groups-rings-fields

[tough raven]

Note: this forms part of the proof that the (prime) spectrum of a commutative ring is sober.

[gusty halo]

so lets first assume choice

suppose I is not prime, so we have ab in I but a,b not in I

if I is not radical, we are done

so suppose it is radical

first note that $I \subsetneq (I:a) = (I:a^2)$. Indeed, we always have $I \subseteq (I:a) \subseteq (I:a^2)$ Since $b \in (I:a)$ but $b \notin I$ the first inclusion is strict.

[cloud walrus]

Phil

[gusty halo]

Then if $c \in (I:a^2)$, i.e. $ca^2 \in I$, we also get $(ca)^2 = c^2a^2 \in I$ so $ca \in \sqrt{I} = I$ and thus $c \in (I:a)$

[cloud walrus]

Phil

[gusty halo]

this shows the first claim $I \subsetneq (I:a) = (I:a^2)$

[cloud walrus]

Phil

[gusty halo]

now I claim that $I = (I : a) \cap (I + \langle a \rangle)$

[cloud walrus]

Phil

[gusty halo]

and then you can use $I = \sqrt(I) = \sqrt{(I:a) \cap (I + \langle a \rangle)} = \sqrt{(I:a)} \cap \sqrt{(1+\langle a \rangle)}$ to get that I is not irreducible

[cloud walrus]

Phil

[gusty halo]

So for the second claim, clearly the inclusion $I \subseteq (I:a) \cap (I + \langle a \rangle)$ holds

[cloud walrus]

Phil

[gusty halo]

now suppose $c \in (I :a) \cap (I + \langle a \rangle)$, so $ca \in I$ and $c = f + ra$ for some $f \in I$ and $r \in R$

[cloud walrus]

Phil

[tough raven]

But then ra^2 in I?

[gusty halo]

yep

and then $(ra)^2 \in I = \sqrt{I}$ so $ra \in I$

[cloud walrus]

Phil

[gusty halo]

thus $c \in I$, and we are done

[cloud walrus]

Phil

[gusty halo]

and i think this should also work with your definitions instead of the usual radical, so without choice

[gusty halo]

do people here generally not assume choice or what?

[hidden haven]

The fact that this has to be said is what's

[gusty halo]

well they explicitly talked about choice in their question

so i thought i'd mention it

[hidden haven]

ye

[tough raven]

That's a me thing

Just curious whether you actually need it

[hidden haven]

where have you been raghuram

why no activity

[tough raven]

College busy
This server very busy

[hidden haven]

[gusty halo]

hm i feel like the server feels more dead than a few months ago

might be me tho

at least the topology channel seems alot more quiet

[hidden haven]

because Toki is no longer asking questions in that channel

was solo carrying that channel before

[gusty halo]

ah and also it was merged with geometry before

[hidden haven]

True

[gusty halo]

good riddance though, i decided this semester maybe geometry is not my thing afterall

had to drop algebraic geometry cause too hard

[hidden haven]

same 🤝

[lavish gale]

toki blew through intro topology lol

[gusty halo]

course here in bonn is literally insane

[hidden haven]

toki is now studying steenrod algebras and stuff

[tough raven]

I muted to stop myself reading everything
So I wouldn't know

[gusty halo]

cool

[hidden haven]

I can't even understand the doubts anymore

[lavish gale]

i remember toki asking about definition of continous function in topological sense
and next time i saw toki, she/he was asking bout algebraic topology

[gusty halo]

i mean you kind of dont need to know much about pset topology

i knew a lot of pset topology beforehand

[hidden haven]

ye he skipped most of munkres, jumped straight to hatcher

[gusty halo]

but never had a course on anything algebraic

in october i had my first encoutner with covering spaces and fundamental group etc

also never had of homology etc before

[hidden haven]

phil please dont be an undergrad

[gusty halo]

but now im suddenly talking about hurewicz and homotopy theory in my courses

[gusty halo]

no i started my masters in bonn this semester

[hidden haven]

ok nice

[gusty halo]

this week we started doing infinity categories in topology

based

[hidden haven]

bruh

very cool

[lavish gale]

maybe
i have not read algebraic topo so i dont know

[hidden haven]

only 1 thing to do then

[gusty halo]

[lavish gale]

highschoolers here are scary

[gusty halo]

true

i wish i had thought of doing something productive during highschool

but alas

it never even crossed my mind

i never even though about studying mathematics

[lavish gale]

same

[gusty halo]

and started with CS instead

then i did both

[hidden haven]

I was learning cat theory right after finishing my ug last year

and first year UG was explaining yoneda lemma to me

[gusty halo]

yeah that was also pretty late for me, i only started learning category theory in october like a week before classses started

big mistake

[hidden haven]

That first year UG was Raghuram

damn

[lavish gale]

started with physics
gradually got into math
Now i like math more

[hidden haven]

I knew a little bit before that

[gusty halo]

literally every course here just assumed you are comfortable with adjunctions limits colimits etc

[hidden haven]

Like universal properties etc

and definition of adjunction

But spent the summer reading mac lane

Almost finished it

[lavish gale]

damn 😂

[hidden haven]

[gusty halo]

oh nice

i read riehl till adjunction now

planning to learn about monads now

[hidden haven]

Nice

[gusty halo]

wondering whether i should also go ever mac lane at some point

[hidden haven]

Have you learned about string diagrams yet

[gusty halo]

i really want to

[hidden haven]

You should before monads

[gusty halo]

but i just cant get my head around them

like i actually tried

but too smallbrained

[hidden haven]

Did you try that arxiv paper

[gusty halo]

yeah i have it

[hidden haven]

category theory using string diagrams

[gusty halo]

dan marsdens

[hidden haven]

ah

Where did you get stuck?

[gusty halo]

i just dont know how people can parse these diagrams

[hidden haven]

Horizontal composition of natural transformations?

[gusty halo]

like i technically know the rules

[hidden haven]

ah you do 2 examples and you get used to it

I had the same problem

I came up with my own formal rules for the manipulation

[gusty halo]

but in the pictures everything is just wiggled around and imo not really recognizable anymore

[hidden haven]

And proved that those work

oof

[gusty halo]

hmm

[hidden haven]

The only reason I had to do this

was that the arxiv paper doesn't say things formally

And the papers that said things formally were first defining topological graphs and talking about their homeomorphisms

Which didn't seem like a good sign for what was to come

[gusty halo]

like for example with the adjunctions you should have well-defined wiggles

but then it starts looking like this

[hidden haven]

Lmao

[gusty halo]

and im just like "where did my cups and caps go"

[hidden haven]

That is also a well defined wiggle

That's just a natural transformation from H to G'KF

Not the one in the adjunction

A separate one

and you use the adjunction cups and caps to manipulate this

[gusty halo]

oh that makes sense actually

i also felt like the string diagrams for monoids or monads are easier

but maybe thats just me

[hidden haven]

Maybe because there's only one category there

So you don't need to worry about the colors

Btw which book did you read AT from

I need books to supplement May, because I am finding it very difficult lol

[gusty halo]

nothing really, i learned mostly from lectures and their exercises

[hidden haven]

I see

[gusty halo]

and lately a LOT of nlab articles

[hidden haven]

Damn nice

Going from being scared of nlab to being able to read nlab is such a great feeling

[gusty halo]

a friend showed me how with these string diagrams for monoids in moidal categories you can show that the [0] in the augmented simplex categoriy is the universal monoid in a monoidal category

[hidden haven]

Oh damn I should try that

[gusty halo]

in the sense that for any monoid in a monoidal category there exists a unique monoidal functor to it from this simplex category sending [0] to that monoid

[hidden haven]

Ye

[gusty halo]

was pretty cool

[hidden haven]

Clerk talks a lot about that property

[gusty halo]

simplicial sets are nice

[hidden haven]

I am reading about them right now

I will agree

[gusty halo]

clerk said he wants to talk about dold kan

im hyped for that

[hidden haven]

Yes

Same

[tough raven]

When I do this

[hidden haven]

The problem was to identify the abelian monoid of the natural transformations from the identity functor to itself for the case of Set, Grp, and Ab

And you and shamrock helped me with that

[chilly ocean]

Guys what is the best book to study abstract algebra? I would like to do it for an hobby so i'm more interested in theory than exercises

[frail perch]

Whats your background and what specifically interests you? Abstract algebra is a big field

[hidden haven]

There should be some book reviews in one of the pinned messages in #book-recommendations

[delicate orchid]

open #groups-rings-fields
set theory

[lethal dune]

category of sets

[wooden ember]

just a question about this one part of my lin alg exam i wasnt too sure about: how would i should that the vector space of quaternions is of dimension 2 over R[Z] where R[Z] = RId + RZ for Z a quatenion (using representation of quaternions as two by two complex matrices)

i just said that it's of dimension 4 over R and so is R[Z] (when Z isnt a scalar matrix), so it must be of dimension 4/2 over R[Z] but i had no way of justifying that directly

i thought this might have to do with field extension degrees but H isnt a field

[hidden haven]

where

R[Z] is a 2 dimensional division algebra over R, and a subalgebra of H. The field extension degrees you are talking about works for division algebras

[wooden ember]

and how would you show it?

[hidden haven]

which part?

[wooden ember]

i mean i was thinking this but i dont remember the proof for the multiplication rule lmao

[hidden haven]

The 2 dimensional algebra over R?

[wooden ember]

no this i know

[hidden haven]

You take a basis of larger thing over middle thing

Another basis of middle thing over smaller one

Show that the pairwise products give a basis for large over small

[wooden ember]

pairwise products?

as in of elements of small -> middle with middle-> big?

[hidden haven]

a_i*b_j

where {a_i} and {b_j} are the bases

[wooden ember]

right okay

okay so i see how to show that's linearly independent

just gotta show it's generating now

[hidden haven]

write the a_i in terms of the b_j or the other way around depending on what is a basis of what

[wooden ember]

oh yeah nvm

i was getting confused and multiplying by the wrong stuff

yeah okay i see

well let's hope my "we saw it in exercises" passes

had like 5m left i couldnt be asked to properly think it through

[hidden haven]

[wooden ember]

just analysis to go and then im exam free

and i can get back to hatcher

[hidden haven]

[wooden ember]

that emoji is cursed

[hidden haven]

[coral shale]

Let $R$ be a non-trivial ring. Is it possible to have an isomorphism between $R$ and $R^n$ for some $n > 1$?

[cloud walrus]

Shuri2060

[gusty halo]

what kind of isomorphism

module?

or ring?

[delicate orchid]

I assume they mean a ring isomorphism

[chilly ocean]

take R=Z^omega maybe?

would that work?

[gusty halo]

yes, i think so

[hidden haven]

Yes

[chilly ocean]

[white nymph]

in this exercise (from dummit and foote), in part a.) are they saying that it is enough to show that gKg^-1 = K in order for K to be in the center? if so, why? is it because we think of the group elements sorta as a basis... so if it commutes with all basis elements it commutes with the group ring?

[upbeat swift]

A is the set of all x ∈ R such that x ≠ 0,1. G = {ε, f, g}, where f(x) = 1 /(1 − x) and g(x) = (x − 1) /x. I have to show that G is a subgroup of S_A, and write the table of G. A little help with this?

[woven delta]

$x\in$?

[cloud walrus]

Emma

[upbeat swift]

Oops sorry

In R

Forgot to type that

Real numbers

[delicate orchid]

and G is a group under function composition or?

[upbeat swift]

A is a subset of R and G is a set of permutations of A

[delicate orchid]

ah I see so you're permuting the elements of A by those functions

nifty

[upbeat swift]

Mhm

I just don't know how to start this problem

[delicate orchid]

I think a good way to start is to show that those functions actually do permute A -> A

[white nymph]

(before it gets too buried, if anyone has help for my question above - im still interested)

[delicate orchid]

for the first part you can use the property that for all $k \in K$ and $\forall g \in G$, $g^{-1}kg \in K$ by definition of a conjugacy class, similar to what the hint is telling you. Then you use properties of multiplication in the group ring to show that this element must be fixed under conjugation by elements in $g \Rightarrow$ commutes with all $g \Rightarrow K$ is in the centre

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

ok back to this

[upbeat swift]

Would I have to show that G is a subset of $S_A$? So I'd have to check for closure, identity and inverse?

[cloud walrus]

Moon Child

[delicate orchid]

yup

[upbeat swift]

And if this passes all, it's a subgroup of Sa

[delicate orchid]

I mean and associativity but function composition is always associative

yeah

keep in mind that $\epsilon(x) = x$

[upbeat swift]

Yeah, reason I didn't say associativity

[cloud walrus]

Wew Lads Tbh

[upbeat swift]

So with each function, I'm showing closure and all

For f(x) and g(x) that was given

[delicate orchid]

no no you need to show that your little set {e, f, g} is closed under function composition

so f(f(x)), f(g(x)), etc. are all in that set

[upbeat swift]

Ohh

[delicate orchid]

cause that's the set you actually want to be a group

who cares about A (that guy sucks)

[upbeat swift]

Haha true

So I'm not computing any of the compositions?

[delicate orchid]

I mean you have to

you need to check that they're in the set

for example (spoiler alert) f(f(x)) = g(x)

[upbeat swift]

So I can check each one like, f(g(x)), g(g(x)), g(f(x)) and f(f(x))?

[delicate orchid]

yup

and you already know that the identity is in your set cause it's the first element so that one's done

[upbeat swift]

Ohh

Okay, let me try this real quick

[potent briar]

sorry for bothering in this channel,

I have 2 vector spaces, one of the form $(-\alpha_1, \alpha_1 + \alpha_2, \alpha_1 - \alpha_2, \alpha_2)$ and another of the form $(\beta_1, 0, 0, \beta_2)$
I'm trying to the find the intersection of these two spaces (which might just be a set) so I tried writing that the intersection of the space are all the vectors of the form (a, b, c, d), where

$a = -\alpha_1 + \beta_1$

$b = \alpha_1 + \alpha_2 = 0$

$c = \alpha_1 - \alpha_2 = 0$

$d =\alpha_2 - \beta_2 = 0$

[cloud walrus]

vik

[delicate orchid]

what's your reasoning behind these equations?

[potent briar]

i mean i need to find the intersection, how else could I do it

[white nymph]

i may be missing something... but what i was wondering is why an element of RG commuting with all the group members implies that an element commutes in \emph{RG} (since in my mind elements of RG look like sum r_i g_i

[delicate orchid]

oh shit is your ring not commutative?

[white nymph]

it is

[delicate orchid]

what definition are you using of the centre?

cause mine is just elements r such that gr = rg for all g in the underlying group

[white nymph]

commuting with all elts of RG im assuming

hmm maybe i should look into that

[delicate orchid]

well I suppose since your ring is commutative all the ring-based operations automatically commute, so it's just dependant on the group operation?

here's the sketch proof I have in my head

[white nymph]

i figured out the problem given that it is enough to show that commuting with group elements implies in center

i guess i was tripped up on defintion perhaps

[delicate orchid]

maybe but I think it holds true for your definition as well

[upbeat swift]

So I got f°g to be the identity
f°f to be g
g°f to be the identity
And g°g to be f
We know the identity is in the set. And we know that f and g are each other's inverse?
And because we got e, f, and g, G is closed under the set A

So it's a subgroup if S_A

Of*

[delicate orchid]

$let K = k_1+k_2+...+k_m$ and $r = \sum r_ig_i$
$Kr = k_1r+k_2r+...+k_m =\sum k_1r_ig_i+\sum k_2r_ig_i+...\sum k_mr_ig_i =\sum r_ik_1g_i+\sum r_ik_2g_i+...\sum r_ik_mg_i$
as $k_i, ; k_j$ are all conjugate we have $gk_i = k_jg$, and so $Kr = \sum r_ig_ik_{j_1}+\sum r_ig_{j_2}k_2+...\sum r_ig_{j_m}k_m = \left(\sum r_ig_i\right)k_{j_1}+...\left(\sum r_ig_i\right)k_{j_m} = rK$ as$K$ is the sum of the entire conjugacy class, these two sums are equal (conjugation is bijective)

[delicate orchid]

yup! exactly what I got

[cloud walrus]

Wew Lads Tbh

[upbeat swift]

Awesome

[delicate orchid]

and now you know the table too ;)

[upbeat swift]

Yes sir, thank you very much

[delicate orchid]

although I knew the table from the start using the cheeky fact that the only group of order 3 is C_3 ;)

[upbeat swift]

[candid kiln]

[delicate orchid]

half of the subscripts in this are wrong but I hope I got the point across btw

put the j's under the k's instead of the g's

[potent briar]

guys can someone help me understand the difference between external and internal direct sums?

(maybe in general, but I'm studying vector spaces)

as I understand it, an external direct sum is when you have 2 spaces U and V, and define a new one W whose elements are all the ordered pairs (x, y) with x in U and y in V

and an internal direct sum is strictly in the context of a big space and contained subspaces

[delicate orchid]

yeah that sounds right

[potent briar]

but

[wooden ember]

They’re essentially the same thing in the sense that the inclusion of your vector spaces in your external direct sum are in internal direct sum within the new bigger space

[delicate orchid]

the internal and external sums are isomorphic

[potent briar]

but in the case of internal sums can you establish, like, isomorphisms between your vectors to vectors of lower dimension

[wooden ember]

Wdym

[delicate orchid]

do you mean homomorphisms

[potent briar]

so that you can do the "operation" of making the ordered pairs

[wooden ember]

I’m not sure what you’re saying

[potent briar]

the span of (0, 1) and (1, 0) are two disjoint subspaces right?

[delicate orchid]

disjoint other than (0, 0) yes

[potent briar]

but I can't use the external direct sum definition

i can't make a new space like (a(1,0), b(0,1))

[wooden ember]

Vectors of the external direct sums aren’t “of dimension 2” if that’s what you mean: you may make pairs but the dimension of the space is dimV + dimU

[potent briar]

right...

[wooden ember]

The point is the internal direct sum of the span of those two vectors is isomorphic to the external direct sum of the underlying field with itself

[delicate orchid]

but this is a new space if you take (1, 0) and (0, 1) to be the basis vectors

nvm you changed it

[wooden ember]

As a vector space

[potent briar]

whaat

[wooden ember]

It’s the same kind of parallel you see with products of groups

[delicate orchid]

yeah I was gonna bring up internal and external products of groups

[potent briar]

well i haven't had a formal maths education so i havent done that much more algebra

[wooden ember]

Like the “external” direct product is really the same thing as the product HF when the intersection is trivial and both normalise each other

Ah okay

What’s important in both these concepts is that in a sense you are looking at the smallest structure containing two substructures

You can either construct this directly by starting with the substructures

[potent briar]

right I guess my fundamental problem is that when I see commas

[wooden ember]

External sums

[potent briar]

i immediatly think # of dimension

the span of (1,1) is of dim 1 not 2

[wooden ember]

Or find that you have an example of the bigger structure, find the two substructures in it and call the bigger structure their internal sum because it’s isomorphic to the external sum of these substructures viewed disjointly

[wooden ember]

Yes

The span of a set of size n is always of dimension at most n

[potent briar]

yes I know this

hmm right so if two subspaces direct sum to a containing space

they must be of (dim of big space) / 2

?

[chilly ocean]

How would I approach b? Idk how to visualize the n and how if j and k divides it does anything

[chilly ocean]

Hmmm

Ill come back to this after lecture

[desert dome]

Suppose for a in K, f(x) is the minimal polynomial of a over k, deg(f) = d and [K:k] = n and r is a root of f(x) in algebraic closure of k. I want to show that there are n/d different sigma in Sigma that maps a to r.
Now I have [k(a):k] = d and [K:k(a)] = n/d and I want to apply Galois correspondence theorem, but I don't see the correct way to apply. Like I don't see how |Gal(L:k(a)) : Gal(L/K)| = n/d helps. Could anyone give me a hint?

[chilly ocean]

sorry I am too high to help lol but just wondering is this for a second undergrad course in algebra? I'm hoping that I didn't miss the boat with a bad class cause we didn't go as hard as the problems you post 💀 like if I go to grad school eventually will I be weak af

[chilly ocean]

meant to reply to this lul whoops

[desert dome]

This is the second class for a grad algebra series

[chilly ocean]

ok ok so I'm not doomed, but it's still tough 😂 good to know! and sorry to interrupt

[desert dome]

Np np

[chilly ocean]

Why do you do PI^j

When we are give n is divisible by n

[chilly ocean]

i dont understand what you told me for help on solving part b of this problem

@chilly ocean

multiple of

okay lemme try

I don’t see how this would equal e

Is there another step

@chilly ocean

[chilly ocean]

where exactly?

[desert dome]

Suppose $\Sigma$ is the set of field monomorphisms from K to $\overline{k}$ that fix $k$. For $\alpha \in K$, suppose $f(x)$ is the minimal polynomial of $\alpha$ over $k$ and has degree d. Also, [K:k] = n. We can observe that $\prod_{\sigma \in \Sigma_{k(\alpha)/k}} (x-\sigma(\alpha)) = f(x)$, but how to show that $\prod_{\sigma \in \Sigma_{K/k}} (x-\sigma(\alpha)) = (f(x))^{n/d}$?

[cloud walrus]

Virginia

[desert dome]

We also have $k \subset K \subset L$ and $L/k$ is Galois. not sure if this is related

[cloud walrus]

Virginia

[tall jay]

This exercise is from Hungerford’s book, and I was wondering if someone could point me towards a good direction. It hinted that it uses the fact that $$ab=ac \implies b=c \quad \forall a,b,c \in G$$

[cloud walrus]

beeswax

[tall jay]

And I proved the implication already

[gritty sparrow]

each row is just the collection of elements of the form ag where a is fixed and g is from the group G. Is any two elements of the row are the same then ag=ah for some g and h, but then by what you showed h=g so they must have been in the same position in the row. Similarily for the collumns. You can show that everything will appear by considering a(a^-1g) for all elements g

[tall jay]

I'll work w/ this real quick and come back

[tall jay]

Oh, so the first 2 sentences is basically half the proof, huh?

I guess I was just unsure of how to structure the proof

[gritty sparrow]

I think I have some idea of how to do this: Basically given any map sigma from k(a) to the algebraic closure we need to show that is has exactly n/d extensions to K. That will basically prove it. Now let [L:K]=e. I'll show we have ne/d extensions to L first. So first of all fix some extension t of sigma (such a thing exists since we are mapping to some algebraically closed field) and wlog we can assume that in fact t maps L into L itself (infact t is an automorphism as L is algebraic over k(a) ) since L is normal so t will anyway map onto some copy of L in the algebraic closure. Now extensions of sigma to L are in correspondence with extensions of L fixing k(a) by mapping g->tg for any g in Gal(L/k(a)). So that proves my first claim as |Gal(L/k(a)|=ne/d. Now i will show that any extension of sigma to K has e extensions to L. The proof is pretty similar actually, and so because |Gal(L/K)|=e we are done. Now to calculate the extensions of sigma to K we simply have to do (en/d)/e =n/d.

[gritty sparrow]

ye pretty much

[upbeat swift]

Can I use the definition of a permutation? Or is there another way?

I'm doing 1-3

[rustic crown]

to show it's a permutation you need to check the definition >.<

[upbeat swift]

Well for me, the definition of a permutation is a bijection from set A to itself.

[rustic crown]

yep!

so is it clear why f_{a, b} is a bijection on R?

[upbeat swift]

So to prove the first one, I'd have to show that ax+b is bijective - injective and surjective?

[rustic crown]

yee

[upbeat swift]

It passes the horizontal line test, and doesn't have a slope of zero

[rustic crown]

well you would still need to "prove" that it passes the test

that test is like a visual thing, so just seeing that it works for graph you're able to see isn't enough

[upbeat swift]

Ah okay

[rustic crown]

i probably am being nit picky tho

to show that a function is injective just start with f(x) = f(y) and deduce that x = y

[coral shale]

nah. You should show it algebraically.

[upbeat swift]

Alright

[upbeat swift]

Hello, hope you're having a good day

Or morning, or night

[coral shale]

I am gn now. lol 👋

[upbeat swift]

[rustic crown]

[upbeat swift]

And this is because A → A?

[rustic crown]

nah, that's just the definition of an injective function.

for a function to be injective you don't need the comain and codomain to be the same

[upbeat swift]

Ah true

I did some algebra to show it's injective

got x=y

[rustic crown]

for surjectivity we need to show that given any y in R, there is an x in R such that f_{a, b}(x) = y

so can you tell what this x should be in terms of y?

[upbeat swift]

(y-b)/a?

[rustic crown]

yep!!

notice that in both these we really needed a != 0

in the first one we had ax+b = ay+b, and cancelling those a can't be done if it was zero

[upbeat swift]

Ah yes

Okay

[tall jay]

The symmetry group for this would just have the flips across each of the black axes, as well as two rotations, right? And those two rotations being 0,180 degrees?

[rustic crown]

are the h and j axes perfectly diagonal?

wait, even in that case i don't think that will be a symmetry

[tall jay]

So, it's pretty much 2 equilateral triangles facing eachother

Oh

[upbeat swift]

I need a quick verification

[tall jay]

Ur good, sorry I interrupted that

[upbeat swift]

No problem, I was working on this lol

[upbeat swift]

Does this seem right?

[rustic crown]

yea i think you're just missing the vertical flip

[tall jay]

Which would be taken care of with composing (horizontal flip) o (rotation) right?

[rustic crown]

yep

so two rotations and these horizontal and vertical flips

[tall jay]

4 elements only?

I was expecting more

I thought it was going to be either 2n or n! bc of the cardinality of Permutation & Dihedral groups

[rustic crown]

dihedrals are for regular polygons which give them lots of symmetries
that two triangles fused together is more like a rectangle, which should intuitively have fewer symmetries than a square

[tall jay]

Ah I see

So, just r0, r1, v, h

Where v and h are the vert and hor flips

Oh nice, thank u!

So, it should be something like this

[rustic crown]

right

[rustic crown]

(also maybe one thing to say, if we assume that the image is drawing to scale, then those diagonals have slope 1/2 and not 1/sqrt(3), which means it's not really equilaterals)

[tall jay]

How does one denote equal lengthed sides again? Just like a perpendicular line in each midpoint?

[rustic crown]

you usually put some sort of mark in the middle of the segments

it can be a single smol segment, a double one if you wanna indicate two pairs of equals sides, maybe some other squiggly mark if you need more

[subtle ivy]

another way to look at this is that the symmetry group of the "sides" is the same as the whole shape, and since these "sides" form "half" of a square, their symmetry group must be a subgroup of order 4 of D_8. order properties lead you to fill in the correct amount of rotations and reflections then

it's not exact, but at least a helpful example. there are situations later on in the study of space groups which use the idea of fitting one shape into another to relate their symmetry groups.

[tall jay]

Hmm interesting

[desert dome]

thank you so much! I am still a bit confused on why |Gal(L/k(a))| / |Gal(L/K)| = n/d means there are n/d such extensions?

[gritty sparrow]

basically every extension of sigma to K has e extensions to L. Also every extension to L comes from extending some extension to K (just restrict that extension to K). So basically the maping (extensions to L)->(extensions to K) which is defined by restriction has the property that it is surjective and the preimage of any extension to K has e elements. This means that e*|(extensions to K)|=|(extensions to K)|

[desert dome]

I am not sure if I follow. like the automorphisms in Gal(L/k(a)) and Gal(L/K) both fix a. suppose r is a root of the minimal polynomial f(x) of a over k. how could we say there are n/d different sigma such that sigma(a) = r? a is fixed in both of them, how could their quotient give us that info

[gritty sparrow]

That's why I first set up a bijection between Gal(L/k(a)) and sigma such that sigma(a)=r

and after that I show that any extension of sigma to K has e further extensions to L

after that I show that there is a correspondence between extensions of any injection from K to K closure to L and Gal(L/K) (i didn't do this but it is entirely the same proof as the first correspondence)

[desert dome]

I see. thank you so much!

[dusty sapphire]

True or false? And how to approach?

[cursive holly]

wondering if the proof logic for this is correct:
let's write our cyclic finite group as $G = H_Z$, where $Z$ is finite. we can apply the theorem that if $G$ is a finite cyclic subgroup of order $n$ then there exists a subgroup of order $m$ where $m \in \mathbb{N}$ $\iff m|n$. we know that Z only has a finite amount of factors, so therefore every finite cyclic group has a finite amount of subgroups

[cloud walrus]

alias

[cursive holly]

the conclusion seems correct but i'm skeptical about the logic

[hidden haven]

Hint, ℝ × ℝ can be thought of as the ring of functions from a 2 element set to ℝ

[hidden haven]

This is fine if you use the stronger version of the theorem that also tells you that when m | n, there is exactly one subgroup of order m, because if there were many for each factor then you could lose finiteness

But there's a simpler argument

Not relying on cyclicity

Any finite group will have finitely many subgroups because it has finitely many subsets

[cursive holly]

ahhh right, that makes sense

haha i always overthink, that's a pretty intuitive way to think about it

thanks!

[dusty sapphire]

still not getting

from R×R to R there is no injective function but how to solve in this

[hidden haven]

Not to R

I am claiming that there is an injection to C(R)

[dusty sapphire]

next

i didn't get a map

[hidden haven]

So first prove isomorphism R x R ~ C({0,1})

Then embed C({0,1}) into C(R)

[dusty sapphire]

yes they are similar i can do this but for ring homomorphism it is same?

[hidden haven]

This is a ring isomorphism yes

[dusty sapphire]

okay then

Thank you

hey but answer is false🙂

[hidden haven]

bruh moment

[dusty sapphire]

is there any ring homomorphism othere then (a,b)-->a or b

[rustic crown]

there is a non-trivial idempotent in R x R namely e = (1, 0), notice e^2 = e

but there aren't any such in C(R)

[dusty sapphire]

yep

[rustic crown]

if a continuous function f : R --> R is idempotent, then f(x) = 0 or 1 at each value

by continuity f = 0 or f = 1

[dusty sapphire]

so (0,0), (0,1), (1,0) and (1,1) have only two choices right

[rustic crown]

you should be able to do something with idempotents again
e will be mapped to something which satisfies x^2 = x, inside R only such are 0 and 1

e = (1, 0) and f = (0, 1) say
then there are 4 options where these can go

e+f = (1, 1), so the sum of their images need to be 1

[dusty sapphire]

got it

Thanks

[rustic crown]

does it work tho >.<?

[dusty sapphire]

yep it will work

[rustic crown]

i don't see it immediately but if you say so

[dusty sapphire]

how to find number of ring homomorphism from $ \mathbb{Z}[x,y] $ to $ \mathbb{F}_2[x]/(x^3+x^2+x+1)$

[rustic crown]

is texit sad?

that thing is just (x+1)^3

[dusty sapphire]

Not working

[dusty sapphire]

yep

[cloud walrus]

det

[dusty sapphire]

space mistake

[rustic crown]

since the right is commutative ring, we could just choose the image of x and y arbitrarily and that should give a ring map

size of the right is like 8 right

so 8^2 = 64 ig

[dusty sapphire]

yes in answer 2^6,

so there is two ways to it

Thanks

[rustic crown]

[cursive holly]

what does the set $\left{ \sigma \in S_4 | \sigma(3) = 3 \right}$ actually denote?

[cloud walrus]

alias

[cursive holly]

i'm seeing it in this and has no clue what it actually means

[hidden haven]

Permutations that fix 3

[chilly ocean]

[blazing parrot]

Hi anyone?

[cursive holly]

i still don't get it

[delicate orchid]

this mf lookin awfully cyclic lemme tell you what

[chilly ocean]

you have bijective functions of {1,2,3,4}, how many of those have property f(3) =3

[devout crow]

if $a^m = b^{-1}$ then $b = a^{-m}$

[cloud walrus]

∧res

[cursive holly]

only 1 right? because it's bijective

[delicate orchid]

not quite

a bijective function from {1, 2, 3, 4} with f(3) = 3 is basically all the permutations of {1, 2, 4}

[chilly ocean]

ye if f(3) = 3 you have to figure out what can 1 map to - 3 choices, then for f(2) you have remaining 2 choices and for f(4) only 1 choice left

[cursive holly]

yeah, the answer is 6

at least in the explanation i saw - my issue is just that i'm REALLY fucking stupid and have no clue what the set means

[chilly ocean]

set is a set

[delicate orchid]

it's the set of "all permutations of 4 elements such that 3 is fixed", is how I'd read that notation

[cursive holly]

OHHHHHH, yeah it just clicked now for me haha. and from there we just find the possibilities

[barren sierra]

Stuck on where to start with b

I want some sort of relation between d and [G : H cap K] probably

I know [G : H][H : H cap K] = [G : H cap K]

and knowing me there's just one small leap in logic I need to make from that but I'm not sure

cause that says mn' = [G : H cap K]

I guess from part a that gives me mn' <= mn

[wooden ember]

use the fact that $|HK| = \frac{|H||K|}{|H\cap K|}$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[barren sierra]

I was going to be doesn't that not hold if H, K are infinite?

or well

it holds but not in any way that's useful persay

[wooden ember]

oh nvm finite index

[barren sierra]

yea

[wooden ember]

yeah my bad

[barren sierra]

that was my first thought as well

[G : H] and [G : K] are finite

but that doesn't tell me anything about H and K themselves

[G : H][H : H cap K] = [G : H cap K]
so I do get [G : H][H : H cap K] = [G : K][K : H cap K]. So mn' = nm'

tho that isn't exactly enlightening

[wooden ember]

yeah you do get that but im struggling to squeeze more out of it

[barren sierra]

same

[wooden ember]

probably some argument via actions im not seeing

[barren sierra]

I think I need something concrete about [G : H cap K]

[wooden ember]

So i want to say ive got it but i probably dont

Consider $x := |G:N_G(H\cap K)|$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

this is $|N_G(H\cap K): H\cap K|\cdot mn' = |N_G(H\cap K): H \cap K|\cdot m'n$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

wait im an idiot

this wont work

[barren sierra]

What is N_g

[wooden ember]

i was about so say H and K normalize H n K which they clearly dont have to

[wooden ember]

normalizer

oh but instead of N_G(H n K)

we cant just consider <H,K>

then the index |<H,K>: H n K| >= m'n'

[wooden ember]

and you can use this to conclude

[wooden ember]

except this is wrong of course and you should be taking a fraction not a multiplication lol

but you get the idea

you get $x \leq \frac{mn'}{m'n'} = \frac{m}{m'} = \frac{n}{n'}$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

and so since $x\geq 1$ this gets you the result

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

wait

god im tired

[wooden ember]

this wont work it would be lcm(m'n')

which might still get you the result since lcm(m'n') = m'n'/(m',n') 🤔

youd end up with m'x < m (m',n')

so if you can show (m',n') divides x you're fine

idk if that's true though

im too tired for this lol good luck

gotta take a break

[tall jay]

Im doing this exercise from Hungerford's book and was wondering what other information I have to fill in the other blank spots

I can guess all of them with pattern matching, but I'm not sure which other Group axioms I can use to justify

[hidden haven]

Try proving that every element appears exactly once in each row and column

[tall jay]

Sorry, I just saw this. My discord stopped working for some reason

[tall jay]

OHHH, I see it. I proved it in a previous exercise. Thanks!!!

Well, not really prove, but like started proving

[hidden haven]

Same

It's broken today

[barren sierra]

Is there a good strategy for coming up with presentations?

I need to come up with a presentation for S_4 with 2 elements

I know (1 2) and (1 2 3 4) generate the whole group

but with what relations is what is giving me a bit of a think

clearly we have x^2 and y^4

but how do I know "that's all"

[hidden haven]

That's not all because it's still infinite

(xy)^n are all still distinct

[barren sierra]

yea

so I'm playing around with combinations of that

xy = (2 3 4) so (xy)^3 is another

so then like

my question is more abstract than this example

[hidden haven]

Right

[barren sierra]

just like in general how do I know I have "enough"

[hidden haven]

So you have < x, y >

[barren sierra]

cause I knew x^2 y^4 wasn't enough since I didn't have any way for those 2 element to play with each other per say

[hidden haven]

The free group

[barren sierra]

yea

[hidden haven]

Which surjects onto S_4

[barren sierra]

and then I take the quotient of hte normal closure of the relations

[hidden haven]

Yes

[barren sierra]

and get an isomorphism hopefully

[barren sierra]

yea

[hidden haven]

So you have this induced map

[barren sierra]

yea that's just first isomorphism theorem

[hidden haven]

Yep

You get surjectivity of this just from the fact that the original map was surjective

You need injectivity, and this is very annoying in general

But for finite groups, we can do this by counting

So you have got a surjection from the quotient to H

That means quotient is ≥ H in cardinality

Now you prove the other inequality, which will prove that the surjection must be a bijection

And to prove the other inequality, you will write down enough elements to make sure that everything in the free group lies in the equivalence class of 1 of those

So for example if you were quotienting by the simpler relations x², y², (xy)³

Then you could prove that
1, x, y, xy, xyx, xyxy, xyxyx, yx, yxy, yxyx, yxyxy
cover everything

You do this by taking any element of the free group and showing that it can be reduced to one of these forms (usually by induction on length)

So for S⁴, you would want to write down all the elements in terms of these 2 elements x and y

[hidden haven]

And then that will be your list as in here

[barren sierra]

but my question is

[hidden haven]

And you'll know that you have enough relations when your list is exhaustive

[barren sierra]

oh hm

so like

get what relations i think I need

attempt the proof

and then if I need more

add it?

[hidden haven]

Yes

[barren sierra]

ok cool

[hidden haven]

Also keep in mind that here we don't claim that all of these elements are distinct

That is a very hard thing to prove

We just need the list to be exhaustive and of cardinality that of the given group

[barren sierra]

how do we know exhaustive if we don't know that they're distinct

like what's "stopping" me from just making <x^2, x^4, x^6, ...>

obviously that's wrong but showing that my list is exhaustive in general sounds rough

[hidden haven]

Exhaustive as in everything in the free group is equivalent to at least one of these

It is enough because we already proved surjectivity

And here we are only trying to prove that the quotient is smaller than or the same size as the given group

[barren sierra]

got it

[barren sierra]

so like

from first isomorphism theorem we get that the free group is isomorphic to the image of phi

wait hm

don't I just need to show that phi is surjective?

Oh but then I need to show that ker(phi) = normal closure of the relations?

I'm just lost as to where injectivity comes in

[hidden haven]

Yes exactly

[barren sierra]

one direction is trivial

it's the other inclusion that sucks?

[hidden haven]

The first isomorphism theorem (the general version) says that there is an induced map from the quotient iff the subgroup you are quotienting by is contained in the kernel

And this is injective iff kernel = that subgroup

[hidden haven]

Here we just use the first version to get the induced surjective map because yes

The equality is impossibly difficult

So we just show that this surjection is also injective through this counting in the finite group case

[barren sierra]

what that ker(phi) is a subset of the normal closure?

hm ok

[hidden haven]

Did I say the opposite thing

Kernel should contain the subgroup you quotient by

Ye that is what I said

[barren sierra]

no I just don't see right now why that inclusion direction is difficult

[hidden haven]

Ah

[barren sierra]

tho tbf I haven't done it, it just seems easy in my head (as I type up the first part lol)

cause clearly the normal closure is a subset of the kernel, that's done already

[hidden haven]

Yeah it is one of those things that seems easy

[barren sierra]

(I've shown a similar result in a prior HW I can just leverage that)

yea it does lol

maybe I'll just try it later when I have time

[hidden haven]

Ye

[barren sierra]

that's probably the best thing

[hidden haven]

It is very hard tho lol

Arguing by this map being an injection is the best method that I know

[chilly ocean]

guys to prove if something is a group under an operation i just prove

closure, identity, inverse, and associativity right?

hnmm

would i just grab arbritary elements

[stone fulcrum]

Depends on the group

[chilly ocean]

one sec

i know what to do but i dont know What i actually need to do this

[stone fulcrum]

Do you know what an element of Z(p) looks like?

[chilly ocean]

yes

its a rational number where the prime number p doesnt divide the denominator

so if it was z4

[stone fulcrum]

Take two general elements, add them. The addition should also be a member of Z(p). Prove this.

[barren sierra]

4 isn't prime

[chilly ocean]

1/3 would be and elment

oh shit

LOOOOOOOL

[barren sierra]

lmao

[chilly ocean]

lets take Z3

1/17 would be part of that set

right?

[stone fulcrum]

Right

[chilly ocean]

can i call those two general elements a, b?

[barren sierra]

might be useful to call them a/b and c/d

[stone fulcrum]

I think, more interestingly, 1/8 would as well

[barren sierra]

but yes they're arbitrary

[chilly ocean]

ight

so taking a/b , c/d element of Zp

adding them

uh oh

he dropped the devestation

emoji

[delicate orchid]

bro...

your b turned into a d....

[chilly ocean]

O Hshit

mb

there ya go

[delicate orchid]

yes! so true!

[proud bear]

,ti moldi

[cloud walrus]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[chilly ocean]

that would satisfyclosure right

[delicate orchid]

ok now use the properties of primes and other assorted integers to show this is in Z_(p)

[proud bear]

moldi go to sleeeeeeep

[chilly ocean]

hmm

[delicate orchid]

idk I'd explicitly state that "from the definition of a prime number, p|bd => p|b or p|d so p cannot divide bd"

where | means divides

idk how universal that notation is

and then the numerator is clearly an integer so it's closed

[chilly ocean]

okay

[delicate orchid]

@hidden haven is it past your bedtime, buster?

[chilly ocean]

we want p not to divide bd

due to the def

[delicate orchid]

yup

[chilly ocean]

coolio

[delicate orchid]

now find inverses and identity

[hidden haven]

No sir

[chilly ocean]

okay

[hidden haven]

It's 2pm I'm not breaking any rules

[delicate orchid]

ahhh just realised why we localise to the compliment of a prime ideal, it's so we have 1/1

how have I only just realised that lmfao

[proud bear]

lol

[proud bear]

i thought it was because the complement is multiplicatively closed

[delicate orchid]

I mean so is an ideal

[chilly ocean]

for inverse cant i just get elements a/b and b/a

wait no

its addition

cant i just get a/b and -a/b

[delicate orchid]

yup

[chilly ocean]

how do i formalise that tho

like get 2 elements?

or like

[delicate orchid]

find the identity, show those two fuckers add to give you the identity

:packwatch: dub epic win etc.

[chilly ocean]

the identity would be 0

and just say

a/b + (-a/b) = identity

?

[delicate orchid]

then literally just add the fractions

[hidden haven]

You can localise at any set because localisation is defined using a universal property that is preserved by taking multiplicative closure of the set that you localise at it's just that the construction becomes a lot neater when the set is already multiplicatively closed

[delicate orchid]

I see the words "universal property" I stop reading moldi you know this

[hidden haven]

I didn't know this

Does that mean

You ignored 90% of all my messages

[delicate orchid]

I know localisation as "define new ring.... funny a/b = c/d <=> exists a unit u such that u(ad-bc) = 0.... b d in multiplictive set,"

wait

so lemme guess

the universal property guarantees the existance of some map from R^2 to R_(p)

[proud bear]

a/b+(-a/b)=(a-a)/b=0/b=0 so yes

[delicate orchid]

is moldi latex-ing a diagram

[hidden haven]

Localisation of R at S is a ring R' with a map i: R → R' which maps every element of S to a unit such that whenever there is a map j: R → R" with this property, there's a unique map k: R' → R" such that j = ki

[chilly ocean]

yessir

and now associativity

grab 3 elements

[delicate orchid]

this is actually a 😵‍💫 moment

ok I had to read it a couple of times but I follow it now

[proud bear]

,tex\begin{tikzcd}R\arrow[r,"i"]\arrow[rd,"j"']&R'\arrow[d,"k"]\&R''\end{tikzcd}

[delicate orchid]

holy shit that's impressive

[hidden haven]

What a god

[proud bear]

Lol

[delicate orchid]

also on the topic of associativity I literally just say "inherited from Z"

or "inherited from the parent ring" etc. like 99% of the time because I hate proving associativity

[chilly ocean]

got it got it

[delicate orchid]

the existence of this unique map is the universal property right?

just to double check my intuition from the definition of the tensor product is right

[hidden haven]

The universal property is this factoring property of the map

[delicate orchid]

by factoring you mean they compose nicely?

[hidden haven]

Like if some map R → R' satisfies that statement

We say that it satisfies this universal property

Existence is not guaranteed

I can make up whatever universal property I want

Doesn't mean that there exists something that satisfies it

So it has to be constructed to prove existence

But universal properties always guarantee uniqueness up to isomorphism

[delicate orchid]

I mean the existence of the map k is the universal property

not i

[hidden haven]

Ah yes uniqueness and existence both

[delicate orchid]

got it

so it is just like the tensor product

hmm

[hidden haven]

Yes

They are both universal properties

[chilly ocean]

just like iso theorem

[delicate orchid]

I mean, I should've expected it to be - in both you're quotienting all pairs (a,b) by an equivalence relation

[delicate orchid]

it's just a different relat-

omg it is the first iso theorem

holy SHITTTTTTTTTTT

cause the equvialence you're quotienting by in the first iso is given by the cosets of the fuckin subgroup

[hidden haven]

[chilly ocean]

https://tenor.com/view/cat-scared-shocked-funny-animals-gif-14395038

[delicate orchid]

if only there was some field of maths that could rigorously link these ideas

[hidden haven]

Perhaps you would like some cat theory 🙈

[delicate orchid]

and there it is

[hidden haven]

The sales pitch

[delicate orchid]

https://tenor.com/view/cat-running-away-escape-getaway-bye-gif-16631286

me when category theory

[hidden haven]

🙈

Just draw arrow bro

[delicate orchid]

yup

just so happens the arrow looks like this

$\rightarrow$ category theory

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

which is a very scary diagram

[hidden haven]

[delicate orchid]

also I love how #chill and #groups-rings-fields just legit swapped places during this convo

[hidden haven]

The important thing is that everyone here enjoyed learning about cat theory and has decided to continue learning it on their own 😌

[delicate orchid]

[hidden haven]

That's the spirit

[delicate orchid]

I have actually done a little bit of independent reading on basic finite category stuff

[hidden haven]

What is finite category stuff though

[delicate orchid]

I cannot remember for the life of me

not much past the definitions though

[hidden haven]

I see

[delicate orchid]

I remember seeing cartesian product being defined

[hidden haven]

Very nice

That is cool

It starts slow because you start by redoing stuff you know

And suddenly you've developed tools that let you learn new stuff much faster

Anyway gn it's past 5am

[chilly ocean]

WAIT m oldi

[delicate orchid]

I knew it

[hidden haven]

[chilly ocean]

good night

[delicate orchid]

how wholesome

[hidden haven]

Good night catfan

[chilly ocean]

[delicate orchid]

wow and I thought I liked the first iso theorem before

[barren sierra]

So I have that aba^-1 = b^2 and bab^-1 = a^2. Then the first relation implies aba^-1b^-1 = b. But ba^-1b^-1 = a^-2. So then a^-1 = b.

So that was my work for showing a^-1 = b (which then gets me a = b = e shortly after)

is that good enough?

or is there something more

[delicate orchid]

that looks right

[barren sierra]

So I don't need to show anything else?

No map from a free group or anything?

[delicate orchid]

I mean you could find a map if you wanted to ig

[barren sierra]

But do I have to?

That's my question cause I honestly don't know if I have to or if what I wrote is sufficient

[gritty sparrow]

what you did was sufficient (atleast if your grader is not a pedant). If you are really worried just add in a line about how every element is a product of a sequence of a's and b's hence a and b being e implies all elements are

[cyan raft]

Determine the order of the group with presentation $$\langle a,b,c|ab=c^2a^4, bc=ca^6, ac=ca^8, c^{2018}=b^{2019}\rangle$$

[cloud walrus]

meg chmonkeynumber3fan

[cyan raft]

please help. i have no idea how to do this at all.

[cyan raft]

...anybody here?

[final oasis]

what is representation theory about

I have read an intro chapter and was wondering what's the big picture idea

[rustic crown]

trying to understand something by acting it on something else (usually vector spaces)

the first something can be groups, rings, algebras, lie algebras, ...

[final oasis]

the symmetric square and alternation square are reminiscent of Fourier analysis stuff, how does that get involved here

[rustic crown]

idk analysis >.<

[final oasis]

characters

[rustic crown]

what's this symmetric and alternation square?

you mean those character tables?

[final oasis]

like direct sum of odd and even functions

[rustic crown]

taking symmetric and exterior powers gives you a way to find more representations of (say a) group

the usual questions to answer are:

1. what are the simple representations
2. what the indecomposable represntations
3. can we classify finite dimensional representation

and finding these simple representation can be hard, you need a place to look for

there was like a theorem that if you have a faithful representation then all the simple representations are hiding inside tensor powers

which could turn out to be pretty useful for small examples

[final oasis]

what types of info does these give of the group

[rustic crown]

i don't think i know too many applications... but i've heard using representation theory you can prove things like every group of order p^aq^b is solvable

[final oasis]

cool thank you

[lavish nexus]

Given two fields A and B

[next obsidian]

I'm sorry, the answer is no

[lavish nexus]

if [A:B]=p^k
can I find for every s< k a subfield C [A:C]=p^s

lol

[next obsidian]

I think so but I have to do a thinky

[chilly ocean]

Are $A,B\in k$?

[cloud walrus]

Arr0w_04

[next obsidian]

A and B are just fields

A containing B

[lavish nexus]

any field

[chilly ocean]

Oh sorry misread the question.

[next obsidian]

So let's say A = B(alpha)

then alpha's minimal polynomial is degree p^k

if you look at B(alpha^{p^r})

no

this won't work

hmmm

[lavish nexus]

specifically someone raised a point with this proof

https://media.discordapp.net/attachments/932835767854694491/936093699249680424/Screen_Shot_2022-01-26_at_10.01.55_PM.png

which seems to rely on that fact

[next obsidian]

wtf are K' and L'

[lavish nexus]

corresponding subgroups in Gal(E/F)

[next obsidian]

I see

Also where do you think it's using the fact?

[chilly ocean]

Where did you even get that photo from? A paper of some kind? A textbook?

[lavish nexus]

Say you have [L:K] to be p^k but no intermediate subgroups

To induct need to be able to reduce it to p

[next obsidian]

https://math.stackexchange.com/questions/746246/is-there-always-an-intermediate-field-between-non-prime-field-extension

it doesn't look like this is true

[lavish nexus]

some lecture notes

[next obsidian]

also I was wrong about being able to refine the field extensions

I thought maybe there'd be a counterexample over Q with a quartic polynomial as I thought about it longer, but couldn't come up with an example

[rustic crown]

it should be true if the extension is galois >.< (and the galois group is solvable)

[south temple]

guys, if a ring contains only a finite number of distinct ideals, then it must be Noetherian right?

[rustic crown]

yea ig

[south temple]

since Noetherian rings have the property that any increasing chain of distinct ideals ends

[rustic crown]

yep, so you can't have a strictly increasing infinite chain

[rustic crown]

I don't know if there's a proper way to do this. Like I've heard this problem is undecidable, but that doesn't mean there aren't nice algorithms for special cases. anyway, so idea is to just simply the relations as much as you can.

look at abc, we get abc = c^2a^4c = aca^6
using the third relation c^3 a^32 = c a^14 so c^2 = a^(-18)

plug it in the first relation ab = a^(-14) so b = a^(-15)

now since ac = ca^8 we get ac^2 = ca^8c = c^2a^64
so a^63 = 1

c^2 = a^(-18) gives c^14 = 1 and b = a^(-15) gives b^21 = 1

the last relation can be now written in an easier way c^2 = b^3 (as 2016 is divisible by both 14 and 21)

writing c^2 and b in terms of a gives a^(-18) = a^(-45) which gives a^27 = 1, and by a^63 = 1 we have a^9 = 1

so c^2 = a^(-18) = 1 and b = a^(-15) = a^3

we now have relations a^9 = c^2 = b^3 = 1 and b = a^3, we need to still understand how a and c behave together so the third relation says that ac = ca^(-1) which is same as (ac)^2 = 1

so we proved that old relations imply these new ones, it's easy to check that these new relations suffice

so the group is
<a, b, c|a^9 = b^3 = c^2 = (ac)^2 = 1 and b = a^3>

the generator b is useless, a already does its job. so the group is actually isomorphic to <r, f| r^9 = f^2 = (rf)^2 = 1>
which is the presentation of D_9 so has 18 elements.

[cyan raft]

oh gosh det thank you so much (even though i haven't read any part of your answer at all)

[prisma shuttle]

hey guys in the first sol in this thread

https://math.stackexchange.com/questions/60969/every-nonzero-element-in-a-finite-ring-is-either-a-unit-or-a-zero-divisor/60974#:~:text=Thus every nonzero element of,zero-divisor nor a unit.

can someone explain why the surjectivity part would fail if the ring was not finite

[chilly ocean]

because then injectivity doesn't imply surjectivity

[gritty sparrow]

Take Z and the element 2 for example

(Z being the integers)

[rustic crown]

who's chmonkeynumber2fan?

[gritty sparrow]

Somebody at some point was, I think senku

[next obsidian]

Why are you chmonkeynumber1fan

[gritty sparrow]

I can't explain it, I guess I was just born like that

[next obsidian]

[cyan raft]

hi chmonkey!

[cyan raft]

senku still is number 2

[lethal dune]

who's number 0

[next obsidian]

Liquid

[kind temple]

what is the intuition behind commutators

like what do they do/what’s the motivation

[next obsidian]

In a group?

[kind temple]

sure. group, ring, does it matter what setting?

[chilly ocean]

commutator measures commutativity

[rustic crown]

gib me motivation too

what's so nice about the jacobi identity in the definition of lie algebras?

[chilly ocean]

derivations

are cool

[rustic crown]

i mean it felt a little weird, the book said we like the operation xy-yx, then said we can abstractly describe such a system even when x*y doesn't make sense, and for that we have jacobi identity, and later there was a remark that all finite dimensional lie algebras will look like xy-yx anyway

[kind temple]

how so?

[chilly ocean]

elements commute iff their commutator vanishes

[kind temple]

so... how is that measuring commutativity? just telling you what elements commute?

[next obsidian]

Commutative iff the subgroup they generate is 0

[chilly ocean]

perhaps i should have specified commutativity of elements.

[next obsidian]

You can also mod out by them which is basically modding out by every relation to make them commute (you actually have to mod out by the set of all products of them, the set of commutators won’t usually be closed)

And you get what is basically the closest abelian approximation of your group

[chilly ocean]

an example i like is the set \frak{so}(3) of 3x3 skew symmetric matrices

if x and y are skew symmetric matrices, xy might not be one, but xy - yx will

[kind temple]

so the reason i asked is because i had a hw problem which explored the commutator of vector fields v and u

[chilly ocean]

generally, if X and Y are derivations on something, then XY might not be one, but the commutator XY - YX will be one. that's nice.

a classic example being what c^2 is probably about to post

[rustic crown]

right, same was true with other derivations, d1 * d2 might not be but [d1, d2] will be

[lethal dune]

enlighten me as well

[chilly ocean]

the jacobi identity just says [x, -] is a derivation wrt the bracket, for all x in your space

more derivations!

[x, [y, z]] = [[x, y], z] + [y, [x, z]] lmao

[kind temple]

so we were given $v,u\in C^{\infty}(A\subseteq\mathbb{R}^n,\mathbb{R}^n)$ and were asked to consider the map $L_v:C^{\infty}(A,\mathbb{R}^n)\to C^{\infty}(A,\mathbb{R}^n)$ given by $$L_v(f)=\sum_{i=1}^nv_iD_if$$ where $v_i$ is the $i$th component function for $v$

[chilly ocean]

sus bet

[rustic crown]

[kind temple]

um

i actually dont know

[chilly ocean]

strange

[cloud walrus]

c squared

[chilly ocean]

ok

so just the derivation defined by a vector field

how do i prove a homomorphism is isomorphic?

show that it's invertible

[kind temple]

and we were asked to show that there is a unique vector field $w\in C^{\infty}(A,\mathbb{R}^n)$ satisfying $L_v\circ L_u-L_u\circ L_v=L_w$ and to provide a formula for $w$

[cloud walrus]

c squared

[chilly ocean]

so the lie bracket

go to #diff-geo-diff-top

c^2

i'll say something there

How would i
Do it here

ohhhhhhhhhhhh

rip

[lethal dune]

you need to show what values of n gives an isomorphism right?

deduce the kernel

[chilly ocean]

eah

how would that help me

[rustic crown]

it's an abelian group hom, not a ring hom (so multiplication need not be preserved)

[lethal dune]

ok nvm

that's always 0

[chilly ocean]

i still dont know how to approach this

[rustic crown]

it's always injective (for non-zero n), so just need to check for which values of n it will also be surjective. so when will 1 be in the image?

[chilly ocean]

iso means bijective

[lethal dune]

can you show it's onto?

[chilly ocean]

so get an element from the codomain

then work backwards?

[lethal dune]

kinda

[lethal dune]

this also works