abstract algebra is cool but man

[next obsidian]

There’s a tiny tiny bit of subtlety here

Mainly in how you interpret <= when |g| could potentially be infinite

But it works out fine

[frank lake]

that is true

[next obsidian]

The proof here shows that if |g| is finite then |g^-1| is also finite

And that’s symmetric so like it’s true if you flip g and g^-1

This is enoigh to tell you that one is infinite if and only if the other is

Or you can just extend what <= means to include infinity

Which is perfectly fine

[frank lake]

tysm

@next obsidian

I'm going to take a break and come back to this later

😂

[wicked zephyr]

hi

what is the index of a submodule?

I encountered this definition and I´ve never seen it

maybe it has sense when taking about free R-modules

context

[rustic crown]

Index means number of cosets of the submodule

Since everything is abelian, it's just the size of the quotient

[trim grove]

Let $G$ is finite Group , If for any two subgroups $A$ and $B$ either $A \subset B$ or $B \subset A$, then this implies $G$ is cyclic?

[cloud walrus]

TheStudent

[gritty sparrow]

yes i think so, here is a sketch of the proof: consider the maximal cyclic subgroup of G (which exists since G is finite) , say (x). for any y in G we know (y) is a subset of (x) hence y=x^n therefore G is cyclic

[trim grove]

but what if this is contained in some other group?

[prisma shuttle]

wdym?

isn't the only group G?

[next obsidian]

You took it to be maximal

This has to exist because G is finite as Saketh said, if there wasn’t one you’d get an infinite ascending chain of groups each with an element the prior didn’t have, so in particular you’d have infinitely many elements

[trim grove]

Now this is clear to me

[trim grove]

any hint for these 1,3,4?

[hidden haven]

Try proving that non cyclic subgroups of ℚ are dense

That will solve 2 of those options

Idk about the last one